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CORNELL
UNIVERSITY
LIBRARY
MATHEMATICS
Cornell University Libra
elementary treatise on the integral
PEPAAT Hcy ES B2Tin. ay
Wapeasy +
4e02
AN ELEMENTARY TREATISE
ON
THE INTEGRAL CALCULUS.
AN ELEMENTARY TREATISE
ON
THE DIFFERENTIAL CALCULUS,
CONTAINING
THE THEORY OF PLANE CURVES.
BY
BENJAMIN WILLIAMSON, F.R.8.
SEVENTH EDITION.
AN ELEMENTARY TREATISE ON DYNAMICS,
CONTAINING
APPLICATIONS TO THERMODYNAMICS.
BY
BENJAMIN WILLIAMSON, E.R.S.,
AND
FRANCIS A. TARLETON, LL.D.,
Fellows of Trinity College, Dublin.
SECOND EDITION.
AN ELEMENTARY TREATISE
OW
THE INTEGRAL CALCULUS,
CONTAINING
APPLICATIONS TO PLANE CURVES
AND SURFACES,
A CHAPTER ON THE CALCULUS OF VARIATIONS,
NUMEROUS EXAMPLES.
BY
BENJAMIN WILLIAMSON, D.Sc., F.R.S.,
FELLOW AND SENIOR ‘I'UTOR OF TRINITY COLLEGE, DUBLIN.
Sixth Gdition, Bebised and Enlarged.
NEW YORK:
:D. APPLETON AND COMPANY.
1891.
ae
[ALL RIGHTS ‘RESERVED. ]
K
MAAS
Dusuw: Printed at THe University Press.
PREFACE.
Tus Book has been written as a companion volume to my
Treatise on the Differential Calculus, and in its construction
I have endeavoured to carry out the same general plan on
which that book was composed. I have, accordingly, studied
simplicity so far as was consistent with rigour of demonstra-
tion, and have tried to make the subject as attractive to
the beginner as the nature of the Calculus would permit.
I have, as far as possible, confined my attention to the
general principles of Integration, and have endeavoured to
arrange the successive portions of the subject in the order
best suited for the Student.
I have paid considerable attention to the geometrical ap-
plications of the Calculus, and have introduced a number of the
leading fundamental properties of the more important curves
and surfaces, so far as they are connected with the Integral
Calculus. This has led me to give many remarkable results,
such as Steiner’s general theorems on the connexion of pedals
and roulettes, Amsler’s planimeter, Kempe’s theorem,
Landen’s theorems on the rectification of the hyperbola,
Genocchi’s theorem on the rectification of the Cartesian oval,
and others which have not been usually included in text-
books on the Integral Calculus.
A Chay ter has been devoted to the discussion of Integrals
of Inertia. For the methods adopted, and a great part of the
vi Preface.
details in this Chapter, I am indebted to the kindness of the
late Professor Townsend. My friend, Professor Crofton,
has laid me under very deep obligations by contributing
a Chapter on Mean Value and Probability. Iam glad to
be able to lay this Chapter before the Student, as an in-
troduction to this branch of the subject by a Mathematician
whose original and admirable Papers, in the Philosophical
Transactions, 1868-69, and elsewhere, have so largely contri-
buted to the recent extension of this important application
of the Integral Calculus.
In this Edition I have introduced Chapters on Line and
Surface Integrals, and on the Calculus of Variations. In
the latter Chapter I have endeavoured to give an account
of the method for the determination of curves possessing
maxima and minima properties, and have applied it to some
of the more important examples. In the space allotted to
the subject I have only discussed the case of Single Integrals ;
the Student who requires further development is referred to
the works on the subject by Jellett, Carll, and Moigno.
Trinity CoLLEcE,
August, 1891.
TABLE OF CONTENTS.
—_—_—<>—_—_
CHAPTER I.
ELEMENTARY FORMS OF INTEGRATION.
PAGE
Integration, . . é A . 7 ‘ . 7 oa
Elementary Integrals, . : és 3 r : A em 2)
Integration by Substitution, : 3 » 5
a dz
Integration of a+ 2be + on” . . 8
Euler’s expressions for sin @ and cos @ deduced by Integration, . - 10
: a:
Integration of a . . . . - - +4
(x —p) Va-+ 2ba + cx®
dz
fo eee GY Pie ec 16
(@ + 2b% + cx?)2
do de
” ” oos8 and sin é’ . . . . . ° - 1
dé
” ” a+4bcose’ . . . . . . . e « 19
Different Methods of Integration, . , 3 : 20
Formula of Integration by Parts, . : ; . é e ‘ - 20
Integration by Rationalization, ; 3 . : . 23
Rationalization by Trigonometrical Transformations, ‘ 7 i - 26
Observations on Fundamental Forms, . . is . 5 29
Definite Integrals, . “ . : 5 . : 30
. - 36
Examples, . s ‘. s -
Vill Contents.
CHAPTER II.
INTEGRATION OF RATIONAL FRAOTIONS.
PAGE
Rational Fractions, ; a net> Se ee
Decomposition into Partial Hinieltovia, ‘ ‘ : : ‘ 3 - 42
Case of Realand Unequal Roots, . . . « « «© «© « 42
Multiple Real Roots, . 7 7 . . Fs : . . 47
Imaginary Roots, . ‘ . ‘ : ‘ ay ie a - 49
Multiple Imaginary Roots, . : é ‘ ‘i ; a : - 52
Integration of woasec . . 5 : 7 - 56
)™ (a — n
dx
” ” Seay? : . . . . - 58
amldy
” i) Seah . . : 2 3 s - . - 60
Examples, . . - . 3 s . . : : s - 61
CHAPTER III.
INTEGRATION BY SUCCESSIVE REDUCTION.
Cases in which sin @ cos" 6 d@ is immediately integrable, ‘ = . 63
Formule of Reduction for [sin 6 cos"@d@, . . . . F - 66
do
tannd’ ie . . . . ° . ° 72
Trigonometrical Transformations, . . . . ‘ 5 - 73
Binomial Differentials, . : 7 ° , ‘ : 3 e « 95
Reduction of fem=andz,. . . fs . . 5 ee: . 76
5 > J 2™ (loga)"dz, . . s . . 7 . - - 717
Integration of tan” 6 d@ and
ee » fa™cosaxdr, . . . . . . . . - 78
rr > fe costadz, . - A : . . . # om (79
35 », f cosa sin nada, ‘ . . . . . . . 79
Reduction by Differentiation, ‘ < 5 id 3 es f . 80
Contents. ix
PAGE
zm dxz
a + cx)» 2
\eaee
soon (ope
hres oer
| ecencae
Reduction of
a+ 2bu + ex")?
amdz
(a+ ou + 6a2)n? m
4 athe e 22 : ake. 84
Integration of f() =a aos 3 A . : , . . 86
(x) Va + 2b0 + ca?
Examples, . . . zi 5 ‘i F : - : 3 . 89
CHAPTER IV.
INTEGRATION BY RATIONALIZATION.
Integration of Monomials, . é Fe : i é : . o1
Rationalization of F(x, Va+ 2bx + cx*)dx, . i : : : » 92
d.
we eI a ke cage a a
(x) Va + 2ba + ox?
General Investigation, . . . .« «© «© w. «2 w 4 OF
Case of Recurring Biquadratic, : : ; : : - . 101
Examples, . . - . : 4 ‘ : : 3 - - 103
CHAPTER V.
MISCELLANEOUS EXAMPLES OF INTEGRATION.
Integral of some eae Y i . . r F . - 104
acosx+ bsing+e
Differentiation under the sign of Integration, . . Z ‘ . . 107
Integration under the sign of Integration, ; A : a : 109
‘a by Infinite Series, 4 3 7 3 5 2 - . 110
Examples, . , : : 3 ‘ j ‘i ‘ : 3 » 115
x Contents.
CHAPTER VI.
DEFINITE INTEGRALS.
Integration regarded as Summation, ‘ . . .
Definite Integrals. Limits of Integration, . . .
a 7
Values of [ sin" «dz and { cos*adt, +. . 5
0 0
ela
an 43 | sin" cos"xdz, when m and » are integers,
0
Sri AS i e*andz, when nis a positive integer, -
Taylor's Theorem, e 2 : : . :
Remainder in Taylor’s Theorem expressed as a Definite Integral,
Bernoulli’s Series, : : . A : ‘
Exceptional Cases in Definite Integrals, 7 : .
Case where f(x) becomes Infinite at either Limit, .
When f(x) hecomes Infinite between the Limits, . .
Case of Infinite Limits, . ss . .
Principal and General Values of a Definite Integral, .
Singular Definite Integrals, . . . . . °
° f(z)
~-— dz, . A : . é * é # .
I, F(z) ;
w gem
\, Tae —— dz, wheren>m,. . . i . ‘
oc atm 3
f, sa da, ‘ , - 3 3 - ‘
Differentiation of Definite Integrals, . 7 . .
Integrals deduced by Differentiation, . . .
Differentiation of a Definite Integral when the Limits vary,
Integration under the Sign of Integration, . . .
f e* dr, ‘i 7 Z i « é ‘ ‘ 7
0
PAGE
114
115
119
120
128
126
127
128
128
128
1380
181
1382
134
135
138
139
143
144
147
148
151
Contents. xi
AL
= cosmudx . a
\, oo fe es “ee Ga AY ok 1 mh Ge cdots
[, log (sino) ao, eat OO ke
O
Theorem of Frullani,* . i é “ ‘ 7 s : : . 155
Value of {. SS dx, 4 : ‘ . ‘ 3 . 157
Remainder in Lagrange’s Series, . : , . . ‘i : . 158
Gamma-Functions, ‘i : : < . . . . 159
r (m) r (7)
Proof of Equation B(m 1) =>
Petrie a Se ee
a 7 sin 27’
Value of (=) r (=) r (3)...r(*=), ‘ . é ‘ . 164
n n n n
Table of Log F'(¥), “ b : a 7 : : 3 - 169
171
Examples, . . é 7 “ , 2 : . .
CHAPTER VII.
AREAS OF PLANE CURVES.
Areas in Cartesian Coordinates, : é ‘ 3 : 3 - 176
The Circle, . i : e ‘ - 7 5 5 : s . 178
The Ellipse, 3 ‘5 ‘i ‘ - ‘ 5 A ‘ . 179
The Parabola, ‘ é ss 7 - 7 ‘ - : ‘ - 180
The Hyperbola, . ; , . 3 j . 7 7 i . 181
Hyperbolic Sines and Gosines, e : 182
The Catenary, . , : A : 183
188
Form for Area of a Closed Curve, .
The Cycloid, 3 ‘ : Z i : ‘ é - ‘ - 189
* This theorem was communicated by Frullani to Plana in 1821, and published after-
wards in Mem. del Soc. Ital., 1828.
Xi. Contents.
Areas in Polar Coordinates, : e
Spiral of Archimedes, . . s .
Elliptic Sector—Lambert’s Theorem, .
Area of a Pedal Curve, zi .
Area of Pedal of Ellipse for any Origin,
Steiner’s Theorem on Areas of Pedals,
5 on Roulettes, .
Theorem of Holditch, . 7 . .
Kempe’s Theorem, fs . e .
Areas by Approximation, = : .
Amsler’s Planimeter, . e . .
Examples, . P z i 7 is
CHAPTER VIII.
LENGTHS OF CURVES.
Rectification in Rectangular Coordinates,
The Parabola—The Catenary, . .
The Semi-cubical Parabola—Evolutes, .
The Ellipse, EE Ss . .
Legendre’s Theorem on Rectification,
Fagnani’s Theorem,
The Hyperbola, . ¢ 7 - a
Landen’s Theorem, ‘ : 7
Graves’s Theorem, é ‘ is .
Difference between Infinite Branch and ere for Hyperbola,
The Limacon—the Epitrochoid, . .
Steiner’s Theorem on Rectification of Roulettes,
Genocchi’s Theorem on Oval of Descartes,
Rectification of Curves of Double Curvature,
Examples, . . . . s
.
.
PAGE
190
194
196
199
201
201
203
206
210
211
214
218
222
228
224
226
228
229
231
232
234
236
237
238
239
243
246
Contents. xiii
CHAPTER IX.
VOLUMES AND SURFACES OF SOLIDS.
PAGE
The Prism and Cylinder, 2 : . % : . i : - 250
The Pyramid and Cone, . ‘ 7 : 3 ‘ . . 251
Surface and Volume of Sphere, . . . . . 7 i » 252
Surfaces of Revolution, . 3 s . é . : . 254
Paraboloid of Revolution, . . . . y és , s - 256
Oblate and Prolate Spheroids, . . 7 . * . - . 257
Surface of Spheroid, . ° : ‘ - . s . F - 257 |
Annular Solids, . . . ° . . . . . . » 261
Guldin’s ‘Theorems, . ‘ 4 2 : es . ; ; - 262
Volume of Elliptic Paraboloid, . . . . . . 7 - 265
Volume of the Ellipsoid, 5 : : 3 . 3 ‘i . - 266
Volume by Double Integration, . . s : J < 3 . 269
Double Integrals, : ‘ . a 5 3 - E P - 273
Quadrature on the Sphere, . . 3 7 ‘ . “ s - 276
Quadrature of Surfaces, 2 . ‘ Fi 7 * 3 3 . 279
Quadrature of the Paraboloid, 5 * A ‘ e é . - 280
Quadrature of the Ellipsoid, : ° . . . s - 282
Integration over a Closed Surface, 3 . é - 284
Examples, . 3 z 7 - ; : : ° . - 288
CHAPTER X.
INTEGRALS OF INERTIA.
Moments and Products of Inertia, . : 7 és : 7 « 291
Moments of Inertia for Parallel Axes, or Planes, . : . 292
Radius of Gyration, . ‘i , ‘ ‘ ‘ ‘ . 293
Uniform Red and Rectangular Lamina, . : - ‘ ef . 294
Xiv Contents.
Rectangular Parallelepiped, . ‘ :
Circular Plate and Cylinder, .
Right Cone, . 5 : . 3
Elliptic Lamina, . . 3
Sphere, . . . . . 7
Ellipsoid, . : . : a :
Moments of Inertia for a Lamina, .
Momental Ellipse, . . : .
Products of Inertia for a Lamina, . ‘
Triangular Lamina and Prism, ; ‘
Momental Ellipse of a Triangular Lamina,
Tetrahedron, . ; :
Solid Ring, . . .
Principal Axes, . 0
Ellipsoid of Gyration, .
Momental Ellipsoid,
Equimomental Cone, . :
Examples, . . 7 .
CHAPTER XI.
MULTIPLE INTEGRALS.
Double Integration, A - 7
Change in Order of inteeritin, ‘ :
Dirichlet’s Theorem, . F ‘
Transformation of Multiple item .
Transformation of Element of a Surface,
General Transformation for ” Variables, .
Green’s Theorems, . ‘ 7 .
Application to Spherical ee :
Laplace’s Theorem on Expansion by Spherical Harmonics,
Examples, . . . . . ‘
PAGE
295
295
296
296
297
298
299
300
301
302
304
304
305
307
309
3809
310
311
313
314
316
321
324
325
326
332
838
342
Contents.
CHAPTER XII.
ON MEAN VALUE AND PROBABILITY.
Mean Values, c
Case of one independent Visrialile,
Case of two or more Independent Variables,
Probabilities,
Buffon’s Problem,
Curve of Frequency,
Errors of Observation, .
Lines drawn at Random,
Application of Probability to the Dateien of Definite Pibainas
Examples,
CHAPTER XIII.
ON FOURIER’S THEOREM.
Fourier’s Theorem,
Examples, . 3 zi F : 3
CHAPTER XIV.
ON LINE AND SURFACE INTEGRALS.
Integration over a Plane Area,
Stokes’ Theorem, . ‘ ‘
Example in Solid Harmonics,
Lemma on Solid Angles,
Differential Equations connecting Solid daidios,
Neumann’s Theorem, . ee fs
Integrals connected with the ‘ehaony of Atsanition,
xXV
PAGE
346
347
350
352
365
369
374
381
384
387
392
399
401
403
405
405
407
408
409
xv Contents.
CHAPTER XV.
CALCULUS OF VARIATIONS,
Definition of Variation By, . :
Variation of a Definite Integral,
Caseof Vaf(rt, 4,47, ¥), - .
Maxima and Minima, .
Case of Geometrical Pasniotoua ‘
V a Function of y, 7, ¥ solely,
Case where V contains the Limits,
Case of Two Dependent Variables,
Relative Maxima and Minima,
Equations of Condition, A
Case of Arc being the Independent Taviale, é
Maxima or Minima of [uds, . 5 7
Lines of Shortest Length,
The Brachystochrone, S -
Minimum Surface of Revolution, .
Properties of the Catenary,
General Transformation,
Criterion for Maxima and Minima,
Application to V = f(x, y, ¥), 7
Examples, . - : ‘5 b 3
MiscELLANEOus ExaMPLes, : :
PAGE
413
415
417
419
421
422
423
423
426
426
426
426
430
432
436
438
443
444
445
447
449
The beginner is recommended to omit the following portions on the jirst
reading :—Arts. 46, 49, 50, 72-76, 79-81, 89, 96-125, 132, 140, 142-147, 149,
158-167, 178, 180, 182, 189-193, Chapters x., xI., XII., 3
INTEGRAL CALCULUS.
—_—__
CHAPTER I.
ELEMENTARY FORMS OF INTEGRATION.
1. Integration.—The Integral Calculus is the inverse of
the Differential. In the more simple case to which this
treatise is principally limited, the object of the Integral
' Calculus is to find a function of a single variable when its
differential is known.
Let the differential be represented by F(x) dx, then the
function whose differential is F(x) dw is called its integral, and
is represented by the notation
| F(z) da.
Thus, since in the notation of the Differential Calculus we
have
df (a) =f’ (@) da,
the integral of /’(«) dx is denoted by f(a) ; i.e.
[7@ ae =7(0).
Moreover, as f(z) and f(x) + C (where C is any arbitrary
quantity that does not vary with ) have the same differen-
tial, it follows, that to find the general form of the integral of
J’ (a) dz it is necessary to add an arbitrary constant to f(z) ;
hence we obtain, as the general expression for the integral
in question,
[@ae-s@) +6 (1)
[1]
2 Elementary Forms of Integration.
In the subsequent integrals the constant C will be omitted,
as it can always be supplied when necessary. In the appli-
cations of the Integral Calculus the value of the constant is
determined in each case by the data of the problem, as will be
more fully explained subsequently.
The process of finding the primitive function or the inte-
gral of any given differential is called integration.
The expression F(x) dv under the sign of integration is
called an element of the integral; it is also, in the limit, the
increment of the primitive function when 2 is changed into
x + dx (Diff. Cale., Art. 7); accordingly, the process of inte-
gration may be regarded as the finding the swm* of an infinite
number of such elements.
We shall postpone the consideration of Integration from
this point of view, and shall commence with the treatment of
Integration regarded as being the inverse of Differentiation.
2, Elementary Integrals.—A very slight acquaint-
ance with the Differential Calculus will at once suggest the
integrals of many differentials. We commence with the
simplest cases, an arbitrary constant being in all cases under-
stood. ,
On referring to the elementary forms of differentiation
established in Chapter I. Diff. Cale. we may write down at
once the following integrals :—
| eae = : |= a (a)
m+ 2” (m—1)a"
dx
[= bos ©. ®)
| sappanac. Oe mie | a (0)
m m
dx dx
lors = tan 2, |= =- cots. (a)
* Tt was in this aspect that the process of integration was treated by Leib-
nitz, the symbol of integration f being regarded as the initial letter of the word
sum, in the same way as the symbol of differentiation @ is the initial letter in
the word difference.
Fundamental Forms. 3
de I 2
| Tag ny en's (7)
(9)
These, together with two or three additional forms which
shall be afterwards supplied, are called the fundamental* or
elementary integrals, to which all other forms,t that admit
of integration in a finite number of terms, are ultimately re-
ducible.
Many integrals are immediately reducible to one or other
of these forms: a few simple examples are given for exercise.
Examp.es.
| 7 Ans, — *.
az ie
2. le ” 2 we
tan # dz. 9 ~ log (cos 2).
I
» ob log (a + ban),
a : tan"! (+) :
a+ bx?" ” J ab a
9» Sec 0.
5. | wae ny Vie
I
eat da, ” a en,
* The fundamental integrals are denoted in this chapter by the letters a, 5, ¢,
&c.; the other formule by numerals 1, 2, 3, &c. ae
t+ By integrable forms are here understood those contained in the elementary
portion of the Integral Calculus as involving the ordinary transcendental func-
tions only, and excluding what are sc all and Hyper-Elliptic functions.
la
4 Elementary Furms of Integration.
da 3
9. \F Ans. - =.
2 ak
2
Io. [= » ne”.
n-1
a”
Il. | oe M x log (% —- a).
uw—a
3. Entegral of a Sum.—It follows immediately from
Art. 12, Diff. Calc., that the integral of the sum of any number
of differentials is the sum of the integrals of each taken sepa-
rately. For example—
f(4a™+ Bu" + Ca + &e.) dv=A fa” da+ BS ade + Of at dae+ &e.
Agim » Bam ; On"
mt+iI n+iI r+itI
+ &e. (2)
Hence we can write down immediately the integral of any
function which is reducible to a finite number of terms con-
sisting of powers of x multiplied by constant coefficients.
Again, to find the integrals of cos’*dz and sin’wdx; here
| costede=| I + COS 24 pe 2 ‘ sin ms (3)
2 2 4
| site | Ee cre ot seo an (4)
2 2 4
A few examples are added for practice.
EXAMPLES.
(1 -— 2)2de igi
I, j—. Ans. ge aa
(% — 2) dz _ 4
2. == é arr
| an/ @ OME Te
3. f tan?adx =f (sec? # — 1) de. » tang—-x,
Integration by Substitution. 5
sin(m+n)x sin(m —n)ax
: cos mx cos nx dx. ; A -
4. Jf cos mz cos nz ns aie) ne
ee fies jp Cs ee ae
2(m — n) 2 (m+n)
a+24 ? x Ss
6. | {eteae. 55 aT ~S/@— a
_ Multiply the numerator and denominator by «/a + @.
eno! 3
1. fa/az+adz. Ans. 2 (@ +a) — 2a (ea)
a:
8. as 3 = (@ + ie 2’) ‘
J arar fa 3a
Multiply the numerator and denominator by the complementary surd
Satra—V/z.
at bx bz ab’ — ba’ aes
| the. Ans, rua —Fa — lee (a’ + B'x).
at+tbe 5b ab’ — ba’
Here w4ee TE (a + Ox)”
4. Integration by Substitution.—The integration of
many expressions is immediately reducible to the elementary
forms in Art. 2, by the substitution of a new variable.
For example, to integrate (a + bx)” dx, we substitute s for
a+ ba; then dz = bdz, and
sds os" (a + bah
[@ eae -\"5 —(n+1)b (n+1)b°
Again, to find
x de
(a + bx)”
we substitute s for a + ba, as before, when the integral be-
comes
z{" - aee
6 3”
I I 2a % a
a ~ 8 fe =3)8"9 (n—2)e* (n-3) a)
6 Elementary Forms of Integration.
On replacing s by a + be the required integral san be ex-
pressed in terms of x.
The more general integral
x™ de
\e + bx)”
where m is any positive integer, by a like substitution be-
comes
1 | (s -— a)" dz
pen | get
Expanding by the binomial theorem and integrating each
term separately the required integral can be immediately
obtained.
Again, to find
da
a” (a + be)”
we substitute z for - + 6, and it becomes
£3 qmntn-l gn ?
I | (3 — b)™" dg
which is integrable, as before, whenever m+ is a positive
integer greater than unity.
Thus, for example, we have
da _ 1 x
lz +b) a 8 (; + a}
It may be observed that all fractional expressions in which
the numerator is the differential of the denominator can be
immediately integrated.
For we obviously have, from (6),
[fede _
i f(z)
og f(z). (s)
, dx
Integration of Foe 7
Examp.es.
_Sin de log (a + 6 cos 2)
a (= Ate b :
dx I a\*
2, es - sin (=
IFS ae (:)*
1 dz I
3: j Ogu. ” = (log a).
dz
4. | alone » log (log 2).
P i xv dz log (a+ br) 302+ 4abx
(a + bays" ” os 28 (a + ba)e”
6 j dx 2, a+bz a+ 2be
; x (a + ba)? a ee aa (a + bz)
y j adz 2(a+ ba)? 2a (a + bz):
. (a qs bayi" ” 3 = a °
8 j zdx 3(a+ bx)8 3a(a + bx)i
. (a + ba)8 ” seOt~t~—<“is—stSC‘“
9 | dx 2 ae (2 -a
eS arf rae — ae "a a”
Assume 2a” — a? = 2, then adw = zdz, and the transformed integral is
2dz
J a? + ae
dee
5. Integration of ——..
“-a
. I I I I
Since —— = — san
v’-a@ 2ale-a a«ta/’
da I L-a
we get -=— ‘ h
8 IE -@ 24a L+a (%)
This is to be regarded as another fundamental formula
additional to those contained in Art. 2.
da
8 Integration of a ole eee
In like manner, since
aa eee eae
da I w-a
we have = lo ( 6
lexjerm ssp o
EXAMPLEs.
dz I %- 3
I \e55 Ane GUE
dx I z-3
eae 0 ye rae
az a
* ape ae ” rary
- 2
4. |= : pet OEM
eas w/3 «+73
dz
6. Integration of —————..
5 a+ 2b4 + cx?
This may be written in the form
eda .
(cx + 6)? + ae — B°’
or, substituting s for cx + b,
dz
s+ac—b*
This is of the form (/) or (2) according as ac — 8 is positive
or negative.
Hence, if ac > 6? we have
dx 2 I ee b
a+2be+ce | /ao_# See (7)
(p + gu) dx
Integration of a oie
If ae < 2,
{ dz 7 1 peeve (8)
Ja+2bu+ cx »,/R ge om +b +/P a0
This latter form can be also immediately obtained from (6).
In the particular case when ac = 0’, the value of the inte-
gral is j
-1
cx +b
(p + qx) da
. Integration of ————..
7 . a+ 2ba + ca
This can at once be written in the form
gq (b+cx)dx pe-—gd dx
= + '
Cat 2bu + cx e at 2be+ ca?
The integral of the first term is evidently
a 2
= log (a + 2b + cx’),
while the integral of the second is obtained by the preceding
Article.
For example, let it be proposed to integrate
(w cos 0 - 1) dw
a — 20 cos 0+1
The expression becomes in this case
cos 8 (2 — cos 0) dx sin’ Odx .
z’—2xcos0+1 («# - cos@)*+sin?9’
hence
BE ea ee
| Sae? 2 log (a? - 22 cos 8 + 1)
xz — cos
— sin 6 tan? —
sin 0
10 Elementary Forms of Integration.
When the roots of a + 2b¢ + cx* are real, it will be found
simpler to integrate the expression by its decomposition into
partial fractions. A general discussion of this method will
be given in the next chapter.
ExaMP.Les.
da 2041
I. \ “ Ans. —. tan ( =).
I+a+ a? V3 V3
{ dau z ie 22-1 ae -1tV75
Ze . ae
1+2—2? ee V5 5 af sf
ax SF
o |e ” = log (==).
di
4. |= ae » tan"l(z + 2).
dx I 5ut+2
fans tage
= Fares gg
dz I 1 + 23
6. \ ae 6 j los (: = =).
( edz I 1 vt—3
a 28 — at 6° 9 30°°8 Natya]
dx
cea -1 ox
8. j oe x, tan} (2-1).
8. Exponential Walue for sin 9 and cos §.—By com-
paring the fundamental formule (7) and (A) the well-known
exponential forms for sin 6 and cos @ can be immediately
deduced, as follows:
Substitute ¢,/— 1 for # in both sides of the equation
oe Soe + const. ;
r-@ 2 8\t—e me
and we get
I ae apes pera + const. ;
1+2 - a Ola sea
Exponential Forms of sin 0 and cos 0. 11
I ; ( tte -1
infot OO saan
Now, let z = tan 0, and this becomes
am oe ( +> 1 tan ‘) + const.
-1
1 ~ /-1 tan 0,
When @ = o, this reduces to o = const.
or, by (f), tan*z = ) + const.
p=
2
Hence V7 = eae feet = (cos 6 + /— 1 sin 6),
cos 9 — ./—1 sin@
or eV = cos 0 + /—1 sin 6,
eV = cos 0 — \/—1 sin 8.
dx
g. Integration of
Jat &
Assume* VUe+ePH8-2,
then we get + @ = 8 — 222,
dz
s-@ 8’
hence (8 — #) dz = zda, or
: dee ds ae
ve | Fea [F = lo82 = tog (2+ fa ta’), (i)
This is to be regarded as another fundamental form.
By aid of this and of form (e) it is evident that all ex-
pressions of the shape
dx
/ a+ 2bx + cx?
* The student will better understand the propriety of this assumption after.
reading a subsequent chapter, in which a general transformation, of which the
above is a particular case, will be given.
12 Elementary Forms of Integration.
can be immediately integrated; a, b, c, being any constants,
positive or negative.
The preceding integration evidently depends on formula
(t), or (é), according as the coefficient of x is positive or
negative.
Thus, we have
d.
lS bes (ae + 0+ Ale rae) 79)
A a+ 2ba + cx
(ree (11)
¢ being regarded as positive in both integrals.
When the factors in the quadratic a + 2b2 + cx’ are real,
and given, the preceding integral can be exhibited in a
simpler form by the method of the two next Articles.
dx
V @ = «)(@ — B)
10. Integration of
Assume e-a=2", then de = 22d;
dx
% = 2d8 3
J # —-a :
ad
hence : ae
eM) Janae
| dz | dz
J (@ — a)(a = B) Veta-B
= 2 log (2 + /#* +a — 8), by (‘),
de
| FeeaeTH MSR. Oe)
or
Exponential Forms of sin 0 and cos 0. 13
dx
11. Integration of = ————_..
Sema Baa)
As before, assume 2 - a = g*, and we get
dau Zz 2dz
V(@-a)(B-2) fB-a-2
Hence, by (e),
dz sic Je = @
TS- 2 sin = (13)
Otherwise, thus :
assume 2 =a cos’) + 8 sin’@,
then B -«=(B — a) cos’9,
dx = 2((3 — a) sin 0 cos 00;
= = 2d0;
h ces Gores Se
ma J (@ = a(R -2)
vw-—a
12. Again, as in Art. 7, the expression
(p + qu) de
/ a+ 2b% + cx
e~a=( —a)sin’6,
and
can be transformed into
q (b+ cx)dx , be- 9b da
C fa + 2bx + cx? C /a+ abe + cx
and is, accordingly, immediately integrable by aid of the
preceding formule.
14 Elementary Forms of Integration.
EXxaMPLes.
I. \__ Ans. 2 log (\/2 +f — 4).
=
é 35. 2s
7 —
3. y= PP 2sinl)/z—I.
4. aS ih log (20 +1 +2\/1+@ +0).
5 | [tgee J G@+t aet d) + (a- 8) log (\/atat/atd)
Multiply the numerator and denominator by »/z + a.
6. j eee Ans. sin“! pad a.
Jems We
(—*_. a FET
J (a+ bx) (a’ — b’a) WAT ab! + ba’
8. Show, as in Art. 8, by comparing the fundamental formule (e) and (i),
that
c0s0 +4/—1 sing = 1,
d.
13. Entegration of SS
(a — p)/ a + 2ba+ cx
Let # - p= *, then
at page te ,
L-p gz s
. l dx -| — dz
"" \ (@ —p)/a+2be + ca J /as?+ 2bs(1 + pe) +e(1 + ps)?
eee
Weewrree td
where @=0c, V=b+ep, C=at 2bp t+ ep’
The integral consequently is reducible to ( 10), or (11), ace
cording as ¢’ is positive or negative.
daz
Integration of ———. 15
. of (a + cx’)a
Examples.
d:
I. | eS dns. > cos () :
ar/ x — & a x
7.
da r+ a
» 5 ha es),
a ea &
3 dx es
lage fF : r+ a"
dz I “
4. |=. Ans. —~ log (——_ 7.) 7
ar/ at 2bx + ox? fa a+ but/ar/ at 2bu + 0x
dx I bz —a
5: | ey a ees Ans, — sin7l (=) .
US cx? + 2ba —a Va arf ac+
4 =
6. | canes, * ait sin! )
G+ a)fiten—e /2 Lae
= | —— - sin-! ( Eye )
(I+ a) f/ita—a2 . (c+ 2)V5
14. The transformation adopted in the last Article is one
of frequent application in Integration. It is, accordingly,
worthy of the student’s notice that when we change # into
I de dz si : 1 dz dz
— we have — =——; and, in general, if a” =-, —=-—.
s a 8 ze NB
These results follow immediately from logarithmic differ-
entiation, and often furnish a clue as to when an Integration
is facilitated by such a transformation.
For example, let us take the integral
| da
a(a+ ba")
Here, the substitution of - for a” gives
£ dz
njaz+e
16 Elementary Forms of Integration.
The value of which is obviously
1 I a”
cle log (az + 8), or F log (- " a) :
Again, to integrate
da
arf ac” +b
I ; :
assume «” = =, and the transformed integral is
2.8
0) /a+ be
This is found by (e) or (7) according as 0 is positive or
negative.
dat
. Integration of ———.
~ E (a + cu?)i
I .
Let # = S and the expression becomes
edz
(az® + c)**
the integral of this is evidently
alas + oP ala + oak
dx x
| eran” aera “a
16. To find the integral of
dx
(a+ 2b + cx*\¥
Hence
This can be written in the form
dx
(ac — 0 + (cw + by}
which is reduced to the preceding on making cv + b = 8.
dé
Int q a
ntegration of sn 0" 17
Hence, we get
| da = b+ cx
(a+ 2be + cx*)8 ~ (ac— 0) (a + 2ba + cx*)¥ (15)
Again, if we substitute - for 2,
ade b - dg
——_———,, becomes —,—_—_—_,
(a + 2ba + cx’)! (as + 2bz + ¢)?
and, accordingly, we have
| ada —_ a+ be
(a+ 2ba + cx*)§ (ae — B)(a + 2bu + ca*)P
Combining these two results, we get
| (p+ ge)de _ bp - ag + (p ~ bq) @ (16)
(a + 2ba + ca*)§ (ac — b)(a + 2bx + ca”)?
do
17. Integration of ——. and —...
7 6 sin 6 cos 0
It will be shown in asubsequent chapter that the integra-
tion of a numerous class of expressions is reducible either to
do ‘ :
that of any ot of aca accordingly propose to inves-
tigate their values here. For this purpose we shall first find
; a0
the integral of ano eoee.
nm
dé cos’? d(tan 8) |
Ener sin@cos@ tand tan@ ’
a6
consequently | a coed log (tan 6). (17)
[2]
18 Elementary Forms of Integration.
Next, to find the integral of
dé
sin 0”
This can be written in the form
do
9
2 sin— cos —
2 2
and, by the preceding, we have
d0 0
lan = log (tan 3) (18)
Again, to determine the integral of = we substitute
os 8
dp
- ¢ for 0, and the expression becomes a : the integral
of this, by (18), is
— log (tan 3} or log (cot $) or log ot ( - 5). 7
Accordingly, we have
| = log feot (= - -)| = log lean (= + al (19)
This integral can also be easily obtained otherwise, as
follows :—
\So- (oF -[“S?
cos 8 cos’ 6 cos’@ *
Let sin 0 = x, and the integral becomes
de 2M (S*2)- 21 1+sin 0
jr-#@ 2 8\r—-a@/° 2°38 od)
The student will find no difficulty in identifying this
result with that contained in (19).
do
Integration oat See 19
dé
a+bcos?
This can be immediately written in the form
do
18. Entegration of
>
(a + 6) cost S + (a — 6) sin’?
soot dd
or —_—_—_*——__:
a+b+(a—0) tants
on substituting s for tan? this becomes
2da 2 ue
a+b+(a—b)zx* a * a
Consequently, by Ex. 6, Art. 2, we get ba
(1) when a > 8,
dé
laces" WS ont) feo tans ta a (20)
(2) when a < 8, oe hate (h),
~ de ; i a
lima (2)
| eat f
If we assume a = b cosa, we deduce immediately from
the latter integral
‘ a-0
| dd I oe
————— = - log ——: Po
2
cosa +cos@ sina |
cos
The integral in (20) can be transformed into
| do I = {ote cost
——_—; = ——— cos! }—__——__}.
a+beosO ,/@_—% (a+ cosd
[24] 7
20 Elementary Forms of Integration.
In a subsequent chapter a more general class of integrals
which depend on the preceding will be discussed.
19. Methods of Integration.—The reduction of the
integration of functions to one or other of the fundamental
formule is usually effected by one of the following methods :—
(1). Transformation by the introduction of a new va-
riable.
(2). Integration by parts.
(3). Integration by rationalization.
(4). Successive reduction.
(5). Decomposition into partial fractions.
Two or more of these methods can often be combined
with advantage. Itmay also be observed that these different
methods are not essentially distinct: thus the method of
rationalization is a case of the first method, as it is always
effected by the substitution of a new variable.
We proceed to illustrate these processes by a few ele-
mentary examples, reserving their fuller treatment for sub-
sequent consideration.
20. Integration by Transformation.—Examples of
this method have been already given in Arts. 4, 10, &c. One
or two more cases are here added.
Ex. 1. To find the integral of sin*# cos*# dz.
Let sin # = y, and the transformed integral is
3 5 SaaS, Ind
2(1 — y?)di -| 2 -| ya See ee
[ro -v)a-|ray-|ya-E-£- 2 -
edz
Ex. 2. \ae
Let e* = y, and we get
(_ 3 5
| ae tan y = tan“ (e*).
21. Integration by Parts.— We have seen in Art. 13,
Diff. Calc., that
d(uv) = udv + vdu ;
hence we get
uv =f udv + f vdu,
or (ud = wo — f vdu (22)
Integration by Parts. 21
Consequently the integration of an expression of the form
udv can always be made to depend on that of the expression
odu.
The advantage of this method will be best exhibited by
applying it to a few elementary cases.
ada
gf toa
=a@sintat /1 — 2.
Ex. 1. |siwtede = esin'e—[
Ex. 2. iE log x de.
2
Let a=loga,v= =, and we get
2 2
[etog 2a ~ ERB Efe = 2 (loge - 2)
2 a) @ 2 2
Ex. 3. | e* wdx.
ett
Let o= th, =e, then
az ax ax
[era ae =( 2 }
a a a a
Ex. 4. | e* sin max dz.
Let sin ma = u, — = », then
é e gin mz m
|e sin mx dx = =a m fee cos mada.
e cos mx
m ;
+—|e“ sin ma da.
a a
Similarly, [es cos max da =
.
22 Elementary Forms of Integration.
Substituting, and solving for { e* sin ma dz, we obtain
; e** (asin maz — m cosma
e* sin mada = Se ee (23)
ae+m
In like manner we get
e* (@ cos ma + m sin ma
e™* cos mx dx = eG contin mh signe) . (24)
atm
Ex. 5. |v a+ x dx.
Let /@ +e =u, then
| Ver F ae 2 /ere -| ae
JV +a
We) 2
also |[Vereae= a(S + |=.
J a+e Ja +e
Hence, by addition, and dividing by 2,
72 yk 2
| Jere ae - wes + < log (e+ Ja +2). (25)
Ex. 6. es (a + A/a + a) de.
Here [tog (w+ A/a £0) dx = x log (w+ Se + a).
7 vdxz
l\7-z
=alog (a+ f/eta)— feta’. (26)
Integration by Rationalization. 23
EXAMPLEs,
I [ eto ade MO :
5 gs e ns. Ei (iog2——".).
I
2. | tan“ 2 dz. » & taney — = log (1 + 2”).
a
3 | 2 tan? edz. »» «tan x + log (cosz) — 2
sin! x dx ‘ew sina 1
. = — 72
an ” Jip ee x),
Let z= sin y, and the integral becomes
Yee = ,
J a cos?y | y a (tan y) = y tan y + log (cos y).
ys | er 2? da. yy 6% (a? — 2% 4 2).
22, Integration by Rationalization.—By a proper
assumption of a new variable we can, in many cases, change
an irrational expression into a rational one, and thus inte-
grate it. An instance of this method has been given in
Art. 9:
The simplest case is where the quantity under the radical
sign is of the form a + dx: such expressions admit of being
easily integrated.
For example, let the expression be of the form
x” da
(a + bx)?
where 2 is a positive integer. Suppose a + bx = 8’, then
28dz 2-a
da = 5? and # = b
making these substitutions, the expression becomes
2 (2? — a)"ds
en
24 Elementary Forms of Integration.
Expanding by the Binomial Theorem and integrating the
terms separately, the required integral can be immediately
found. It is also evident that the expression es can
(a + bx)¢
be integrated by a similar substitution.
vm dg
5 ti f ——
23. Integration o (a4 ca"?
where m is a positive integer.
sdz v-a
Let a + cz = 3’; then adz = a ge = ; and the
transformed expression is
(3? - a)™ dz
cml
This can be integrated as before. It can be easily seen
gem
that the expression is immediately integrable by
r
(a + cx*)3
the same substitution.
A considerable number of integrals will be found to be
reducible to this form: afew examples are given for illustra-
tion.
EXAMPLes.
3 — 42\3
t. { sii ia dg BEE omy,
Vt -x2 3
dz 25 228 —_——
Ze | ——. ” —-—+2; wherez=1/i1+a%.
Si+e 5 3
Bdx — (24 + 30x?)
| = erent
(a + cx?)? 30° (a + cx?)
24. It is easily seen that the more general expression
S (x*) xdaz
sf @ + cu?
where f(x) is a rational algebraic function, can be ration-
alized by the same transformation.
2 : fj dx
integration of (A+ Cx") (a+ ca?) 25
a 8 I :
Again, if we make x = — the expression
&
daz
x" (a + cx?)
transforms into
lds |
(az? + c)h’
and is reducible to the preceding form when n is an even posi-
tive integer.
Hence, in this case, the expression can be easily integrated
by the substitution (a + cx’)? = ay.
It will be subsequently seen that the integrals discussed
in this and the preceding Articles are cases of a more general
form, which is integrable by a similar transformation.
EXxaMPLEs.
az SV -1I ni 7
il ae Ans. 3a (2a + 1}.
2 4
jana 8 JAE) 4+}.
26 (1 + 2°)A 15% a xt
dx
25. Iniegration of (4+ Oa")(a + ca")P
As in the preceding Article, let (@ + cx’) = az, or
a+ ca’ =2°2': then, if we differentiate and divide by 22, we
shal} have
d.
edz = sda + xz dz, or eee se
UB o-%
dx dz
a poe” (27)
"* (a+ cn) cs?’
and the transformed expression evidently is
a
(Ac — Ca) — Az®
26 Elementary Forms of Integration.
This is reducible to the fundamental formula (A), or (/),
according as is positive or negative.
- Ca
A
I is (4s + cx”) + arf Ac — es (28)
2/ A (Ac — Ca) J Ala + ca?) — #/ Ac — Ca
Ac - Ca
A
cC¢-
A
Hence, (1) if zh
>o, the integral is easily seen to be
(2). If < 0, the value of the integral is
I i e/ Ca— Ac
/ A(Oa- Ac) / A(a + cx”)
ExaMpe.es.
dz af 2
1 eee +R aE Ans. Fr tan! (5 = =)
ang ee
(3 + 427) (4- 327) Wi VW 12 = gx
j dx Hie a 3+ 4a + 50
3 (4 — 32°) (3 + 4a?)8 ” 20 Sata + 4u* — 5%
26. Rationalization by Trigonometrical Trans-
formation.—It can be easily seen, as in Art. 6, that the
irrational expression ,/a + 2bx + cx* can be always trans-
formed into one or other of the following shapes:
(1) (a?—s")4, (2) (a? + 2°), (3) (8 - @)AS
neglecting a constant multiplier in each case.
Accordingly, any algebraic expression in # which con-
tains one, and but one, surd of a quadratic form, is capable
of being rationalized by a trigonometrical transformation :
the first of the forms, by making z =a sin 8; the second, by
z=atan@; and the third, by z = a sec 0.
Rationalization by Trigonometrical Transformation. 27
For, (1) when s= a sin 0, we have (a’ - s*)4 = a cos 0, and
dz = a cos 0d0.
(2). When s = a tan @,.... (a°+ #)4=a sec 0, and
ad@
cos’ 0"
3). When s =a secO,.... (s°- a’)! = a tan 6, and
dz = a tan 0 sec 00.
A number of integrations can be performed by aid of one
or other of these transformations. In a subsequent place this
class of transformations will be again considered. For the
present we shall merely illustrate the method by a few ex-
amples.
EXAMptes.
; ( du
i aw ( + 2°)¥
Let z = tan @, and the integral becomes
sin?@ 6) sin?@0—Sts SiN x
je ode = {2% 6) I Vita
i dz
2. 7
(a? - 22)2
Let =a sin 6, and we get
=| a6 tan @ =
a} cos*@ a gt / gt
This has been integrated by another transformation in Art. I5.
dx
. jaan Caan
Let # = sec 0, and the integral becomes
sin@cos@ @
—— +7:
| cos? @d@; or, by (3) Art. 3, :
accordingly, the value of the integral in question is
fet
— +: - sec"!z.
2x
28 Elementary Forms of Integration.
ee tan” 2 dx
4. <<
(1 +2)?
Let 2 = tan 6, and we get
78 (a cos @ + sin 8)
\ cos 66% d0; or by (23), It+@
-1 21
dz ettan 2% (a + 2) eatan 2
Hence | = + __—.
{I+ 23 = (1 + @)(1 + 2)
a
5. { ae sins ( = ) 3
at+aZz
* = sin? 6, or « =a tan?9, and the integral becomes
x
af@d(tan?6), or af@d(sec?@): (since sec?@ = I + tan?9),
Integrating by parts, we have ;
J 0d (sec? 0) = @ sec? 6 — f sec? 6 dd = @ sec? @ — tan 6:
hence the value of the proposed integral is
(a + 2) tan} (2) — (ax).
It may be observed that the fundamental formule (e) and (/) can be at once
obtained by aid of the transformations of this Article.
S 27. Remarks on Integration.—The student must
not, however, take for granted that whenever one or other of
the preceding transformations is applicable, it furnishes the
simplest method of integration. We have, in Arts. 9 and 13,
already met with integrals of the class here discussed, and
have treated them by other substitutions: all that can be
stated is, that the method given in the preceding Article will
often be found the most simple and useful. The most suit-
able transformation in each case can only be arrived at after
considerable practice and familiarity with the results intro-
duced by such transformations.
By employing different methods we often obtain integrals
of the same expression which appear at first sight not to
agree. On examination, however, it will always be found
that they only differ by some constant ; otherwise, they could
not have the same differential.
Observations on Fundamental Forms. 29
28. Higher Transcendental Functions.— Whenever
the expression under the radical sign contains powers of x
beyond the second, the integral cannot, unless in exceptional
cases, be reduced to any of the fundamental formule; and
consequently cannot be represented in finite terms of a, or of
the ordinary transcendental functions: i.e. logarithmic, ex-
ponential, trigonometrical, or circular functions. Accord-
ingly, the investigation of such integrals necessitates the
introduction of higher classes of transcendental functions.
Thus the integration of irrational functions of x, in which
the expression under the square root is of the third or fourth
degree in w, depends on a higher class of transcendentals
called Elliptic Functions.
29. The method of integration by successive reduction is
reserved for a subsequent place. The integration of rational
fractions by the method of decomposition into partial frac-
tions will be considered in the next chapter.
30. Observations on Fundamental Forms.—From
what has been already stated, the sign of integration ({) may
be regarded in the light of a question: i.e. the meaning of
the expression f F(x) dx is the same as asking what function
of w has F'(«) for its first derived. The answer to this ques-
tion can only be derived from our previous knowledge of the
differential coefficients of the different classes of functions, as
obtained by the aid of the Differential Calculus. The number
of fundamental formule of integration must therefore, ulti-
mately, be the same as the number of independent kinds of
functions in Algebra and Trigonometry. These may be
briefly classed as follows :—
Pp
(1). Ordinary powers and roots, such as 2”, x7, &e.
(2). Exponentials, a”, &c., and their inverse functions;
viz., Logarithms.
(3). Trigonometric functions, sin z, tan 2, &c., and their
inverse functions; sin“, tana, &e.
This classification may assist the student towards under-
standing why an expression, in order to be capable of inte-
gration in a finite form, in terms of 2 and the ordinary
transcendental functions, must be reducible by transforma-
tion to one or other of the fundamental formule given in
30 Elementary Forms of Integration.
this chapter. He will also soon find that the classes of in-
tegrals which are so reducible are very limited, and that the
large majority of expressions can only be integrated by the
aid of infinite series.
The student must not expect to understand at once the
reason for each transformation which he finds given: as he,
however, gains familiarity with the subject he will find that
most of the elementary integrations which can be performed
group themselves under a few heads; and that the proper
transformations are in general simple, not numerous, and
usually not difficult to arrive at. He must often be prepared
to abandon the transformations which seemed at first sight
the most suitable: such failures are not, however, to be con-
sidered as waste of time, for it is by the application of such
processes only that the student is enabled gradually to arrive
at the general principles according to which integrals may be
classified.
Many expressions will be found to admit of integration
in two or more different ways. Such modes of arriving at
the same results mutually throw light on each other, and will
be found an instructive exercise for the beginner.
31. Definite Integrals.—We now proceed to a brief
consideration of the process of integration regarded as a sum-
mation, reserving a more complete discussion for a subsequent
chapter.
If we suppose any magnitude, wv, to vary continuously by
successive increments, commencing with a value a, and termi-
nating with a value 3, its total increment is obviously repre-
sented by 3—a. But this total increment is equal to the sum
of its partial increments; and this holds, however small we
consider cach increment to be.
This result is denoted in the case of finite increments by
the equation
B
= (Au) = 6 -a;
and in the case of infinitely small increments, by
[ia ee . (30)
a
Definite Integrals. 3l
in which (9 and a are called the limits of integration : the
former being the superior and the latter the imferior limit.
Now, suppose « to be a function of another variable, 2,
represented by the equation
u=f(e):
then, if when a = a, u becomes a, and when z = b, w becomes
3, we have
a=f(a), B =/().
Moreover, in the limit, we have
du =f" (a) dx,
neglecting* infinitely small quantities of the second order
(See Diff. Cale., Art. 7).
Hence, formula (30) becomes
[7@ 4-70-70) (31)
in which 8 and a are styled the superior and the inferior limits
of x, respectively.
3b
Tt should be observed that the expression | J’(«) da, re-
&
presents here the dimit of the sum denoted by = (/"(z) Az),
when Az is regarded as evanescent.
In the preceding we assume that each element /’(x) dv is
infinitely small for all values of x between the limits of inte-
gration a and 0; and also that the limits, a and 8, are both
finite.
A general investigation of these exceptional cases will be
found in a subsequent chapter: meanwhile it may be stated,
reserving these exceptions, that whenever /(z), i.e. the integral
of f’(x) dx, can be found, the value of the definite integral
[7@ae is obtained by substituting each limit separately
4
* In a subsequent chapter on Definite Integrals a rigorous demonstration
will be found of the property here assumed, namely that the sum of these
quantities of the second order becomes evanescent in the limit, and consequently
may be neglected. Compare also Art. 39, Dif Cale.
32 Elementary Forms of Integration.
instead of x in f(x), and subtracting the value for the lower
limit from that for the upper. :
A few easy examples are added for illustration.
ExaMrzes.
1 iy ;
* ie ‘i ——
I. \, da, ns. rae
ee
iB
2. \ sin 6d0 » «iT.
0
ao dx Tv
‘ ee ” 4a
z
4 ‘e sin? x dz. 3 a
0 4
T
a is F wT oT
: 4. sintx dx. Shei,
- Ne We Bea a
Tr
6. | sin? ¢ da. pres
0 2
7 f, és I
1 x ” a
1 dz T
| : ae
ol +4%+2 3/3
Tv
z 2.4
costa dx. cr
? I OSS
ee {, adz ry
; 26+ 22 3 08 3+
fe f ax
. a) %
a &— a)(8 - 2)
See Art. 11.
TT
ae
12. J xz sin # dx. wo
0
7 dx
13. | ——., where a>. oo. eee
o @+ beosax Jee
; 7 dx v
- I I — 2a cosa+a* 7 Tog
y Change of Limits. 33
32. Change of Limits.—It should be observed that it
is not necessary that the increment dz should be regarded as
positive, for we may regard w as decreasing by successive
stages, as well as increasing.
Accordingly we have
("7 de = sla) 40) =-[ Pear (52)
v
That is, the interchange of the limits is equivalent to a change
of sign of the definite integral.
Also, it is obvious that
[oleyie= |" ocayae + | prayers
and so on.
Again, if we assume z to be any function of a new variable
2, so that $(x)dx becomes (z)ds, we obviously have
(n=l im (33)
le is
where Z and z are the values which s assumes when X and
% are substituted for x, respectively.
: ‘ da
For example, if = atans, the expression @+ ai be-
COS & a and if the limits of # be o and a, those of
comes
s are o and a Consequently
T
q a a2 ‘ coss ds = :
Ha FO Ty
Also, if we substitute a — z for 2, we have
i (a)da = -f g(a — s\ds = i, g(a — 8)dz.
[3]
384 Elementary Forms of Integration.
Since neither w nor s occurs in the result, this equatior
may evidently be written in the form
U" (a) de = i g(a - a)de. (34)
For example, let (x) = sin”#, then @ (= - *) = cos", and
we have
2 j
| sina dx = | cos"x dz.
0 0
And, in general, for any function,
Tr Tv
[[Acine)ae = [/A(eos a\da. (35)
T T
33. Walues of| sin max sinnrda, ana| cosma cosne da.
0 0
Since
2 sin mz sin nx = cos (m — n) x - cos (m+n) a,
and
2 COs mx cos nz = cos (m — n) # + cos (m+n) 2,
we have
| ; : sin(m—n)e sin(m+n)x
sin me sin nada = —_———__—_ - —______,
2(m — n) 2(m +n)
and | cos ma cosnede — am — ")# sin (m +) @
2 (m —n) 2 (m+n)
Hence, when m and » are unequal integers, we have
T wT
sin ma sin nedz = 0, and | cos ma cos nxdx = 0. (36)
J0 0
When m =, we have
iy I — cos 2nw e sin 2nx
sin? na da = |\_—————. dx = — -
2 2 4n
vr
. T . .
os | sin’ ne da = as when ” is an integer.
0
Definite Integrais. 35
In like manner, with the same condition, we have
[eos nedx ==, (37)
0 2
Again, to find the value of
B SL erannEEEEEEREEEEnEIETEEEEEEEEEEEEEEE
ie V (e—a) (8-2) de.
Assume, as in Art. 11, 7 =acos’@ + B sin’ 9; then, when
Tv
0 =0, we have «=a; and when O=—,e=B.
Hence, as in the article referred to, we have
wT
B 2
if / (#—a) (B —2)dax = 2 (B - a)? [sin 6 cos’0.d0.
Also 2 [jsiw Ocos’0.d0 = 3 [sin 2 6d0
0 Oo
= 2 |"sine 9p ze z
0 8
% i (@= a) (B=«) de = (B-a)* (38)
(e]
36 Examples.
EXAMPLEs.
(1 + cos x) dx I I
Ans. — - -——_——..
(2 + sin x)§ 2 (x + sin x)?
x sin « dz, 9 sins — ecosz.
I-
2log (I+ 2) - 2.
I+2 ”
(a@ + dgn)met
aoen- sae ea Saas
4. (a + ban)mgr) de, a ae
eda 2 I
5: @ =- ” 3@+e)
(1 a tans” pi HR ee
rama
ere
a exes
\ Jove
base=e
| crates ame
© \eaoiers
7. 1» 28in! Za
Js
8 edz I 16 (S aay
o+e—2° » 9 8 \aa)°
I b
— tan! (=
* a nee + a sin?” ” ab tan (5 tan *) :
tanadx I .
a+btae PS log (a cos? # + bsin? x).
log «) d ;
Ts jeter » sin(loga@).
12. Show that the integral of & can be obtained from that of amdzx.
gmt] _ gmtl
Write the integral of adz in the form eae and, by the method of
indeterminate forms, Ex. 5, Ch. iv. Diff. Calc., it can easily be seen that the
true value of the fraction when m+1=o0 is log (: ), or log x, omitting the
arbitrary constant.
13. feo” sin max cos nz daz.
This is immediately reducible to the integral given in formula (23).
d 445 tan —-
14. \—Sar 6 Ans. > tan ( =]:
5+4sme% 3 3
Examples.
tan” 2
15. | daa Ans.
: (a+ 2s
16. \ a(a + x)bda. »
dx
17. ———_.. 4
| (a+ bx2)3
Let a + b2? = 2%
j% + q cos x)dz
18. TT
a+bcosz
This is equivalent to
37
entan” 2 (ax -1)
(I oe + a)h
3(a + 2)8 (4x - 3a)
4-7
2a + bx
B(a+ ba)"
ee dz
b b a+bcosz’
and accordingly can be integrated by Art. 18.
verde e
mp: | afar Ans. ae
a dx I
20. | ice ” 5 tan! (a?).
2 dx 2
21. ————., » —:
(a+ ba2)8 3a(at b22)8
PA dx o({ren 1+a@5-1
| ES Oh 5 ae ad
Let 84+1=2%
| ax La sa 1am —
23. 4, its integral is Jee
log (« + 6/2? — 1), &e.
I
SJB —@
sin-1y; and when a < 4, it is
Examples.
25. Deduce Gregory’s expansion for tan-!x from formula (f).
When # < 1, we have
1
Seer =I-et+aet— xt &.;
1+2'
5 yl
# tate = [ae et ee Se
I+e@ 3.05 =#7
No constant is added since tan“! # vanishes with «.
26. Deduce in a similar manner the expansions of log (1 + x), and sin"! x.
de
. Find the int —_———————.
27. Find the integral of Serv EET
This can be reduced to the form in Art. 18, by assuming : = cota, &.
dz
28, \ a
(a+ ba) 4/1 + x2
I a+ be
Ans. log { as
A a+ ee
This can be integrated either by the method of Art. 13 or by that of Art. 23.
a: n
29. j=. Ans. = seo (=) 2
es
a at a dx
1
3°. 5 log 2.
cos az *
Aly
3t. fea, » log (x + v2
i dx I
37 J, @+30)8 ” 3
mae
33. {7 a — «de. Be
2 2
34 J" eversin: (2 ) ‘9 gna
0 4
rT
3. Poe hog 2
Te Mg oe
T
2 2 I
= tan-1{_),
ol eeaae wae Bog (5)
( 89 )
CHAPTER II.
INTEGRATION OF RATIONAL FRACTIONS.
34. Rational Fractions.—A fraction whose numerator
and denominator are both rational and algebraic functions of
a variable is called a rational fraction.
Let the expression in question be of the form
aa™ + ba + cx™ + &e.,
2" 4+ Wa" + x? + &e
in which m and 7 are positive integers, and a, 6,...a',0,...
are constants.
In the first place, if the degree of the numerator be
greater than, or equal to, that of the denominator, by division
we can obtain a quotient, together with a new fraction in
which the numerator is of a lower degree than the deno-
minator: the former part can be immediately integrated by
Art. 3. The integration of the latter part in general comes
under the method of Partial Fractions.
35. Elementary Applications.— Before proceeding to
the general process of integration of rational fractions, we
propose to consider a few elementary examples, which will
lead up to, and indicate in what the general method really
consists.
We commence with the form already considered in Art. 7;
in which, denoting by a, and a, the roots of the denominator,
the expression to be integrated may be represented by
(p + gx)de
(@ - ai)( — az)’
Assume
p+gx _ A ” A,
(e—a)(e—-m@) w—-a, wap
40 Integration of Rational Fractions.
Multiplying by (w - a) (# — a:) we get
Pp + qu = (Aiaz + Aya;) + (A + A,) Ho
Hence, we get for the determination of 4, and A, the
equations
p=-Am- Aw, g= A+ A;
whence we obtain
AaB te 4 B+ gee
a; — a2 a, — ae
Consequently
| (p + ga) de - £28 dx - pete | dex
(w—- a1)(@ — a2) a@—-a,)%-ay, a, — a2 &L— Ae
a3 - {(p + gas)log (#- a) - (p+ gas) log (ea).
ai — a2
In like manner
pan A, A;
es a
(@ -—ai)(@-—a@) @-a @- a,
where A, and A, have the same values as above; hence
| (p + ga’) dx ~22 80 da eal dz
a
(@- 1) (@-—m@) a-aje-a a-aje-a;
But each of the latter integrals is of one or other of the
fundamental forms (f) and (4) of Chapter I.; hence the
proposed expression can be always integrated.
Again, let it be proposed to integrate an expression of
the form
(p + gu + ra’) dx
(w — ay) (@ - az) (a — as)
We assume
pt get ra A, A, A;
= + +
(a — a,)(v — az) (# — as) w- ay, t— Ag e—- as
.
>
Elementary Applications. 4}
then clearing from fractions, and identifying both sides by
equating the coefficients of 2”, of x, and the part independent
of x, at both sides, we obtain three equations of the first
degree in A,,.A.,-As, which can be readily solved by ordinary
algebra ; thus determining the values of A,, A., A; in terms
of the given constants.
By this means we get
F (p + qe + ra”) dx = Ail da = + |
% — a) (# — a2) (w@ - as x
= A, log (x = a) + A, log (w = a2) + A, log (@ = as).
da
wv - a3
+ A,|
@— a2
We shall illustrate these results by a few simple examples.
ExampPtes.
(@—1) dx 2 3
I. Ans. —log (% ~ 3) 4+- —log (# + 2).
lecsiee sree 8) + oe et 2)
2 (SR ig (@ + yeep (w-1
. e+ ae Z arg 8 3 4 8 ).
{ dx 1 “—-t r Sy
3: a ‘3 ak age ante,
Hpk ie Raed ©
4. = ” re g tau >
(==. I, gat
5+ a 3G 8 Par
(30? — 2) dx I g—2
6. ere ea 35 log (32) + tants,
oe I I I
Ts \ Aas 6 g log et log (# abs Meet 3
Here the denominator is equal to x (# — 2)(% + 3); and we have
w+a-—t Ay Ag Ag
tH tH
a(v~—2)(@+3) %@ 4-2 £43
42 Integration of Rational Fractions.
hence e+e —1 = Ai (2? + & — 6) + Aa t+ 3) + Asx (x — 2);
.. the equations for determining 4), 42 and -A3 are
4, + 424+ 43=1, 41+ 342-243=1, 6A, =1,
whence we get
I
e 43=-.
I
A, = Ag = =
3
I
2
Ans. x? + log (x? + 2+ 1).
8 (ete 1) dz
3 w+e4 .
We now proceed to the consideration of the general
method, and, as it is based on the decomposition of partial
fractions, we begin with the latter process.
36. Partial Fractions.—The method of decomposition
of a fraction into its partial fractions is usually given in
treatises on Algebra; as, however, the process is intimately
connected with the integration of a large class of expressions,
a short space is devoted to its consideration here.
For brevity, we shall denote the fraction under con-
f(z)
g(@)’
Let a, az, as... an denote the roots of g(x); then
o() = (@ — a) (@ - az)(# - a3)... (@ — an). (1)
There are four cases to be considered, according as we
have roots, (1) real and unequal; (2) real and equal; (3)
imaginary and unequal; (4) imaginary and equal.
‘We proceed to discuss each class separately
37. Real and Unequal Roots.—In this case we may
sideration by
assume
TAO) = ts aa at Hes 2 eh Fx. (2)
$ (x) Ga Ca, C= oe v= ay
where A,, A, .... A, are independent of # For, if the
equation be cleared from fractions by multiplying by 4),
on equating the coefficients of like powers of w on both
sides we obtain » equations for the determination of the »
constants A,, 4. ... An.
Real and Unequal Roots. 43
Moreover, since these equations contain A,, A, &c., only
in the first degree, they can always be solved : however, since
the equations are often too complicated for ready solution,
the following method is usually more expeditious :—
The question (2), when cleared from fractions, gives
J (@) =.Aj(@ - ay) (@- as) « « (@— an) +A2(@—ay)(@ — a3) .. (@ — an)
+ &e. + An (e—a1)(@- a2)... (@- ana);
and since, by hypothesis, both sides of this equation are
identical for all values of z, we may substitute a, for z
throughout; this gives
J (a1) = Ai (ai — a2) (a1 ~ a3)... (a1 — an),
_ f(a)
= ¢ (a)
In like manner, we have
As a J (a) Aes: F (as) A _ fF (an) (3)
© g(a) 8 (as)? g(a)’
Hence, when all the roots are unequal, we have
PAE) PAB 8 Ta) Fal Te) 28 8 a
9 (2) F ¢ (a1) v— ay 2 ¢ (az) wv — ay ¢’ (an) @- dn
Accordingly, in this case
F(a) ma F(a) v-a J (a2) 3
lia” MG ee) gy ee ase
oe 5 log (# — an). (s)
‘The preceding investigation shows that to any root (a),
which is not a multiple root, corresponds a single term in the
integral, viz.
F(a)
Seay oe a)
44 Integration of Rational Fractions.
one which can always be found, whether the remaining roots
are known or not; and whether they are real or imaginary.
38. Case where Numerator is of higher Degree
than Benominator.—It should also be observed that even
when the degree of z in the numerator is greater than, or
equal to, that in the denominator, the partial fraction cor-
responding to any root (a) in the denominator is still of the
form found above.
For let
LO 2g
Cea
(2)
where Q and R denote the quotient and remainder, and let
& = be the partial fraction of = corresponding to a single
XG :
root a; then, by multiplying by ¢(x) and substituting a in-
stead of a, it is easily seen, as before, that we get
_f(a)
¢'(a)
For, example, let it be proposed to integrate the ex-
pression
aidx
2 — 2a — 54 +6
Here the factors of the denominator are easily seen to be
2-1, +2, andw- 33
accordingly, we may assume
a ay e+P+ Z + z G
oe — 20-50 +6 e @-l @+2 £2-3
To find a and 8, we equate the coefficients of z‘ and 2° to
zero, after clearing from fractions: this gives, immediately,
a= 2,and PB =9.
Again, since p(x) = a — 22° — 5x + 6, we have
(x) = 32° — 4x - 5.
Real and Unequal Roots. 45
Accordingly, substituting 1, - 2, and 3, successively for w
in the fraction
5
xz
3a? — 4x — 5°
we get
I 2 243
0 re a
and hence
x ae I 32 243,
e@ooe—5et+6 | 9 Gea) ig(@42) 10(@-3)'
: x de e , log (# - 1)
te a ee
32 243 -
- ise (e@+2)+—— log (# - 3).
39. Case of Even Powers.—If the numerator and
denominator contain x in even powers only, the process can
generally be simplified; for, on substituting z for 2’, the
fraction becomes of the form
43)
9(2)
Accordingly, whenever the roots of ¢(s) are real and
unequal, the fraction can be decomposed into partial fractions,
and to any root (a) corresponds a fraction of the form
fa) 3
g(a) s-a
The corresponding term in the integral of
f(#)
$(2’)
is obviously represented by
F (a) da
9 (a) | aa
daz
46 Integration of Rational Fractions.
This is of the form (/) or (4), according as a is a positive
or negative root.
The case of imaginary roots in $(s) will be considered in
a subsequent part of the chapter.
It may be observed that the integrals treated of in Art. 5
are simple cases of the method of partial fractions discussed
in this Article.
EXAMPLEs.
{ (20 + 3) de
w+ a — 20°
Here the factors of the denominator evidently are x, —1,and2%+23; we
accordingly assume
20 +3 A B Cc
so + :
w+ a2? — 24 Go Bot a+2
Again, as p(x) = 4° + @? — 27, we have (x) = 3a? + 2x —2;
I 2 SEES
*" gla) 3at@+ar—2
Hence, by (3) we have
Bia, Bae, (en
2 3
6 ?
consequently
(24+ 3)de 3 5 t '
(a Se Se 1) é log (# + 2).
: ax
. Jarmery
Here
I I I I
@PLAC+A P-P (3 +B gy ays
hence the value of the required integral is
I I £ I 2
—— Ji tan (=) — = tant (=) §.
(a — &) {5 ten (5) oo (7)
var
\erea
Multiple Real Roots. 47
Substitute z for 2? and the transformed integral is
j I dz
2 (z+ a) (z+ b)
Consequently the value of the required integral is
I 1 (3+)
2 (a — 3) °C Nea al”
(32? — 1) dx
4 ( Ans. 3@ + 11 log (% — 2) — 2 log (x - 1).
(2? — 3) dx I i 5
= lease ” ge et) + laste abt ee a as
(2@ + 1) da I 3
. |r n 5 loge + log (w+ 1) — 2 log (x + 2).
vdz a x >
ee laa » ~—— tan Care 78 (==).
8 | dz(a' + b'x”)
; ala + bar) *
Let wa
z
40. Multiple Real Roots.—Suppose ¢(x) has r roots
each equal to a, then the fraction can be written in the shape
I (2)
(x — a)"P (2)
In this case we may assume
J (2) NM, MM, _ih
P
(«@ —a)"'b (a) ~ (@ =a)" "tesa?" ia we
where the last term arises from the remaining roots.
For, when the expression is cleared from fractions, it is
readily seen that, on equating the coefficients of like powers
at both sides, we have as many equations as there are
unknown quantities, and accordingly the assumption is a
legitimate one.
48 Integration of Rational Fractions.
In order to determine the coefficients, Ih, Ih, &e. .. . Il,
clear from fractions, and we get
f(#) = My(«) + (# - a)b (2) + Uy(a - a)*p(w) +&e.... (6)
This gives, when a is substituted for z,
f(a) = Myla), or I, = e (7)
Next, differentiate with respect to z, and substitute a
instead of # in the resulting equation, and we get
F(a) = Mrf'(a) + Iep(a) 5 (8)
which determines J/,.
By a second differentiation, I, can be determined; and
so on.
It can be readily seen, that the series of equations thus
arrived at may be written as follows—
F(a) = My(a),
faa Ma (a) +1.Mab(a),
S’(a) = My" (a) + 2. thy (a) + 1.2. Ua),
f(a) = M"(a) + 3. Mab'"(a) + 2.3 D(a) + 1.2.3. Mah(a),
fie(a) = Mrp?(a) + 4. Map") + Serna a) + 2.3.4.Map(a)
1.2.3.4. Mp(a),
in which the law of formation is obvious, and the coefficients
can be obtained in succession.
The corresponding part of the integral of
S (x) de
(z — a)" (2)
evidently is
M,-1 1 U,_. 1
Hog (2-0) ~ Ft 5 Oe Gea (9)
If ¢(z) have a second set of multiple roots, the cor-
responding terms in the integral can be obtained in like
manner.
Imaginary Roots. 49
41. Imaginary Roots.—The results arrived at in
Art. 37 apply to the case of imaginary, as well as to real
roots; however, as the corresponding partial fractions appear
in this case under an imaginary form, it is desirable to show
that conjugate imaginaries give rise to groups in which the
coefficients are all real.
Suppose a + b,/—1 and a- 6./—1 to bea pair of con-
jugate roots in the equation ¢(#) =0; then the corresponding
quadratic factor is
(e-a)?+0?; which may be written in the form a+ pa+g.
We accordingly assume
g(a) = (2° + pe + g) (2),
and hence
S(t) LIe+M P
o(z) w@+pe+gq Q
where represents the portion arising from the remaining
Le+
roots, and —
e+ pet
atb/f-1.
Multiplying by ¢(x) we get
J (#) = (Le +) p(x) + (@ + pe + 9) é p (a). (10)
If in this, - (px +g) be substituted for 2’, the last term
disappears; and by repeating the same substitution in the
equation
S (2) = p(2)(Le + Ht),
it ultimately reduces to a simple equation in w: on identify-
ing both sides of this equation, we can determine the values
of Z and I.
42. In many cases we can determine the coefficients Z,
more expeditiously, either by equating coefficients directly,
or else by determining the other partial fractions first, and
subtracting their sum from the given fraction.
It will also be found that the determination of many
[4]
is the part arising from the roots
50 Integration of Rational Fractions.
integrals of this class can be much simplified by a trans-
formation to a new variable, or by some other suitable
expedient.
Some elementary examples are added for the purpose
of illustration.
EXAMPLEs.
aan
te le +2) (1+ 2)"
Assume
x A Le+ iM
(i+a)G4e) Ise) 14e
clearing from fractions, this becomes
a= A(1+a?) + (Le + M)(1 +2).
Equate the coefficients, and we get
L+A=0, L+M=t1, A+M=o0.
Hence
and accordingly
x : 1. ITI+2%
(1+ %) (1+ 2) 2t¢e* 214a'
j adz tie 1 +2? gay Ly
O17 oe Se >—3| + — tana.
(arte) 48) (re a] "2
dz
Ze
+28
Let
consequently, 4 = - by formula (3). Substituting and clearing from fractions
we have
Z=t-e+e?4+3(Le+ M)(1+2);
hence, dividing by 1 + 2, we have
2-2=3(I«e+ MM.
Imaginary Roots. 51
Consequently
dz _ if dz + 1{/Go8e
len ele gii-a+e*
I I I 2n-1
= Hog (1 + 2) — glog (x ~ # + 2#) +7 tan ( =).
3 V3 V3
dz I I+at+a I 4 (/2%+1
.t |= Ane. og (a) tm (==).
This can be got from the last by changing the sign of #.
dx
at J I-a
In this case we have
I I I I
—a=5(-—s ta):
I-z 2\r1-2 I+2
aidz I (a4 — 1)? I (20+
Se \a— Ans. qhe( | tee { ;
Let # = z, and the integral becomes
=| 2d2
or
vdz
‘ leery
Assume
eo AB det ih
(e-1)?(@? +1) (v—1) got 1+e
To find Z and ¥, clear from fractions, and by Art. 41 the values of Z and M
are found by making 2?=~ 1 in the following equation:
= (Le+ M)(2-1%
This gives immediately L =— -, M=0.
Again, by Art. 40, we get immediately 4 = *.
To find B, make z = 0 in both sides of our identity, and we get
o=A-B+U; B= Ass.
[4a]
52 Integration of Rational Fractions.
Finally
2 _iro41 rr I @ ,
(@—12@ 41) 2(z—1* 2e-1 21422’
wdz Io. I I
| — > = Ht — 1) ——log (#? + 1).
jeer apa te og (% — 1) 4 g ( )
7 dz
; lagna
Here the denominator is easily seen to be 23(#—- 1)(# + 1)*(2? +1), and the
expression becomes
{ dz
®@-he+i@+iy
Assume # = *, and the transformed expression is evidently
fe 28 dz
(2 — 1) (2 + 1)?(22 +1)
The quotient is easily seen to bez — 1; and, by the method of Art. 38, we may
assume
28 anes A Ba (4 iz+M
(@-1)@4+1"@+HD goa @+ipP er” eer
Hence (Arts. 37, 40), we have
I
> B=--.
4
A=
Col =
Next, Z and Mf are found by making 2? = — 1, in the equation
& = (Lz + M)(z-1)(2 + 1)?5
of P= 2224+ M)@4+ 1) =2{Le2 + (2+ U)z+ HY},
which gives
L+M=0, L-M=-%;
& t= = 7 cee
4 4
In order to find the remaining coefficient C, we make z = 0, when we get
o=-1-A+B4 O44; w= 8
Multiple Imaginary Roots. 53
hence we have
28 = 1 I 9 2-1 |
Geta TB ei) gee * ern) en)!
: a8ds ae eh >To (@-1)+
“J gse—+ Try
9 sgh 2 stan=l
, + 3 log @ +1) 3 bee (& rae ee Ze
Hence
j dz I 1
Bake, x a @+1
x84+a7—at—28) 224 x 4(@+ 1)
I—2 I I
= log ——-= + log —— + ~ tan-)-
gS Tae te 2 + {ee
“z—TI I
+3 @—-1
j (3 + 1) da
I
G_fesa Ans. . log
43. Multiple Imaginary Roots.—To complete the
discussion of the decomposition of the fraction a, suppose
the denominator ¢(z) to contain r pairs of equal and imaginary
roots, i.e. let the denominator contain a factor of the form
{(@ -— a)? + &}"; and suppose p(x) = {(# — a)? + 8°)" g(a)
In this case we assume
S (2) _ Lett, | Lert,
((e- a)? + B}"gi(z) ((w@-a)?+ Br ((w—a)? +}
ao. Tet h P
(ways B* Gia)
the remaining partial fractions being obtained from the other
roots.
There is no difficulty in seeing that we shall still have
as many equations as unknown quantities, Z,, 1%, In, Mh,...
when the coefficients of like powers of # are equated on both
sides.
To determine L,, Mh, L., &e.; let the factor (x- a)? +B
be represented by X, and multiply up by X7, when we get
F(z) =< r-1 Px,
Sg) Tie + Mat (Law + Ma) + &e. + (Ly + Uy) X aa (11)
54 Integration of Rational Fractions.
The coefficients Z, and WW, are determined as in Art. 41.
To find Z, and I,; differentiate with respect to x, and sub-
stitute a + b,/ — 1 for w in the result, when it becomes
af Fl)
. iz ole Taso 2 aitae eae:
where x = a + Bal
Hence, equating real and imaginary parts, we get two
equations for the determination of LZ, and If. By a second
differentiation, Z; and If; can be determined, and so on.
It is unnecessary to go into further detail, as sufficient has
been stated to show that the decomposition into partial frac-
tions is possible in all cases, when the roots of ¢(#) =o are
known.
The practical application is often simplified by transfor-
mation to a new variable.
44. The preceding investigation shows that the integra-
tion of rational fractions is in all cases reducible to that of
one or more fractions of the following forms:
da da (A + B)dx (Le + M)dx
a-a@ (w©-ay” (@—ay+b? {(@-a)?+ Bb)"
The methods of integrating the first three forms have been
given already. We proceed to show the mode of dealing
with the last.
45. In the first place it can be divided into two others,
L(x - a)dx és (La + M) dx
{((@-aP+B\" {w-al+ Br
The integral of the first part is evidently
eRe sei sas
2a(r—aj{(w@-aP + Bye
‘To determine the integral of the other part, we substitute
s for « — a, and, omitting the constant coefficient, it becomes
dz
[ (2° + BY"
Multiple Imaginary Roots. 55
Again
dz (v+B-2)dz 1 dz -7 2 dz
le + 6)" al (x? + by" “Ble + Bo BI e+ Rr
But we get by integration by parts
ede | ads dl I )
leap ~ Pee 2r- ml . (ew +B) *
. I dg
"2-8 +E” ar Sle +Pyrr
Substituting in the preceding, we obtain
dz ar-3 ds Z
lem ~ r= ee BP aGsne ese |)
This formula reduces the integral to another of the same
shape, in which the exponent r is replaced by r-1. By
successive repetitions of this a the integral can be re-
duced to depend on that oa Gay
The preceding is a case a the method of integration by
successive reduction, referred to in Art. 19. Other examples
of this method will be found in the next Chapter.
The preceding integral can often be found more expedi-
tiously by the following transformation :—Substitute 6 tan 0
for z, and the expression ———. becomes, obviously,
dz
(s° + 8%)"
=| cos”"6 dé.
br- 1
The discussion of this class of integrals will be found in
the next Chapter.
(w?)da
46. We shall next return to the integration of Tey
which has been already considered in Art. 39 in the case
56 Integration of Rational Fractions.
where the roots of ¢(s) are real. To a pair of imaginary
roots, a + b,/— 1, corresponds a partial fraction of the form
(Az? + B) dx (Az? + B) de
(a? — a)? + 6” at — 20x +o”
where ¢? = a? + b.
In order to integrate this, we assume a = ¢ cos 2g, when
the fraction becomes
(Ax? + B) dx
at — 2a°¢ cos2g + 6
The quadratic factors of the denominator are easily seen
to be
a — 20 cos p + ¢, and a? + 22/c COs + 6
Accordingly we assume
Ac’ +B = Ia+M z Te + WM’
wv —2e%ecos2pte? a — 20/6 Cos p +¢ a + 20/6 cos p +o
hence it can be seen without difficulty that
gape 3) Vawus,
4c cos } 2¢
and after a few easy transformations, we find
(Az? +B)de — Ac—B ie @ — 24 fC COSPte
pas oe eae 8 @ + 2ar/c cos +e
+ Zs acs tan (2am = 2)
se 4 sin ¢ of e-a
oe ¢
vy dx
ti ii CC.
Ais BMEERCON OS Gaerne ar
This expression can be easily transformed into a shape
Integration of G 2
@ =a @— "
which is immediately integrable, by the following substitu-
tion :—
Assume # — a = («— 6)8; then
_a-bs, (a - djs _a-b _(a—b)ds_
x 6 @-a@=+——~, «-b dz
I-38’ reg? I-33 (1-3)?
and the expression transforms into
(1 ~ ayers
(a i periee
Expand the numerator by the Binomial Theorem, and the
integral can be immediately obtained. (Compare Art. 4.)
For example, take the integral
dx
(x — a)*(x- 6)
Here the transformed expression is
| (1 - 38)'dg
(a — 6)4x??
or
1 (fr 3 nr (# :
aay (-2+3-2)e- Sp - 38+ 3 logs +— .
Substituting —— for z, the integral can be expressed in
terms of z.
gem dz
(a+ ca*)*
where m and n are integers.
Let a + ca* = 2, and the expression becomes
(s - a)™ds
2emtl gn ,
a form which is immediately integrable by aid of the Bino-
mial Theorem.
48. Integration of
58 Integration of Rational Fractions.
It is evident that the expression is made integrable by the
same transformation when x is either a fractional or a nega-
tive index.
It may be also observed that the more general expression
2
fee can be integrated by the same transformation, where
J (2?) denotes an integral algebraic function of 2”.
EXAMPLES.
xidx at ee
> / aor tt log (a? — 27).
| (a — x2)?" ane 2(a? — x?) a a 8 }
oda I Z a
| (a + cx)" 9 40? (a + cx)? * 6c? (a + cx)
wda I I I
‘ ——:: —— - a t= log (a? + 1).
3 leas uaa ea og (a + 1)
Ge = ¥ ees
Tp
. Integration of i
49 a > ex /
where 7 is a positive integer.
Suppose a an imaginary root of v” - 1 = 0, then it is evi-
dent that ais the conjugate root: also, by (3), the partial
fraction corresponding to the root a is
I
na” (a - a)’
eh (w — a)"
If to this the fraction arising from the root a be added,
we get
I( a at I w(ata")—2
= -+— Ee Se Nd eee.
nlx —a ga? nla —-(at+ejet i
But, by the theory of equations, a is of the form
2kr . 2kr
cos —- +Y-18smM ae 5
2 ti
de
Integrati
ntegration oa =a
59
where & is any integer;
igen
n
Hence, if 6 be substituted for = the preceding fraction
becomes
2 x cos 0-1
n° a —20c0s0+1°
The integral of this, by Art. 7, is
cos 6 2 sin 8 x — cos 0
eee, = ie a a)
A oe 22 cos 0+ a”) 2 tan ( in 8 )
There are two cases to be considered, according as n is
even or odd.
(1). Let = 27: in this case the equation 2” — 1 = 0 has
two real roots, viz., + 1 and—1; and it is easily seen that
dx I G@-r 1 ker kn,
\aoa- pees ; + 5, 2 008 — log (1-20 cos — + a*)
; i 0 NO Ce
2 3en te | (13)
r r
sin — zy
r
where the summation represented by = extends to all integer
values of & from 1 tor —1.
(2). Let n = 27 + 1, we obtain
da log (#-1 I 2kr 2khar
= Bay) + 2 cos log| 1-2acos +0
erel oy art+i 2ar+i 2ar+i 2r+I
x
ahr
2 aha Sra ee
- Ssin tan? | -—_—__——. ], (14)
2ar+i ar+i . 2kr
sin
2r+ I
60 Integration of Rational Fractions.
where the summation represented by 3 extends to all integer
values of & from 1 up to r.
ml Jon
, where m is less than 7 +1.
I
50. Integration of zm
As before, let a be a root, and the corresponding partial
m1 a”
na” (# — a) bi n(a@ — a)
arising from the conjugate roots, a and a™, is
1/ a” rn a? 2 ela" a") = (er eae)
n\e-a w-a) wn @-(ata')e+1 °
traction is ; hence the partial fraction
2 2cosmO — cos(m — 1)0
n @—20¢c080+1 ’
where @ is of the same form as before.
The corresponding term in the proposed integral is easily
seen, by Art. 7, to be
x—cos@
sin 8
(15)
I 5
7 1008 m0 log (a? — 27 cos + 1) — 2 sin m0 tan™
By giving to & all values from 1 tom — 1, when ” is even, and
from 1 to > when n is odd, the integral required can be
written down as in the preceding Article.
(Ss
ae
a
(4+ Bel\de
a(a + bx*)
ay ie
dz
waar
bees
ery
joe vdz
J
| S29. (2% — 5)de
» (ener
dz
s § x(a + ban)?
du
liza + bane"
Let a + ba” = xz, and the transformed expression is —
xdz
10, oe
aa I
(e+ 3)(@+*
Examples. 61
Examp.es.
Ans. slog (2 us *) :
2 a+ 4
» 2log(#—2) + log(x+1).
Ba - Ab :
a log (a + bx*).
V2
” oat gO al:
g+1° 3
tan-! ee =) :
I-2?
2%
2 Re B+
»
1 lo wa ett I
= 4/2 af 241 aaa
7: II “+i
eel og [ee
" 2@+D 4 os (=)
I
pace x
2a (a + ba)" 22 °° \a 4 ba8)
lo (5 ae
ela a) + na(a + bar)
I
na
(2 — 6)rdz
narar
Ans. slog (a? +1) ~=log (e+ 1) + * tana.
II. lace » log ed pe Aten dig a
A+ 484 5074+ 4044 25 w+ 25 5 (a+ 2)
12. Apply the method of Art. 47 to the integration of a
The transformed expression is — =
td Ans. ; we - = lo; : + =.
13. | aa"
62 Examples.
14. Prove that
d: m+n-2 qe
| ee transforms into — j ae
an (I — a) gm :
if we make 4 = ‘
I+z
dx I = I x
sl ear oes in-— 1 ie
5 | sane ehoad ans jy 2 a—-6b Mg 08S
b
ape log (@ + 8 cos 2).
Multiply by sin z, substitute « for cos z, and the integral becomes
— du
j (1 — #)(a + bu)
dz «
s ————_—_—_—. = --1 =4 2
16 | sanepaes Ans. = og sin = og cos = = 10g (3 + 2 e082),
— 2) d: 2
17. ( (1 aS rn an log (“*5*'),
a(L+ 224 a4) 2 B/ 3 4 z
Let 2? =-, &e.
z
18. Prove that
j des Pape yy anes) Lig (1 ~ 22 008 =" a2)
ish oe 2n 2n 2n
(2% — 1)
x — cos -——__—
ts (2k —1)9 pes 2n :
+ -— sin ag ee ee 7
sin
2n
where & extends through all integer values from I to ”, inclusive.
6. j dz _log(1 +a) 1 3 008 =)" tog ( 1 -acos-=— 0, 1)
tt+aeml oo am+r anti an+1 2an+1
2k—1
(Gk 1) 2 — cos CEH
2 2 I)
—— in —————. tan 4 ee
eae an+tI . (2k —1)7 :
sin -————_—_
2n+1
where & assumes all integer values from I to # inclusive.
on~
63 )
CHAPTER III.
INTEGRATION BY SUCCESSIVE REDUCTION.
51. Cases in which sin”@ cos"0d0 is immediately In-
tegrable.—We shall commence this Chapter* with the dis-
cussion of the integral
f sin” @ cos” 06;
to which form it will be seen that a number of other expres-
sions are readily reducible.
In the first place it is easily seen that whenever either m or
nis an odd positive integer the expression sin”@ cos"™d0 can
be immediately integrated.
For, if n = 2r + 1, the integral becomes
fsin”@ cos” 6d0, or, f sin” @ (cos’ 0)" d(sin 6).
Ii we assume wv = sin 0, the integral transforms into
fami — 2)" de; (1)
and as, by hypothesis, 7 is a positive integer, (1 — 2°)” can
be expanded by the Binomial Theorem in a finite number of
terms, each of which can be integrated separately. In like
manner, if the index of sin 0 be an odd integer, we assume
x= cos 0, &e.
A few examples are added for the purpose of making the
student familiar with this principle.
* Tt may be observed that a large number of the integrals discussed in this
Chapter do not require the method of Successive Reduction: however, since
other integrals of the same form require this method, it was not considered
advisable to separate the discussion into distinct Chapters.
64 Integration by Successive Reduction.
Examptes.
3
I. [sineeo. Ans. ee cos.
ind
2. [costa ao. sind - 2 ag
3 5
cos!®@ —cos8 6
sin? @ cos’ 6d@. at ees *
3 | » “10 8
— ji 083 8
_ cos?0 ® icone oe
= 3 zsin?@ 2 sin¥a
5 ay sin @ cos? a0. 3) ue
3 7
sin? @d6 2 costa ore
° lee = 5 :
cos6 d@ , ee
: : 49 —3sinéo.
7 | sinto » 351080 rae 6
52. Again, whenever m + n ts an even negative integer
the expression sin” @ cos” d0 can be readily integrated.
For if we assume x = tan0, we have
(ola 4 sin @ = eee and d@ = : ae
fi+a fite +a”
and the expression transforms into
a™ dx
mtn
(1 + 2”) ?
Hence, if m+n =-— 2r, this becomes
x™(1 + 2”)""da,
a form which is immediately integrable.
Cases in which sin™@ cos" dO is immediately Integrable. 65
sin?0d0
cos’® *
Take, for example, |
Let w = tan 6, and we get
tan?@ = tan®6
+ =e
|x + x”) dx, or
dé
Next, to find |sxpeoro"
Making the same substitution, we obtain
{[f + w)*de
x
Hence, the value of the proposed integral is
tan‘é
+ tan’6 + log (tan 8).
dé
sin?@ cosi9”
Again, to find |
(1 + a*)dz
xi
cordingly the value of the proposed integral is
Here the transformed expression is » and ac-
2 2
* tani@ — —~—
5 tan30 fanl’
In many cases it is more convenient to assume 2 = cot 6
For example, to fina | at
sin‘
Since d(cot @) = - _ if cot @ = 2, the transformed
integral is
3
-| (1 + #")da, or - cot w
The following examples are added for illustration :—
(5]
66 Integration by Successive Reduction.
EXAMPLES.
sin3 @ do ‘Bibs tan‘ @
do 2tan3@ tan¢
Bacen: 9» tan@+ ee
3 5
de tan? 6
————— —— + log (tan 6).
sin 6 cos3@ a 2 alngven 9)
. : 2 4
2 cos? @ ” 3 tan? 6.
dé 8
° i — 8 cot 20 — — cot?29.
° sin‘ @ cos!@ ” cot 26 3 cot’ 20
d
6. a
j sin! 6d 2
2
95 2 tanto (« Ps *).
sin! @ cos! 0
When neither of the preceding methods is applicable, the
integration of the expression sin” @ cos"@d0 can be obtained
only by aid of successive reduction.
We proceed to establish the formule of reduction suitable
to this case.
53. Formule of Reduction for sin” @ cos” 6d0.
| sin” @ cos" 0d0 = | eos 6 sin™ Od (sin 0):
consequently, if we assume
sin”!
u=cos*™9, v= — ;
M+I
the formula for integration by parts (Art. 21) gives
cos" O sin"! n-—
- me = : mm n-
| inna cos"§ d@ = ——————__ + i [sin 70 cos? 4.d6. (2)
mti
Case of One Positive and One Negative Index. 67
In like manner, if the integral be written in the form
- | sin” @ cos” 0d (cos 6),
we obtain
m—-1
nti
sin” 4 cos”! 8
n+1 u
| sin” cos"0d0= | sin™@ cos"?6d0 — (3)
It may be observed that this latter formula can be de-
rived from (2) by substituting = ¢ for 0, and interchanging
the letters m and n in it.
54. Case of one Positive and one Negative Index.
—The results in (2) and (3) hold whether m or n be positive
or negative; accordingly, let one of them be negative ( sup-
pose), and on changing x into — n, formula (3) becomes
| sin” 9) sin” @ _m—i1sin"?6 (A)
cos" (w= 1) cos™"0— nn — ‘lees :
in which m and » are supposed to have positive* signs.
By this formula the integral of onto is made to de-
pend on another in which the indices of sin 6 and cos@ are
each diminished by two. The same method is applicable to
the new integral, and so on.
If m be an odd integer, the expression is integrable im-
mediately by Art. 51. If mbe even, and even and greater
than m, the method of Art. 52 is applicable; if m =n, the
expression becomes f{ tan”@d0, which will be treated subse-
quently ; if 2 < m, the integral reduces to that of sin™” 0 d0.
ost” 6 ;
Again, if » be odd, and > m, the integral reduces to |;
* The formule of reduction employed in practice are indicated by the capital
letters 4, B, &c.; and in them the indices m and » are supposed to have always
positive signs. By this means the formule will be more easily apprehended
and applied by the student.
[5a]
68 Integration by Successive Reduction.
and if n < m, it reduces to —— The mode of find-
ing these latter integrals will be considered subsequently.
Again, if the index of sin 0 be negative, we get, by
changing the sign of m in (2),
n 2 fs: n-2
(= 0 cos”! @ n t [oe 9 6 (B,
sin"0-— (m—1)sin™'@ =m-iJjsin™?6 ~
We shall next consider the case where the indices are
both positive.
55. Emdices both Positive.—If sin”6 (1 - cos’6) be
written instead of sin” @ in formula (2), it becomes
cos" @ sin™*! @
m+iI
{ sin”@ cos” 6 d@ =
cos” @ sin™*! 9
get | sin” @ (cos? 6 - cos”) d0 =
m+i os
eee {sin 0 cos’*6d0 — maa | sin" cos” 6 d0:
mM+tI M+
hence, transposing the latter integral to the other side, and
dividing by —, we get
2-1 iymr —
[ sin" 6 cost 0 = oo a eae | sinm 0 cos*0.40. (0)
m+n m+n
In ike manner, from (3), we get
m—t
m+n
sin” @ cos”6
m+n
[sine 0 cos” 6 d@ = [sin 6.cos” 6 dO —
By aid of these formule the integral of sin” 0 cos” d6 is
made to depend on another in which the index of either
sin 9, or of cos @, is reduced by two. By successive appli-
cation of these formule the complete integral can always be
found when the indices are integers.
Indices both Negative. 69
56. Formule of Reduction for sin” 0 d0 and cos"0d6.
These integrals are evidently cases of the general formule
(C) and (D) ; however, they are so frequently employed that
we give the formule of reduction separately in their case,
re _ sin cos™10) n-1 a
{ cos PS = | cos 60. (4)
| sinnaag S26 sinh" 1) sinmtgao. (5)
The former gives, when z is even,
se ates! cos’ 6 + &e.)
| cose 6.0 = (cost 6 (apa - cos” 8
(n - 2)(n - 4)
, @- lm - 3)(@—- 5)..- 19 (6)
n(n—2)(n—4)...2 °
A similar expression is readily obtained for the latter
integral.
EXaMPLes.
sin@ cos@/.
I. { sinto ae. Ans. — pacer (sinro + 3) + 30.
so gina sin@cos@/sin‘@ sin?@ 1 0
2. [ cos 6 sin‘ 6 dé. 7) 3 e 3 16"
sin @ cos°6 5 5/.
3. { costo a. 53 nO EU (conte + 8) + 5 (sino con e).
57. Indices both Negative.—It remains to consider
the case where the indices of sin 9 and cos @ are both
negative. :
Writing — m and — instead of m and 2, in formula (C),
it becomes
dd -1I mtd i do ;
| sin”0cos"@ (m+n) cos’"@ sin”™0 © m+n) sin@ cos"”6’
70 Integration by Successive Reduction.
or, transposing and multiplying by “— =,
dé I , mee i __ 40
| samo cos™?9 (+1) cos*Osin™ 9 +1 Jsin™O cos"6
Again, if we substitute n for x + 2 in this, it becomes
dO O I
sin”Ocos"@ (mn — 1) cos”7@sin”"0
(2)
nm—-1 jsin™@cos”*0"
minnel dé
+ ————
Making alike transformation* in formula (D), it becomes
| dé 7 -1
sin”@cos"@ (m — 1) sin”™6 cos”
7 dé
+ ————
sin”@ cos”@"
(F)
m-TtI
In each of these, one of the indices is reduced by two
degrees, and consequently, by successive applications of the
formule, the integrals are reducible ultimately to those of
one or other of the forms ae or aa : these have been
cos@ sin
already integrated in Art. 17.
ao
sin’@ Gos"
important that they are added independently, as follows :—
The formule of reduction for
are so
* It may be observed that formule (B), (D), and (F) can be immediately
obtained from (A), (C), and (£), by interchanging the letters m and n, and
remount : soos :
substituting > — > instead of @. For, in this case, sin 0, cos 6, and d@, transform
into cos ¢, sing, and — d@, respectively.
Application of Method of Differentiation. 71
{< _ sin 8 p22] do
cos" (n — 1) cos"? n — 1) cos"? (7)
[Be 7 — cos6 | dé 3
sin’@)= (n- 1) sin" © n — 1 Jin"? (8)
It may be here observed that, since sin?™ + cos’0 = 1, we
have immediately
{ do [a a
sin"0 cos"O J sin™=0cos"6 * | sin™@ cos™=0° (9)
and a similar process is applicable to the latter integrals.
This method is often useful in elementary cases.
EXAMPLEs.
f do ee I See
sin@cos?@ J cos?@ sind cos) 8 “3
| dé a jae +] Le)
sinédcos'@ } cosé sin cos?”
and is accordingly immediately integrated by the last.
d@ cosé I @
—.: . — ——— + — log tan-.
| sin?@ dns 2sin? 6 ag Boe ng
de I cose 3 @
, ——. -— = log tan -.
4 | sin’ @ cos? @ ” cos@ 2sin26 * 2 8 2
58. Application of Method of Differentiation.—
The formule of reduction given in the preceding Articles
can also be readily arrived at by direct differentiation.
Thus, for example, we have
d (= 5) msin™ 9 n sin™@
= +
d@ \ cos”6 cos"0 cos"
and, consequently,
| sin”™*!@ 1sin”9 m™ eS
cos™ 9° ~—s xn cos?@— xn J cos"? 0
This result is easily identified with formula (A).
72 Integration by Successive Reduction.
Again,
d
0 (sin@ cos"8) = m sin” cos”*"9 — n sin” cos".
If we substitute for cos*0 its equivalent cos’"0 (1 —sin’6),
we get
d,.
76 (sin”6 cos"6) = m sin” cos""6 — (m+n) sin™16 cos”"0 ;
hence we get
gh “
| sin” cos*""9 d0 = — ee gaa [sin cos" 6 0,
m+n m+n
a result easily identified with (D).
The other formule of reduction can be readily obtained
in like manner.
59. Entegration of tan”0d0 and ae
These integrals may be regarded as cases of the preceding :
they can, however, be arrived at in a simpler manner, as
follows :—
Since tan?@ = sec?@ — 1, we have
| tan”0 d0 = | tan”’@ (sec’@ -— 1) dO = | tan™9 d (tan 0)
t
- | tan”*@ d@ = me | tan d@. (10)
n—-I
By aid of this formula we have, at once,
tan”"@ tan”°@ tan”5@ e
n-1 n— 3 n—- 5
| ten". 20 = &o. (11)
(1.) If x = 2r + 1, the last term is easily seen to be
(— 1)" log (cos 6).
(2.) If n= 27, the two last terms may be represented
by (- 1) (tan 6 - 6).
Trigonometrical Transformations. 73
In a similar manner we have
d0 ~ (scot | do = -1 _ do ia
| tan"@ J tan" tan”*9 (n— 1) tan""0 Prec (
EXampiezs.
tan3
tan46 dé. Ans. ~~ — tan@+ 6.
I Ss
fan ae ee
ans@ ” “4 tant@ + Ttant@
+3
cot*e 49. ee a + coto +9.
a \ aw = —* _ 4 log (sin 6).
60. Trigonometrical Transformations.— Many ele-
mentary integrations are immediately reducible to one or
other of the preceding formule of reduction by aid of the
transformations given in Art. 26. For example, if we
m
assume @ = a tan 0, the expression transforms into
a+ a)
sin” 6 cos”-”-? dO (neglecting a constant multiplier).
In like manner, the substitution of a sin @ for x trans-
a™-"+1 sin” @ 0
forms the expression nto : and, if
P (a? - «3 cos” ,
: .,. cos” d
a = asec @, the expression ~ transforms into ————_—
are sin” 6
e—w)t
(neglecting the constant raultiplier).
A similar transformation may be applied in ‘other cases.
. a” de
For example, to find the integral of Geos
let xz = 2a. 8in?@, then dz = 4a sin 6 cos 0 dé,
and the transformed integral is
ant gf | sin Ode:
accordingly the formula of reduction is the same as that in (5) ;
74 Integration by Successive Reduction.
ExaMPtLes.
4 : 1-2
I. ae Ans, 3 eee (3 + 22%).
(1 — 2) 2 8
2 \ da 1 1 fi-# /f/i-#
z a =a » 308 ie 242
\ dx £ Pe
+ \@rer BETA het oF
[22 a At eine
4. (@ + 2) ” 2(@ 4a) 12 o—atan’ }-
i, (20x — a2) (2 4 32 rein |
5. Ga a2 ” ax — x?) Bk eS + 3a*sin” 77.
The integrals considered in this Article admit also of
a more direct treatment. We shall commence with the
following :—
. A a a 2 :
61. Cases in which ————— is immediately inte-
grable. (a + cx’)#
We have seen, in Art. 48, that the proposed expression is
integrable immediately when m is an odd positive integer.
Again, when m is an even integer, if we assume a + cx’
= «2°, the transformed expresssion 1s
n-m-3
-(s?-c) * ds
n-m-1 .
a 2 gr
This is immediately integrable when » - m-—1 is even
and positive, i.e. when m is either an even negative integer,
or an even positive integer, less than n — 1.
2-3
(s? — ¢)* dg
= becomes - ———j—-—,
For example, ————.
(a + cx)?
mT and
a 2 gr
accordingly is always integrable by this transformation,
since m is an odd integer, by hypothesis.
;
Binomial Differentials, 75
ExaMPLes.
x cx
" hes + x2) ae (a + cays fH ~3(a+ my
es { wdz Pd { 1 ex?
(a+ cas?) Oe (a + ca2)® (3 5 (a+ ox?) ;
= edz — (202 + 322)
(a+ ai a 3 (a2 + a)i ;
z ( dz
; a (a+ on)
The differentials considered in this Article are cases of a
more general class called binomial differentials.
62. Binomial Differentiais.—Expressions of the form
x™(a + bx" \? dx,
in which m,n, p denote any numbers, positive, negative, or
fractional, are called Binomial Differentials.
Such expressions can be immediately integrated in two
cases, which we proceed to determine by transformations
analogous to those adopted in the preceding Article :—
ak XL
(1). Let a+ be =; then 2 = (: \
b
a-)
and de = 3 (25*)" ds;
met y
hence a™(a + ba”)Pde = oe
nb”
+1. eae
Consequently, whenever oe es positive integer, the
transformed expression is immediately integrable after ex-
pansion by the Binomial Theorem.
76 Integration by Successive Reduction.
(2). Again, if we substitute ; for 2, the differential
becomes
— 7 -™? (ay + 0)? dy.
This is immediately integrable, as in the preceding
_ —(mpt+m+t). sao iste ;
case, whenever =e et) is a positive integer; i.e. when
m+iI
+p isa negative integer. In this latter case the inte-
gration is effected by the substitution of s for ax” + 6.
Examp.es.
5 2(1 + 23)) (8 — 2
| a dno 2 =
2 { us pan
" Jagat ” G+)
dx (1 + a4)t
3. | aq ar y ee,
dx 2a
iz + oye e (r+ at
‘When neither of the preceding processes is applicable, the
expression, if p be a fractional index, is, in general, incapable
of integration in a finite number of terms. Before proceed-
ing with this investigation we shall discuss a few simple
forms of integration by reduction, involving transcendental
functions.
63. Reduction of | one a" dat,
where 7 is an integer,
Integrating by parts, we have
aerm Hm
i pmax a: i i. n=l pmax .
- en” da [= em dv (13)
By successive applications of this formula the integral
is made to depend on | e”"® da, i.e. on —.
Reduction of { x” (log x)” da. 77
mae
Again, to find (5 dx.
Assuming u =e”, v = ae and integrating by
parts, we have
{= - — em m (54 (ea
a (n-s)a na) a” "a
By means of this the integral is reduced to depend on
| e"™ de
a
The value of this integral cannot be obtained in a finite
form; it however may be exhibited in the shape of an
infinite series ; for, expanding e”” and integrating each term
separately, we have
| e”" cee mx m x°
aT
[no 2g 223
= log 2+" + + &e. (15)
The integral of a*#"dz is immediately reducible to the
preceding, since a® = e”!8%, Consequently, by the substitu-
tion of log a for m in (13) and (14), we obtain the formule
of reduction for
| a* x" de and | = da.
In like manner we have immediately
fetarde = eta" +n feta de. (16)
64. Reduction of {2 (log 2x)" dz.
Let y = log z, and the integral reduces to that discussed
in the last Article.
The formula of reduction is
wt (log x)” __ fh [= (log #)*""da. (17)
m+ m+i
[em (log a)" de =
78 Integration by Successive Reduction.
EXAMPLES,
2 2,1
I. [ sends. Ans. a perpar=2ye 3 ; \.
a a a a
a! log” 1
2. [80g a) de. » {ogy -“8 + 3}.
et dx ex (1 I ~| I {=
(>. » -“latmts oS. i 2°
65. Reduction of = {x" cosaxda.
e"sinaz n :
Here | x” cos ax da = ———— - “| a" sin axda ;
a a
again
; zw cosaw n—I
fom sin az dx = — ————— + —— | 2” cos aw da,
hence
x” (av sinax + ncosar) n(n—1)
a
[x cos axdz = [em cos az da.
ae
The formula of reduction for « sin avdz can be obtained
in like manner.
Again, if we substitute y for sin“, the integral
f (sin) dx
transforms into
Sy” cosydy,
and accordingly its value can be found by the preceding
formula.
EXAmp.es.
I, [ 9 c0s edz, Ans. «3 sin « + 322 cosa —3.2.e8in%—3.2.1.c082.
2. iE sin dz.
Ans, —- cosx+ 4 sina+4.3.¢ cos4-4.3.2.u8n4%—4.3.2.1.cosz.
Reduction of { cos™x sin nada. 79
66. Reduction of [{e*” cos"xdz.
Integrating by parts, we get
cos" e"= 0 .
[e= cos"a dae = —— + “| e* cos” sin adx.
a
Again,
| 6 cos” sin ada
Cc a Sie Gy ‘
ge {cos*a# — (n — 1) cos"*xsin’ x} dx
e* cos*esing (n-1)( ., ‘A n
————e + er e* cos” *a da — a e* cos" ada :
substituting, and solving for f e*” cos"adx, we get
e* cos" 1x4 (a cosa +nsin
| et" eos” dx = ee ere)
C+n®
n(n — 1)
n=) e** cos” x dz. (18)
“+n
The form of reduction for e” sin”xdz can be obtained in
like manner.
67. Reduction of f{cos”z sin nada.
Integrating by parts, we get
5 cos”# cosnz mM s
[ cosm sin nada = — — "a mi cos”“'x cos nx sin x da:
replacing cos nz sin by sin nx cos # — sin (x — 1) a, after one
or two simple transformations we get
ioe cos” COS Nx
cos™# sin nada = — ——————_
m+n
cos” a sin(n — 1) ada. I
| (n-1) (19)
The mode of reduction for cosa cos nvdz, sin™x cos nadz,
and sina sin nadz can be easily found in like manner.
80 Integration by Successive Reduction.
Exampies.
‘ ersinz, , 2e%r
1 | eo sin*adz. Ans. (a sing — 2cosz) + ————5.-
44+ a a(4+a*)
: | fot panied cos’%cos4x cos@%.cos3% cos 2x
. n oa SN de
in 4x da. 3 7: ai
e? :
3- je cos? a dz. » — — (cos?# — sin 2” + 2). *
5
68. Reduction by Differentiation.— We shall now
return to the discussion of the integrals already considered in
Arts. 60 and 61; and commence with the reduction of the
expression ee This, as well as other formule of re-
(a + ca’)
duction of the same type, is best investigated-by the aid of a
previous differentiation.
Thus we have
d ea”
sees m-1 2\k FL = m—2 2) 3
at (a + cx’) (m - 1)a™* (a + en”) Gea
_ (m— 1) a" (a+ ca’) + ca™
. (a + ca*)2
_ (m— 1) aa 2 mex™ |
(a+ cx)” (a+ cn)?
hence, transposing and integrating, we obtain
ade a (a+ cx") (m — pal a™ de
la + ca)a me me (a + cx*)¥ 28)
By this formula the integral is reduced to one or more
dimensions; and by repetition of the same process the ex-
pression can be always integrated when m is a positive
integer.
The formula (20) evidently holds whether m be positive
x” da
I ——.. 1
Reduction of | a+ oo 8
or negative; accordingly, if we change m into — (m - 2), we
obtain, after transposing and dividing,
2) 2 3.
| dx (a+ cx)t — (m el da (oat
a™(a + ca) (m—1)axe™ ~ (m—1)ala"™ (a+ ca)
69. More generally, we have
£ {a1 (a + cx)”} = (m—1) x” (a + cx)" + 2ncx™ (a + cx”)
= (a+cn")" {(m—-1)ax™* + (m+ an—1)cx™).
Hence
a” (a + cx?) "dee = oli ead
(m+ 2n-1)e
(m aa 1)a m-2 2\ n-1
- Goel (a + cx*)""dx. (22)
Consequently, when m is positive the integral can be
reduced to one lower by two degrees. If m be negative,
the formula can be transformed as in the preceding Article,
and the integration reduced two degrees.
We next proceed to consider the case where n is negative.
Pp 8
Oo. Reduction of ae ae
. eductlh
7 (a + cx)”
m and n being both positive.
a” dx = wee ada
Here > a fe G+ oy"
wa ade _
eee le + x?)® *s
or ee ee v%,
2(n — 1)e(a + cx?)
and we get
a de an m1 a” de on
lear ~ 2 (n—1)e (a + ca*)™ +3 (n - Sal Gea 3)
8
82 Integration by Successive Reduction.
By successive applications of this form the integral admits
of being reduced to another of a simpler shape. We are not
able, however, to find the complete integral by this formula,
: r
unless when x is either an integer, or is of the form = where
r is an integer.
7 a” dx
71. Reduction of
(a+ 2bx + cx")
By differentiation, we have
s {a""\(a + 2ba + cx*)4} = (m— 1) a" (a + 2bu + cx*)A
a™\(b+ cr) — (m—1) aa? + (2m—1) ba™* + mex” |
(a+2bx+ cx)s — (a+ 2bu + c2")2 :
oe x de _ a (a+ 2bu + cx?)
ot (a + 2bx +ca*)t me
(2m—1)b a” de (m-1)a a dar (24)
me (a+ 2bx + cx?)a me J(a+2bx+cxu*)® +
This furnishes the formula of reduction for this case: by
successive applications of it the integral depends ultimately
on those of
ade dx
(a+ 2bx + cx*)8 ane (a+ 2bat cx*)®
These have been determined already in Arts. 9 and 12.
‘ 3 da
Again, the integral of Fates can be reduced to
the preceding form by making x = ,
72. The more general integral
a” da
(a+ 2bu + ca)”
admits of being treated in like manner,
a” da
(a + 2ba + cx*)"”
Reduction of | 83
For if a+ 26x + cx* be represented by 7, we have, by
differentiation,
a aon _ (m —1)e"* 2(n—- 1)e"7 (6 4+ ez)
Tr Te qT
_ (m— 1) a? (a+ 2be + ca") - 2(n- 1) 2! (b + cn)
= 7
_ (m-1) au 7 2b(m-n)a"* (2n—m —1) cu
7 ee i te i :
Hence, we get the formula of reduction
\=- - Fa 2(m—n)b [a dx
7” ~ (an—m-—1)eT™ ° (2n—m-1)e} 7
‘g (m-1)a (7. (25)
(2n-m-—1)e} I
By aid of this, the integral of tha when m is a positive |
integer, is made to depend on those of - and * Again,
it is easily seen that the integral of —~ is reduced to that of
de
To for
ade f(b +ex)de bf de
Tee) ae ere
-1 b( dx
= ———___ - -|=. 6
2(n—-1)e7T™ le Ge)
[6a]
84 Integration by Successive Reduction.
dz
. Reducti £) —————__.-
pase la + 2b + cx’)®
In order to reduce | we have
ee
= (Gr) _ ¢ _ 2n(b + cx)?
dz\ T* Le Ee
ce 2n(ac—b*) ane a2n(ac—8) (2n—-1)e
— pat pap
dx b+ cx an —1)e | dx
Hence | = mae bE + Sle (27)
By aid of this formula of reduction the integral of a can
be found whenever » is an integer, or when it is of the form
- (r being an integer).
dx
; i £ | —.——
74. Reduction o | (a+b cos a
when 1 is a positive integer.
U-a
Let Pap ehane then Shane, cos # = —
dx
Again, by differentiation, we have
@ (sin 2| _ Cosa (n-1)b sin’e
alee. ya U*
_ C08 e (n—1)b (m — 1)d cos?a |
~ wi v7 u” 2
substitute a “ for cos «in the numerators of these fractions,
and we get
d | I 4 , @m=1)b _ m-1 2(n—1)a
dz Um ~ b0™ 607 UU” 60" a 607
(n-1)@ —(n-2) ie (2n-3)a_ (n—-1)(@- 6)
bU" ~ b0 bu 60” &
a dx
] ——_—. 85
Reduction of | (Ga bose
Hence, transposing and integrating, we get
dae — bsine n (2n -—3)a da
\R-@e 1)(@-B)U™ (n- Mio | Ds
n-2 da
wane sa TF a
By this formula the proposed integral can be reduced to
depend on
=
a + bcosa’
the value of which has been found in Art. 18.
75. The integral considered in the last Article can also
be found by aid of a transformation, whenever a is greater
than 0, as follows :—
da _ dx
(a + bcos)” {(@ + b) cos? - + (a-6) sine"
2% n
de (: + tan’ =) dx
(4 cos’ * + Bsin’ a) (4 + B tan? sy
2 2 2
(where 4 =a+6, B=a- 0).
|A
Next, assume tan = = A tan ¢, then
(: + tan? ‘) da = 2 lz (1 + tan’) dg:
86 Integration by Successive Reduction.
and we get
(: + tant) dee jess do
Ey 22-2
A+ Btan’ a eae
_ 2(Boos’¢ + A sing)” * do
. (4B)
Hence, replacing 4 and B by a + } and a — 6, we get
dz _. ((@— boos 29)" do
lg +bcosaz)" : | (?@-B)ra © (29)
When n is a positive integer, the integral at the right-
hand side can be found by expanding (a - 0 cos 29)"", and
integrating each term separately by formula (4).
Again, if in (29) we make b=acosa, and 2$=¥y, we
obtain
de I
| (1 + Cos a cos a)” . mar aE coon mace g\"" ayy (39)
a x
where tan” = tan “tan —.
2 2 2
Hence, if we take o and “ as limits for z, we have
2 dz ;
| aeweoea 7 sin?”q = Ge cos a cos y)" dy.
J (2) de
$ (a)/a+ abe + ca*
We shall conclude this Chapter with the discussion of the
above form, where f(x) and ¢(«) are supposed rational alge-
braic functions of x.
If f(z) be of higher dimensions than ¢(z), the fraction
may be written in the form
J (2) R
—— =Q+—W.
¢ (2) ¢ (2)
76. Integration of —
Integration of PSE 87
a) fa + 2bx + ca
Again, since Q is of the form p + gz + ra’ + &e., the inte-
we can be found by the method of
gration of ————___
a+ 2be + ca?
Art. 71.
The fraction 5 can be decomposed by the method of
partial fractions (Chap. IT.). To any root a, which is not a
eer and the
corresponding term in the expression under discussion is
A de
(e — a) /a+ 2b0 + cx*
The method of integration of this has been given in Art. 13.
Next, to a multiple root correspond terms of the form
Bda
(@ — a)" a+ 2bx + ca
This is reducible to the form of Art. 71 on making
multiple root, corresponds a term of the form
@-a= =. Again, to a pair of imaginary roots corresponds
an expression of the form
(le + m)dx
{(@ — a)? + B*} a + 2ba + cx
If s be substituted for 2 — a, the transformed expression
may be written
(Lz + I) dz
(Oop) /d4 see 8
where ZL, Wf, A, B, C, are constants.
To integrate this form; assume* z = 6 tan (0 + y), where
* For this simple method of determining the integral in question I am
indebted to Mr. Cathcart.
88 Integration by Successive Reduction.
6 is a new variable, and y an arbitrary constant, and the
transformed expression is
{LB sin (0 + y) + cos (0 + y)\d8
B/ A cos' (+) + 2B cos(0 + y) sin (0 + y) + Cp? sin’*(8+ 7)
Again, the expression under the square root is easily
transformed into
4{4 + OB? + (4- OB?) cos 2(0 + y) + 2BB sin 2(0 + y)}
= [4+ Cf? + cos 20 {(A — CB?) cos 2y + 2BB sin 2y}
+sin 20 {2B cos 2y — (A - Cp?) sin 2) |.
Moreover, since y is perfectly arbitrary, it may be assumed
so as to satisfy the equation
2B cos 2y — (A - Cf’) sin 2y = 0, or tan 2y = AOR
and consequently the proposed expression is reducible to the
form
(L’ cos 0 + M’ sin 6)d0
WP + Qcos20
(in which L’, If’, P and Q are constants), or
I'd (sin 6) M’d (cos 6)
/ P+Q- 2Qsin?6 "ca 2Q cos? @”
each of which is immediately integrable.
Examples. 89
Examr.es,
i { cos®@ sin20d0. Ans. — : cos 9,
ae ‘
2. | sin?@ cos?@ dé. es ae ee
3 5
3 sin®@ cos56 dé. : ? cos? Z
. > & 4008 20 —— cos 20+ © cost 20}.
4 3 5
coe ae 8
4. (= 5 i + cos@ + log (tan :) 5
cos‘ 6 d@ 3 1 3
. sg —2 —_ -=
5 | ant” 93 (cos 6 : cos OF 36 5 8 tan (5).
6. j— ” (4-2 4 40 41) a
(I eae 5°3 3 (1 +28
1 =
} [aa + ban)P dx. ,, (a + ban) Pl {(p + 1) bun — a}
n(p + 1)(p + 2)0?
- o( ‘
8. { e* cos? x dx. a ee {3(sin2— cos.) + coste (3 sin # — cos x).
it { do = A do
ea sin™@cos"@ sin™1@ cosn"1@ j sin™-? @ cos" @’
determine the values of 4 and B by differentiation.
(2? — a)dx
10, aor ar
\ sin?@ d@
Ir,
6
——- Ans. 2 tan-—- 6.
(1 + cos 6)? 2
sin dp
cos2n-m ?
sin6 d@
(I + cos 6)"
12. Prove that the integral | transforms into 271 |
where 0 = 29.
90 Examples. ;
- | pt. ; a
a (a + b cos x)* a eae
—dsinz 2a : a-—b\} * «@
BIG pe —~* tan { ($55) stom:
- “(Pat bose) * (ewe (lard 2
oF neh ‘ . tan —
Loess a eye ‘cos 6d0 ws pin. BL ; 2\.
14. | eeecear Ans. ©) =" ~ = ten1| —]}.
(5 + 4 cos 6)? 9 5G O08 27 NN 3
eae Se ae (7 ae Ae SS
BE ay Bae:
v,
TS (sin le) dy = {(sina)*- 4.3 - inca) 32 T}y.
& +44/1 —2? ae sine (in) - - 3 2}.
a whe. Ey fo) Le os
J (cos x) dz 2
16. Prove = eae 14, that any expression of the oe Gt b cosa) 18
* gapable of ‘being integrated when f (cos x) consists of integral pone of cos #.
7. Show, i in like manner, that the expression re
a
floose, sit aides 1S Se 3°
Ort coke, : ieee
™~ cm ie
ae aN
can be ieereree when F (cos # # sin 2) consists “only of fiitegral powers of cos
and sin 2. 2 i : eG, es te
gen He Be ERIE a : ey
(A + Ba + Cx?) de ae
- ee P log e
: 7 a iF (a+ Bala + ba. oz2)> : ees - sole tee ee h
A es
find the values of P, Q, and R.
oe _ a Page Get -(a°—b)sin 39
Fg . | (@'cos?6-+ 0 ain? 6)? ~, rd aad) ‘e -.-44ab)?
” et * ie ay : 5 Bae ® Se
sitions teas o= /? tang. ipa ae eT ES ots
a ae ot : rads
is integrable in finite
At
zo. Find the values of » for ae
Tz qn a an
ms %
ean $e
terms.
21. Prove that eo eigtge Pi
7 dz
| o (1+ cosa Cost.
ee ee Pe
bg Got ge G6 See as Pretag rete On fovste le
UR Pe ee
& pertire Sf Med 2g trheon. tk] = BR, Jon
Ke Who oo (ME He forvent Uf tam otk
ns we he
2 ih Yxlé Blaky Ce Let
hen Ate Miaf Phe SO 6 easpakn
beg VIS Pus was ean Be
‘ Aaapeens tlh & : ea
oy Bodh Pi 2: Sao
lrk{b2/ Loy oii Ag few api. -
lgney richer K-& fox Bt
<. wey
e a “= -atZc pug
«- eee = 4, i 3
(EY Say ee
42 yg IE
Sete ee
bt OOS 2 fan ou
Prvte Cres pee ay
PEGE a
7 es et nim H2
a aes SP Posse
sul i nl, OeL. Car
ie ta ea Ws te i eeay f
( 91 )
CHAPTER IV.
INTEGRATION BY RATIONALIZATION.
77. Integration of Monomials.—TI{ an algebraic expres-
sion contain fractional powers of the variable x it can
evidently be rendered rational by assuming z = z", where n
is the least common multiple of the denominators of the
several fractional powers. By this means the integration of |
such expressions is reduced to that of rational functions.
For example, to find
| (1 + wt) dex
1+ a
Let x = s‘, and the transformed expression is
(1 + 2)dz
| r+?
Consequently the value of the integral is
$
7 + 2a! — gat + 4 tan (et) — 2 log (1 + 2%).
Again, any algebraic expression containing integral
powers of x along with irrational powers of an expression
of the form a + dz is immediately reduced to the preceding,
by the substitution of s for a + bz.
EXAMPLEs.
: =
I. \ 7 = Ans. ae * [50° + 60? + 8 + 16].
a-1 ‘
\ ad 2 (2a + ba)
(a + b2)8 me J at be
3. a » log (e@t+/e-1)- ? tan (eter),
“I
etl v3 v3
92 Integration by Rationalization.
78. Rationalization of F (2, ,/a + 2bx + cx*)de. It
has been observed (Art. 28) that the integration, in a finite
form of irrational expressions containing powers of x beyond
the second, is in general impossible without introducing new
transcendental functions. We shall accordingly restrict our .
investigation to the case of an algebraic function containing
a single radical of the form / a + 2bx + cx*, where a, b, ¢ are
any constants, positive or negative.
Integrals of this form have been already treated by the
method of Reduction (Art. 76). We shall discuss them here
by the method of rationalization.
ote da
The expression* ‘—— ————____—-
$(%) ,/a+ abe + cx
tional in several ways, which we propose to consider in
order :—
(1). Assume fat 2barce=s-—a2 fe. (1)
can be made ra-
Then a+ 2be = 8" - 2a2,/c; .*. bde = dz — fc (ads + edz),
or de(b+%./c) = ds(z —2/e) = deat 2ba + cx;
dx dz
Ss Se 2
fatr2ber+ce bt+sJfe @)
2-4
Baer
This substitution obviously renders the proposed ex-
pression rational ; and its integration is reducible to that of
the class considered in Chapter IT.
Also (3)
* Tt will be shown subsequently that the integration of all expressions of
the form
F(z, Va + 2bx + cx?) dx
is reducible to that of the above when F is a rational algebraic function.
It may also be observed that, in general, the most expeditious method of in-
tegration in practice is that of successive Reduction (Arts. 71, 72, 76).
|
Rationalization of F(x, /a+ 2bx + cx’) de, 93
When 3 = o, we get
d. d
a ee, and w = (see Art. 9).
/a+ecn s/c cn
By aid of the preceding substitution the expression
dx
Oo (Art.
Conver tinae 13)
transforms into - ‘
s° — 2ep o/c — a — 2b
dz
F le to find Ss eS
or example, Gea
s-—1 dx 2dz
Here w= and Soy ale dc ee
28” (p+qx)/i+e ge + 2pe—q’
{ dee Seat ae (Ete “£).
“S(ptge) fit ae Spr¢ e get pt/p+ge
When the coefficient ¢ is negative the preceding method
introduces imaginaries: we proceed to other transformations
in which they are avoided.
(2). Assume* /a+ 2ba + ca = /a + as. (4)
Squaring both sides, we get immediately
2b+ cx = 28/4 + #27;
*, da(e — 8°) = 2ds(,/a+ az) = 2ds/a+ 2bu + ca.
di di:
Hence eo ‘ (5)
fat 2bat+er®? ce-#
* This is reducible to the preceding, by changing x into ; and then em-
ploving the former transformation.
94 Integration by Rationalization.
And t= 2(ev/a - 8) (6)
c-2
This substitution also evidently renders the proposed
expression rational, provided a be positive.
For example, to find
| dz
af 1-2
Assume \/1 —- 2 =1- xz, and we get
dae dg Laufiae
lon — 2 -|F rege = log ($==* \
(3). Again, when the roots of a+ 2bx + cx’ are real, there
is another method of transformation.
For, let a and £ be the roots, and the radical becomes
of the form
V¢(e-a)(~ - B), or e(v—a)(B - 2),
according as the coefficient of a is positive or negative.
In the former case, assume /z—a=s,/z — B, and we
get
_a— Be | _a-B. , de _ 2%dz
=~ ge hence #- 6 = ——3; “e-p 1-8
Accordingly
da = da _ 2 de (7)
V/¢(#—a) (2-2) s(e-B)/e /er-s
In the latter case, let ./7 —a =84/, B - #, and we get
a + Bs?
1+3??
t=
dz 2 dg
Vc (ea) (8-2) . fe +s (8)
and
Rationalization of F(a, / a + 2b« + cx*) da. 95
For example, the integral
| dat
(p +92) f1-a
transforms into
| 2ds
(p+g)s+p-g
: 2-1
on making w =
S+1
The student can compare this method of integrating the
preceding example with that of Art. 13, and he will find no
difficulty in identifying the results.
It may be observed that in the application of the fore-
going methods it is advisable that the student should in each
case select whichever method avoids the introduction of
imaginaries.
Thus, as already observed, the first should be em-
ployed only when ¢ is positive: in like manner, the second
requires a to be positive; and the third, that the roots
be real.
_ Itis easily seen that when a and ¢ are both negative, the
roots must be real; for the expression
eee 2 ee _ A\2
af be 2bx — cx’, or Jo ae oe PF
is imaginary for all real values of x unless 8? — ac is positive ;
i.e. unless the roots are real.
Accordingly, the third method is always applicable when
the other two fail.
From the preceding investigation it follows that the
expression
F(a, ./ a+ 2ba + cx”) dx
can be always rationalized; F' denoting a rational algebraic
function of x and of ,/a + 2bx + ca.
96 Integration by Rationalization.
EXAmMPLes.
| dz I eae 2-2 -2@
TF Ans. ———___—_ —
(2 + 32) 4-2 es pA ae —a
2 { dz
Ji@+e a tae
Assume ¢ = (a? + 2?) + 2, and we get for the value of the proposed integral
2a?
52-25
5a
3 [ae at/ar at Ans, 22 +af2te 2.
3 Vg4/2H + fate 2+ a2
{= {(@ + 0)h + w}nde.
Making the same assumption as in Ex. 2, the transformed expression is
—@)™ (a +27) de
2mtl gm-nt2 ’
which is immediately integrable when m is a positive integer.
de [rt ath + at (rg ok bat
+ J {(1+22)h—a}o" a 2(n +1) - 2 (m — 1)
2\} nd:
6. { {(1 ~ a z, - {(r + 2) + ofr
dx
Z \ FSS Vw
Let fa + 26a + ca® + afe= z, then, as in Art. 78, we get
Sa oy
Vat2batex® bts ge
hence the proposed expression transforms into
dz
bi e/a
-. &e.
General Investigation. 97
79. General Investigation.—The following more
general investigation may be worthy of the notice of the
student.
Let R denote the quadratic expression a + 26x + ca’;
then, since the even powers of »/F are rational, and the odd
contain RF as a factor, any rational algebraic function of x
and of RB can evidently be reduced to the form
P+Q/SR
P+Q/R
where P, Q, P’, Q are rational algebraic functions of x.
On multiplying the numerator and denominator of this
fraction by the complementary surd P’ - Q’ \/R, the deno-
minator becomes rational, and the resulting expression may
be written in the form
M+N/R,
where If and XW are rational functions.
The integration of Mdzr is effected by the methods of
Chapter IT. i
NR dx
Also { Rdx = love ;
which is of the form
| S(e)de
(2) /a-+ 2bx + ca
Let, as before, /a+ 2bu + ca = V ¢(@ — a)(# - 8), and
A + 2us + v2”
N+ 2p'8+0's"
substitute instead of #, when the radical becomes
Vef{a—ar’t 2 (u—apm')2 + (v— av’) 27} {A — BA +2 (mB) 2+ (v— By) 27}
Ao + 2p'e + v2, (
9
Again, if the quadratic factors under this radical be made
each a perfect square, the expression obviously becomes
rational.
[7]
98 Integration by Rationalization.
The simplest method of fulfilling these conditions is by
reducing one factor to a constant, and the other to the term
containing 3”.
Accordingly, let
A - aX’ =0, fe — an’ = 0, pw - Bul =0, v— Pv =03
or w=0, w =0, A= aN’, v= By’.
On making these substitutions the expression (9) becomes
3 a , 122
(B = a)a/— Nv while z = ee
Vave Nave
In order that »/— cd’v’ should be real, X’ and v’ must have
opposite signs when ¢ is positive, and the same sign when ¢
is negative.
It is also easily seen that without loss* of generality we
may assume X’=1, and vy =+1.
- Bs
=, and when
=
. "ye a
Hence, when ¢ is positive, we get # = :
2
c is negative, x = —_
These agree with the third transformation in the preced-
ing Article.
More generally, when the factors in (9) are each squares,
we must have
(u — ap)’ — (A — ad’) (v — av’) = 0,
or we Av + (Av’ + vr’ - 2up’) a+ (u” = vy) a’ =0, (10)
and a similar equation with 3 instead of a.
Moreover, by hypothesis, a satisfies the equation
a+ 2ba+ca=0.
igi "a
* For the substitution of y* for a transforms
an’ + Br's? , a+ By?
7 into ee
A+ VE I+y
", &.
General Investigation. 99
Accordingly (10) is satisfied if we assume the constants
A, pw, &c., so as to satisfy the equations
2
w—-Av =a, Nv+dAv-2pp'=2), w?-Nv' =e. (11)
Again, solving for s from the equation
a(X’ + 2p’s + v's”) =A + 2us + v3’, (12)
we obtain
(v-av’) s+ w— ap! =/ p? — dv+ (Av + N= 2p’) w+ (ue- Ny’) a?
=./a+ 2bu + cx. (13)
Also, by differentiation, we get from (12),
(A' + 2's + v's") da= 2{ut vs—a(y'+r’z)}ds
=2/a+ 2ba + cx ds;
~ a 2dz
" Sas aberca N + 2p'a + v's"
(14)
Now, since we have but three equations (11) connecting
d, w, &e., they can be satisfied in an indefinite number of
ways.
We proceed to consider the simplest cases for real trans-
formations.
(1). Let a be positive, and we may assume vy = 0, and
pw = 03 this gives
w= if th Av’ = 2b, Nv =e.
Again, without loss of generality, we may assume v’=— 1,
which gives
A =- 2b, XN’ =c; whence # = aaa,
da 2d
a+ 2be+ co e-8
These agree with the results in (5) and (6).
[7a]
and
100 Integration by Rationalization.
(2). In like manner, if ¢ be positive we may assume
v=0, w=0, and v=1,
which gives
w=V/c, A=-a, and N=26;
2-4 dx dz
“= ————, and ———______- ss
2(b + 2/c) /a+ 2b2 +00 b+a/e
as in (2) and (3).
It may be observed that since these results do not contain
the roots a and (3, they hold whether these roots be real or
imaginary; as already shown in Art. 78.
It is easily seen that if we make u =o, and yp’ = 0, we
get the third transformation.
80. If the expression to be integrated be of the form
Se) de
J a+ 2bu + cx
where f(x) is a rational algebraic function of 2, it is often
more convenient to proceed as follows :—
The substitution of s -: for # transforms the proposed
f(s - :) dz
into —_., where a’=
Jf a + cx
If the even and odd powers be separated in the expan-
ac —
sion of f (= - : it can plainly be written in the form
p(s") + ab(2"),
and the proposed integral becomes
| ¢ (3°) ds a ap (8) dz
Sioa life
The former of these is rationalized (Art. 24), by making
Jd + cz = ys, and the latter by making /@ + cs? = y,
Case of a Recurring Biquadratic under the Radical Sign. 101
It may be observed that in general the expression
S (@) = dx
(22) o/a + ox
is also made rational by the transformation
fa + cH = wy.
81. Case of a Recurring Biquadratic under the
Radical Sign.—As the solution of a recurring equation of
the fourth degree is immediately reducible to that of a
quadratic, it is natural to consider in what case an Elliptic
Integral (Art. 28), in which the biquadratic under the radi-
cal sign is recurring, is reducible by the corresponding sub-
stitution.
Writing the expression in the form
(a) dx oe (2) dw
Sa yee Se cf CU
J a+ abs + cu? + 2ba + ant 2 Ja(ers z)+20(a+ 5) +e
? I ; re
and, assuming # + ria the radical becomes ,/az* + 262+ ¢— 24;
:
and also z (« - =) = dz.
x
Consequently, in order that the transformed expression
should be of the required type, it is obvious that ¢(z) must
be reducible to the form
eee
( ~2\7(e + 2\ar
a+ 2bx + cx + 202° + ax'
___ SF (8) de
J/ as’ + 2b8+e - 2a
In this case
transforms into
102 Integration by Rationalization.
In like manner, the expression
€ +2\7( ~2)ae
/ a+ 2ba + ca® — 2a + ant
transforms into [5——“—— Sed —— by the assumption
az’ — 2be+2ate
a--=8.
@
When 6 = 0 the expression can in some cases be reduced
by assuming either
I I
e+ or #-—=8.
a a
Examples.
/
a — 1)d. 2 I
las zee dei ee
aft + of x
(2 + 1)de wt —144/i+ at
» log ———_—_—_—_-
oe &
1-2
. = . is sin7! 2),
ays Fi I+ 2%
1+2? de I J tt att arf/2
Se Wi — log —___—_
I~ 224/14 at 2 1-2
This and the preceding were given by Euler (Cale. Int., tom. 4):
connexion, however, of their solution with the method of recurring artes
does not appear to have been pointed out by him.
4_ 1d 2
er \ (2 de tne, WEE +1
a At ee %
Let e+ oa &e.
6 j— "4
al (@ + at + 1) (a + Bu + 1)
WEES ETE LETIES
Ans. 2 log =
(1-2 )dz ‘ ( x )
——=_=—. Ans. sin 3
| aS 1+ 2
yt rs y . -UTY¥Y) 1 aan ge
=<[\F : ? 424 _,
K | Le. k= aoe ~ ep} ¥
ty = (1- a4, “AOS ng 4 ae ZOD ay
fide GF
= ob ~ ata les
[/-4O) 4
la Ga
C- A-00o/ &
2% Pee
ef ay 2 / a
- (/— gt Ge
ett Z
[-~ 4% Bf nyt. vfs vad
ag T W-2
nw.
/— 24) %
Cro AF
| EZ’.
bate
" &¥ etre fas P . he ag
~ rp en
PO PEM G te hn the
Lo “ f-a8
eat Ma ape
(AE fin poke
te (a. 9ftera — £6 4 /-AY ahs
5 ee ‘wee
: ong (/- a**G 4 feng Lg fe i
¥ 3 . B
i? pee fo
wane
f-he EO) ae
: }
—_~_—-
Examples. 103
adx 3 (252 — 3a) a
\ er Ans. = (a + bz)a.
9. f= de re ijog Vite tattoo
I afi +a?+ at V3 I — 2?
ax . ind x
ec loamy ? en ram)”
Assume e= (I+ aj sin 6, &e.
»
az {
Ir eee SEE ete oT ee in-!
; — eens .
le + a?n){(1 + 22a)” — a2}h as aaa
‘3 j a dx
: (I+ a)i+(+a)¥
Assume I+2=28,
f a? dz
3 Ja-aatey
4 2 #4
Ans. J jg eas ct "
4 2 I-22 4 2 2/2
+ a4) d.
ve jee ) si Ans. =
(1 —2)8 (1 — 24)s
eee
as
s sa © Mdns, bs ag p( Cet evs), : tan" ave
oe cu Bayh 2 et ,-# 2/2 Sit at
i 1-2 os dt
I+ 200+ 2 ,/7 4200 + 2ba® + 2008 + oh re
17 a go ss
1+ ax EEE
nae 2 2 ad
ge 2e=0) a) 4TH 2008 + ata , when ¢ >a.
Faaa5™ I + ax?
I cit (2CE9), when o> 0
Jaga I+az
( 104 )
CHAPTER V.
MISCELLANEOUS EXAMPLES OF INTEGRATION.
(A cosv+ Bsinx + C) dx
82. Integration of G
acosa+Osing+c
: : du
Let acosz +d sing +c=4, then -asinw + 6 cose = re
Next assume
F du
A cose + Bsing +O = Aut pT + Dy
and, equating coefficients, we have
A=)a+ pd, B=Xb- ua, C=det+v.
Solving for A, pu, v, We get
_ Aa+ Bb Ab - Ba v= co 44s Boje
Pape? ON eae? e+e ¢
Hence
(A cosa + B sina + C) dex
acosx + bsing+e¢
z (Aa + Bb)w | Ab - Ba
eae agp 18 (4 cos a+ 6 sing + ¢)
(a? + b?) C—-(Aat Boje dz
at 3 .
a’ + 0 ae ees
The latter integral can be readily found; for, if we make
a=rcosa,b=rsina, we get
acosx+ 6 sinw = r (cose cosa +sin# sin a) = rcos (# — a).
J (cos 2, sin x) dx
acos’a+bsineg+¢
Integration of 105
On making 2 ~ a = 0, the integral reduces to the form con-
sidered in Art. 18.
As a simple example, let us take
(= B tan x) dx
a+btanz
ie A+ Btanz Acose+ Bsine
a+btane acose+b sing’
and we evidently have
[A+B tone) ae (da+ Bb)e Abd- Bay bei
at+btane @4+8 gap Ce One oene)
J (cosa, sina)de _
83. Integration of ,
3 ° acos#+obsing +c’
where fis a rational algebraic function, not involving frac-
tions.
As in the preceding Article, assume x = 9+ a, and the
expression becomes of tke form
¢ (cos 8, sin 6) dO
Acos0+B ~
Again, since sin?@ = 1-cos’6, any integral function of sin 8
and cos # can be transformed into another of the form
. (cos @) + sin 0 ¢2(cos 8).
Accordingly, the proposed expression is reducible to
g.(cos 0)d0 2 $2(cos 0) sin 00
Acos#+B Acos6+B
The latter is immediately integrable, by assuming
Acos0+ B=s.
To integrate the former, we divide by A cos 0+ B, and
integrate each term separately.
106 Miscellaneous Examples of Integration.
84. Integration of
J (cos x) da
(a; + , cos 2) (a2 + b, cos)... «(dn + By C08 @)’
where f, as before, denotes a rational algebraic function.
Substitute s for cos « and decompose
J (3)
(ay + b18) (a2 + baz)... «(dn + On8)
by the method of partial fractions: then the expression to be
integrated reduces to the sum of a number of terms of the form
dz
A+ Bcos2’
each of which can be immediately integrated.
EXxaMP_es.
tain z
dx I r+sing 3 2
“ I. j ——_—_—_——.._ Ans. — log | ———_}- = tan"! \ —— }.
cos “(5 + 3 cos 2) 10 t-sinz/ I0 2
dx
' 2. | ataeedomay when a > 6.
pee 6-—acosz _ B souk (ee)
(?-0?)sing (a - #2 a+bcosz!
dz tanz 3b w a\ BP dz
nme | scat booas a eee (+2)+5| pee
85. Integration of { f(x) + (2) je da.
The expression e* Pdz is immediately integrable whenever
P can be divided into the sum of two functions, one of which
is the derived of the other.
For, let P= f(e) +(e),
then fetPda = fet f(a) dx + f ef" (a) de.
Differentiation under the Sign of Integration. 107
Again, integrating by parts, we have
ff (2) de = f(x) & - f ef (a) dx
Accordingly,
(f(x) + f'(a)} & de = e* f(a).
For instance, to find
» #
fe G +aP cen
I I
Here oF ar a ea
(1+a@)? 1+a@ (1 +2)’
consequently the value of the proposed integral is aay
Lote
ExampLes.
| e* (cos xv + sin x) dx. Ans, e* sing.
I +e" ©
. fet z 9 log.
sf w+ a z—I
ee ——.
(e+1)? 1 ar
I - 2\2 ez
. e i —.
eS | (3) # 7 1+ 2
86. Differentiation under the Sign of Integra-
tion.—The integral of any expression of the form 9(-, a) dz,
where a is independent of 2, is obviously a function of a as
well as of a.
Suppose the integral to be denoted by F(z, a), i.e. let
F(a, a) = $(2, a) da,
then © (F(ea)} = 9(2 9).
108 Miscellaneous Examples of Integration.
Again, differentiating both sides with respect to a, we
have, since # and a are independent,
a. F(a,a)_ d. o(#, a)
dadz - da 2
or (Art. 119, Diff. Calc.),
d (ad. F(a, a) _@. $(#, a)
z\ da )- da
Consequently, integrating with respect to x, we get
d. F(z, a) _ d. $(a, a)
da al a
Le. pax a) dx - | 29 dg. (1)
In other words, if
u=Jo(a, ade.
du (do
then ao \Z da,
provided a be independent of ; in which case, accordingly, it
is permitted to differentiate under the sign of integration.
By continuing the same process of reasoning we obviously
get
in, nm
du -| o(x, a) die, (2)
da" da”
where u = { ¢(z, a)dz, a being independent of #.
For example, if the equation
[ eae = ee
a
Integration under the Sign of Integration. 109
be differentiated » times with respect to a, we get
feo (ate
ne (eat
(See Art. 49, Diff. Calc.).
Again, in Art. 21 we have seen that
| Pipes o= (a sin ma — m COS mit)
m + a?
Accordingly,
n ax s. is
Penge) oe ee
da m + a?
We now proceed to consider the inverse process, namely,
the method of integration under the sign of integration.
87. Integration under the Sign of Integration.—
If in the last Article we suppose ¢(x, a) to be the derived
with respect to a of another function », ie. if
$ (2, a) = o>
then » =f (a, a) da.
Also by the preceding Article we have
@ (fae) i \Z ie -| 6 Sae2 Fea
Hence [eae -|Fe, a) da. -
In other words, if
F(2, a) = fea a) da,
110 Miscellaneous Examples of Integration.
ea. [Fe Sane: fc [o(, a) da] de. (3)
It may be remarked that the results established in this
and in the preceding Article are chiefly of importance in
connexion with definite integrals. Some examples of such
application will be given in the next Chapter.
88. Integration by Infinite Series.—It has been
already observed that in most cases we fail in exhibiting the
integral of any proposed expression in finite terms. In such
cases, however, we can often represent the integral in the
form of a series containing an infinite number of terms.
An example of an integral exhibited in such a form has
been given in Art. 63.
The simplest mode of seeking the integral of /(x)\dx in the
form of an infinite series consists in expanding f(z) in a
series of ascending powers of w, and integrating each term
separately: then if the series thus obtained be convergent, it
represents the integral proposed.
It can be easily seen that if the expansion of f(x) be a
ee series, that of | f(x)dx is also convergent.
or let
S (@) = + e+ e+... dna” + &e,
then
ae axe Aye
[7@)de = ae +4 a w+...
m+
Now (Diff. Cale., Art. 73), the expression for f(x) is
convergent whenever “:” is less than unity for all values
nl
of » beyond a certain number; and the latter series is con-
“So ‘ye less than unity, under the same
vergent provided.
M+ I Any
conditions.
Accordingly, the latter series is convergent whenever the
former is so.
Integration by Infinite Series. 111
EXxaMptLes.
a: tro 1.32% 1.3.5 x8
t | oi pete ee ge eS Tie
Yi-@® 1 26 2.411 2.4.6 16
coe in? int
W012 { Ma 2fiine (1 te, Sey.)
J sine a 5 2.4 9
ean —g)c gn
a eg Re ge +e.)
qm+n T.2.¢g° m+ 2n
o> | (I + cor) gamle =am (= +
89. Expansion of | log (1 + 2 mcosx + m’) da.
We shall conclude by showing that the integral
[log (1 + 2mcosa + m’) dx
can be exhibited in the form of an infinite series.
For we have
I + 2mcose@ + m= (1+ me*) (1 + mee’),
Hence
log (1 + 2mcose + m*) = log (1 + me) + log (1 + me*”)
t= gta 0 7 aS WG
=m (e+ e*") Sr ("1 + '*"1) + Ke.
m ms
= 2 | moose —— p82 arn 008 3a — &e. }.
Accordingly
: sin 2” sin 3a
ftog(+2m cost?) ema m sin e—m? 552" 4m? er -) (4)
This series becomes divergent when m is greater than
unity. Inthat case, however, the corresponding series can be
easily obtained.
112 Miscellaneous Examples of Integration.
outs eo
For I + 2m-cosie + mt =m*( 1 + \Q+ )
m m
and accordingly
cos# COS 2” COS 3a
log (1 + 2mcosx +m?) =2logm+2 (= = tt -&e.)
Consequently, when m > 1, we have
4 sing sin27 sin3e
log (1+2mcos2+m’)du=2xlogm+2| —-——; +; }-
m
From the above it is easily seen that the integral
flog (1 + acosz) dx
can be exhibited in the form of an infinite series when a is
we have
less than unity: for making a = eal 5
1 +m
log (1 + acosz) = log (1 + 2m cos + m?) — log (1 + m’).
The relation between m and a admits of being exhibited
in a simple form; for let a = sin a, and we get m = tan .
Making this substitution in (4), we get
| log (1 + sin a cos 2) dx = 2x log (cos ‘)
2?
+2 (tan *sine — tan? tO” + &. ), (3)
2 2
Examples. 113
EXampres.
(2.008 # + 3 sin x) de Ans. © — log (30082 + 2 sin 2).
3cosx+2sinz I 13
tan“ (tan 6/2).
1 — sin‘ 2 2/2
bea, | a - a “tan + :
et (m8 + @ + I) da ea
{ 3. a+ » af tee aoe
I
it 4. Seah mains 7g 08(t + 0080) + Tog (1-008 4) ~ 2 log (1~ 2 c08 0).
0 . 6 =e
sin ¢ tan Sao 1~ sin ‘ 2 sin? 4 I
le 5. ; =log + —= log ;
08 1+ sin > 2 V2 sins — 1
6. When 2? < 1, prove that
a. @ 1% 6.3989 1.3.5 28
a [nts ee ast,
V+ a T 25 2.40 2.4.6 13
a ye.
fitch 2 2505 2 gon 2.4. 6 13x18
tt 7. Prove that
| dz 5 ae a I.31 Kav 3 og
eat abt+«2 a (b+2)?
=e ib to an ne oe oe
le A {tog (5+) +$ ao Se \,
and determine when the series is convergent, and when divergent.
y 8. Prove that
a «pl + M48
he 4 he sin’ w® «a2 +1? sin” *y
———— sin* wdw = a ‘
2 w+I 1.2 Bt+3
(a2 + 1)(a? + 32) sin? w
1.2.3-4 mts
Substitute w for sin-!z in the expansion of OE Dig Cale., Art. 87), &e.
Aw —rAw A sin’*? ala? 22) si Bk
eos [ant ote 4 AQ? + 2%) sin!
2 I +2 1.2.3 pt+4
A(A? + 22) (a2 + 42) sin”? wy
1.2.3-4.5 pmpt+6
[8]
+ &e.
( 114 )
CHAPTER VI.
DEFINITE INTEGRALS.
go. Integration regarded as Summation.— We have in
the commencement observed that the process of integration
may be regarded as that of finding the limit of the sum of
the series of values of a differential / (x) dz, when x varies by
indefinitely small increments from any one assigned value to
another.
It is in this aspect that the practical importance of inte-
gration chiefly consists. For example, in seeking the area of
a curve, we conceive it divided into an indefinite number of
suitable elementary areas, of which we seek to determine the
sum by a process of integration. Applications of finding
areas by this method will be given in the next Chapter.
We now proceed to show more fully than in Chapter I.
the connexion between the process of integration regarded
from this point of view and that from which we have hitherto
considered it.
Suppose ¢ (#) to represent a function of # which is finite
and continuous for all values of x between the limits X anda;
suppose also that X — a is divided into xz intervals 2, — %,
%— 1, W3— %,... X—a#,4; then by definition (Diff. Calc.,
Art. 6), we have
$ (21) — $(@)
% — &%
= $ (%)
in the limit when m = 2; accordingly we have
(#1) — (ao) = (a1 ~ 0)("(we) + &0),
Limits of Integration. 115
where « becomes infinitely small along with x, - a. Hence
we may write
(#1) — (#0) = (1 — a) {p" (wo) + eo},
(2) — $ (m) = (@2 — %1) (p’(a) + 4},
G(s) — (2) = (@s — a2) ($"(m2) + &2},
$(X) — b (@n1) = (X - ans) {$' (ena) + ena},
where &, 4... 1-1 become evanescent when the intervals are
taken as infinitely small.
By addition, we have
$(X) — p (ao) = (#1 — %) G’ (ao) + (@ — m1) g(a) +...
+ (X- Ln-1) # (@n-1) + (a = £0) Eg + (a2 — a) Et... (X - &y1) En-1-
Now if n denote the greatest of the quantities &, &,.. . &a1)
the latter portion of the right-hand side is evidently less
than (X — x,)y; and accordingly becomes evanescent ulti-
mately (compare Diff. Cale., Art. 39).
Hence
¢(X) — $ (a) = limit of [ (a1 - 20) $’ (a) + (t2- m1) o’(a1) +...
+ (X- a2) ¢'(#na)], (1)
when » is increased indefinitely.
This result can also be written in the form
$(X) — $ (m) = B9"(@) dey,
where the sign of summation = is supposed to extend through
all values of # between the limits x and X.
gl. Definite Integrals, Limits of Integration.—
The result just arrived at, as already stated in Art. 31, is
written in the form
x
P(X) ~ Flas) = |" 7") de (2)
0
where X is called the superior, and a the inferior limit of the
integral.
[8a]
116 Definite Integrals.
Again, the expression
x
| (x) dx
a)
is called the definite incegral of o(~)dx between the limits
and X, and represents the limit of the sum of the infinitely
small elements ¢ (x) dz, taken between the proposed limits.
’ From equation (1) we see that the limit of
(0,24) fan) + (0 — 1) fan) +0 + (XK An) Ca),
when 2, — 2% %:—%,...X—4y1 become evanescent, is got
by finding the integral of /’(x) dx (i.e. the function of which
7’(a) is the derived), and substituting the limits z, X for # in
it, and subtracting the value for the lower limit from that for
the upper.
If we write x instead of X in (z) we have
fO=fe)\= i, P(e) de, (3)
in which the upper limit* 2 may be regarded as variable.
Again, as the lower limit 2 may be assumed arbitrarily, (a)
may have any value, and may be regarded as an arbitrary
constant. This agrees with the results hitherto arrived at.
In contradistinction, the name indefinite integrals is often
applied to integrals such as have been considered in the pre-
vious chapters, in which the form of the function is merely
taken into account, without regard to any assigned limits.
As already observed, the definite integral of any expres-
sion between assigned limits can be at once found whenever
the indefinite integral is known.
A few easy examples are added for illustration.
* The student should observe that in (3) the letter x which -tands for the
superior limit and the x in the element f’(x)dxz must be considered as being
entirely distinct. The want of attention to this distinction often causes much
confusion in the mind of the beginner.
Elementary Examples. 117
Exampres.
d bn+l — qnek
1. { an da. Ans. eee ONS
a A+1
Tv
qsin @d@ =
# [c= cos? 6° ” fa -1.
3. i Fae ” +a (/ 2-3.
|
. °c da i
. 0 @t+ae Oa
5 ie =
0 aS Phat
me a
6. | ear dx (a positive). oo
o a
7 = = -
. ee 7? 2sing
8 (Ss eM
0 sean ” sin ¢
* m
: e-9" gin max dx. =
|, 9 ar + me
a
10. | €9% cos mx dz. » =F
9 at +m
——— _, when ac ~ 8 is positive.
ae (” eee J a —
92. To prove that
os eae, ee ig Feta oD
\\* es w= |9 (1a) de aei)..(atm—1)
when mand n are positive, and m is an integer.
118 Definite Integrals.
The first relation is evident from (34), Art. 32.
Again, integrating by parts, we have
m—tI
jen (1 — a)" da = = (1 -a)™1 + [anc — a)" dz,
Moreover, since » and m — 1 are positive, the term
z”(1 —a)"" vanishes for both limits ;
1 7p
2 | xo (1— a)" de = “| an (1 — )™? da.
0 0
The repeated application of this formula reduces the in-
1
tegral to depend on | #*"~*dz, the value of which is op
0
Hence we have
” iglat - £6223 enn G= 7)
he ee) nO ea cas = 8) (4)
This formula, combined with the equation
1
1
| a (1 —2)"" dx = | w™ (1 — x)" da,
e 0
shows that when either m or » is an integer the definite
integral
| x (1 — a)" dx
0
oan be easily evaluated.
When m and n are both fractional, the preceding is one
of the most important definite integrals in analysis.
Woe purpose in a subsequent part of the Chapter to give
an investigation of some of its simplest properties.
Examples.
1
I. ( (1 — xi dz. Aus. re
“0 +7611. 13
218
1
‘ 4(1 — x)tde. ae
2 [2c a)t da ee a ee
Elementary Examples. 119
wv Tv
2, 2
93. Walues of | sin” «dz ana | cos* edz.
0 0
One of the simplest and most useful applications of
definite integration is to the case of the circular integrals
considered in the commencement of Chapter III.
We begin with the simple case of
T
i sin” a da.
0
If in the equation (Art. 56)
[sine eee eBay | simmer
we take o and * for limits, the term aa vanishes
for both limits, and we have
if sin"a dx =" —* [sine de,
0 n 0
Now, if » be an integer, the definite integral can be
easily obtained; its form, however, depends on whether the
index n is even or odd.
(1). Suppose the index even, and represented by 2m,
then
Tv wv
2, 2m-1f?.
sin?” eda = sin?” x dx.
‘ 2m fo
Similarly,
% uw
a 2m — Bs
sin?" ¢ da = 3} sin?m-4 dr ;
0 2m— 2},
and by successive application of the formula, we get
Da oe Sahin C3) &
[sia CO ee 2m "2. (5)
120 Definite Integrals.
(2). Suppose the index odd, and represented by 2m + 1,
then
[sine ede = ae [sini
P 2m +1}
Hence, it is easily seen that
i. 2.4.6.... 2m
sin?" ¢ dy = ————__-_—______., (6)
6 3-5.7....(2m+1)
Again, it is evident from (35), Art. 32, that
2 as
| cos” eda = | sin” x dz,
0 0
and consequently (5) and (6) hold when cos # is substituted
for sin a.
y%
wly
94. Investigation of [ sin” x cos" x da.
Qo
v
From Art. 55, when m and » are positive, we have
rT
2, m- 1f2, :
| sin” x cos" a” dx | sin” 2 cus" ada,
0 M+Nno
wT Tv
2, m—-f2.
and sin” x cos"« dx sin” » cos” # dx.
: to mtn)
il
Hence, when one of the indices is an odd integer, the
value of the definite* integral is easily found.
* The result in this case follows also immediately from Art. 92, by making
cos? =z; for this substitution. transforms the integral into
rfl fe
=| (1 —2z)™2 ? dz.
2Jo
Elementary Examples. 121
For, writing 2m + 1 instead of m, we have
Tv
oe 2m @
[ sin?! y cos" a dx = miei sin”! x cos" # dx.
No 2m+n+1]}o
Hence
Bs
[[sineme x cos" a dx
: =
2m(2m—2)....2 3
= ( ) = sing cos" x dx
(2m+n+1)(2m+n—-1).... (n+3) J,
__ 2-4-6... (2m) (7)
(n+1)(n +3)... (m+ 2m+1)
In like manner,
= =
[[sintma cosa de = 7 F [sinvns cos" x da.
4 2(m+n)),
Hence
x
2 i
| sin?” 2 cosa dx = ees (1): ‘sina de
0 (2m+ 2)... (2m+2n) },
1.3.5...(2n-1).1.3.5...(2m-1) @
= z ° oe (8)
2.4.6. . 2. . 1 2.) (2m+ta2n) 2
in which m and n are supposed both positive integers.
Many elementary definite integrals are immediately re-
ducible to one or other of the preceding forms.
For example, on making x = tan 0, we get
Tv
0 0
: .. (2n-3) - (9)
tae
2 bn OSes QR 2h”
S|32 nln
Similarly, by 2 =a sin 0, | a" (a? — #)? de transforms into
0
Tr
te
qm | sin” @ cos” 6 d@.
0
122 Definite Integrals,
m
a ate
In like manner ( (2ax ~ x)* dz,
0
on making x = a (1 — cos 9), becomes
v
am | * sin 0 0.
0
The expressions for these integrals, when m and n are
fractional in form, will be given in a subsequent Article.
EXAMPLEs.
wT
os 2
I, [Fsinte costa x. Ans. pio ae
0 3°-5-7-11
E
2 [sint cost x de. 3 Bae ee se.
0 9.19. 29.39.49
13
1.2.3.6. (m — 1)
2 en tm-1 2n-1 Sy Sea nat GEST Per EL
3: [osint» woosin la dx. mn. (mt)... (nt m—1))
1 2420-4 ai
4. [rand ” Pog Gees Gr)
j 3-5-7--.- (2%+1)
: i anda 1.3-5++-(2%—1) @
* Je/ina® Racha. vem
i. zentl dy 2) SAO wens 2n
s O/T a G 3.5 .7+-5 (241)
7. Deduce Wallis’s value for m by aid of the two preceding definite integrals.
a App ees OSD
= \ (a + b22)*" Sig ep eves we VJ abe
when » is an odd integer.
9. | “2 (2ax— a?) P de. en
° / / be }
/ ie
Elementary Examples. 123
95. Walue of | ¢* 2" dz, when nis a positive integer.
0
In Art. 63 we have seen that
[ere du =-— ea + n| et a" da.
Again, the expression = vanishes when # = 0, and also
when # =oo (Diff. Cale., Art. 94, Ex. 2).
ao @o
Hence | e* a" dx =n | e* a daz. (10)
0
0
@
Consequently | CPO dpa 16 2s 3 a5 (11)
0
Many other forms are immediately reducible to the pre-
ceding definite integral.
For example, if we make x = az we get
7 PD! Bley att
[evade = (12)
0
in which a is supposed to be positive.
1
Again, to find | a™ (log «)"dx; let « = ¢*, and the in-
0
tegral becomes
123 eM
— jy)” —(m+})8 on = (27)? ——
(— 1) [ie a” dz = (-1) (a iy
Since log # = - log (=), this result may be written in the
form
1 n
| (log =) pee ees (13).
‘ x
(m +1)"
124 Definite Integrals.
The definite integral | e*2"dz is sometimes known as
The Second* Eulerian Integral, and is fundamental in the
theory of definite integrals. Being obviously a function
of n, it is denoted by the symbolI’(n), and is styled the
Gamma-Function.
It follows from (10) that
T(n+1)=nF(n). (14)
Also, when n is an integer we have
PUGH TS 7 38) ate (15)
Again, when ~ is less than unity, we have
=1t+@+a+ae+ &;
I-2@
dx
I1-@
i ie El Ei fad
SS Sb SS eh Ss poe
\ 2h og? 6°
(by a well-known result in Trigonometry).
In like manner we get
1 1
és ( log x -| loga(r+@tars.. dite
a 0
‘log a dx a
> i+@ 12”
An account of the more elementary properties of Gamma-
Functions will be given at the end of this Chapter.
1
* The integral { am1(1 -2)*1 dz, considered in Art. 92, is sometimes called
J0
the First Eulerian Integral; we shall show subsequently how it can be exe
pressed in terms of Gamma-Functions.
Ewvampies. 125
ExaMpies,
1 1\ jn
I {, flee (=) }" ae ANB TBs Bose se:
0 x
coud Peden
Zz [orende. ” Tog ayer
1 log a
|, oS
1 dz (log x)?” I I
4. aes ar bag Gnn [rece ae].
l dz I+e@ r
5: ( = 18 (; **). Ans, z
96. If u and v be both functions of x, and if v preserve the
same sign while x varies from a to X, then we shall have
x x
[ uvda = v | 0 det,
7% xo
where U is some quantity comprised between the greatest and the
least values of u, between the assigned limits.
For, let A and B be the greatest and the least values pf
u, and we shall have, when »# is positive,
Av > w > Bo;
when » is negative,
Av < w < Bo.
Consequently, for all values of x between 2 and X the
expression wrdx lies between Avdx and Bovdz, and accord-
ingly, if the sign of » does not change between the limits,
x “xX “x
| uvdz lies between A | vdx and B | vdz,
Zo
% zo
which establishes the theorem proposed.
126 Definite Integrals.
Cor. If f(x) be finite and continuous for all values of x
between the finite limits x and X, then the integral
xX
[. sea
will also have a finite value.
For, let A be the greatest value of f(x), and B the least,
XK
then { (a) dx evidently lies between the quantities
0
x x
4{ dz and BI dat;
0
“ [. 7@) de > B(X — a) and < A(X - #).
97. Waylor’s Theorem.—The method of definite inte-
gration combined with that of integration by parts furnishes
a simple proof of Taylor’s series.
For, if in the equation
Xth
fein Qe i te) de
,, We assume = X +h —2, we get du = — dz, and also
ies Pages [ f(X+h—a)de;
x 0
h
P(X +h) -f(X) = | f(X+h-s)ds.
oO
Again, integrating by parts, we have
[7@&+a-s) ds = sf’ (X+h—2) + | of (X+h-s)de,
Hence, substituting the limits, we have
he
”
| f'(X +h—a) de = A(X) + f of"(X +h) ds.
Taylor’s Theorem. 127
In like manner,
of’ (X+h-s)ds =" /"(K+h-s) +fEr f'(X+h-2)ds
which gives
h h? h 2?
\, of (X+h-s) ds= sf) + Sf (e+ h—-3#)ds;
0
and so on.
Accordingly, we have finally
X+h) A(X) +S r(E)+* pws
=e )
2 ds
16
mer (8)
+[ pie
This is Taylor’s well-known expansion.*
98. Remainder in Taylor’s Theorem expressed
as a Definite Integral.—Let &, represent the remainder
after n terms in Taylor’sseries, then by the preceding Article
we have
ght da
ant a —
There is no difficulty in deducing Lagrange’s form for
the remainder from this result.
For, by Art. 96, we have
= [FOX +h-2) (17)
h n-1 n
Rez ul ee pe,
0
1.2.3... (m-1) 1.2...”
where U lies between the greatest and least values which
7)(X + h -—s) assumes while s varies between o and 4h.
* The student will observe that it is essential for the validity of this proof
(Art. 90), that the successive derived functions, f’ (x), f(x), &c., should be
finite and continuous for all values of # between the limits X and X +2.
Compare Articles 54 and 75, Diff. Cale.
128 Definite Integrals.
Hence, asin Art. 75, Diff. Cale. (since any value of z between
o and h may be represented by (1 — 0) #, where 0>0 and <1);
we have
hn
= ——— _ fi)
dn ae ee
where 0 is some quantity between the limits zero and unity
99. Bernouili’s Series.—I{ we apply the method of
integration by parts to the expression /(x)dz we get
| £0) av = ap (@) - [af eae:
A {-7@) dz = Xf (X) - [re x de.
In like manner,
2 x is
[re ada = ~ 7® =, r" (a) x sd
I. Te2
x 2 3 x 3
fre s-re-pre ss,
1.2 14.223
and so on.
Hence, we get finally
[ree =4 9m) - 7)
15243
f'(X)—- &e.... (18)
Compare Art 66, Diff. Cale., where the result was obtained
directly from Taylor’s expansion.
100. Exceptional Cases im Definite Integrals.—
In the foregoing discussion of definite integrals we have sup-
posed that the function /(«), under the sign of integration,
has a finite value for all values of 2 between the limits. We
have also supposed that the limits are finite. We purpose now
to give a short discussion of the exceptional cases.* They may
* The complete investigation of definite integrals in these exceptional cases
is due to Cauchy. For a more general discussion the student is referred to
M. Moigno’s Calewl Intégral, as also to those of M. Serret and M. Bertrand.
Exceptional Cases in Definite Integrals. 129
be classed as follows :—(1). When f(x) becomes infinite at
one of the limits of integration. (2). When f(r) becomes
infinite for one or more values of x between the limits of
integration. (3). When one or both of the limits become
infinite.
a
In these cases, the integral | f(x)dvx may still have a
0
finite value, or it may be infinite, or indeterminate: depend-
ing on the form of the function f(x) in each particular case.
The following investigation will be found to comprise the
cases which usually arise.
101. Case im which /(x) becomes infinite at one of
the Limits.—Suppose that f(c) is finite for all values of «
between a and X, but that it becomes infinite when x = X.
The case that most commonly arises is where f(x) is of
(ec) de
=, in which y(«) is finite for all values
the form
between the limits, and » is a positive index.
Let a be assumed so that (wv) preserves the same sign
between the limits a and X; then
[. le) de -|" L(x) dx +f" W(x) dx
a,(X— 4)” Je (X- 2)” Ja (X - a)”
The former of the integrals at the right-hand side is
finite by Art. 96. The consideration of the latter resolv@g™
into two cases, according as n is less or greater than unity.
(x). Let x < 1, and also let A and B be the greatest
and least values of (x) between the limits a and X: then,
by Art. 96, the integral
¥ (ade ,. x dz i dx
ee A ;
i (x -2) lies between iF (Xap and B .x-a
Moreover, since < 1, we have evidently
A Ge. ea
ee I-n ?
and consequently, in this case, the proposed integral has a
finite value.
[9]
130 Definite Integrals.
(2). Let » > 1, and, as before, suppose A and B the
greatest and least values of ¥(z) between a and X; then
XW(a)dz .. = de x dex
I (X-a)* lies between 4 [ (x-ap and Bl (K-ay"
Again, we have
{ dx = I
(Xa @- (kay
Now roae becomes infinite when xz = X, but has a
finite value when x =a; consequently the definite integral
proposed has an infinite value in this case.
dx :
When » = ules = -— log (X -2). This becomes
infinite when x = X; and consequently in this casé also the
proposed integral becomes infinite.
The investigation when f(z) becomes infinite for 2 = a
follows from the preceding by interchanging the limits.
102. Case where /(z) becomes infinite between
the Limits.—Suppose f(z) becomes infinite when 2 = a,
where a lies between the limits 2 and X; then since
[_perae=["7e)ae+ [7a
the investigation is reduced to two integrals, each of which
may be treated as in the preceding Article.
(2)
eae
the last Article, that ! J (x) dz has a finite or an infinite
%
value according as is less or not less than unity.
The case in which f(x) becomes infinite for two or more
values between the limits is treated in a similar manner.
Hence, if we suppose f(z) =
it follows, as in
Case of Infinite Limits. 131
For example, if
J(u) = %, SF (a2) =, . « - (an) = &,
where @,, @ . . . lie between the limits X and 2; then
xX a a x
| F(a) ae=| * ¥(2) | ” f (0) de + &e. | F (2) dr,
X % a ay
each of which can be treated separately.
103. Case of Infinite Limits.—Suppose the superior
limit X to be infinite, and, as in the preceding discussion, let
f(z) be of the form a va where y)(z) is finite for all values
of 2.
As before, we have
x
[7 (x) da = 2 (wv) da + i f(a) de
The integral between the finite limits 7, and a has a finite
value as before. The investigation of the other integral con-
sists again of two cases.
(1). Let 2 >1, and let A be the greatest value of (7)
between the limits a and o, then sn
le) de is less than 4 ae :
Pa — a)" a (@— a)”
x dx 1 I I
But i (w - a)" er le - a) a (X - =I
The latter term becomes evanescent when X= o : accord-
ingly in this case the proposed integral has a finite value.
In like manner it is easily seen that if n be not greater than
unity, the definite integral
\ie=a
[9a]
132 Definite Integrals.
has an infinite value ; and consequently
| * Wa) de
« (w-a)”
is also infinite, provided ~() does not become evanescent for
infinite values of 2.
Hence, the definite integral
f b(2) dx
ay (@ — a)”
has, in general, a finite or an infinite value according as n is
greater or not greater than unity : Y(x) being supposed finite,
and 2, being greater than a,
If X become — o, a similar investigation is applicable; for
on changing # into - w, we have
-X
[/ peyare-[_ re a)ae
“no
in which the superior limit becomes o.
104. Principal and General Values of a Definite
Entegral.—We shall conclude this discussion with a short
account of Cauchy’s* method of investigation.
Suppose /(#) to be infinite when # =a, where a lies be-
x
tween the limits x, and X; then the integral | J (#) dx is re-
garded as the limit towards which the sum °
le (a) da + [. JS (x) dex
Zo a+ve
approaches when « becomes evanescent; » and v being any
arbitrary constants.
* This and the four following Articles have been taken, with some modifica-
tions, from Moigno’s Calcul Intégrai.
Principal Value of a Definite Integral. 133
This value depends on the nature of f(x), and may be
finite and determinate, or infinite, or indeterminate.
If we suppose p = v, the limiting value of the preceding
sum is called the principal value of the proposed integral ;
while that given above is called its general value.
x
For example, let us consider the integral | =.
“9
x x = me
Here | Ge iat [| o+( "SI
7 ax ve x To a
But [St (= =)
Also, making # = - 8,
(c-fos)
ty & zs %
x
Accordingly, the principal value off is log (=) while
ae) 0
its general value is log =) + log (f). The latter expres-
0
sion is perfectly arbitrary and indeterminate.
« da
Again, let us take | =
—% @
x x “pe
As before, { a limit [| oe +{" I
rye ve & ay v
x “pe
But | gran zie] GS-25;
PP ee oa [-+2-5-2|
ae pe ve X |
Consequently, both the principal and the general value of the
integral are infinite in this case.
134 Definite Integrais.
In like manner,
x
LS = limit of ( - oe + .. - =)
Hence the general value of the integral is infinite, while
its principal value is : = - me :
It may be observed that the principal value of
x dr. 1 x da
ze 18 equa to| =
a
fe x
2 %
This holds also whenever f(x) is a function of an odd
order: i.e. when f(- x) =- f(a).
For we have
LC I (@) de = r F(a) de + ie f(a) de.
a,
But fe Hte)de=—| #(-a)ae = [°r- wae
0
%
Accordingly, if f(- v) = —f(x), we get
/ ic J (a) dz = 0.
: Again, if f(x) be of an even order, ie. if f(- 7) = f(a), we
ave
X
0
-[P pe)de=[° (7) +4(- 2) (19)
: fava 2 "#(«) de.
x,
—X 0
105. Singular Definite Integral.—The difference
between the general and the principal value of the integral
considered at the commencement of the preceding Article is
represented by di ai
‘at Le $e
Poses,
atve
in which f(a) =o, and ¢ is evanescent.
Infinite Limits. Example. 135
Such an integral is called by Cauchy a singular definite
integral, in which the limits differ by an infinitely small
quantity. The preceding discussion shows that such an in-
tegral may be either infinite or indeterminate.
106. Infinite Limits.—Ii the superior limit be infinite,
we regard | J () de as the limit of [7 (x) de, when « becomes
evanescent. : :
I
Also | J (2) de = limit of | r _/ («) dz when ¢ is evanescent.
pe
In the latter case the value of the definite integral when
u = v is, as before, called the principal value of
[7 (a) da.
In this we assume that f(x) does not become infinite for
any real value of 2.
107. Example.—Suppose aa to be a rational algebraic
fraction, in which f(x) is at least two degrees lower in # than
F(x), and suppose all the roots of F(x) = o to be imaginary,
it is required to find the value of
* (2)
ir Fe) da.
From the foregoing conditions it follows that om cannot
become infinite for any real value of x: accordingly the true
value of the integral is the limit of
I
* SG)
when « vanishes.
136 Definite Integrals.
S(e)
F(«)
thod of partial fractions, and let
To find this value, suppose decomposed by the me-
AaB yas AGB /=i%
e-a-bf/-~1 «£-atb/-1
be the fractions corresponding to the pair of conjugate roots
a+ b/—1 anda-b/~1, of F(z) =0;
and
then the corresponding quadratic fraction is the sum of
AEB 1
Pia be ow
ALB f=a
a-a-—b/-1
2A (x — a) + 2Bb
and
i.e.
(c-a)? +B
: Bod. -
Again \ece = 2B taxn(® b *) :
a [os = 27B when « vanishes.
we
2A (x — a)dx
Also SS = Alog {(# - a)? + 8};
: ve oA (vw - a)dx _ pw? (1 — ave)? + Bye?
ge i (@—-a)* +6 =A log S (1 + Que)? + Bu?
» pe
= 2Alog®, when ¢=0
Investigation of | Fade 437
% {2A (w—-a)+2B\ dz
Hence | : rae ari =2A tog(#) +27B. (20)
Now, suppose F(x) to be of the degree 2n in x, and let the
values of A and B, corresponding to the n pairs of imaginary
roots, be denoted by 4,, A,,... An, and B,, B.,... Bn, re-
spectively ; then we have
i coe 2(Ai+ Ag te wet 4,) log (*)
I
ua
+ 27(B, + Bot... + B,).
Again, since (7) is of the degree 2 — 2 at most, we have
A,+A,+...+ An =
For, if we clear the equation
J(@) _ 2Ai(e-a) + 2B - 2An(% — ay) + 2Budn
FF) (w-a)+be ene (@ — ay)? + Dn®
from fractions, the coefficient of 2*”" at the right-hand side is
evidently
2(4,+ 4,+...+ An);
which must be zero, as there is no corresponding term on the
other side.
Accordingly we have, in this* case,
(45 ye = an(Bi+ Bat aw Bek (21)
* It may be observed re when f(x) is but one degree lower than F(z),
the principal value of (. dz is still of the form given in (21).
Fe
138 Definite Integrais.
We proceed to apply this result to an important example.
em
wv” dx zie
108. Walue of | area when m and 7 are Positive
ol+z
Entegers, and n> ™.
Let a be a root of 2” + 1 = 0, and, by Art. 37, we have
Again, by the theory of equations, a is of the form
pi (2k+1)7 ee (2k + ie
2n 2n
in which k is either zero or a positive integer less than n;
am! = cos (2k + 1)0+4/— 1sin (2k + 1)8,
where 0= aS E
2n
Hence B = = ; and accordingly we have
Bit+ B+... + By=— {sin O + sin 30+... +sin (2n —1)6}.
To find this sum, let
S=sin#+sin30+...+sin(2n- 1)0;
then
28 sin 0 =2sin’9+ 2 sin @sin 30 +...+ 2 sin Osin(2n—1)6
= 1- cos 2 + cos 20 — cos 40+... + cos(2m — 2) 0 — cos 2n0
. ° T
= 1- cos 2n0 = 2sin’nO =2 sin?(2m-+ 1) ses
I I
2 S=— = \ .
sin@ . (2m + 1\7
sin ——_——
2n
2m
Value | —; dz. 139
0
x
I+a
Accordingly, we have
i a2” dr T
ee ee ee
Ta . (2m+ 1)
to n sin (Git tbe
2n
Hence, by (19),
i
2 amd, 1f°? 2" de o« I (22)
————_—_ = -— Se OO OO —————.. 22
ol ta™ 2) e1+e" an. (2m+ 1)7
sin. ———————-
2n
We now proceed to consider the analogous integral
oem
| —,,, where m and 2, as before, are positive integers,
ol -2%
and n>m.
cy 2m
:0g. Investigation ot] ap
20
0 zc
We commence by showing that
A
7 a ae
This is easily seen as follows :
| * dx | . dx | ” da
2 at
oi -2 ol -2# 1I-
Now, transform the latter integral, by v= *, and
“de (° ds [' oc [* a ,
|: fer Joie s” J tae’
7 2. e
oe ? I os ee ~~"
Again, proceeding to the integral
(=
209
pI -2
we get
140 Definite Integrals. ‘
we observe that 1 + 2 and 1 — 2 are the only real factors of
1 — 2”, and that the corresponding partial quadratic fraction
in the decomposition of
ae I
—,, is ——_;
1-2" n(1 — a’)
Consequently, the part of the definite integral which corre-
sponds to the real roots disappears.
Moreover, it is easily seen that the method of Arts. 107
and 108 applies to the fractions arising from the n — 1 pairs
of imaginary roots, and accordingly
| ori = 2m (Bi + Bat... + Bus)
where B,, B., ... Bn have the same signification as before.
Again, singe the roots of #*” — 1 = 0 are of the form
k ‘
oi a fOr an,
n n
it follows, as in Art. 108, that
B,+ Bot+.. + Bai = = [sin 20+sin 40 +...+sin 2(n-1)6],
_ (2m + 1)z
where 6 , 28 before.
Proceeding as in the former case, it is easily seen that
sin 20+ sing0+...+sin2(n-1)0
cos 8 — cos (2n- 1) 0 cot 2m tt '
= 7 a 7
2sin 0 an ty
“ 2"de a 2m+1
Hence mm =~ cot 75
I -v" n 2n
“mde om of ett
= 2
ol -xv” an 2n (23)
Examples. 141
Again, if we transform (22) and (23) by making "=
2m +1
and a= » we get
“3% dg 7 “27 dz
=——, = 7 cota. (24)
oit+s sinar ol - 8
The conditions imposed on m and n require that a should
be positive and less than unity.
Moreover, since the results in (24) hold for all integer
values of mand n, provided n > m, we assume, by the law of
continuity, that they hold for all values of a, so long as it is
positive and less than unity.
110. The definite integrals discussed in the two preceding
Articles admit of several important transformations, of which
we proceed to add a few.
For example, on making wu = s* in (24), we get
“du ar “du
i=- i 5 = am cot am.
ol+ue singr ol — Us
If = 7, these become
° du T ° du T ot = o
= =— eot — 2
ot tur wom ls r r 5
rsin—
r
where r is positive and greater than unity.
Again
“atde [) a®de ; * a dee
lee [tee ree
Now, if in the latter integral we make 2 = *, we get
“ade [(° sds (tarde.
gioet jase erga?
: (= (a=
2 2
oite J, 1+2
dx. (26)
142 Definite Integrals.
Moreover, from (22), when x is less than unity, we have
° ot dan 7
la +e nw (27)
2 cos —
2
Accordingly
: a+ om de i Tv (28)
o@tat go
~ Nn
2 cos —
2
In like manner, it is easily seen that
(f¢ 7 nT
— =—tan a (29)
0 &—-x a 2
It should be noted, that in these results n must be less than
unity.
Again, transform (28) and (29) by making 2 =e and
nr =a, and we get
im e% 4 gt I a oe et — ea I a
——. dz =~ sec-, ——, dz = — tan-. (30)
oer +Ee™ 2 2 9 em — Ee 2 2
We add a few examples for illustration.
EXamp.es,
a d.
1 | & S Ans, —
° (an — anyn nsin=
n
i. dx wv
o (2? + a”) (x? + 8) ” 2ab(a + by
7 dz T
S j ol — ae
vl
4. j tan"@d0, where m lies between+ Iand-1. ,, ———.
0
Differentiation under the Sign
tt —, where n > m. Ans,
[pee
0 2* + a" x
ul 6. E (eax + ean) (eb= ott eb) ay
0 ene + e-me am
of Integration.
143
wv
—-———— |
mr
2” COS —
2n
a b
2 cos ~ cos —
2 2
cosa + cosd°
sin 3
woo
© (esx ax\ (gba _ ¢-bx
| (e9% 4 ear) (ebm — ¢ ) ae 2
0 ent — g-me cosa + cosb”
It should be observed, that in these we must have a + 6 < 7.
“ 8. Hence, when 4 < a, prove that
oftest
e+e ) cos 5
a fi [. = ~ cos ax dx —Cecniaee
Oe ee erer rrr
4 5; i su 2 = sin az dz =: es
(( 9 ; oF ae, Ans. m cot am —~.
o I-24 a
111. Differentiation of Definite Integrals.—It is
plain from Art. 86 that the method of differentiation under
the sign of integration applies to definite as well as to in-
definite integrals, provided the limits of integration are
independent of the quantity with respect to which we dif-
ferentiate.
On account of the importance of this principle we add an
independent proof, as follows :—
Suppose w to denote the definite integral in question, i.e.
et
“= {9 a)da,
where a and 4 are independent of a.
To find a let Aw denote the change in w arising from the
a
change Aa in a; then, since the limits are unaltered,
144 Definite Integrals.
Au [16 a + Aa) - $(2, a)} dex;
yi (estes oil a
veer, Ne Aa
Hence, on passing to the limit,* we have
du ae [ ado (x, a)
da |, da es
Also, if we differentiate times in succession, we ob-
viously have
du (> d"o(a, a)
da = i da” at
The importance of this method will be best exhibited by a
few elementary examples.
112. Imtegrals deduced by Differentiation.— If
the equation
. I
| e% dz = —
a a
be differentiated » times with respect to a, we get
o
Pee 3D
NM p-aAt laos "aaa eae
[we dx = “i s
as in Art. 95.
Again, from the equation
° dx _wi
o @+a 2a”
we get, after n differentiations with respect to a,
i de =o w1.3-5...(2m—-1) 1
0
ze ‘
Oa 25 A Oawe Oe are
which agrees with Art. 94.
* For exceptions to this general result the student is referred to Bertrand’s
Caleul Intégral, p. 181.
Differentiation under the Sign of Integration. 148
Again, if we take o and & for limits in the integrals (23)
and (24) of Art. 21, we get
(31)
@ 2
a ‘ m
e* cos ma dz = =—., e sin ma dx = ———..
0 a+m J, a +m
Now, differentiate each of these n times with respect to a,
and we get
ei aXe fe
ak an w= (— ef os coe
[ e-% a” cos ma dit = (— 1) (Z) (a )
_([m-cos(nt+1)0 4.,
ne?
(@ + m*) >
| 2. sin (n+1)6
mI (32)
| eo go" gin ma da =
; (a +m?) ®
where m=atan@. (See Ex. 17, 18, Diff. Calc., pp. 58, 59.)
Next, from (24) we have
ie a da
= 7 cotar.
o 1-2
Accordingly, if we differentiate with respect to a, we have
° #™ log x dx cd
° 1-@ sin? ar
Again, if the equation
p I
n=) =
ly dy a
0
be transformed, by making ye, it evidently gives
o (a+ bay" nab
[10]
| Sgt dy I
146 Definite Integrals.
Now, differentiating with respect to a, we have
“ade _ I
o (a+ ba) n(n + 1)ab™
If we proceed to differentiate m — 1 times with regard to
a, we have
i ada | 1.2.3... (m-—1) 1
Jo (a+ bx)™™ = n.(n+1)(n+2)...(n+m—1) ° anh”
113. By aid of the preceding method the determination
of a definite integral can often be reduced to a known integral.
We shall illustrate this statement by one or two examples.
Ex. 1. To find
a
"log (1 + sina cos)
i: cos #
dz.
Denote the definite integral by wu, and differentiate with
respect to a; then
du | * cos adz
7 =7 (by Art. 18).
> 1+ sina cosx ~
Hence, we get
dv log (1 + sin a cos 2)
cos x
=a.
oO
No constant is added since the integral evidently vanishes
along with a.
2 p-Ak oo
e* sin mx
Ex. 2. “= ———— dz.
0 z
In this case
du 3 a
= =| e&* cosmadr = =—,3
dm Jy Crm
: dm m
“. w@=a) ——, = tan? (—).
Ia+m a
No constant is added since wu vanishes with m.
Case where the Limits are Variable. 147
Ex. 3. Next suppose
je 2 mp2
| log (1 + aa) 2
1+ 02
du 2aax° dx
Here da | (1 + a?a)(1 + 0? x")
_ot ° 2ada _ ° 2adu
Pell le Bee ly pas
= ee ona es
G=-0Gb 5 Bas oy
Tv da T
oe U ~F | op F log (a + 2) + cont.
To determine the constant: let a= 0, and we obviously
have wu =o.
Consequently, the constant is - . log b;
_— beter), 2 a+b
oo | SEO de =F log (“F),
The method adopted in this Article is plainly equivalent
to a process of integration under the sign of integration.
Before proceeding to this method we shall consider the case
of differentiation when the limits @ and 6 are functions of
the quantity with respect to which we differentiate.
114. Differentiation where the Limits are Wa-
riable.—Let the indefinite integral of the expression
(2, a)dx be denoted by F(a, a); then, by Art. 91, we have
w= | $l 0)de =F, a) - Fle 0)
; du d. F(b, a)
‘ad db
[10 a]
= (6, a);
148 Depenite Integrals.
du dF(a,a)__
and eee (a, a).
Again, taking the total differential coefficient of w re-
garding @ and 6 as functions of a, we have
du_ [ dolee)g dude , du da
da |, da dbda dada
’ d(x, a) db da
-| Pie de + $(b, «) 5 - $(4 a) 7 (33)
du d
By repeating this process, the values of >? = , &e., can
be obtained, if required.
115. Integration under the Sign of Integration.—
Returning to the equation
“= [oe a)dx,
where the limits are independent of a, it is obvious, as in
Art. 87, that
| wa = i | o(2, a) aa da,
provided a be taken between the same limits in both cases.
If we denote the limits of a by a and a, we get
[° uda “(TP Gi, a) aa dar,
or il $(«, a) as | 2 {[ is $(2, a) as] de. (34)
0
This result is easily written in the form
an J& a,
i i emis i [9 a) dade. (35)
Integration under the Sign of Integration. 149
These expressions are called double definite integrals, as in-
volving successive integrations with respect to two variables,
taken between limits.
It may be observed that the expression
i i (a, a)deda
a, Ja
is here taken as an abbreviation of
[een
in which the definite integral between the brackets is sup-
posed to be first determined, and the result afterwards
integrated with respect to a, between the limits a, and a.
The principle* established above may be otherwise stated,
thus: In the determination of the integral of the expression
o(x, a)de da
between the respective limits a, a, aNd ar ar, we may effect the
integrations in either order, provided the limits of x and a are
independent of each other.
In a subsequent chapter the geometrical interpretation of
this, as well as of a more general theorem, will be given.
‘We now proceed to illustrate the importance of this
method by a few examples.
116. Applications of Integration under the Sign {.
Ex. 1. From the equation
1
[we - 2
0 a
1 pa, ‘a da (2)
a di ae | — =log\—}.
i i: 7 ao a S \ag
* It should be noted that this principle fails whenever (2, a), or either of
its integrals with respect to a, or tox, becomes infinite for any values of z anda
contained between the limits of integration. The student will find that the
examples here given are exempt from such failure.
we get
)
150 Definite Integrals.
1 gal gyaomi a
Se -g(4)
| » loge se a,
Again, if we make x = ¢* in this equation, we get
iS 7% — E742 _ a,
| : =. = log (*).
Ex. 2. We have already seen that
Tlence
| &€~ cos mrdz =
0
a? + m*
Hence
. Or . ada
| {[ om da | cos max daz -| Tae
; ay a, @ +m
I lo a,’ + m?*
28 a? +m)
ent — eto 1 ay +m
or ———— cosmadx = — log | —— }.
° x 2 a +m
Ex. 3. Again, from the equation
7 : m
| e~ sin madx = ———,,
5 a’ + m
we get
"ar sin madade =| Be
0 Ja, ay a +m
© pnt _ p-a,t
ot | eee sin mrdz = tan (2) - tan (=\ “
0 x m m
Compare Ex. 2, Art. 113.
Tf we make a = 0 and a,=© in the latter result, we
obtain
ee
sin mr T
dv = -—.
0 x 2
Value of | eda. oe
Ex. 4. To find the value of
| e* dx.
0
Denoting the proposed integral by &, and substituting
ax for x, we obviously have
| evade =k;
0
Ss | eo (42")q dye = ke,
0
Hence
| | €@04)adadz = x evda =k’.
oJo o
But 2
| cada = +— 39
7 21+¢@
i * dat yas « © 242
ool apres eg ee
Hence
| eM de = b= =x. (36)
0
This definite integral is of considerable importance, and
several others are readily deduced from it.
117. For example, to find.
Here
Again, let s = <, and we get
152 Depimte Integrals.
a? a?
"29 ade [" @--
e —=|e dz =u;
a
0 0
du 4
. = =—-2u; hence u= Ce.
da
To determine C, let a= 0, and, by the preceding example,
Tv
wu becomes ra
Consequently :
[Fae = Li gre, (37)
Again, to find
(B) we [, e008 2ba de.
0
Here
du
ee ee 9 2 s .
ab 2 [ evr x sin 2badx
But, integrating by parts, we have
ev sin 2b
2 | eo? sin 2beadz =— ,
a
26
+= \" cos 2ba dx;
a
3 ;
i 2b
# | c= sin 2baadz = 7 i e*= cos 2badz.
0 ; 0
Hence
du 2bu du 2bdb
a a a
2
Hence u=Ce”,
v
Also, when 6 = 0, u becomes a
a
32
. [a cos iieteel oe. (38)
0
Examples. 153
Again, if we differentiate » times, with respect to a, the
equation
Le
at 2s
\,e ™ 2f/a
and afterwards make a = 1, we get
() [evenae = 1.3. 5---(2-1) ope
grt
Next, to find
A
(D) |
We obviously have
“cos max da
o lt+e
I
+ 2?’
2 | a e7 (142) dq =
0 I
” cosmadax
Bs 2| | ae" ("42") cos mx dada -| —;.
0 Jo o I+”
But, by (38), we have
Ja
o
2| e"** cos mxdx = —— € 4a?’
0 a
cae | ne de | * cos mx da
0 o it+2
Hence, by (37), we have
“cosmadx ow
2
= —m
eae a (39)
Again, differentiating with respect to m, we obtain
=aee (40)
8
asinmaedx 37
o I+2 2
154 Definite Integrais.
Examptes.
i, eat 4 eax I a
Bag. Ans. — sec? -.
jo em — Em 4 2
ie wT Pyare ae
o a [Sete te (Z% wy Yog (tan 2),
, tk a
it
a l+t 2
when a>oand<1.
(RUS dz 1 ar )
a 0 «mi-s lgz » 08 \ Sin am)”
wT
2 dz 1/r
. log (I + cos 6 cos x) ——. ee ge ee 2 ie
x OS E B (r+ : "eae ae (= )
= 22+ B
“ 5+ i cos # log (75) az. » 7 (c= - 8) A
_ 13
6 © 2f log zdz an
4 : \, “T4+2" = 4 ott
cos? —
{, S24 Ilea-T
7. 0 670 —e-7 0" il 4ea + r
118. The values of some important definite integrals can
be easily deduced from formula (34), Art. 32.
For example,* to find
wT
fe log (sin 6).
J0
Hie | * top {ain 8)a0 = | * tog (oos 6) a8.
Hence, denoting either integral by u, we have
rT
2u = . {log (sin 6) + log (cos @)} a0
* These examples are taken from a Paper, signed ‘‘H. G.,” in the Cambridga
Mathematical Journal, Vol. 3.
Theorem of Frullani. 155
= i log (sin 20)d6 - * log 2
Again, if s = 20, we have
[tg (sin 20)d0 = an log (sin z)dz
0 0
= =I’ log (sin z)dz + | "log (sin 8) dz;
2
but, since sin (7 - z) = sing,
—— :
| log (sin s)ds = | log (sin s) dz.
T 0
=
Consequently
| "log (sin 26)d0 = i log (sin 6)d0 ;
0 0
Be [0g (sin 0)d0 = - “ log (2): (41)
Again, to find
i 6 log (sin 0)d0.
Here
| @ log (sin 6) 0 = Ie ( - 8) log (sin 6) 0 ;
es i 6 log (sin 0)d0 = = [log (sin 6)d6@ = - © log (2).
119. Theorem of Frullani.—To prove that
lk a de = (0) log (;)
156 Definite Integrals.
v0
Let u -|) oo oo) dz; substitute ax for z, and we get
3
If we substitute d for a, we get
0 xu
h
Pe | g (02) - 9() a,
a h h
= (0) log 5. (42)
h
h
Hence (" #(00)~ 9 (02) 2, -|; (00) de _ 4(0) log (2). (43)
If we suppose 4 =o, we get
[, PEt a= oo og (5), a)
S
provided i, stk dz =o when h =o.
6
For example, let ¢(z) = cos, and, since the integral
&
> cos bx ae
i
evidently vanishes when h = «, we have
al
| cos ax — cos bx aie b
0 a
Theorem of Frullani. 157
Frullani’s theorem plainly fails when ¢(av) tends to a
definite limit when x becomes infinitely great. The formule
can be exhibited, however, in this case in a simple shape, as
was shown by Mr. E. B. Elliott.*
For, in (42) let 4 = ab, and it becomes
i plac)de i ae = $(0) log 8} (45)
0 x
Again, if (00) denote the definite value to which ¢(az)
tends when increases indefinitely, then when h becomes
infinite we may substitute g(0) instead of g(dx) in the
integral
[iota
al>
in which case it becomes
A
5 dx a
- = ( co) log (3).
On making this substitution in (43), we get
{, He) = 90) a, = fo(%) ~ 6(0)} og (5) (46)
0
(c=) |
al>
For example, let ¢(az) = tan™'(az) then we have ¢(0) = 0,
and o( 0) = _
Accordingly we have
A
ie tan" ax — tan" be 7 = a _t 1og (5)
2jaa 2
0 x
* Educational Times, 1875. The student will find some remarkable exten-
sions of the formule, given above, to Multiple Definite Integrals, by Mr. Elliott,
in the Proceedings of the London Mathematical Society, 1876, 1877. Also by
Mr. Lendesdorf, in the same Journal, 1878.
158 Definite Integrals.
119 a. Remainder in Lagrange’s Series.— We next
proceed to show that the. remainder in Lagrange’s series
(Diff. Calc., Art. 125) admits of being represented by a
definite integral. This result, I believe, was first given by
M. Popoff (Comptes Rendus, 1861, pp. 795-8).
The following proof, which at the same time affords a
demonstration of the series, of a simple character, is due to
M. Zolotareff :—
Let z = 2 + y¢(z) ; and consider the definite integral
8, = { o(u) +e — u)" F’(u)du.
Differentiating this with respect to z, we get, by (33),
Art. 114,
di nm nm RY
Te 7 Sma — ¥"{H(@)}” F'(@). (47)
If in this we make n = 1, we get
d
& = y p(2) F(a)+
but 8) = F(s) - F(a) ;
. d. 1
. FG) = FQ) + $(e) P@) +2. (48)
In like manner, making » = 2, we have
P ape ds, |
2m = v'(ge)}* Fe) +S
a YY a a I
dan Hs |@) F@) | Tle a
Substituting in (48) it becomes
Fe) -F@)+49@F@ +4 2 [gorro|+5
Again,
ap
(o@ FQ) +2;
& =
wlS
Gamma Functions. 159
i Ce Ff . : ] I a8,
eS B|vorre + 7.2.3 00"
3p
na = Y" (a) }"L"(x) + ae
I a” sy I = , |
— = x)\" F
1.2...n-1 da 1.2...n dg" (9(@)} (°)
I a” Sn
*T2...” de"
Hence we get finally
FQ) = Fe) +4 9@)F@ + 4 [(@) (e) | + &
+ (E) [iv ewre-apr man 9)
Consequently the remainder in Lagrange’s series is always
represented by a definite integral.
We next proceed to consider a general class of Definite
Integrals first introduced into analysis by Kuler.
120. Gamma Functions.—It may be observed that
there is no branch of analysis which has occupied the atten-
tion of mathematicians more than that which treats of
Definite Integrals, both single and multiple; nor in which
the results arrived at are of greater elegance and interest.
It would be manifestly impossible in the limits of an
elementary treatise to give more than a sketch of the results
arrived at. At the same time the Gamma or Eulerian
Integrals hold so fundamental a place, that no treatise,
however elementary, would be complete without giving at
least an outline of their properties. With such an outline
we propose to conclude this Chapter.
The definitions of the Eulerian Integrals, both First and
Second, have been given already in Art. 95.
The First Eulerian Integral, viz.,
1
| a" (1 — 2)" da,
0
is evidently a function of its two parameters, m and n; it is
usually represented by the notation B(m, n).
160 Definite Integrals.
Thus, we have by definition
[, a (1 — 2)" de = B(m, n).
®
7 (so)
| ea dx = T(p).
The constants m, n, are supposed positive in all cases.
It is evident that the result in equation (14), Art. 95, still
holds when p is of fractional form.
Hence, we have in all cases
I'(p + 1) = pT(p). (51)
This may be regarded as the fundamental property of
Gamma Functions, and by aid of it the calculations of all
such functions can be reduced to those for which the para-
meter p is comprised between any two consecutive integers.
For this purpose the values of I(p), or rather of log I'(p),
have been tabulated by Legendre* to 12 decimal places, for
all values of p (between 1 and 2) to 3 decimal places. The
student will find Tablesto 6 decimal places at the end of this
chapter. By aid of such Tables we can readily calculate the
approximate values of all definite integrals which are re-
ducible to Gamma Functions.
It may be remarked that we have
p being any integer. For negative values of p which are
not integer the function has a finite value.
Again, if we substitute sv instead of 2, where g is a con-
stant with respect to 2, we obviously have
‘ 7B ~m~1 TD (m)
| er" dx = = (52)
* See Traité des Fonctions Elliptiques, Tome 2, Int. Euler, chap. 16.
1s
Cv (1) =I, [ (0) =, T(-p)= co, | ind
Expression for B(m, n). 161
With respect to the First Eulerian Integral, we have
already seen (Art. 92) that
1 : 1
| gery - 2)" ae =| al (1 — 2)" da;
0
*, B(m, n) = B(n, m).
Hence, the interchange of the constants m and n does not
alter the value of the integral.
Again, if we substitute a for v, we get
1 1m m-1
ev (1 — 2)" dx = es
Jerre aero = | ies
= yr dy a
Hence ; Gay™ = B(m, n). (53)
We now proceed to express B (m, n) in terms of Gamma
Functions.
121. To prove that
Bins eee o.
From equation (52) we have
T (m) = i et gi 1 Tp,
Hence
Lo
T (m) et gt = @-3 (142) gmtn-1 ym-1 dz;
10
* Pim) | e* 2) dg = | er) get as a” da.
0 J0 0
[11]
162 Definite Integrals.
But, if s (1 + 7) = y, we get
| e@ (142) gntn-1 dg a ey ym dy = r (m + n) a
0
~ (1 +a)mn?
- g™ de
a eae
ara,
“. T(m) T(n) = (m+ 2) |
Accordingly, by (53), we have
_ Pm) Tn)
Bim, n) = rae: (54)
Again, if m = 1 — n, we get, by (24),
° gl de fa
T(n) P(r - 0) = i ae (55)
If in this n = , we get
)-7
This agrees with (36), for if we make 2’ = z, we get
I
ap TG)
,¢ wa | ets leer (56)
Again, if we suppose in the double integral
{J gn y dx dy
x and y extended to all postive values, subject to the condi-
tion that «+ y is not greater than unity; then, integrating
with respect to y, between the limits o and 1 - 2, the
integral becomes
1T(m) D+ 1)
1 : 1 n =. .
= |.2 Ge n T(m+n+1) » by (54);
7 [Jere any - TEE (57)
in which z and y are always positive, and subject to the con-
dition z+y <1.
Gamma Functions. 163
122. By aid of the relation in (54) a number of definite
integrals are reducible to Gamma Functions.
For instance, we have
= y™ dy = 1 yn" dy f cae yn dy
la tym Jo(r+yy™ Ji(r + yy
Now, substituting ~ for y in the last integral, we get
{, yn dy | 1 gn dx
1a+y™ Journ
Hence
emis ght — T(m) T(n)
[, (+a) Tim +n)" (58)
Next, if we make «= +, we get
(= spp a ee
o(1 +2)" lie + ymin?
ymdy _ _ Pm) P(0) ea
“(ay +6)" ab" T(m + ny 9
Again,* let 2 = sin’0, and we get
1
| a (1 —a)" de = 2| sin?” cos” 6.40 ;
0 0
ie Pecans 2n-1 = T(m) T'(n)
oe [[sn 0 cos’ 0.0 rer (60)
This result may alsv be written as follows :
serterea 0
[[sine0 cost 9d0 = =A AL, (61)
_ ar(? ; ‘)
* These results may be regurded as generalizations of the formule given in
Arts. 93, 94, to which the student can readily see that they are reducible when
the indices are integers.
164 Definite Integrals.
wa “en is
a
Again, if p = g in (61) it becomes
If we make g = 1, we get
7
i sin? 9 d0 =
ia
rh ¢
; ro) = sin 0 cos? 6.d0 = or sin 26a0.
Let 26 = 2, and we have
wT
i sin?! 20d0 = = [sine sdg = \ sin? |g dg
9 2 Jo 0
r(?)
rs __\? | 6
= + ‘)
Hence r(5) r (on) = Va Hp), |
2 2p
If we substitute 2m for p, this becomes
T(m) r(m + ;) = a I'(2 m). (63)
Again, make y = tan’@ in (59), and we get
f sin” cos™0d0 _ —-P(m) Fn) ;
o (asin’@ + bcos’O)™™ — 2a" b" (m+n) (64)
123. To find the Value’ of
ra) (a) a) 1)
n being any integer.
* This important theorem is due to Euler, by whom, as already noticed, the
Gamma Functions were first investigated.
: - 65
Vaue opt (2)r(2)r(2)...r(@—)
nn} \n] \n n
Multiply the expression by itself, reversing the order of
the factors, and we get its square under the form
aa)
that is, by (55),
re
_w . 20. 39 . (n- 1)
sin — sin — sin —... sin ~———
n n n n
To calculate this expression, we have by the theory of
equations
1-9
1-2
wT 4 2 (n-1)r
=( I - 2x%cos— +2" || 1 — 2ecos— +2’ }...{ 1 — 2@c08 +2").
n n n
Making successively in this, 2 = 1, and #=- 1, and re
placing the first member by its true value n, we get
sw Vf ow. Qa\? . (n- 1)r\?
n=(2 sin — }( 2 sin —)...[(2 sin ~——~_},
2n 2n 2n
a\? 2n\ (m — 1)m\?
n = (2 cos —]| 2 cos — |... ( 2 cos —_——~ ],
2n 2n 2n
whence, multiplying and extracting the square root,
ee ee . (n-1)7
n= 2""sin — sin — ... SIN Sire
n n n
Hence, it follows that
n-1
2
r (2) r() seeP (*=*) = se (65)
166 Definite Integrals.
124. To find the values of
a
oO
| é cos bax” da, ana | e* sin ba v™ de.
0 0
If in (52) a—b/- 1 be substituted* for s the equation
becomes
rags i T(m) _ D(m) (a+ b/- 1)"
| (a-b/-1 (4
Let a = (a? + 8°)* cos 0, then d= (a? + a sin 0, and the
preceding result becomes
e- (cos ba + 4/— 1 sin be) a™ de
10
= AND 36 af 4 sin 6)"
(a? + B)
T(m)
(cos m0 + 4/—1 sin m6).
m
(a + b)*
Hence, equating real and imaginary parts, we have
q (a + oF
T'(m)
[ e“= cos baa” da = ee cos ”)
’ (66)
ao ?
| sin bax” da = sin m0 |
: (a? + 0°)? J
in which @ = tax(7)
If we make a =o, 0 becomes = and these formulz become
* For a rigorous proof of the validity of this transformation the student is
referred to Serrett’s Cale. Int., p. 194.
Gamma Functions. 167
| cos baa" de = ae cos =|
0 “on 2
‘ ROR nee (67)
| sin baa! da =~ gin J
0 b™ 2
It may be observed that these latter integrals can be ar-
rived at in another manner, as follows :—
From (52) we have
b r @
T(n) — i i e** a eos be dx;
*, T(n) | = 2 oe | | e*” cos bz a” da ds.
0 & 0 Jo
But, by (32), we have
7. 2
0% =e .
i. e* cos bz dz Page?
=| cos badzs 1 a 2” de
0
se P(n)jo B+ eH
ote =,
T) 5 e053 by (27)
in which x must be positive and < 1.
In like manner we find
* sin bsdg 6" wT
o s* — P(n)
The results in (67) follow from these by aid of the relation
contained in equation (55).
168 Definite Integrals.
EXAMPLES.
m+ti
- T(p+ or (™=*)
t[ am(x apace, Ans. ees)
nv (2 is ii
—_ = ayerda T'(m) I (x)
0} 6(a@+a)mm * am(T+a)™T (mn + 2)
3. Prove that
j. dx ~f. de®
O(n — ath Jor + Ah 2/2
wv
o L I’ (m + 1) cos (» *)
x | cos (82") de. ee
0
f de es r (>)
0/1 — an Be es r(: a
Ss.
© sin bx
6. | = dz. 5
0
v] a
125. Numerical Calculation of Gamma Func-
tions.—The following Table gives the values of log I'(p),
to six decimal places, for all values of p between 1 and 2
(taken to three decimal places).
It may be observed that we have '(1)=I(2) =1, and
that for all values of p between 1 and 2, I'( p) is positive and
less than unity ; and hence the values of log '(p) are negative
for all such values. Consequently, as in ordinary trigono-
metrical logarithmic Tables, the Tabular logarithm is obtained
by adding 10 to the natural logarithm. The method of
calculating these Tables is too complicated for insertion in
ap elementary Treatise.
p Oo 1 2 3 4 5 | 6 7 8 9
1,00 9750 | 9500 | 9251 | 9003 | 8755 | 8509 | 8263 | 8017 | 7773
1.01 | 9. 991529 7285 | 7043 | 6801 | 6560 | 6320 | 6080 | 5841 | 5602 | 5365
1.02 5128 | 4892 | 4656 | 4421 | 4187 | 3953 | 3721 | 3489 | 3257 | 3026
1.03 2796 | 2567 | 2338 | 2110 | 1883 | 1656 | 1430 | 1205 | 0981 | 0775
1.04 0533 | 0311 | 0089 | 9868 | 9647 | 9427 | 9208 | 8989 | 8772 | 8554
1.05 | 9.988338 | 8122 | 7907 | 7692 | 7478 | 7265 | 7052 | 6841 | 6629 | 6419
1.06 6209 | 6000 | 5791 | 5583 | 5378 | 5169 | 4963 | 4758 | 4553 | 4349
1.07 4145 | 3943 | 3741 | 3539 | 3338 | 3138 | 2939 | 2740 | 2541 | 2344
1.08 2147 | I95T | 1755 | 1560 | 1365 | 1172 , 0978 | 0786 | 0594 | 0403
1.09 0212 | 0022 | 9833 | 9644 | 9456 | 9269 ; 9082 | 8900 | 8710 | 8525
1.10 | 9.978341 | 8157 | 7974 | 7791 | 7610 | 7428 | 7248 | 7068 | 6888 | 6709
1.11 6531 | 6354 | 6177 | LLCO | 5825 | 5650 | 5475 | 5301 | 5128 | 4955
1,12 4783 | 4612 | 4441 | 4271 | 4ror | 3932 | 3764 | 3596 | 3429 | 3262
1.13 3096 | 2931 | 2766 | 2602 | 2438 | 2275 | 2113 | 1951 | 1790 | 1629
L.14 1469 | 1309 | 1150 | Ogg2 | 0835 | 0677 | 0521 | 0365 | O210 | 0055
1.15 | 9-969901 | 9747 | 9594 | 9442 | 9290 | 9139 | 8988 | 8838 | 8688 | 8539
1.16 8390 | 8243 | 8096 | 7949 | 7803 | 7658 | 7513 | 7369 | 7225 | 7082
1.17 6939 | 6797 | 6655 | &514 | 6374 | 6234 | 6095 | 5957 | 5818 | 5681
1.18 5544 | 5408 | 5272 | 5137 | 5002 | 4868 | 4734 | 4601 | 4469 | 4337
119 4205 | 4075 | 3944 | 3515 | 3686 | 3557 | 3429 | 3302 | 3175 | 3048
1.20 2922 | 2797 | 2672 | 2548 | 2425 | 2302 | 2179 | 2057 | 1936 | 1815
1.21 1695 | 1575 | 1456 | 1337 | 1219 | 1101 | 0984 | 0867 | 0751 | 0636
1.22 0521 | 0407 | 0293 | 180 | 0067 | 5955 | 9843 | 9732 | 9621 | 9512
1.23 | 9.959401 | 9292 | 9184 | 9076 | 8968 | 8861 | 8755 | 8649 | 8544 | 8439
1.24 8335 | 8231 | 8128 | 8025 | 7923 | 7821 | 7720 | 7620 | 7520 | 7420
1.25 7321 | 7223 | 7125 | 7027 | 6930 | 6834 | 6738 | 6642 | 6547 | 6453
1.26 6259 | 6267 | 6173 | 6081 | 5989 | 5898 | 5807 | 5716 | 5627 | 5537
1.27 5449 | 5360 | 5273 | 5185 | 5099 | 5013 | 4927 | 4842 | 4757 | 4673
1,28 4589 | 4506 | 4423 ; 4341 | 4259 | 4178 | 4097 | 4017 | 3938 | 3858
1.29 3780 | 3702 | 3624 | 3547 | 3470 | 3394 | 3318 | 3243 | 3168 | 3094
1.30 3020 | 2947 | 2874 | 2802 | 2730 | 2659 | 2588 | 2518 | 2448 | 2379
1.31 2310 | 2242 | 2174 | 2106 | 2040 | 1973 | 1007 | 1842 | 1777 | 1712
1.32 1648 | 1585 | 1522 | 1459 | 1397 , 1336 | 1275 | 1214 | 1154 | 1094
1.33 1035 | 0977 | 0918 | o86r | 0803 | 0747 | A690 | 0634 | 0579 | 0524
1.34 0470 | 0416 | 0362 | 0309 | 0257 | 0205 | e153 | o102 | 0051 | ooor
1.35 | 9 eee 9902 | 9853 | 9805 | 9757 | 9710 | 9663 | 9617 | 9571 | 9525
1.36 9480 | 9435 | 9391 | 9348 | 9304 | 9262 | 9219 | 9178 | 9136 | 9095
1.37 9054 | 9015 | 8975 | 8936 ; 8898 | 8859 | 8822 | 8785 | 8748 | 8711
1.38 8676 | 8640 | 8605 | 8571 | 8537 | 8503 | 8470 | 8437 | 8405 | 8373
1.39 8342 | 8311 | 8280 | 8250 | 8221 | 8192 | 8163 | 8135 | 8107 | 8080
1.40 8053 | 8026 | 8000 | 7975 | 7950 | 7925 | 7901 | 7877 | 7854 | 7831
1.41 7808 | 7786 | 7765 | 7744 | 7723 | 7703 | 7683 | 7664 | 7645 | 7626
1.42 7608 | 7590 | 7573 | 7556 | 7540 | 7524 | 7509 | 7494 | 7479 | 7465
1.43 7451 | 7438 | 7425 | 7413 | 7401 | 7389 | 7378 | 7368 | 7357 | 7348
1.44 7338 | 7329 | 7321 | 7312 | 7305 | 7298 | 7291 | 7284 | 7278 | 7273
1.45 7262 | 7263 | 7259 | 7255 | 7251 | 7248 | 7246 | 7244 | 7242 | 7241
1.46 7240 | 7239 | 7239 | 7240 | 7240 | 7242 | 7243 | 7245 | 7248 | 7251
1.47 7254 | 7258 | 7262 | 7266 | 7271 | 7277 | 7282 | 7289 | 7295 | 7302
1.48 7310 | 7317 | 7326 | 7334 | 7343 | 7353 | 7363 | 7373 | 7384 | 7395
1.49 7407 | 7419 | 7431 | 7444 | 7457 | 7471 | 7485 | 7499 | 7514 | 7529
170
p Oo 1/2/s[a4ls | 6|7|s]|9
1.50 | 9.947545 | 7561 | 7577 | 7594 | 7612 | 7629 | 7647 | 7666 | 7685 | 7704
1.51 7724 | 7744 | 7764 | 7785 | 7806 | 7828 | 7850 | 7873 | 7896 | 7919
1.52 7943 | 7967 | 7991 | 8016 | 8041 | 8067 | 8093 | 8120 | 8146 | 8174
1.53 8201 | 8229 | 8258 | 8287 | 8316 | 8346 | 8376 | 8406 | 8437 | 8468
1.54 8500 | 8532 | 8564 | 8597 | 8630 | 8664 | 8698 | 8732 | 8767 | 8802
1.55 8837 | 8873 | 8910 | 8946 | 8983 | 9021 | 9059 | 9097 | 9135 | 9174
1.56 9214 | 9254 | 9294 | 9334 | 9375 | 9417 | 9458 | 9500 | 9543 | 9586
1.57 9629 | 9672 | 9716 | 8761 | 9806 | 9851 | 9896 | 9942 | 9989 | 0035
1.58 | 9.950082 | 0130 | 0177 | 0225 | 0274 | 0323 | 0372 | 0422 | 0472 | 0522
1.59 0573 | 0624 | 0676 | 0728 | 0780 | 0833 | 0886 | 0939 | 0993 | 1047
1.60 1102 | 1f57 | 1212 | 1268 | 1324 | 1380 | 1437 | 1494 | 1552 | 1610
1.61 1668 | 1727 | 1786 | 1845 | 1905 | 1965 | 2025 | 2086 | 2147 | 2209
1.62 2271 | 2333 | 2396 | 2459 | 2522 | 2586 | 2650 | 2715 | 2780 | 2845
1.63 2911 | 2977 | 3043 | 3110 | 3177 | 3244 | 3312 | 3380 | 3449 | 3517
1.64 3587 | 3656 | 3726 | 3797 | 3867 | 3938 | 4or0 | qo8r | 4154 | 4226
1.65 4299 | 4372 | 4446 | 4519 | 4594 | 4668 | 4743 | 4819 | 4804 | 4970
1.66 5047 | 5124 | 520% | 5278 | 5356 | 5434 | 5513 | 5592 | 5671 | 5740
1.67 5830 | 591r | S991 | 6072 | 6154 | 6235 | 6317 | 6400 | 6482 | 6566
1.68 6649 | 6733 | 6817 | 6901 | 6986 | 7072 | 7157 | 7243 | 7322 | 7416
1.69 7503 | 7590 | 7678 | 7766 | 7854 | 7943 | 8032 | 8122 | 8211 | 8301
1.70 8391 | 8482 | 8573 | 8664 | 8756 | 8848 | 8941 | 9034 | 9127 | 9220
1.71 9314 | 9409 | 9502 | 9598 | 9693 | 9788 | 9884 | 9980 | 0077 | 0174
1.72 | 9.960271 | 0369 | 0467 | 0565 | 0664 | 0763 | 0862 | og61 | 1061 | 1162
1.73 1262 | 1363 | 1464 | 1566 | 1668 | 1770 | 1873 | 1976 | 2079 | 2183
1.74 2287 | 2391 | 2496 | 2601 | 2706 | 2812 | 2918 | 3024 | 3131 | 3238
1.75 3345 | 3453 | 3561 | 3669 | 3778 | 3887 | 3996 | 410s | 4215 | 4326
1.76 4436 | 4547 | 4659 | 4770 | 4882 | 4994 | 5107 | 5220 | 5333 | 5447
1.77 5561 | 5675 | 5789 | 5904 | 6019 | 6135 | 6251 | 6367 | 6484 | 6600
1.78 6718 | 6835 | 6953 | 7071 | 7189 | 7308 | 7427 | 7547 | 7066 | 7787
1.79 7907 | 8023 | 8149 | 8270 | 8392 | 8514 | 8636 | 8759 | 8882 | goos
1.80 9129 | 9253 | 9377 | 9501 | 9626 | 9751 | 9877 | c008 | 5129 | Gass
1.81 | 9.970383 | 0509 | 0637 | 0765 | 0893 | Io2zr | 1150 | 1279 | 1408 | 1538
1.82 1668 | 1798 | 1929 | 2060 | 21g9t | 2322 | 2454 | 2586 | 2719 | 2852
1.83 2985 | 3118 | 3252 | 3386 | 3520 | 3655 | 3790 | 3925 | 4061 | 41907
1.84 4333 | 4470 | 4606 | 4744 | 4881 | Sorg | 5157 | 5295 | 5434 | 5573
1.85 5712 | 5852 | 5992 | 6132 | 6273 | 6414 | 6555 | 6697 | 6838 | 6980
1.86 7123 | 7266 | 7408 | 7552 | 7696 | 7840 | 7984 | 8128 | 8273 | 8419
1.87 8564 | 8710 | 8856 | goo2 | 9149 | 9296 | 9443 | 9591 | 9739 | 9887
1.88 | 9.980036 | 9184 | 0333 | 0483 | 0633 | 0783 | 0933 | 1084 | 1234 | 1386
1.89 1537 | 1689 | 1841 | 1994 | 2147 | 2299 | 2453 | 2607 | 2761 | 2915
1.90 3069 | 3224 | 3379 | 3535 | 3690 | 3846 ; 4003 | 4159 | 4316 | 4474
1.91 4631 | 4789 | 4947 | 5105 | 5264 | 5423 | 5582 | 5742 | 5902 | 6062
1.92 6223 | 6383 | 6544 | 6706 | 6867 | 7029 | 7192 | 7354 | 7517 | 7680
1.93 7844 | 8007 | 8171 | 8336 | 8500 | 8665 | 8830 | 8996 | gt61 | 9327
1.94 9494 | 9660 | 9827 | 9995 | 0162 | 0330 |.0498 | 0666 | 0835 | 1004
1.95 | 9-991173 | 1343 | 1512 | 1683 | 1853 | 2024 | 2195 | 236 | 2537 | 2709
“1.96 2881 | 3054 | 3227 | 3399 | 3573 | 3746 | 3920 | 4004 | 4269 | 4443
1.97 4618 | 4794 | 4969 | 5145 | 5321 | 5498 | 5674 | 5851 | 6029 | 6206
1.98 6384 | 6562 | 6740 | 6919 | 7098 | 7277 | 7457 | 7637 | 7817 | 7997
1.99 8178 | 8359 | 8540 | 8722 | 8903 | 9085 | 9268 | 9450 | 9633 | 9816
Examples, 171
ExaMprezs.
a de
I. — : :
ge Ans, af
z. If f(z) = f(a + #) for all values of x, prove that
\7@) de =n I; fle) de,
where ” is an integer.
3 [. dz
: ree e ™.
0 V an — a ”
4 . dz -
1 arf @ = z ” 3
1
5- | sin-) «dz. - os
° 2
6 j, ax v
9 +2)/1+20—-a ” ee
mt dz ; zs, -
: a _ Be ; :
: j a & + 260 + cx” ao — b? being positive. ,, Ws =
8. Prove that
dz = =
J, a@+ 2ba* + oat = af where h =2 (A ae + 6).
é vw
Ans. ——.
- Weer wee ano
wT
= 6
10. Fe
’ oe sin 0
J
J
a rE + - cos?a
.
T
z a (a? + 8%)
12.
0
(@ aes + - (a sin? x + 6? cos?a)* nae
172 Definite Integrals.
a x“ nw es at
13. [ /e— x cos"! 5 a. ant. 16° 4.
41 dz 3 Tv
14. | —_— a> 6 ” =
-1 (a — ba) f1-# V @ — b
15 ( dz i
15. [SSS » ~™
a o/ (¢ — a) (8 — 2)
Va2-b3 24 BY) yd
16. ces eA ” wae
ava V aby = (y? + BP
” sin ax cos bx
17. Show that | -dg = *, or 0, according as @ > or< 4; and
0 x
that when a = 6 the value of the integral is :
1 ab
13. [. erik go eee ab<1. Ans. s te (272)
a J (i —20% 4) (1— 202 +8)’ J ab 1 ab :
Tv
+ tisk a. : log 2 2
19. [* tants ee » 3 gs op
3r
|? sinedz 7 I
20. | ae anne ” — + tan =
o I+ cos*% 4 2
21. If every infinitesimal element of the side ¢ of any triangle be divided
‘by its distance from the opposite angle C, and the sum taken, show that its
value is
A
log (cot — cot =) ‘
2 2
22. Being given the base of a triangle; if the sum of every element of the
base multiplied by the square of the distance from the vertex be constant, show
that the locus of the vertex is a circle.
Tr
7 cos? @ sin@ de aes 1 tance
23 \; 1 + cos?@" " @ 8°
T rare
ia cos?@ sin@da VI+e@ log(et+/t +6)
24. Se a
z 0 4/1 + & cos?@ " 2e 26 :
Examples. 173
25. Deduce the expansions for sin # and cos z from Bernoulli's series.
26, Show that the integral
1
| 2" (log x)™ dx
0
can be immediately evaluated by the method of Art. 111, when m is an integer.
© ton-l
= i tan-! (az) dx
Tv
» eGR, Ans. zee (1 +4).
28, Find the value of
"tog (I - 2acos x + a?) dz,
0
distinguishing between the cases where a is > or <1.
Ans. a <1, its value is o.
» @> TI, its value is 2m log a.
29. If f (x) can be expanded in a series of the form
& + 4 COS% + 4200824 +... +ancosnvt+...,
show that any coefficient after a can be exhibited in the form of a definite
integral.
ANS. an = 2 \77@ cos nx dx.
rJo
30. Find the analogous theorem when f(z) can be expanded in a series of
sines of multiples of +; and apply the method to prove the relation
x=2 (sine -S 4S o0.),
2 3
when « lies between + 7.
31. Prove the relation
Tv
z dé { aa
Esagy reese i =.
es oV sin 6d Eg
sia
32. Express the definite integral
T
S j 2 do
04/1 — w?sin?O
in the form of a series, « being <1.
2 2 2
Ans. = (1+ (;) Ket (=) att (+8) + &.),
2 2 2.4 2.4.6
174 Definite Integrals.
i ‘
z 1 [x
se {? log (1 + cos acos 2) ae hee (= = a).
0 cos & 2\4
oo ee Be
34. i axe 003 budz, where a4 > 0. 99 4
B
° tan ax tan-1p4 ® (a + 8)" }
35- \, agg eee » > log ae oS
=
36. [108 (a? cos? @ + B? sin? 6) dé. » 7 log ae oi
T
2 at+bsin0\ do b . (8
37° (; 8 \7—ban6) ane? » TS Na)"
ls de v
8, j ? lan
JF Ge =2/F-@-| = —,
| / x —
Adding, and dividing by 2, we get
2 2
Emits ERE olde dee
| , 2 Je -@ ec
af x — a
== — “tog (2+ f/x — a").
182 Areas of Plane Curves.
Accordingly, if we suppose the area counted from the
summit A, we have
= b ICS ab Ut ava
APN = 208-8 - 2 hog ( =
Again, since the triangle CPN = $y, it follows that
sector ACP = 2 log (? + z).
2 a b
For a geometrical method of finding the area of a hyper-
bolic sector, see Salmon’s Conics, Art. 395.
130 (a). Hyperbolic Sine and Cosine.—If S repre-:
sent the sector ACP, the final equation of the preceding
Article becomes
ab @ y\ |
= tog (§ + t) 8, (1)
which may also be written
& at y = ef,
a
b
introducing a single letter v to denote the quantity
a 28
log (2+ $) = Fae
Hence, by the equation of the hyperbola, we get
z Y _ ge,
a 6b
Thus, in analogy with the last result of Art. 128, calling the
following functions the hyperbolic cosine and hyperbolic
sine of v, and for brevity writing them cosh 2, and sinh 2,
e+e%=2coshe, e —e° = 2 sinhy, (2)
the co-ordinates of any point on the curve are
28 y 28
x ‘ ‘
a5 cosh » = cosh ee sinh » = sinh oi
The Catenary. 183
We might have treated the matter differently by intro-
ducing the angle ¢ defined by the equation x = a sec ¢, and
therefore y = btan @ (for the geometric meaning of this
transformation, see Salmon’s Conics, Art. 232); whence (1)
may be written*
28 T
we a
oF +0 = Tog tan (= +S),
and we see that the hyperbolic cosine of a real quantity is the
secant, and the hyperbolic sine the tangent of the same real
angle. Also, since
cosh »
sinh 0”
: sinh v I abi I -
sin ¢ =, cos ¢ = —_— = ——, cosec ¢ =
$ cosh v” % cosh v” e sinh »’ %
we can obviously extend the names of the other trigonometrical
functions likewise. Again, putting in (2) for 2, u/ — 1, or
iu, they become, by Art. 8,
i cos w= coshzu, 7% sin w# = sinh zu.
¥
131. Whe Catenary.—lIi an inelastic string of uniform
density be allowed to hang freely from two fixed points, the
curve which it assumes is called the Catenary.
its equation can be easily arrived
at from elementary mechanics, as fol-
lows :—
Let V be the lowest point on the
curve; then any portion VP of the
string must be in equilibrium under
the action of the tensions at its ex-
tremities, and its own weight, W. Fig. 6
Let A be the tension at V; 7 that :
at P, which acts along PR, the tangent at P; PRM = 9.
Then, by the property of the triangle of force, we have
W:A=PM: RM;
“. W=A tang.
* When @ is related to » by this equation, p is what Professor Cayley
(Elliptie Functions, p. 56) calls the gudermannian of v, after Professor Guder-
mann, and writes the inverse equation @ = gdv.
184 Areas of Plane Ourves.
Again, if s be the length of VP, and a that of the portion
of the string whose weight is 4, we have, since the string is
uniform,
W-=A°*;
a
“. S=a tang.
This is the intrinsic equation of the catenary. (Diff.
Cale., Art. 242 (a).)
Its equation in Cartesian co-
ordinates can be easily arrived at.
For, on the vertical through V
take VO = a, and draw OX in the
horizontal direction, and assume
OX and OY as axes of co-ordi-
nates. Let
PN=y, ON=2,
then
di
2 = tan ¢,
we sin oe COS o 5
fee ay Sas
. dy _dyds__ sing dy a
cos’g’ dp cos¢’
Hence y =asecg, «=a log (sec¢ + tang). (3)
No constant is added to either integral, since y = a, and
2 = 0, when ¢ =0.
From the latter equation we get
secg + tan p = e%;
x
I -2
also oe Et aie
Hence, we have
x xz =z
Zz
2se0p = e* +e %, 2tang = e*-e%,
Examples. 185
(+6 c), (4)
ono (#- 64), (5)
In the notation of last Article these equations may be
written
Consequently,
wre NIR
Also
ze s ee
Y — cosh = and — = sinh-.
a a a a
Again, if NZ be drawn perpendicular to the tangent at
P, we have
NL=PN cos; .. NL=a. (6)
Also Phe NE Gang; Phas] PV. (7)
The area of any portion VPNO is
Accordingly, the area VPVO is double that of the triangle
PNL.
EXAmMPLes.
1. To find the area of the oval of the parabola of the third degree with a
double point
2 - 5)?
ay = (e— a)(a- 8) ; F
The area in question is represented by 6 x
2
b
Fall (b — 2) x — ade.
Fig. 8.
8(b —a)®
3-50 °
2. Find the whole area of the curve a*y? = #3 (2a — 2). Ans. wa,
3. Find the whole area between the cissoid #3 = y? (a — x) and its asymptote.
2/8
times the rectangle which circumscribes the oval, having its sides parallel to the
co-ordinate axes.
Let 2 —a@ = 2%, and we easily find the area* to be
* The student will find little difficulty in proving that this area is —_—
186 Areas of Plane Ourves.
Since «—a=0 is the equation of the asymptote the area in question is re-
presented by
@ wide
| o(a— at
Let x = asin?6, and this becomes
nla
2a? i sin‘ @ d@:
hence the area in question is 3 na’.
4. Find the area of the loop of the curve vy
ay? = a4(b + 2).
This curve has been considered in Art. 262, Diff.
Cale. Its form is exhibited in the annexed figure ; and
the area of the loop is plainly
=f A ‘0
or 2 ap
a ew b+ ade.
Let 3 + # = 27, and it is easily seen that the area
in question is represented by
8. 38
ea ; Fig. 9.
3.5.7. a 18-9
5. Find the area between the witch of Agnesi
zy” = 4a" (2a — 2)
and its asymptote. Ans. 4na?.
132. In finding the whole area of a closed curve, such as
that represented in the figure, we
suppose lines, PA, QN, &c., drawn PQ
parallel to the axis of y; then, as- |¥
suming each of these lines to meet ;
the curve in but two points, and
making PM =», P’M = y, the
elementary area PQQ'’P’ is repre-
sented by (y2—- 4) da, and the en-
tire* area by
Oo 8B MN BX
OB’
| (y2 a UP) dix 3 Fig. to.
OB
in which OB, OB’ are the limiting values of z.
* This form still holds when the axis of w intersects the curve, for the ordi-
nates below that axis have a negative sign, and (y2 — yi) a will still represent
the element of the area between two parallel ordinates.
The Ellipse. 187
For example, let it be proposed to find the whole area of
an ellipse given by the general equation
au? + 2hay + by’ + 292 + 2fy+c=o.
Here, solving for y, we easily find
~-"= ; / = ab) a + 2 (if — bg) a+ 7" — be.
Also, the limiting values of z are the roots of the quadratic
expression under the radical sign.
_ Accordingly, denoting these roots by a and #3, and observ-
ing that 4? — ab is negative for an ellipse, the entire area is
represented by
Sif ab =P
CELA Yc —a)(B — #)dz.
To find this, assume ww — a = (8 — a)sin’6 ;
then B - x = (B - a) cos’O,
and we get
i VY (e — a)(B — 2) de = 2(B - a)? [sino cos’ 6 dO
=; (8-4).
: (if — by)? + (f? - be) (ab - 1?)
— 2
Again, (8 -a)?=4. (ab
_ 4b(af? + by? + ch? — 2fgh — abe)
7 (ab — h*)? ;
Hence the area of the ellipse is represented by
a (af? + bg? + ch? — 2fgh — abc)
(ab — #3
This result can be verified without difficulty, by deter-
mining the value of the rectangle under the semiaxes of an
ellipse, in terms of the enefficients of its general equation.
It is worthy of observation that if we suppose a closed
curve to be described by the motion of a point round its en-
tire perimeter, the whole inclosed area ts represented by | ydz,
taken for every point around the entire curve.
188 Areas of Plane Curves.
Thus, in the preceding figure, if we proceed from A to A’
along the upper portion of the curve, the corresponding part
of the integral { ydz represents the area APA’B’B. Again,
in returning from J’ to A along the lower part of the curve,
the increment dz is negative, and the corresponding part
of { ydz is also negative (assuming that the curve does not
intersect the axis of x), and represents the area A’P’ ABB,
taken with a negative sign. Consequently, the whole area of
the closed curve is represented by the integral f ydz, taken
for all points on the curve.
The student will find no difficulty in showing that this
proof is general, whatever be the form of the curve, and
whatever the number of points in which it is met by the
parallel ordinates.
To avoid ambiguity, the preceding result may be stated as
/ follows :—The area of any closed curve is represented by
dz
[vga
taken through the entire perimeter of the curve, the element of the
curve being regarded as positive throughout.
The preceding is on the hypothesis that the curve has no
double point. If the curve cut itself, so as to form two loops,
it is easily scen that | y eas, when taken round the entire
perimeter, represents the difference between the areas of the
two loops. The corresponding result in the case of three or
more loops can be readily determined.
133. Inmany cases, instead of determining y in terms of
, we can express them both in terms of a single variable,
and thus determine the area by expressing its element in
terms of that variable.
For instance, in the ellipse, if we make «=a sing, we
get y = 5 cos ¢, and ydx becomes ab cos’ > d@, the integral of
which gives the same result as before.
In like manner, to find the area of the curve
e.-
Let # = asin’¢, then y = b cos*¢, and ydx becomes
3ab sin’ cos'¢ do :
The Cycloid. 189
hence the entire area of the curve is represented by
T
12ab { sin’? cos'¢ de = 3 wad.
0
ExaMpies.
1. Find the whole area of the evolute of the ellipse
ey 3m (a? — 82)?
a + il I. Ans. a an
2. Find the whole area of the curve aK 7
; ye
2 2 .
@\ tm | (y\i _ 5
Gms (Pre
‘ 1.3.5 ...(2m+1).1.3.5... (241)
{ é Aue Zs4i0 a 6 BS ee Ss irae x2
134. Whe Cycloid.—In the cycloid, we have (Diff.
Cale., Art. 272),
x=a(0-sin@), y=a(1- cos);
ts | vae =@ Kc — cos 9)?d0 = 4a? | sint Sa,
Taking 6 between o and 7, we get 37a” for the entire
area between the cycloid and its base.
The area of the cycloid admits also of an elementary
geometrical deduction, as follows :—
D A
Fig. 11.
It is obviously sufficient to find the area between the
semicircle BPD and the semi-cycloid ByA. To determine
this, let points P and P” be taken on the semicircle such that
arc BP = are DP’: draw MPp and M’P’p’ perpendicula> to
BD. Take MN and M’N’ of equal length, and draw Vg
and W’¢, also perpendicular to BD: then, by the fundamen-
tal property of the cycloid, the line Pp = are BP, and P’p’
=are BP’: .. Pp+ P’p’ = semicircle = za.
&
ith
|
9
196 Areas of Plane Curves.
Now, if the interval ICN be regarded as indefinitely small,
the sum of the elementary areas PpgQ and P’p'7’Q is equal
to the rectangle under JLN and thesum of Pp and P’y’, or to
wax WN.
Again, if the entire figure be supposed divided in like
manner, it is obvious that the whole area between the semi-
circle and the cycloid is equal to za multiplied by the sum of
the elements IZW, taken from B to the centre C, i.e. equal to 7a’.
Consequently the whole area of the cycloid is 37a’, as
before.
The area of a prolate or curtate cycloid can be obtained
in like manner.
' 135. Areas in Polar Co-ordinates.—Suppose the
curve APB to be referred to polar co-ordinates, O being the
pole, andlet OP, OQ, OR represent consecutive radii vectores,
and PL, QM, ares of circles described with O as centre. Then
the area OPQ =OPL+ PLQ; but
PLQ becomes evanescent in com- ie oe
parison with OPLZ when P and Q
are infinitely near points; conse-
quently, in the limit the elemen- Q
2.
tary area OPQ = area OPL = aS ;
vy and @ being the polar co-ordi-
nates of P. %
Hence the sectorial area AOB
is represented by Fig. 12.
| rd,
2) 8
where a and (3 are the values of @ corresponding to the limit-
ing points A and B.
136. Area of Pedals of Ellipse and Hyperbola.—
For example, let it be proposed to. find the area of the locus
of the foot of the perpendicular from the centre on a tangent
to an ellipse.
2 2
Writing the equation of the ellipse in the form, + = i,
the equation of the locus in question is obviously
y* = a’cos’@ + 0b sin? 0.
Area of Pedals of Ellipse and Hyperbola. 191
Hence its area is
2
2 21 pe 2
a | cos*oa0 is = | sintOa0 = ea eae
2 4 4
2
2
sin 6 cos 0.
The entire area of the locus is
Ty. 2
5 (a + 3),
_ The equation of the corresponding locus for the hyperbola
is
rv? = a cos’0 — 8 sin’ 6.
In finding its area, since » must be real, we must have
a’cos’@ — 0’ sin’@ positive: accordingly, the limits for 0 are o
a
-12
and tan a
Integrating between these limits, and multiplying by 4,
we get for the entire area
ab + (a ~ 0°) tant.
In this case, if we had at once integrated between 0=0
and @ = 27, we should have found for the area (a? — b’) .
This anomaly would arise from our having integrated
through an interval for which 7° is negative, and for which,
therefore, the corresponding part of the curve is imaginary.
The expression for the area of the pedal of an ellipse with
respect to any origin will be given in a subsequent Article.
EXAaMPLes.
I. Show that the entire area of the Lemniscate
x” = a? cos 20
is a2,
2. In the hyperbolic spiral
r0=a,
prove that the area bounded by any two radii vectores is proportional to the
difference between their lengths.
3. Find the area of a loop of the curve
a
1? = a cos 8. Ans. <
192 Examples.
4. Find the area of the loop of the Folium of Descartes, whose equation is
B+ y= 3axy.
Transforming to polar co-ordinates, we have
_ 34 cos @ sin @
~ sin’ @ + cos?6"
Again, the limiting values of 6 are o and 33
a” 3 sin? 6 cos*6 d0
+ ae | Ge + ooo
Let tan @ = u, and this expression becomes
oe ae" wu? du _3@
(1+e8P 2°
5. To find the area of the Limacon
r=acosé+b
Here we must distinguish between two cases. i
(1). Let >a. In this case the curve consists of one loop, and its area is
2 2
=| "(a cos 6 + 5)2d0= (#+5)x
2J0 2
2
‘When 0d = a, the curve becomes a Cardioid, and the area =
(2). Let a, andr, + 1, = 2¢ cos 8,
and nr,=C-a@35 ° N-hn= 2/@— ¢ sin? 0.
Hence (r? -— 77) d0 = 4¢ cos 0,/a? — c sin’?0d0; and the
limiting values of 6 are + sin.
Hence the whole area is
sin-? ¢
ze : cos 0 /a? — c sin? 6 0.
[13]
a1a
194
Let ¢ sin @ = a sin ¢, and this integral transforms into
20 |
Again, if the origin be inside, we have ¢ < a, and
I
. (ri? + 1:°) = a@ + & cos 20;
as +177) d0 = [@ + ¢ cos 26) dO = 7a’.
0 0
ela wa
Areas of Plane Curves.
cos*¢ dp = 1a".
The method given above may be applied to find the area
included between two branches of the same spiral curve. As
an example, let us consider the spiral of Archimedes.
138. The Spiral of Archimedes.—The equation of
this curve is r= a0,
and its form, for
positive* values of 0,
is represented in
the accompanying
figure, in which O
is the pole and OA
the line from which
6 is measured. Let
any line drawn
through O meet the
different branches
of the spiral in
points P, Q, R, &e.:
then, if OP=r, and
LPOA=6,wehave,
from the equation
of the curve,
OP =a0, OQ=a(0+ 27), OR=a(0+ az), &e.
Fig. 15.
—
* Tt should be noted that when negative values of @ are taken, we get for
the remaining half of the spiral a curve symmetrically situated with respect to
the prime vector 04.
Areas by Polar Co-ordinates. 195
Hence PQ = QR = &e., = 2am = ¢ (suppose); i.e. the
intercepts between any two consecutive branches of the spiral
are of constant length.
Again, let OQ=7,, OR=r,=7, + ¢, and the area between
the two corresponding branches is
5 |e = 7") do = e[n dO + =| ao
Now, suppose INV and mn represent the limiting lines,
and let (3 and a be the corresponding values of 0; then the
area nVMm will be equal to
c) a0<0 [a9 =£(B - a) (aa + af +6)
=° (8 - a) (OM + On). (0).
If 8 -a=r7, this gives for the area of the portion
between two consecutive branches QH’Q’ and RF’R’, inter-
cepted by any right line RR’ drawn through the pole,
RQ. QR’, i.e. half the area of the ellipse whose semi-axes
are RQ and RQ.
139. Another Expression for Area.—The formula
in Article 137 still holds, obviously, when AB and ad repre-
sent portions of different curves.
It is also easily seen, as in Art. 132, that if a point be
supposed to move round any closed boundary, the included
area is in all cases represented by | a0, taken round the
entire boundary, whatever be its form; the elementary angle
a0 being taken with its proper sign throughout.
Again, if we transform to rectangular axes by the rela-
tions # = rcos0, y=rsin 8, we get
ens dO xdy — yde
tan 6="5 ao a
Hence db = ady - ydx;
[18 a]
196 Areas of Plane Curves.
and the area swept out by the radius vector is represented by
the integral
5 {ety - vin),
taken between suitable limits; a result which can also be
easily arrived at geometrically.
140. Area of Elliptic Sector. Lambert's Theo-
rem.—It is of importance in
Astronomy to be able to express
the area AFP swept out by the
focal radius vector of an ellipse.
This can be arrived at by inte-
gration from the polar equation |
of the curve; it is, however, a
more easily obtained geometri-
cally.
For, if the ordinate PN be produced to meet the auxiliary
circle in Q, we have
Fig. 16.
area AFP = : x area AFQ = *(4cQ - CFQ)
= = (u —e sine), (10)
where u = 2 ACQ.
By aid of this result, the area of any elliptic sector can be
expressed in terms of the focal distances of its extremities,
and of the chord joining them.
For (Fig. 17), let QP re- %
present the sector, and let
FP =p, FQ=p’, PQ=8; then, Q
denoting by wand w’ the eccen-
tric angles corresponding to 4’ é
P and Q, the area of the sector :
QFP, by (10), isrepresented by Bee
b :
= u-—w — e(sinu -sinw’)}.
FMMUNVA
Lambert's Theorem. 197
We proceed to show that this result can be written in
the form
{¢ - 9” - (sing — sin g’)}. (11)
0)
2
where ¢ and ¢’ are given by the equations
; I +p+8 .¢ 1 +p'-98
ats preee cin 2 fete
2 2 a 2 2 a
For, assume that » and ¢’ are determined by the equations
u-w=$- 4%’, e(sinw - sinw’) = sing -sing’. (a)
The latter gives
, ? , ,
_u-w utw i. o- ES
esin ——— cos =sin2—© go :.
2 2 2
utu’ +9"
or by the former, € co8 —-— = C08 ore.
Again, since the co-ordinates of P and Q are a cos u,
bsin u, and acos w’, b sin w’, respectively, we have
& = a’ (cosu — cos wu’)? + 0? (sin wu — sin w’)?
+ tf
.,u-u . wt u+u
= 4s81n? a’ sin? + 5? cos? ——
2 2
ri a
: -u uU+u
= 4a’ sin? (: — e* cos* )
2 2
, ,
~-_ "s +
= 4a’ ane? y sin 2 78 ,
2 2
, #
_ o-oo . ot
8 = 208in® ~ sin 2 ; ¢ = a(cos p’ — cos ¢). (0)
Again, from the ellipse, we have
p=a(l-ecosu), p =a(t—ecosy),
w+u 4—-a
cos
. p +p = 2a —ae(cos u + cosw’) = 2a — 2a€ cos
= 20 ~ 200082 cos £—# = 24 — 4(008@ + cos ¢’). (ce)
198 Areas of Plane Curves.
Hence, adding and subtracting (b) and (c), we get
+ p’ ‘
= os 2 (1 - cos) = asin’,
pte e. 2(1 — cos¢’) = asin’ ©,
which proves the theorem in question.
Consequently, the area* of any focal sector of an ellipse can
be expressed in terms of the focal distances of tts oda 8
the chord which joins them, and of the axes of the curve. .
141. We next proceed to an elementary principle which
is sometimes useful in determining areas, viz. :—
The area of any portion of the curve represented by the
equation
F Oe
G 5)
is ab times the area of the corresponding portion of the curve
F(a, y) =e.
This result is obvious, for the former equation is trans-
formed into the latter, by the assumption = =, ; =y; and
hence ydx becomes aby‘dzx" ;
| uae = as | vac,
the integrals being taken through corresponding limits—a
result which is also easily shown by projection.
2
2
Thus, for example, the area of the ellipse 5 + g =!
* This remarkable result is an extension, by Lambert (in his treatise entitled :
Insigniores orbite cometarum proprietates, published in 1761), of the correspond-
ing formula for a parabola given by Euler in Misceil. Berolin, 1743. It
furnishes an expression for the time of describing any arc of a planet’s orbit, in
terms of its chord, the distances of its extremities from the sun, and the major
axis of the orbit; neglecting the disturbing action of the other bodies of the
solar system.
Area of a Pedal Curve. 199
reduces to that of the circle ; and the area of the hyperbola
2 2
vy
—->- s=I!I
a Bb
to that of the equilateral hyperbola «? — y? = 1.
Again, let it be proposed to find the area of the curve
on? y\? “gt yf
(5+o)-5+4
The transformed equation is
2,2
(r+ yy = TE OF,
or, in polaz co-ordinates,
a cor? = -B* sin? 8
Ue m *
P=
2 2
But the whole area of this (Art. 136) is ~ a - j
2 [? 2
m
Consequently the whole area of the proposed curve is
a 8
- “ab (F Ze +3)
It may be remarked that the equations
2
FG, 2) ae F(x, y) =¢,
represent similar se and their corresponding linear
dimensions are as a: 1. Consequently the areas of similar
curves are as the savin of their dimensions; as is also
obvious from geometry.
142, Area of a Pedal Curve.—If from any point
perpendiculars be drawn to the tangents to any curve, the
200 Areas of Plane Curves.
locus of their feet is a new curve, called the pedal of the
original (Diff. Cale., Art. 187).
Ii p and w be the polar co-
ordinates of WV, the foot of the
perpendicular from the origin 0,
then the polar element of area of
the locus described by WV is plainly ®
2
a and the sectorial area of any
portion isaccordinglyrepresented by Fig. 18.
I
3 | po,
taken between proper limits.
There is another expression for the area of a closed pedal
curve which is sometimes useful.
Let S, denote the whole area of the pedal, and S that of
the original curve; then the area included between the two
curves is ultimately equal to the sum of the elements repre-
sented by NTN’ in the figure.
Hence S, = 8+ BNIN’ = 8+ | PN%dw. (12)
Again, by the preceding,
8:= +[0Nde.
Accordingly, by addition,
see 7 | OPide. (13)
It is easily seen that equation (12) admits of being stated
in the following form :—
The whole area of the pedal of any closed curve ts equal to
the sum of the areas of the curve and of the pedal of its evolute:
both pedals having the same origin.
For, PN is equal in length to the perpendicular from 0
on the normal at P: and hence -PN "dw represents the ele-
Steiner’s Theorem on Areas of Pedal Curves. 201
ment of area of the locus described by the foot of this perpen-
dicular, i.e. of the pedal of the evolute of the original curve.
For example, it follows from Art. 136 that the area of
the pedal of the evolute of an ellipse is < (a — b)*, the centre
being origin.
143. Area of Pedal of Ellipse for any Orisin.—
Suppose O to be the pedal
origin, and OM, OM’ perpen-
diculars on two parallel tan-
gents to the ellipse; draw CV
the perpendicular from the
centre C; let OM =p,, OM’
=D CN=p, OC=c, LOCA
=a, LACN=w; then
Fig. 19.
pi = MD - OD =p - cc0s(w - a),
P2 =p + €C08 (w — a).
Again, the whole area of the pedal is
5 [ws + p2*) dw -["e" + ¢ cos" (w — a)}dw
0 oO
v
3 (a°+B+#0*). (14)
7 [[otaw +e [cos (w — a) dw =
0
0
That is, the area of the pedal with respect to O as origin
exceeds the area of its pedal with respect to C by half the
area of the circle whose radius is OC.
If the origin O lie outside the ellipse, the pedal o msists
of two loops intersecting at O and lying one inside the other;
and in that case the expression in (14) represents the um of
the areas of the two loops, as can be easily seen.
The result established above is a particular case of a
general theorem of Steiner, which we next proceed to
consider.
144. Steiner’s Theorem on Areas of Pedal Curves.
Suppose .4 to be the whole area of the pedal of any closed
curve with respect to any internal origin O, and A’ the area
202 Areas of Plane Curves.
of its pedal with respect to another origin 0’; then, if p and
p’ be the lengths of the perpendiculars from O and O' on a
tangent to the curve, we have
23 ar
Anz |"pdo, A’ =F] "pide.
2Jo 2 Jo
Also, adopting the notation of the last article,
p=p- e cos(w — a) =p—2C0Sw —YSIDW3
where 2, y represent the co-ordinates of O’ with respect to
rectangular axes drawn through 0. Hence we get
ar
A’-A= Al (v7 cosw + ysinw)*dw
0
Qa 2ar
- s| pcoswdw — v| psinwdw.
0
0
ar hs : 29 :
But | cos? w dw = 7, ] sin’? w dw = 7, | SIN w COSw dw =0.
0 0 0
Qa ‘2a
Also, for a given curve, | pcosw dw ana | psinwdw are
constants when O is given. ‘Denoting their values by g and
h, we have
A'- A=" (+ y) ~ go hy. (15)
This equation shows that if O be fixed, the locus of the
origin O', for which the area of the pedal of a closed curve is
constant, is a circle.* The centre of this circle is the same,
whatever be the given area, and all the circles got by varying
the pedal area are concentric.
* It can be seen, without difficulty, from the demonstration given above,
that when the curve is not closed, the locus of the origin for pedals of equal area
is a conic: a theorem due to Prof. Raabe, of Zurich. See Crelle’s Journal,
vol. 1., p. 193.
tie cvident will find a discussion of these theorems by Prof. Hirst in the
Transactions of the Royal Society, 1863, in which he has investigated the corre-
sponding relations connecting the volumes of the pedals of surfaces.
Areas of Roulettes. 203
If the origin O be supposed taken at the centre of this
circle, the constants g and / will disappear; and, in this case,.
the pedal area is a minimum, and the difference between the
areas of the pedals is equal to half the area of the circle whose
radius is the distance between the pedal origins. a
For example, if we take the origin at the centre, the
pedal of a circle, whose radius i is a, is the circle itself. For
any other origin the pedal is a limagon; hence the whole
: : o? : ;
area of a limacon is (a + 3) as found in Art. 136, Ex. 5. _
145. Areas of Roulettes on Rectilinear Bases.
The connexion between the areas of roulettes and of pedals.
is contained in a very elegant theorem,* also due to Steiner,
which may be stated as follows :—
When a closed curve rolls on a right line, the area between
the right line and the roulette generated in a complete revolution
by any point invariably connected with the rolling curve is double
the area of the pedal of the rolling curve, this pedal being taken
with respect to the generating point as origin.
To prove this, suppose O to be the describing point in any
“
QP
Fig 20.
position of the rolling curve, and P the corresponding point
of contact. Let O’ represent an infinitely near position of the
describing point, Q’ the corresponding point of contact, and Q
* See Creiie’s Journal, vol. xxi. The corresponding theorem of Steiner
connecting the lengths of roulettes and pedals will be given in the next Chapter.
By the area of a roulette we understand the area between the roulette, the
base, and the normals drawn at the extremities of one segment of the roulette.
204. Areas of Plane Curves.
a point, on the curve such that PQ = PQ’; then Q is the point
which. coincides with Q’ in the new position of the rolling
curve; and, denoting the angle between the tangents at P
and Q (the angle of contingence) by dw, we have OPO' = dw,
since we may regard the curve as turning round P at the in-
stant (Diff. Cale., Art. 275).
Moreover, QQ’ ultimately is infinitely small in comparison
with QP, and consequently the elementary area OPQ’O’ is
ultimately the sum of the areas POO’ and QO’P, neglecting
an area which is infinitely small in comparison with either of
these areas.
2
Again, if OP =r, we have POO’ = oe, and area QO’P
= QOP in the limit.
Also the sum of the elements QOP in an entire revolu-
tion is equal to the area (S) of the rolling curve. Conse-
quently the entire area of the roulette described by O is
S+tfrdu.
But we have already seen (13) that this is double the area of
the pedal of the curve with respect to the point O; which
establishes our proposition.
Again, from Art. 144, it follows that there is one point in
any closed curve for which the entire area of the correspond-
ing roulette is a minimum. Also, the area of the roulette
described by any other point exceeds that of the minimum
roulette by the area of the circle whose radius is the distance
between the points.
For instance, if a circle roll on a right line, its centre de-
scribes a parallel line, and the area between these lines after
a complete revolution is equal to the rectangle under the
radius of the circle and its circumference ; i.e. is 27a”; denot-
ing the radius by a.
Consequently, for a point on the circumference, the area
generated is 27a’ + ra’, or 37a’; which agrees with the area
found already for the cycloid.
Ix like manner, by ,Steiner’s theorem, the area of the or-
dinary cycloid is the same as that of the cardioid: and the
area of a prolate or curtate cycloid the same as that of a
limacon.
General Case of Area of Roulette. 205
Again, if an ellipse roll on a right line, the area of the
path described by any point can be immediately obtained.
For example, the pedal of an ellipse with respect to a focus
is the circle described on its axis major. Hence, if an ellipse
roll upon a right line, the area of the roulette described by its
focus in a complete revolution is double the area of the auxiliary
circle. Also, the area of the roulette described by the centre
of the ellipse is equal to the sum of the circles described on
the axes of the ellipse as diameters, and is less than the area _
of the roulette described by any other point. (a4
146. General Case of Area of Ronlette.__If the
curve, instead of rolling on a right line, roll on another
curve, it is easily seen that the method of proof given in the
last article still holds; provided we take, instead of dw, the
sum of the angles of contingence of the two curves at the
point P.
Hence the element of area OPO’ is in this case
+ OP'do (1 = } or 5 OPtdu(1 4 2),
2 dw Py
where p and p’ are the radii of curvature at P of the rolling
and fixed curves, respectively.
Hence it follows that the area between the roulette, the
fixed curve, and the two extreme normals, after a complete
revolution, is represented by
S45 | rdw (: + *)
2 p
If a closed curve roll on a curve identical with itself,
having corresponding points always in contact, the formula
for the area generated becomes
|>
S+ rdw.
In this case the area generated is four times that of the
corresponding pedal; a result which appears at once geome-
trically by drawing a figure.
206 Areas of Plane Curves.
Exame.res.
1. If A be the area of a loop of the curve 7™ = a™ cosm@, and 4, the area
of its pedal with respect to the polar origin, prove that
A= (: + o) 4.
2
It is easily seen, as in Diff. Cale., Art. 190, thatthe angle between the radius
vector and the perpendicular on the tangent is m@; and .°. w = (m+ 1)@
Hence, by Art. 142,
pee
7 fr'd@ = (m+ 2)A.
2. Ifa circle of radius 4 roll on a circle of radius @, and if A denote the
area, after a complete revolution, between the fixed circle, the roulette described
by any point, and the extreme normals; and if 4’ be the area of the pedal of
the circle with respect to the generating point, prove that
Aa+ Bb=2(a4+b)A’.
where B is the area of the rolling circle.
3. Apply this result to find the area included between the fixed circle and the
are of an epicycloid extending from one cusp to the next.
147. Molditch’s Theorem.*—Ti a line CC’ of a given
length move with its extre-
mities on two fixed closed
curves, to find, in terms of
the areas of the two fixed
curves, an expression for the
whole area of the curve gene-
rated, in a complete revolu-
tion, by any given point P
situated on the moving line.
Let CP =c, PC’ = ¢, and suppose (a, y,), (#, y), and
(a2, Y2) to be the co-ordinates of the points C, P, and C’, re-
spectively, with reference to any rectangular axes.
Fig. 21.
* This simple and elegant theorem appeared, in a modified form, as the
Prize Question, by Mr. Holditch, under the name of “Petrarch,” in the Lady’s
and Gentleman’s Diary for the year 1858. The first proof given above is due to
Mr. Woolhouse, and contains his extension of Mr. Holditch’s theorem.
ee 2
e AS sens
Holditch’s Theorem. 207
Then, if @ be the angle made by CC” with the axis of y,
we have evidently
a=a%-csn0, y,=y-c cos 8,
m=a“2t+esnd, y=y+e' cos.
Hence we have
yida, = yd« — ¢ cos 0 (dx + yd@) + c cos’ Od6 ;
yr.de, = yde +c cos 0 (dx + yd0) + ¢? cos’ Od6.
Multiplying the former equation by ¢’, and the latter by ec,
and adding, we get
cy, day + cy2dx, = (¢ + ¢)yda+(c+c)ce’ cos0d0;
vw. Ofyida t+ cfyzdit, = (e+ ¢)fydx + (e+ &) ce’ cos’ Odd.
If we suppose the rod to make a complete revolution, so
as to return to its original position, and if we denote by (C),
(C’), (P), the areas of the curves described by the points
CO, C’, and P, respectively, we shall have (since in this case
the angle @ revolves through 27)
[(C) +0e(C") =(e+c)(P)+r(e+ cee,
é(C) + ¢(C’)
e+e¢
This determines the area (P) in terms of the areas (C),
(C’) and of the segments ¢, ¢’.
‘When the extremities C, C’ move on the same identical
curve we have(C) = (C”’), and hence (C) - (P) = mee’.
Consequently, if a chord of given length move inside any
closed curve, having a tracing point P at the distances e and
ce’ from its ends, the area comprised between the two curves is
equal to mec’.
More generally, if the extremities C, C’ move on curves
of equal area, we have, as before,
(C) - (P) = ee’. (17)
Should the extremities, instead of revolving, oscillate
back to their former positions, then (C) = 0, (C’) =0, and
or = (P) + wee’. (16)
|
208 Areas of Plane Curves.
(
“. (P) =— ec’. The negative sign implies that the area is
described in a direction contrary to that in which the rod re-
volves.
Again, if the rod returns to its original position after
n revolutions, the limits for 8 become o and 2n7, and equa-
tion (16) becomes
coy ee) = (P) + nee’. (18)
If (C) = (C”"), this gives
(C) - (P) = nee’. (19)
If the line oscillate back to its former position, without
making a revolution, we have » = 0, and (19) becomes
(C) = (P).
Hence, in this case, if two points describe curves of equal
area, then any point on the line joining these points describes
a curve of the same area.
The theorem in (16) can also be proved simply in another
manner, as follows:—
Let O denote the point of intersection of the moving line
\ CC’ with its infinitely near position ; that is to say, the point
, of contact with its envelope; and let OP =r. Adopting the
| same notation as before, let (O) represent the area of the en-
velope, and it is easily seen that
\
0) ~ (0)=4| (Oop a0=4 |" e- »*a8,
(07) - (0) =4 | (ooyaa=3[ "e+ r)a0,
(P) -(0)= | (OP)*d0 =4] rao;
hence
Qn
(C) +6(0") -(0+¢)(P) -1| (e'(e—r)*+0(e+r)?—(e4 er" d0
eed (e+c)n,
as before.
Holditch’s Theorem. 209
A remarkable extension of Holditch’s theorem was given
by Mr. E. B. Elliott, in the Messenger of Mathematics,
February, 1878.
Mr. Elliott supposed the length of the moving line C’C to
vary, but that it ig in all positions divided in the constant
ratio m:n in a point P.
Then, if C travel round the perimeter of any closed area
(C), and C’ move simultaneously round another area (C’), the
two motions being quite independent and subject to no re-
strictions whatever, except that both are continuous, having
no abrupt passage from one position to another finitely differ-
ing from it, then P will travel simultaneously round the
perimeter of another closed area (P).
Adopting the same notation as before, we have
(m+n)v = may + Nt, (m+n)y = my. + Nr} iy.
(m+n) yda = (my, + ny2) (mda, + nda.)
= my, da, +n’ y,dt,+ mn (y,dx, + y:dr2)
= (m+ n)(my, day + nyz da) —mn (Y2- Yi) a (2-21).
Integrating for a complete circuit, and dividing by (m +7),
we have
mn
m+n
(m-+n)(P)=m (0) + 9(0) ~ ™[-y)d(e—m). (20)
This result is stated as follows by Mr. Elliott :—
Through any fixed point in the plane of a closed area S
let radii vectores be drawn to all points in its perimeter, and let
chords AB, parallel and equal to the radii vectores, be placed
with one extremity A in each case in the perimeter of a closed
area (A), and the other B on that of another (B); then, if
the points .A, B, travel respectively all round the perimeters,
and do not in either case return to their first positions from
the same sides as that eT ia they left them ; and, if
14
210 Areas of Plane Curves.
(C) represent the area described by a point always dividing BA
in the constant ratio m : », then the areas (A), (B), (C), (S)
are connected by the following relation :
m(A) +” (B) mn
q) m+ 1 (m + n)* Sy ae
This follows immediately from (20) by altering the nota-
tion.
Areas described in opposite directions of rotation must be
taken with opposite signs.
For particular modifications in this result, as also for its
extension to surfaces, the student is referred to Mr. Elliott’s
paper ; as also to Mr. Leudesdori’s papers in the same
Journal. ‘o
147 (2). Kempe’s Theorem.— We next proceed to the
consideration of a singularly elegant theorem* discovered by
Mr. Kempe, and which may be stated as follows :—
If one plane sliding upon another start from any position,
move in any manner, and return to its original position after
making one or more complete revolutions; then every point
in the moving area describes a closed curve, and the locus, in
the moving plane, of points which describe equal areas is a circle ;
fe by varying the area we get a system of concentric circles for
oci
This result can be readily de- 2 e
duced from Holditch’s theorem, for s
if we suppose A, B, C, to be three 7
points which generate equal areas; it ae, \
can easily be seen that any fourth
point, D, which generates the same
area, lies on the circle circum-
scribing ABC.
Let AB and CD intersect in P,
then, let (P) represent the area :
described by the point P, as before ; Ha 2s
and » the number of revolutions made before AB returns
to its original position: then we have, by (19), denoting by
o
* Messenger of Mathematics, July, 1878.
Kemve’s Theorem. 211
(C) the common area described by each of the points
A, B, C, D,
(C) - (P) =nwAP. PB,
and, by same theorem,
(C) - (P) =nx CP. PD;
hence
AP.PB=CP.PD;
consequently A, B, C, D, lie on the circumference of the
same circle.
Again, let O be the centre of this circle, and join OP and
OA, then the preceding equation gives
(CO) - (P) = nr(O4? - OF”).
Hence all points which describe an area equal to that of ©
(P) lie on a circle, having O for centre, and OP for radius, |
which establishes the second part of the theorem.
For the effect of two or more loops in the area described
by a moving point see Art. 132.
i48. Areas by Approximation.—In many cases it is
necessary to approximate to the value of the area included
within a closed contour. The usual method is by drawing a
convenient number of parallel ordinates at equal intervals ;
then, when a rough approximation is sufficient, we may
regard the area of the curve as that of the polygon got by
joining the points of intersection of the parallel ordinates
with the curve. Hence, if 2 be the common distance between
the ordinates, and if
Yor Yrs Yrs SC-y Yny
represent the system of parallel ordinates, the area of the
polygon, since it consists of a number of trapeziums of equal
breadth, is plainly represented by
+9 t+ Y.+ &. + Ynap.
[14a]
re
212 Areas of Plane Curves.
Hence the rule: add together the halves of the extreme
ordinates, and the whole of the intermediate ordinates, and
multiply the result by the common interval.
When a nearer approximation is required, the method
next in simplicity supposes the curve to consist of a number
of parabolic arcs; each parabola having its axis parallel to
the equidistant ordinates, and being determined by three of
those ordinates.
To find the area of the parabola passing through the
points whose ordinates are %, 1, ¥2; let y=a+ Bx + yx" be
the equation of the parabola, and, for simplicity, assume the
origin at the foot of the intermediate ordinate y,, then we
have
Yo=a-Pht+yh?, y=a, Y=at BA+ yh’.
Again, the area between the first and third ordinate is
hh s i?
le (a + Pa + yx*)dx = 2h(a +y¥ =)
But y + ¥2 = 241 + 2yh’: hence the area in question is
aan
5 ie + ays + He.
Now, if we suppose the number of intervals n to be even,
and add the different parabolic areas, we get, as an approxi-
mation to the area, the expression
h
3 {Yo+ Yn + 4(m +Ys+ &e. + Yn-1) + 2(Y2 + Yat &e. + Yn-2)}.
Hence the rule: add together the first and last ordinates,
twice every second intermediate ordinate, and four times each
remaining ordinate; and multiply by one-third of the common
interval.
We get a closer approximation by supposing the number
of equal intervals a multiple of 3, and regarding the curve
as a series of parabole of the third degree, each being
determined by four equidistant ordinates. To find the area
corresponding to one of these parabolic curves, let y,, Yay Yo» Ys
be four equidistant ordinates, and for convenience assume
a
C
«
Areas by Approximation. 213
the origin midway between y: and y,; then if the equation
of the parabolic curve be ‘ sage
y=at Bat yx + d2°, Lg et
and the common interval on the axis of x be Monotda by 2h,
we have
Yo = a — 3Bh + gyh? — 2762,
fi =a - Bh + yh? — dh’,
y2=at Bh+ yh? + Ohi,
Ys =a t+ 33h + gyh® + 278h3.
Hence y+¥s=2(at+ oyh’), "+4. = 2(at yh’).
Again, the parabolic area between y and y; is
3h
| (a + Ba + yx? + d2°)dx = 3h(2a + 6yh?).
3h
Substituting in this the values of a and y obtained from
the two preceding equations, the expression for the area
becomes
h
2 (Yot+ Ys + 3 (Yi + Y2)}-
If the corresponding expressions be added together, we
easily arrive at the following rule :*—Add together the first
and last ordinates, twice every third intermediate ordinate, and
thrice each remaining ordinate ; and multiply by 3ths of the
common interval.
It is readily seen that these rules also apply to the ap-
proximation to any closed area, by drawing a system of lines,
parallel and equidistant, and adopting the intercepts made by
the curve instead of the ordinates, in each rule.
Since every definite integral may be represented by a
* This and the preceding are commonly called ‘‘ Simpson’s rules’? for cal-
culating areas; they were however previously noticed by Newton (see Opuscula.
Method. Diff, Prop. 6, scholium) as a particular application of the method of
interpolation. By taking seven equidistant ordinates, Mr. Weddle (Camb. and
Dub. Math. Jour., 1854), obtained the following simple and important rule for
finding the area:— To five times the sum of the even ordinates add the middle ordi-
nate and all the odd ordinates, multiply the sum by three-tenths of the common
interval, and the product will be the required area, approximately. The proof,
which is too long for insertion here, will be found in Mr. Weddle’s memoir :
and also, with applications, in Boole’s Calculus of Finite Differences. The student
is referred to Bertrand’s Cale. Int., 1. 1, ch. xii., for more general and aceurate
methods of approximation by Cotes and Gauss.
214 Areas of Plane Curves.
curvilinear area, the methods given above are applicable to
the approximate determination of any such integral.
In practice the accuracy of these methods is increased by
increasing the number of intervals.
149. Planimeters.—Several mechanical contrivances
have been introduced for the purpose of practically estimating
the area inclosed within any curved boundary. Such instru-
ments are called Planimeters. The simplest and most elegant
is that of Professor Amsler of Schaffhausen. It consists of
two arms jointed together so asto move in perfect freedom in
one plane. A point at the extremity of one arm is made a
fixed centre round which the instrument turns; and a wheel
is fixed to, and turns on the other arm as an axis, and records
by its revolution the area of the figure traced out by a point
on this arm. From its construction it is plain that the re-
volving wheel registers only the motion which is perpendi-
cular to the moving arm on which it revolves.
In the practical application of the instrument it is neces-
sary that the two arms, CA and AB, should return to their
original position after the tracing point B has been moved
round the entire boundary of the required area.
We shall commence by showing that the length registered
by the wheel while B has moved round the entire closed area
is independent of the wheel’s position on the moving arm;
i.e. is the same as if the wheel be supposed placed at the joint.
To prove this, suppose P to represent the point on the
arm at which the centre of the .
revolving wheel is situated. Let
A’B’ represent a new position of
AB very near to AB, and P’ the
corresponding position of the
point P. Draw PN perpendicular
to A’B’; then PV represents the
length registered by the wheel
while the arin moves from AB to
the infinitely near position A’B’.
Next, draw 4 NV’ perpendicular,
and AL parallel, to A’B’.
Let PN = ds’, AN’ =ds, AP =e,
PAL=d; then PN=PL+AN’,
or ds’ = ds + edd.
B
Amsler’s Planimeter. 215
Now, if we suppose AB after a complete circuit of the
curve to return to its original position, we have obviously
=(dp~)=0; and therefore & (ds’) = & (ds), i.e. the whole length
registered by the revolving wheel at P is the same as if it
were placed at A.
Next, let x and y be the co-ordinates of B with respect to
rectangular axes drawn through C, and let AC = a, AB = b,
ZL ACX = 0; and suppose ¢ the angle which BA produced
makes with the axis of ; then we shall have
x=acos0+bcosg, y=asinO+bsing.
Hence xdy - ydv = a’ dO + bd + ab cos(O — 9) d(0 + ¢).
Also ds = AN’ = AA’ sin AA’N = ad cos (0 - 9).
But 6+ =20-(0-¢);
.. ab cos (0 - ¢)d(0 + ¢)
= 2ab cos(@ - p)d0 — ab cos(0 — 9) d(0 - 4)
= 2bds — ab cos(0 - $) d(6 - ¢).
Consequently
ady —yde=a dO + b’ dy + 2bds — ab cos(0 - ¢) d(0- ¢).
But, by Art. 139, the area traced out by B in a complete
revolution is represented by 4 | (edy - ydz) taken around the
entire curve.
Also, since AC and AB return to their original positions,
the integrals of the terms a’d6, bd and ab cos (0 — ¢) d(0-¢)
disappear; and hence the area in question is equal to 6S, where
S denotes the entire length registered by the revolving wheel.
On account of the importance of the principle of this in-
strument, the following proof, for 8
which I am indebted to Prof. Ball,
based on elementary geometrical
principles, is also added.
Let CO, A, B represent, as before,
the positions of the fixed centre, the
joint, and the tracing point, respec-
tively ; and suppose & to represent
the position of the roller, or revolv-
ing wheel; then draw CP and RS Fig. 24.
perpendicular to AB.
216 Areas of Plane Curves.
Let AC=a, AB=b, AR=1, BC=r.
Now, if the instrument be rotated about C through an
angle @ without altering the angle CAB, it is easily seen
(eee the circumference of the roller is rotated through an are
represented by
a oe
PR.O= (14977 —")o,
— 2b
Again, if the instrument be rotated about S through a
small angle the roller does not revolve.
Hence a curve can be drawn through B,
such that, if the tracing point B be
moved along it, the roller will not
revolve.
Now, let Au, A’p’ be the two adjacent
circles described with C as centre, and
suppose aa’ and 3)’ two adjacent non-
rolling curves, such as just stated: and : :
suppose the tracing point B to move Big. 25,
round the indefinitely small area aa’ BG: then the arc through
which the roller has turned is represented by
(5 + =") 30 7 (45 +B-(r+ “ 36
2b 2b
= 189 _ aren of SOP
since a8 = 760; and or = ad’ sin B.
Now suppose the instrument works correctly for the area
dXa’a, then it will work correctly for the area AX’B’B; for,
start from a to X, N’, a’, then the area adda’ must be regis-
tered, since the roller does not turn in moving from a’ to a;
proceed then from a’ to (3’, B, a, then, by what has been just
proved, the area a’3’Ba will be added. Hence the instrument
will work correctly for the strip ANp'p.
Again, suppose the instrument works correctly for the
area Aup, then it will work correctly for \’y’p; for suppose
we start from A to p, wu, and back to A: then start from A to
Amsler’s Planimeter. 217
wu, mu’, A’ and A; the two journeys from \ to uw and » to A
will neutralize each other, and it follows that if the instrument
works correctly for the area Aup, it will work correctly for
the area X’u’p: hence, if the instrament works correctly for
any portion of the area, however small, it works correctly for
the entire area.
The student will find a description of Amsler’s Planimeter,
with another mode of demonstration, in a communication by
Mr. F. J. Bramwell, C.E., to the British Association.—See
Report, 1872, pp. 401-412.
218 Examples.
EXAMPLEs.
1. Find the whole area between the curve
ary? + ab? = ary?
and its asymptotes. Ans. 2rab.
z. Find the whole area of the curve
8a?
aty* = a4 (a? — 2), os 7 .
3. Find the whole area of the curve
a\? (y\8 3
(=)'+ (j)'=" Riga
4. Find the whole area included between the folium of Descartes
B+ y—3ary=0
Ans. 3a
and its asymptote. .
5. In the logarithmic curve y = a*, prove that the area between the axis of
«and any two ordinates is proportional to the difference between the ordinates.
6. Find the area of a loop of the curve
2
r =a cos 26. Aig,
Find the area of a loop of the curve
r=acosné + b sinné. oy (a2 +8?) =
The equation of the curve may be written in the form
raVf/ert & cos (n@ + a),
8
; and consequently its area can be found from the preceding
where tana = — —
a
example.
8. Find the area of a loop of the curve
r? = a® cos nd + 8 sin nd. Ans. LOsE
n
Examples. 219
g. Find the area of the tractrix.
The characteristic property of the tractrix is that the intercept on a tangent
to the curve between its point of contact and a fixed right line is constant.
Denoting the constant by a, and taking the origin O at the point for which
the tangent 0.4 is perpendicular
to the axis, we have, P being A
any point on the curve
PT =a, PN=y, ae?
SE
OV te Ping
dz VJ ae—y¥ oS
. Ydt =— VS @& —ydy.
Hence the element of the area of
the tractrix is equal to that of Fig. 26.
a circle of radius a.
It follows immediately that the whole area between the four infinite branches
of the tractrix is equal to ra”. This example furnishes an instance of our being
able to determine the area of a curve from a geometrical property of the curve,
without a previous determination of its equation.
If the equation of the tractrix be required, it can be derived from its differ-
ential equation
A @ = pay
dz=—- eee
from which we get
at fax =a log
That the equation of the tractrix depends on logarithms was noticed oy
Newton. See his Second Epistle to Oldenburg (Oct. 1676). This was,
believe, the first example of the determination of the equation of a curve by
integration ; or, what at the time was called the inverse method of tangents.
10. If each focal radius vector of an ellipse be produced a constant length c,
show that the area between the curve so formed and the ellipse is me(2d + ¢),
6 being the semi-axis minor of the ellipse.
11. Find the area of a loop of the curve 7 = a” cos 0.
fur)
r(-+-
2 7”
B/p
8. ——
re
n
12, If a right line carrying three tracing points 4, B, C, move in any manner
in a plane, returning to its original position after making a complete revolution ;
and if (.4), (B), (C) represent the entire areas of the closed curves described by
the points 4, B, C, respectively, prove that
BOC x (A) + GA x (B) + AB x (€) +7. AB. BC. CA =0,
in which the lmes 4B, BC, &c., are taken with their proper signs; i.e.,
AB =— BA, &e.
a+ fe-—¥
5 :
An,
220 Examples.
13. A, B, C, D, are four points rigidly connected together, and moving in
any way in a plane; if they describe closed curves, of areas (4), (B), (C), (Dj,
respectively; and if x, y, 2, be the areolar coordinates of D referred to the
triangle ABC, prove that
(D) = (A) + y(B) + 2(C) + 0*.
where ¢ is the length of the tangent from D to the circle circumscribed to the
triangle 4BC. Mr. Leudesdorf, Messenger of Mathematics, 1878.
This follows immediately : for let P be the point of intersection of the lines
AB and CD, then, by (18), we get a relation between (4), (B), and (P); and
also between (C), (D), and (P). If P be eliminated between these equations we
get the required result.
14. Show that a corresponding equation connects the areas of the pedals of
any given closed curve with respect to four points A, B, C, D, taken respectively
as pedal origin. Mr. Leudesdorf.
15. If a curve be referred to its radius vector » and the perpendicular p on
the tangent, prove that its area is represented by
I \ prdr
2) moe
16. A chord of constant length (c) moves about within a parabola, and
tangents are drawn at its extremities; find the total area between the parabola
and the locus of intersection of the tangents.
2
me
Ans. —.
2
17. From the centre of an ellipse a tangent is drawn to a semicircle
described on an ordinate to the axis major; prove that the polar equation of the
locus of the point of contact is
# a? 5?
” B4 (@ +P) tanto?
and that the whole area of the locus is
T ab
18. Apply the three methods of approximation of Art. 148 to the calculation
Ld:
to 6 decimal places of the definite integral \ — adopting = as the common
: 0
interval in each case. Ans. (1), 693669. (2), .693266. (3), .693224.
The -eal value of the integral being log 2, or .693147, to the same number
of decimal places.
1g. Prove that the sectorial area bounded by two focal vectors r and >” of a
parabola is represented by
a y(eeey (ara,
where ¢ is the chord of the arc, and a the semiparameter of the parabola.
Examples. 221
2 yf
20. Show that the whole area of the inverse of the ellipse “5 + x =Tis
represented by
ak’ I I I 1\ /a2 p*
nls at (a- 5) (S-5)}
('- 5-3)
where a, 8, are the co-ordinates of the origin of inversion, and / is the radius of
the circle of inversion.
a1. A given arc of a plane curve turns through a given angle round a fixed
point in its plane; what is the area described ?
22. Given the base of a triangle, prove that the polar equation of the locus
of its vertex, when the vertical angle is double one of its base angles is
ra 2 cos 26 + £)
"2 cos @
Hence show that the entire area of the loop of the curve is 30/3
4
23. Ois a point within a closed oval curve, P any point on the curve, QPQ’
a straight line drawn in a given direction such that QP = PQ’ = PO; prove that
as P moves round the curve, Q, Q’, trace out two closed loops the sum of whose
areas is twice the area of the original curve. Camb. Trip. Exam., 1874.
24. Prove that the area of the pedal of the cardioid r = a(1 — cos @) taken
with respect to an internal point at the distance ¢ from the pole is
: (5a8 — 2a¢ + 20”). (Ibid., 1876.)
25. The co-ordinates of a point are expressed as follows:
30 _ 36
Yat
;
find the equation of the curve described by the point, and the area of the portion
of the plane inclosed thereby.
(283
CHAPTER VIII.
LENGTHS OF CURVES.
150. Length of Curves referred to Rectangular Axes.
The usual mode of considering the length of a curve is by
treating it as the limit of a polygon when each of its sides 1s
infinitely small. If the curve be referred to rectangular axes
of co-ordinates, the length of the chord joining the points
(2, y) and («+ dz, y+ dy) is ,/dz* + dy*, and, consequently, if
s represent the length of the curve measured from a fixed
point on it, we shall have ds = ,/dz? + dy’, or, integrating,
on fr e(W) a, (1)
taken between suitable limits.
The value of wu in terms of x is to be got from the equa-
tion of the curve, and thus the finding of s is reducible to a
question of integration.
The determination of the length of an arc of a curve is
called its rectification.
It is evident that if y be taken for the independent variable
we shall have
dx\?
8 -| 1+ (=) dy.
Again, when x and y are given functions of a single va-
riable g, we have
dx\* (dy?
(a) * Ge) }
|
In each case the form of the equation of the curve deter-
mines which of these formulz should be employed.
The Catenary. 223
The curves whose lengths can be obtained in finite terms
(compare Art. 2) are very limitedin number. We proceed to
consider some of the simplest cases.
151. The Parabola.—Writing the equation of the
V +.
parabola in the form y? = 2ma, we get da mee 2 fa,
24 bon AK dy m nm
I ae m4 : 4 *
Hence = a(|Ve + mdy. ae
The value of this integral can be obtained from that of
the area of a hyperbola (Art. 130), by substituting y for a,
and m? for — a’. Cae ween ee
Thus we have >
pam om + of Pam
IVY +™ tog (4 y \
Ve
i 1
: (2)
2m m
the arc being measured from the vertex of the curve.
152. Whe Catemary.—The equation of the catenary
(Art. 131), is
cues
y=S(¢ +eé )
d 1/2 -2
Z-3(¢-«*)
a gaia et euge oe
de iS da®) 2 , Fig. 27.
Hence
I 2 I oie afz aie
8 ~5 [eae t le *dy= “(a —@é *) + const.
2 2 2
If s be measured from the vertex V, we have
a7 = om
Ts *);
2
the same result as already arrived at in Art. 131.
Again, since PL = PV,and NLis constant, it follows that
the catenary is the evolute of the tractrix (see Ex. 9, p. 219).
224 Lengths of Curves.
153. Semi-cubical Parabola.—The equation of this
curve is of the form ay’ = 2°.
wt dy 3 (a\b ds _ ga\t
hence ye He S(, Boe)
4 : 3
3.8 =((: +2) dz = a (1 +2) + const.
4a 27 4a
If the arc be measured from the vertex, we get
poe (2 + ae }.
27(\ "4a
The semi-cubical parabola is the first curve whose length
was determined. This result was discovered by William
Neil, in 1660.
154. Rectification of Evolutes.—It may be noted
that the rectification of the semi-cubical parabola is an
immediate consequence of its being the evolute of the ordinary
parabola (see Diff. Cale., Art. 239). In like manner the
length of any curve can be found if it be the evolute of a
known curve, from the property that any portion of the are
of the evolute is the difference between the two corresponding
radii of curvature of the curve of which it is the evolute.
For example, we get by this means the lengths of the
cycloid, the epicycloid and the hypocycloid.
Again, since the equation of the evolute of an ellipse is
(ax) + (by) = (@ - &)f,
the length of any arc of this curve can be at once found.
This can also be readily got otherwise; for, writing the
equation in the form
a\i 2
(2) *(B)
and making 2 = a sin’¢, we get y = (3 cos*¢, and
ds = (dx* + dy’)? = 3 sin $ cos g(a’ sin’¢ + (3 cos’ g)4d¢
3(a? sin’ + 3? cos’p)#
2(a° — p")
d(a? sin’p + 3? cos’¢).
(oo \ ? os
4 om \ =
4
bin
\
+
oN
Examples in Rectification. 225
Hence
_ (@ sin’ + 6? cos’g)?
a’ — (3?
If the arc be measured from the point x = 0, y = B, we
get the constant
ae (a? sin?¢ + B? cos’ )3 - B®
- p and s ae i :
If a = B, the expression for ds becomes 3a sin p cos pd¢ ;
+ const.
hence we get s = 3 a sin’¢, the arc being measured from the
same point as above.
EXAMPLEs,
1. Find the length of the logarithmic curve y = ca*.
dz 6b I
l = 1 Cy _—=- =>.
Here og y=xzlogat+loge; ae where 4 lea
B+ y?)idy yay B dy
Hence e=] ( ={ |
y (P+ yb Jy (P+ 9?)
24 y2)h
= (07 + yb +b log ae.
2. Find the length of the tractrix.
‘Here, by definition (see fig. 26), we have PT’ =a;
p ds a
-. sin PIN = , hence a = >
~8Se o{[2-- a log y + const.
If the arc be measured from the vertex A, we get
a
arc AP =a log (<) 3
3. Find in what cases the curves represented by amy" = a+” are rectifiable.
Here we have
226 Lengths of Curves.
2m
2 Bn : :
ee , and making 1 + d# ~ = 2°, this becomes
2m
na"
Substituting 8 for
feel,
n g? — 1\ am r
=— — de,
nb | ( 6 ) fae
This expression is immediately integrable when — is a positive integer.
Hence, if ZS =r, we see that curves of the form ay*" = «r+! are rectifiable.
Again, if = be a negative integer, the expression under the integral sign
becomes rational, and can accordingly be integrated. This leads to the form
y =ax, Accordingly, all curves comprised in the equation ay™ = z™*! are p
rectifiable, being any integer. (Compare Art. 62).
aa
155. Whe Ellipse.—The simplest expression for the are
of an ellipse is obtained by taking x = a sin ¢, whence
y= b cos p, and ds = (a* cos’ + b’ sin’ 9)! dg ;
8 -| (a cos’ + b sin’ $)2d¢.
It is often more convenient to write this in the form
g= af (1 - & sin’¢)*d¢, (3)
e being the eccentricity of the ellipse.
It may be observed that ¢ isthe complement of the eccen-
tric angle belonging to the point (2, y).
The length of an elliptic quadrant is represented by the
definite integral
Rly
a| (1 - & sin’¢)# do.
We postpone the further consideration of elliptic ares to
a subsequent part of the Chapter.
156, Rectification in Polar Ce-ordinates.—TIf the
curve be referred to polar co-ordinates we plainly have (Diff.
Cale., Art. 180) ds’ = dr? + °d@’ ; hence we get
> dr? \3 7? dO?2\3
s= iG + #) d0, ors= [(: + ) dr. (4)
LA (\ wo Shu
a — ’
Os eC A
ms e
A'aln OF
“yyy I fie Y
is i.
a
i
?
~~
ain ¥(A4 L)
» ¢
Rectification in Polar Co-ordinates. 227
For example, the length of the spiral of Archimedes, 7 = a8,
is given by the equation
=+| (r+ @)h
aK + @)sdr.
Comparing this with the formula (2) for the parabola, it
follows that the length of any arc of the spiral, measured
from its pole, is equal to that of a parabola measured from its
vertex.
EXxaMPLes.
1. Cardioid, r= a(I + Cos 6).
qi a
Here = =—a@ sin 0, and hence
s= af {(I + cos 6)? + sin?6}#d9 = 2a | cos * a = 4a sin + constant,
The constant becomes zero if we measure s from the point for which @ = o.
2. Logarithmic spiral, r= as.
: I
Here, if b= inne we get
r.
me n8; de s=| (1 4 Bde = (1 + BE (m — 10).
,
To
Accordingly, the length of any arc is proportional to the difference between
the vectors of its extremities; a result which also follows immediately from the (
property that the curve cuts its radius vector at a constant angle.
3. rm = am cos me.
di
Taking the logarithmic differentials, we get = =— tan m0;
* cal = sec m@
** 7d i
del
Hence s=a | (cos m6)” dd.
Or, writing ¢ for m0, E
a
a m
8 =| (cos ¢) = dp.
a 3 r
This is readily integrated when 7, 8 an integer (see Art. 56).
[15 a]
228 Lengths of Curves.
Whatever be the value of m, we can express the complete length of a loop of
the curve in Gamma Functions. For if we integrate between o and we ob-
viously get the length of half the loop.
Hence the length of the loop (Art. 122) is
I
afn* (5) ;
m m+
(e)
157. Formula of Legendre on Rectification.—
Another formula* of considerable utility in rectification fol-
lows immediately from the result obtained in Art. 192, Diff.
Cale. For, if this result be written in the form
| d(s—¢
US ~ 9 _ p, wo got s— t= piu. (5)
Consequently, the total increment of s —- ¢ between any two
points on a curve is equal to { pdw taken between the same
two points.
For example, in the parabola we have p = —, and
@w@
hence
dw
s-t=al
COS w
=a log tan ( + “) + const.
If we measure the arc from the vertex of the curve, and
observe that ¢ = o this gives
w
a sinw
cos? w
wT w
+a log tan (= + =)
The student can without difficulty identify this result with
that given in Art. 151.
* This theorem is due to Legendre. See TZraité des Fonctions Elliptiques,
tome ii., p. 588.
Fagnani’s Theorem. 229
It should be observed that when the curve is closed, its
whole length is, in general, represented by
20
| pdw.
0
Equation (5) furnishes a simple method of expressing the
intrinsic equation of a curve, when we are given its equation
in terms of p and w.
For, if p = f(w) we have
2 + {paw he [rt0) iis (6)
taken between suitable limits.
158. Application to Ellipse. Fagnani’s Theorem.
In the ellipse we have
p = @ cos’w + 8 sin’w.
Hence, measuring the are
from the vertex A, and observ-
ing that in this case PIV is to be
taken with a negative sign, we
have
Fig. 28.
arc AP + PN =|" (a? cos*w + 3? sin*w)} dw,
0
where a = 2 ACN.
But, in Art. 155, we have found that if ¢ be measured
from the vertex B, the arc is represented by
| (a? cos*p + 0° sin’9)}d¢.
Consequently, if we make 2 BOQ =a=2 ACN, and draw
QM perpendicular to the axis major meeting the curve in P’,
we shall have
are BP” = are AP + PN,
or, taking away the ccmmon are PP’,
BP - AP’ = PN. (7)
230 Lengths of Curves. ‘
This remarkable result is known as Fagnani’s Theorem*,
and shows that we can in an indefinite number of ways find
ve arcs of an ellipse whose difference is expressible by a right
ine.
We add a few properties connecting the points P and P’
in this construction.
EXAmMpPteEs.
1. If (2, y) and (z’, y’) be the co-ordinates of P and P’, respectively ; prove
the following :—
oe
Q). py=* =, (2). PN=P’N'’, (3). ON. CN’ = CA. CB,
(4). CP? + CN”? = CA? + CB? = CP” + CN?
2. Divide an elliptic quadrant into two parts whose difference shall be equal
to the difference of the semiaxes.
This takes place when P and P’ coincide; in which case CV = VY ab, and
PN=a~-d,
We shall designate the point so determined on the elliptic quadrant as Fag-
nani’s point.
3. Show that if a tangent be drawn at Fagnani’s point, the intercepts
between its point of contact and its points of intersection with the axes are
respectively equal in length to the semi-axes of the ellipse.
4. If the lines PN and P’N’ be produced to meet, show that they intersect
on the confocal hyperbola which passes through the points of intersection of the
tangents to the ellipse at its vertices. Show also that this hyperbola,cuts the
ellipse in Fagnani’s point.
* Fagnani, Giornale de’ Letterati d’ Italia, 1716, reprinted in his Produzioni
Matematiche, 1750. It may be noted that if we integrate the equation of Art.
116, Diff. Cale., taking the angle C as obtuse, and adopting zero for the lowest
limit in each integral, we obtain
a Oo
| ie sade [vt 1 ain®b db
¢ es
ma | we = Painted + Hsin a ain b sino,
where & is defined by the equation sin C= % sinc, and a, 5, ¢ are connected by
the relation
cos ¢ = cosa cos) — sina sin b4/1 — k* sin2e,
This equation furnishes a relation between three elliptic arcs, from which
.Fagnani’s theorem can be readily deduced, as well as many other theorems con-
nected with such arcs. See Legendre, Fone. Ellip., tomei., ch. g.
“A
~
Py
ee
kag tbe = 1
lotr flown =p ;
os lait 2 2h g we po
uae ae a p.
From Hobrhol oLirp Aea~ 46K
a 40- G's ws or wit fs 490
femter tl w' £6 =40
Tope Pym Cag Fa pi Open Pe
Nerrret lt [ow ference of 0, fC 6. fx G
i ge Ben mtay Sad P/p = Che hats F fg
ss
The Hyperbola. 231
6
The equation of PN is
asin + y cos@ = 4/ a? sin’@ + 5? cos?6,
and that of P’N’ is
wcoos@ ysing
+ =I
a 6
If we eliminate 6, we get 4
2
2
i =a-5,
which represents the hyperbola in question.
SY
159. Whe Hyperbola.—In the hyperbola we have
p= cos’ w — 0’ sin’ w.
Hence, measuring the arc from the vertex A of the curve,
we find, since w is measured below the axis,
PN - AP = i, (a? c0s'w — Bsin?w)tdw, (8)
0
where a = 2 ACN.
As we proceed along the hyperbola
the perpendicular p diminishes, and
vanishes when the tangent becomes
the asymptote.
Moreover, as the limit of w in this
case becomes tan™ i it follows that the
difference between the asymptote and
the infinite hyperbolic are, measured
from the vertex, is represented by the
definite integral
Fig. 29.
a
tan-1—
| ? (@cos’w — sin’ w)? dw.
- ;
® EXAMPLES,
N
1. If a>, prove that
S(a@+ bcos o)idp
is represented by an elliptic arc, and that the semiaxes of the ellipse are the
greatest and least values of (a + 4 cos )}.
2. If a <3, prove that
S(a+ dcosp)idp
is represented by the difference between a right line and a hyperbolic arc.
232 Lengths of Curves.
160. Landen’s Theorem on a Wyperbolic Are.—
We next proceed to establish an important theorem, due to
Landen ;* namely, that any are of a hyperbola can be expressed
in terms of the arcs of two ellipses.
This can be easily seen as follows:—In any triangle,
adopting the usual notation, we have
c=acosB + bcos A.
Now, representing by C the external angle at the vertex
C, we have C = A + B, and hence
ed = (acosB + bcos A) dA + (acosB + bcos.A) dB.
Consequently, supposing the sides a and 0 constant, and
the remaining parts variable, we have
[eae =| acosBdA + | bcos AdB + 2asin B + const.,
or
[Va +6? + 2abcos Gao=[v a’ — b*sin?A a4+|/B - asin’ B dB
+ 2asin B + const. (9)
Now, if we suppose a > 8, | r= ina dA represents
(Art. 155) the are of an ellipse, of axis major 2a and eccen-
tricity . Also |v 0 — a sin’B dB represents (Art. 159) the
difference between a right line and the arc of a hyperbola,
whose axis major is } and eccentricity ;
Again, ,/a? + 0 + 2ab cosC'= Jee - bytsint (a+ By cost”,
* Landen, Philosophical Transactions, 1775; also, Mathematical Memoirs,
1780.
Landen’s Theorem on a Hyperbolic Are. 233
and consequently the integral
|v a + 6? + 2ab cosOd0
represents an arc of the ellipse whose semiaxes are a + b and
a—Oob.
Hence, Landen’s theorem follows immediately.
It should be noted that the limiting values of A, B and
C are connected by the relations
asinB = bsinA, and C= 4+ B.
Again, if we suppose the angle 4 to increase from o to 7,
the external angle C will increase at the same time from
o to z, while B will commence by increasing from o to a,
and afterwards diminish from a to o( where a= sin” j
Moreover, in the latter stage bcos A is negative, and dB also
negative, consequently the term 6 cos A dB is positive through-
out the entire integration ; and the total value of
| / 0 — asin’ BdB is represented by 2 I. / 0 — asin’ BaB.
Hence, substituting ¢ for and integrating between the
limits indicated, we get, after dividing by 2,
a (a + b)* sin?’ + (a — 5)’ cos*p}2dp
-['@ — Bsin’A)MdA + [ie — asin BAB. (10)
0 0
Accordingly, the difference between the length of the asymp-
tote and of the infinite are of a hyperbola is equal to the differ-
ence between two elliptic quadrants. This result is also due to
Landen.
We next proceed to two important theorems, which may
be regarded as extensions of Fagnani’s theorem.
234 Lengths of Curves
161. Theorem* of Dr. Graves.—If from any point
P on the exter‘or of two confocal ellipses, tangents P7' and
PT’ be drawn to the in-
terior, then the difference
(PT + PT’ - TT’) between
the sum of the tangents
and the are between their
points of contact is con-
stant.
For, draw the tangents
QS and QS’ from a point
Q, regarded as infinitely
near to P, and drop the
perpendiculars PN and Fig. 30.
QN’; then, since the conics are confocal, we have
£ PQN =2 QPN’; «. PN’ = QN.
\ Als, P?=7R+RN=TR+RS+ SN = TS+SN
- TS + SQ- QN. 4
Tn like manner ges i oe jhe ;
PY’ = PN’ + 8’Q- T'S’;
PL 4 PT = Q8+ Q8'+ TS- T'S,
or PT + PT - TT = Q8+ QS’ - 88". }
Hence, PT + PT’ - TT’ does not change in passing to
the consecutive point Q; which proves that PT + PI’ - TT’
has a constant value.
a,
1 nw
* This elegant theorem was arrived at by Dr. Graves, now Bishop of Limerick,
for the more general case of spherical conics, from the reciprocal theorem, viz. :—
If two spherical conics have the same cyclic arcs, then any arc touching the
inner will cut from the outer a segment of constant area. (See Graves’ transla-
tion of Chasles on Cones and Spherical Conics, p. 77, Dublin, 1841.)
It should be remarked that the theorems of this and of the following article
were investigated independently by M. Chasles. The student will find in the
Comptes Rendus, 1843, 1844, a number of beautiful applications by that great
geometrician of these theorems, as well to properties of confocal conics, as also
to the addition of elliptic functions of the first species.
Theorem of Dr. Graves.
235
This value can be readily expressed by taking the point
at B’, one of the extremities
of the minor axis of the
exterior ellipse. Let D be
respectively.
Let CA =a, CB =4,
B
the point of contact of the B
tangent drawn from B’, and N
drop DM, and DN perpen- a
dicular to CA and CB, |
CA’=d’, CB’ =’, e the eccen- et
tricity ‘of interior ellipse. Bie, ae
Then, by Art. 155, the length of are
BD = aK ~ ésin’¢)!d¢,
where
a. DU. | CH CB _&
““ CB CB CB
Again,
BD = BN’ + DN? = (0 — bcosa)’ + @ sin’a
, & Be
-(¥-F) eels)
hence
BD -2 fF =a’ sina.
Consequently we have
BD - BD =¢d sina - al" ~ ésin’¢)4d¢,
0
Hence, in general,
PI + PI’ - TT’ = 2d sina- za (1 — e sin’ $)2dp, (11)
}
where a = cos? (>)
236 Lengths of Curves.
The analogous theorem, due to Professor Mac Cullagh,
may be stated as follows :—
162. Theorem.—lIf tangents PT, PT’ be drawn to an
ellipse from any point on a con-
focal hyperbola, then the differ-
ence of the tangents is equal to E
the difference of the arcs 7'’K and T eo
KT’.
The proofisleft tothe student, ar
and is nearly identical with that a ie
given for the previous theorem.
This result still holds when
the tangents are drawn from a
point on an ellipse to a confocal
hyperbola, provided that the tan-
gents both touch the same branch
of the hyperbola; as can be seen
without difficulty.
As an application* we shall prove another theorem of
Landen; viz., that the difference between the length of the
asymptote and of the infinite branch of
a hyperbola can be expressed in terms
of an are of the hyperbola.
For, let the tangent at A meet
the asymptote in D, and suppose a
confocal ellipse drawn through D.
Then, regarding DT as a tangent to
the hyperbola, it follows, by the
theorem just established, that the
difference between DT and KT is
equal to the difference between DA
and AK.
Consequently the difference be-
tween the asymptote C7 and the
hyperbolic branch AT is equal to
DA+DC-2KA. Consequently the Rigs 3
required difference is expressible in
terms of given lines and of the hyperbolic are 4K.
Fig. 32.
* I am indebted to Dr. Ingram for this application of Professor M‘Cullagh’s
theorem.
The Epitrochoid and Hypotrochotd. 237
We next proceed to consider two important curves whose
rectification depends on that of the ellipse.
163. The Limagon.—From the equation of the limacon,
r=acos0+, wo got = — asin 6,
and hence
ds = (a? + 6? + 2ab cos 0)2d0;
me [{e + bY? cost? + (a —by* sint Sl a.
Accordingly, the rectification of the limacon depends on
that of the ellipse whose semiaxes are a + 6 and a — b.
164. Whe Epitrochoid and Hypotrochoid.—The
epitrochoid is represented by the equations (see Diff. Calc.,
Art. 284)
2 = (a+ bd) cos - ¢ os *6,
y= (a+ 8) sind - eosin 2**9,
Hence
ae fsin 0 - Fin “=" ol,
= (a+ d) foos 0 — 5 cos “Gl.
Scalia and adding we get
ds’ fa+b(., , ad
(3) (S aaa +¢ - 2b0 008 FI
. _a+b 2 elt c5 ad
sat lfs +¢ 2be cos da.
a
Hence, substituting — for 0, we get
= HELA (6 + a sinty + 0 ~ 0) cost gag.
f
238 Lengths of Curves.
Consequently the length of an are of the epitrochoid is equal
to that of an ellipse. :
The corresponding form for the hypotrochoid is obtained
by changing the sign of 0.
165. Steiner’s Theorem on Rectification of
Roulettes.—If any curve roll on a right line, the length
of the are of the roulette described by any point is equal
to that of the corresponding are of the pedal, taken with
respect to the generating point as origin.
For (see fig. 20, Art. 145), the element OO’ of the roulette
is equal to OPdw.
Again, to find the element of the pedal. Since the angles
at NW and WN’ are right, the
quadrilateral NN’TO is inscri-
bable in a circle, and consequently
NW’ = OT sin NON’. But, in
the limit, WV’ becomes the ele-
ment of the pedal, and OT becomes
. OP: hence the element of pedal
is OPdw; consequently the ele-
ment of the pedal is equal to the
corresponding element of the Fig. 34.
roulette; .°. &c.
We proceed to point out a few elementary examples of this
principle. In the first place it follows that the length of an
are of the cycloid is the same as that of the cardioid; and
the length of the trochoid as that of the limacon. Again, if
an ellipse roll on a right line, the length of the roulette
described by either focus is equal to the corresponding are of
the auxiliary circle.
Moreover, it is easily seen, as in Art. 146, that, if one
curve roll on another, the elements ds and ds’, of the roulette,
and of the corresponding pedal are connected by the relation
ds = as (i + 6
Pp
In the case of one circle rolling on another, this relation
shows that the arcs of epicycloids and of epitrochoids are
proportional to the ares of cardioids and of limagons, which
agrees with the results established already.
x
f
(mer
166. Oval of Descartes.—We next proceed to the
rectification of the Ovals of Descartes, some properties of
which curves we have given in chapter xx., Diff. Cale.
The curve is de-
fined as the locus of
a point whose dis- 4
tances, rand 7’,from
two fixed points are
connected by the P/
equation
mr + lr =d, !
A B F c/\D FF,
ies
Oval oy Descartes. 239
where /, m, ad are
constants.
For convenience
we shall write the
equation inthe form
mr + Ur’ = ne, (12)
where c is the dis-
tance between the a
fixed points. BAA
The polar equation of the curve is easily got. For, let F
and F, be the fixed points, and 2 /, FP = 0, then we have
r? = 7 4 6 — 2recos 8;
also from (12),
Pr? = (ne ~ mr)’,
hence the polar equation of the locus is readily seen to be
m—-Posd ,w—P
wae + * gp (13)
vr — 2re
For simplicity we shall write this in the form
v— 2rQ+ C=0. (14)
Solving this equation for r, we get
r=2+4/Q?- 6, or FP, = 0+/Q'- 0, FP=2-./0'- 6.
It can be seen without difficulty that, so long as /, m, » are
real and unequal, the curve consists of two ovals, one lying
inside the other, as in the figure.
240 Lengths of Curves.
Again we get from (14), by differentiation
d
(r - Q)dr =ro'dd, where o =F;
, . aah
(Ge Mo ae ee
rd) r-Q ,/Q?-O rao r= 6
@?+0?-C E
go
the upper sign corresponding to the outer oval, and the lower
to the inner.
Hence the difference between the two corresponding
elementary arcs is equal to
2f Q? + Q* — Cdé, or, 2/ a + 2ab cos 0 + b — Cb,
(writing Q in the form a+b cos@); this plainly represents
the element of an ellipse. Consequently, the difference
between two corresponding arcs of the ovals can be repre-
sented by the arc of an ellipse. This remarkable theorem is
due to Mr. W. Roberts (Liouville, 1847, p. 195). Some years
after its publication it was shown by Professor Genocchi
(Tortolini, 1864, p. 97), that the arc* of a Cartesian is ex-
pressible in terms of three elliptic arcs.
In order to establish this result we commence by proving
one or two elementary properties of the curve.
Suppose a circle described through F, F,, and P; and let
PQ be the normal at P to the oval, meeting the circle in Q,
and join FQ and FQ; then let 2 FPQ = w, and A PQ =’;
dr i
: ? : ‘
and since m ae 1 mo we have /sinw’ =m sinw;
. FQ: FQ=l:m.
* For the proof of this theorem given in the text I am indebted to Mr.
Panton.
The Cartesian Oval. 241
Also, since mr +r’ = ne; and (by Ptolemy’s theorem)
FP .F.Q+FP.FQ = FF,. PQ,
FQ_F@_P@
we have
Hence, denoting the common value of these fractions by
u, we have
FQ=lu, FQ=mu, PQ= nu.
Again
, 2
muse. eo =. sy Sy eae
rag sf PE J/ Q? + Q?- C
Hence the first term in the expression for ds in (15) is
equal to
_ 7
ade e mn-i 008 On
cosw me — 7° COS w
Again, let ZFPF,=y, c4PREC=¢,
and we have the two following relations between the angles
0, o, p:
g=O+y, Jsin0+msing =n siny. (16)
Hence
dp-d0=dp, 1cos0d0+mcospdp =n cosydyp ;
*. (mn —P cos 6)d0 = m(n + loosp) dg — n (m+ 7 cos pb) dp,
or
2
mn —1 0088 ig _ 4, Mtl eosg 5 ag FOES fi)
COS w COs w COs w
Again, from the triangle PQ, we have
r cos w = PQ + FQ cos ¢ = (n + J cos $)u;
n +1 cos ——
. eet Pie + 2/n cos ¢.
c08 w
[16]
242 Lengths of Curves.
In the same manner it can be shown that
m+ loos _
os VP +m + 2lm cos Wp.
COS w u
Hence we have
[2< = won| VE 2in cos o dp
cosw mF
ne
a alV 2? +m + 2hn cos pay. (18)
Each of these latter integrals is represented by the are of an
ellipse, and, accordingly, the are of a Cartesian Oval is
expressible in the required manner.
It should be noted that the limiting values of 0, , andy
are connected by the relations given in (16).
Again, it can be shown without difficulty that the axes of
the ellipses are the lines (AB, CD), (AC, BD), and (AD, BC),
respectively : a result also given by Signor Genocchi. First,
with respect to the ellipse whose element is »/Q? + Q? — 040,
it is plain that its axes are the greatest and least values of
2,/0?+Q”- C, or of 2\/a? + b+ 2ab cos 6 — C; but these
are 2,/(a + b)*— © and 2,/(a — 6)? — C, which are plainly
the same as the greatest and least values of PP,; and, con-
sequently, are AB and CD.
Again, from the equation mr + i’ = ne, we get
tim
In like manner,
fone
c+m
Again, since we get the points on the outer oval by
changing the sign of J, we have
(n -2)e
m—-t
» £D=
.
’
Rectification of Curves of Double Curvature. 248
and, consequently,
2ne 2ue
a2 “m—? BG l+m
2me(n + 2) Bp a 2mele= 4) .
au m—T ? nar?
but these are readily seen to be the values for the axes of the
ellipses in (18).
It should be noted that if we substitute in (15) the values
for a and 4, the expression for the element ds becomes of the
following symmetrical form :
ds= (Font+ 2incos dp -—<—, / Pm 2lm cosy dp
2 ik oF amin 003 000. (19)
We shall conclude the Chapter with a brief account of
the rectification of curves of double curvature.
167. Rectification of Curves of Double Curvature.
If the points in a curve be not situated in the same plane, the
curve is said to be one of double curvature. The expression
for its length is obtained in an analogous manner to that
adopted for plane curves; for, if we refer the curve to a
system of rectangular axes in space, and denote the co-ordi-
nates of twoconsecutive points by (a, y, 2), (xv+dx, y+dy, s+ ds),
we get for the element of length, ds, the value
ds = / da? + dy? + ds’.
The curve is commonly supposed to be determined by the
intersection of two cylindrical surfaces, whose equations are
of the form
I(, Y) = 0, o(@, 8) =0.
From these equations, if oy and S be determined, the formula
of rectification is
[fe (8)
[16 a |
(=) en (20)
244 Lengths of Curves.
When z is taken as the independent variable, this formula
becomes
a da\* fdy\\* , .
r= [{r+(2) +(%) dz;
the limits being in each case determined by the conditions of
the question.
The simplest example is that of the helix, or the curve
formed by the thread of ascrew. From its mode of generation
it is easily seen that the helix is represented by two equations
of the form
z 3 &
= a e05(3), y=asin (5).
CE a PN Od Bete
deb” i) dsb °° \B)’
2 2
ees as =(1 + rr) ors= (: + mes
the are being measured from the point in which the helix
meets the plane of zy.
This result can also be readily established geometrically.
ence
Examp es.
1. Find the length of the curve whose equations are
aw od
Y= Fe oe
ae at \h a eo :
Here r=((1+5+3) doa |(1 +) deme Faetes
the arc being measured from the origin. ; .
This is a case of a system of curves which are readily rectified ; for, in ge-
neral, whenever
dy\? . de
da} da’
ee dy? dN} dz
we have (: 14+) =(: +E):
and therefore ds=da+dz, or s=% +4 + const.
Rectification of Curves of Double Curvature. 245
ai
Thus, if y = f(x) be one of the equations of a curve, we get ~ = f'(x), and
hence, if a second equation be determined from the equation
the length of the curve is represented by x + 2+ const.; the value of the con-
stant being determined by the conditions of the problem.
For instance, if y = a sin z, we get f’(x) = @ cosa, and
a costa; . = (« + cos sin z)
dz 2 Baap :
Hence the length of the curve of intersection of the cylindrical surfaces
a
y=asing, Bao le + toss sit 2)
is z+ 2; the length being measured from the origin.
oes 28
z. y=2/ az —2, seen: =. Ans. s=uty—zs.
2 2 Go ge . .
a ene ie #), the length being measured from the point
a
of intersection of the curve with the plane of zy.
(@ +2)
@
Ans, s= (2? — a”),
246 Lengths of Curves.
EXAMpPLes.
1. Find the length of any arc of the catenary
al 2
a (c+),
and show that the area between the curve, the axis of z, and the ordinates at
two points on the curve, is equal to a times the length of the arc terminated by
those points.
z. In any curve prove that s = | = and hence find the length of a
72 et p
parabolic arc.
3. Show that the integral —- may be represented by an arc of
VY bx? — a — 2
a circle, and find the limiting values of # for its possibility.
2 9 2
4. Show that the length of an elliptic arc is represented w| J a sey
where a is the semiaxis major, and ¢ the eccentricity.
5. Express the length of an elliptic quadrant in a series of ascending powers
of its eccentricity.
2 @2 4 2 66
Ans, {: - (;) es (3)'¢- (33) =~ be.
2 2j/ 1 2.4] 3 2.4.6) §
6. Prove that the integral of
a — x
a dx
V (a = B)(a? = 23)
can be represented by an arc of the ellipse whose semiaxes are a and B.
7. Show that the rectification of the sinusoid y = 6 sin 2 is the same as that
of an ellipse.
8. Prove that the whole length of the first negative pedal of an ellipse, taken
with respect to a focus, is equal to the circumference of the circle described on
the axis minor as diameter.
g. Show that the length of an arc of the curve r = a sin ”@ is equal to that
of an arc of the ellipse whose semiaxes are a and na.
10. If, from the equation of a curve referred to rectangular co-ordinates, we
form an equation in polar co-ordinates, by taking r = y and rd@ = dz, then the
lengths of the corresponding arcs of the two curves are equal, and the area f yda
of the former curve is equal to the corresponding sectorial area of the latter.
11. Prove that the difference between the lengths of the two loops of the
limacon r = a cos @ + 6 is equal to 84: @ being greater than 0b.
12, Being given three points 4, B, C on the circumference of an ellipse,
show that we can always find, at either side of C, a fourth point D such that the
difference between 4B and CD shall be equal to a right line.
Examples. 247
13. If a circle be described touching two tangents to an ellipse and also
touching the ellipse, prove that the point of contact with the ellipse divides the
elliptic arc between the points of contact of the tangents into two parts, whose
difference is equal to the difference of the lengths of the tangents (Chasles,
Comptes Rendus, 1843).
14. Prove that the entire length of any closed curve is represented by
= taken round the entire curve ; p being the radius of curvature at any
p
point, and p the length of the perpendicular from any fixed point on the tangent.
ds @t+t1
em +1 *
ae be the equation of a curve, prove that a me and
hence rectify the curve.
16. Calculate approximately, by the tables of Art. 125, the whole length of
a loop of the curve er, cos +9,
5
Here, by Ex. 3, Art. 156, the required length is
15. Ifev=
AD eu a
aan Ne =" \ 8:
Jaf ee or 20\/ 7 mia
vr) GQ)
8 3
Hence, taking logarithms, and observing that 3 = 1.625, and 2 = 1.125, we
get as the required approximation a x 3.29483. The figure of this curve is
exhibited in Art. 268, Diff. Calc.
17. In a Cartesian Oval whose two internal foci coincide, prove that the
difference of the two ares, intercepted by any two transversals from the exter-
nal focus, is equal to a straight line which may be found. [The above curve
is the inverse of an ellipse from a focus.]—Professor Crofton, Educ. Times,
June, 1874.
From (13) Art. 166, it follows, making » = m, that the equation of the
limagon, in this case, is
age
72 4 aro ESO oo,
which is of the form
r? + 2r(a cos @ — B) + (a — B)* =o.
Hence, by (15), the difference between two corresponding elementary arcs is
— @
4r/ aB cos 7 2.
Consequently, if 6; and 62 be the values of 9 for the two transversals in
question, we get the difference of the corresponding arcs
asf. 9 . Ah
= 8,/ a8 (sin - sin).
Also, it can be readily seen that the distance between the vertices of the
limacon is 4r/ oB; ~. &.
248 Lengths of Curves.
2 yp
18. Show that the length of an arc of the ellipse = + 5 = 1 is represented
by the integral
aoe { ees
(a? cos?@ + 5? sin?a)?
2 B2
This result is easily seen, for we have ds = pd, and p = ae 3 te Ge,
DB
19. Show, in like manner, that the length of a hyperbolic are is represented
by
ans i do
(a? cos? 9 — 8? sin? @)?
20. Hence prove that the integral
dz
| (a — b22)8 (a’ —2'22)8
is represented by an elliptic are when ab’ > ba’, and by a hyperbolic arc when
ab’ < ba’.
21. Prove that the differential of the are of the curve found by cutting in
the ratio m : 1 the normals to the cycloid
y=a+bcoosu, c=au+ bsinu,
is sl (a + nb)? + 4nad sin? = du.
22. Each element of the periphery of an ellipse is divided by the diameter
parallel to it: find the sum of all the elementary quotients extended to the entire
ellipse. Ans. 7.
23. In the figure of Art. 158, ifa= 2 ACN’, and B= £ BCN, prove that
tana _ tan B
a 6°
24. Find the length, measured from the origin, of the curve
as
g = a2 (I — e).
Ans. s=a log (<=) - a,
a-&
25. Find the length, measured from ¢ =o, of the curve which is represented
by the equations
az = (2a — b) sing — (a — 4) sin’ ¢,
y = (2b ~a) cos @ — (5 — a) cos*.
Ans. 8=} (a+ b)p +3 (a — b) sing cos¢.
26. Prove that the sides of a polygon of maximum perimeter inscribed in a
conic are tangents to a confocal conic.—Chasles, Comptes Rendus, 1845.
27. To two ares of an equilateral hyperbola, whose difference is rectifiable,
correspond equal arcs of the lemniscate which is the pedal of the hyperbola.
Ibid.
Examples. 249
28. The tangents at the extremities of two arcs of a conic, whose difference
is rectifiable, form a quadrilateral, whose sides are tangents to the same circle.—
Lhid.
2g. In an equilateral hyperbola prove that
rds = ha? d(tan 26),
and hence show that {rds taken between any two points on the curve is equal to
the rectangle under the chord joining the points and the line connecting the
middle point of the chord with the centre of the hyperbola. Mr. W. 8. M‘Cay.
30. If
z+2 be 2—28
Fee? 8 ae
be any point on a curve, show that the arc is the integral of
=
dz
— (M. Serret
a f/2 Jf Ite 4°
What curve do the equations represent ?
31. Through any point in a plane two conics of a confocal system can be.
drawn. If the distance between the foci be 2c, and the transverse semi-axes of
these conics be yu, v, prove the following expression for any are of a curve
dp? dy?
ds? = (pe — y*) {a - oI.
32. Prove that the following relation is satisfied by the u and y of any point
on a tangent to the ellipse for which yu has the value mi:
dp + dy aes:
VP) (mt) (A F(a)
33. The arc of the envelope of the right line x sina — y cosa= f(a) is the-
integral of (f(a) +f” (a)) da. (Hermite, Cours a’ Analyse.)
34. Thearc of the curve in which ? + a # — 24x = 0 and 22 —b? 2? 4 2b4=0
intersect, if a? = 1 + 0%, is
V2 = dds (Ibid).
WATC — az) (2— bz)
35. Show that the are of the curve = + ¢ =1 depends on an integral of
the form
{a V a (1 +2)* + (1 —2)*, where hae 2.
36. Show that rectification may, in general, be reduced to quadratures as.
follows :—
Produce each ordinate of the curve to be rectified until the whole length isin
aconstant ratio to the corresponding normal divided by the old ordinate, then.
the locus of the extremity of the ordinate so produced is a curve whose area is in
a constant ratio to the length of the given curve.
By this theorem Van Huraet rectified the semi-cubical parabola nearly simul-
taneously with Wm. Neil.
( 250 )
CHAPTER IX.
VOLUMES AND SURFACES OF SOLIDS.
168. Solids. The Prism and Cylinder.—The most
simple solid is the cube, which is accordingly the measure of
all solids, as the square is that of all areas. Hence the
finding the volume of a solid is called its cubature. Before
proceeding to the application of the Integral Calculus to
finding the volumes and surfaces of solids we propose to show
how, in certain cases, such volumes and surfaces can be found
from geometrical considerations. In the first place, the
volume of a rectangular parallelepiped is measured by the
continued product of the three adjacent edges; and that of
any parallelepiped by the area of a face multiplied by its
distance from the opposite face.
Again, the volume of a right prism is measured by the
product of its altitude into the area of its
base. For example, the volume of the right D
prism represented in the figure is mea- ¢. x
sured by the area of the polygon ABCDE, i
multiplied by the altitude 44’. Again,
since each lateral face, AB B’A’ for ex-
ample, is a rectangle, it follows that the
sum of the areas of all the faces (exclusive |
of the two bases), ie. the area of the sur- D
face of the prism, isequal to therectangle A
under the altitude and the perimeter of B
the polygon which forms its base. Fig. 36
This and the preceding result still hold ae
in the limit, when the base, instead of a polygon, is a closed
curve of any form, in which case the surface generated is
called a cylinder. Hence, if V denote the volume of the por-
tion of a cylinder bounded by two planes drawn perpendi-
cular to its edges, h its height, and A the area of its base, we
get V = Ah.
The Pyramid and Cone. 261
Again, if = denote the superficial area of a cylinder,
bounded as before, and S§ the length of the curve which forms
its base, we have 3 = Sh.
169. The Pyramid and Cone.—If the angular points
of a polygon be joined to any external point, the solid so
formed is called a pyramid. Any section of a pyramid by a
plane parallel to its base is a polygon similar to that
which forms the base, and the ratio of their homologous
sides is the same as that of the distances of the planes from
the vertex of the pyramid. Hence it follows that pyramids
standing on the same base, and whose vertices lie in a plane
parallel to the base, are equal in volume. For, the sections
made by any plane parallel to the base are equal in every
respect ; and, consequently, if we suppose the pyramids
divided into an indefinite number of slices by planes parallel
to the base, the volumes of the corresponding slices will be
the same for all the pyramids; and hence the entire volumes
are equal.
Also, if two pyramids have equal altitudes, but stand on
different polygonal bases, the volumes of the pyramids will
be to each other in the same proportion as the areas of the
polygonal bases. For, this proportion holds between the
areas of the sections made by any plane parallel to the base ;
and consequently between the slices made by two infinitely
near planes.
Again, the pyramid whose base is one of the faces of a
cube, and whose vertex is at the centre of the cube, is
the one-sixth part of the cube; for the entire cube can be
divided into six equal pyramids, one for each face. Hence,
denoting the side of a cube by a, the volume of the pyramid
3
in question is represented by a3 i.e. by the product of the
area of its base into one-third of its height.
Now, if we vary the base, without altering the height,
from what has been established above it follows that the
volume of any pyramid is the area of its base multiplied by
one-third of its height.*
* This demonstration is taken from Clairaut’s El/iens de Géometrie. The
student is supposed familiar with the more ancient proof, from the property that
a triangular prism can be divided into three pyramids of equal volume.
252 Volumes and Surfaces of Solids.
If the base of the pyramid be any closed curve, the solid
so formed is called a cone; and we infer that the volume of a
cone is equal to one-third of the product of the area of tts base
into tts height.
If the base of a pyramid be a regular polygon, and the
vertex be equidistant from the angular points of the polygon,
the pyramid is called a right pyramid.
In this case each face of the pyramid is an isosceles triangle,
whose area is the rectangle under the side of the polygon
and half the perpendicular of the triangle. Hence the
surface of the pyramid is equal to the rectangle under the
semi-perimeter of the regular polygon and the perpendicular
common to each face of the pyramid.
Again, if we suppose the number of sides of the regular
polygon to become infinite, the pyramid becomes a right
cone; and we infer that the entire surface of a right cone is
equal to the rectangle under the semi-cireumference of its
circular base and the length of an edge of the cone.
Hence, if a be the semi-angle of the cone, / the length of
an edge, and 7 the radius of its base, wo have r = /sin a, and
the surface of the cone is represented by 72? sin a.
If aright cone be divided by two planes ABC, DEF,
perpendicular to its axis, asin figure, the 0
part intercepted by the planes is called a
truncated cone.
The surface of a truncated cone is
easily expressed; forif OA =1, OD=1,
the required surface is 7 sina (J? — 1”),
or w(J— 0’) (2 + 7) sina.
Now, if the circular section LUN be
drawn bisecting the distance between
ABC and DEF, the circumference of the /,
circle LUN isw (2+ 1’) sina. Hencethe 4(
surface of the truncated cone is equal to
the rectangle under the edge 4D and the Fig. 37.
circumference of LIZN its mean section.
170, Surface and Wolume of a Sphere.—To find the
superficial area of a sphere; suppose a regular polygon in-
scribed in a semicircle, and let the figure revolve around the
diameter AB; then each side of the polygon, PQ for
example, will describe a truncated cone.
Surface and Volume of a Sphere. 253
Now, from the centre C draw CD perpendicular to PQ,
and construct, asin figure; then, by the preceding Article,
the surface generated by PQ is
equal to 27 PQ. DI. Q
Again, by similar triangles, P, JN
we have DC: DI= PQ: MN;
“, PQ. DI= DC. MN.
Accordingly, since the per-
pendicular CD is ofsame length |
for each side of the polygon, the 4 MIN ¢ B
surface generated by the entire Fig. 38.
polygon in a complete revo-
lution is equal to 27 CD : AB = 4m R’ cos ; where » repre-
sents the number of sides of the polygon, and R the radius of
the circle.
If we suppose » to become infinite, the solid generated
by the polygon becomes a sphere; and we get 47 #’ for the
entire surtace of the sphere. Hence, the surface of a sphere
is equal to four times the area of one of its great circles.
Again, it is easy to find the surface generated by any
number of sides of the polygon. Thus, for example, that
generated by all the sides lying between the points A and Q
is plainly equal to 2r CD. AN.
Hence, in the limit, the surface generated in a complete
revolution by the arc AQ is equal to 27. AC. AN. Sucha
portion of a sphere is called a spherical cap.
Again, suppose the points A and Q connected ; then, since
AQ = AB. AN, it follows that the area of the spherical cap
generated by the arc AQ is equal to the area of the circle
whose radius is the chord AQ.
The volume of a sphere is readily found from its surface ;
for we may regard the volume as consisting of an infinitely
great number of pyramids, having their common vertex at
the centre, and whose bases form the entire surface. But the
volume of each pyramid is represented by the product of one-
third of its height (i.e. the radius) by its base. Hence the
entire volume of the sphere is one-third of its radius multi-
plied by its surface, i.e. i RB.
254 Volumes and Surfaces of Solids.
EXAMPLES.
1. Ifa sphere and its circumscribing cylinder be cut by planes perpendi-
cular to the axis of the cylinder, prove that the intercepted portions of the
surfaces are equal in area.
2. Prove that the volume of a sphere is to that of its circumscribing cylinder
in the proportion of 2 to 3: and that their surfaces also are in the same propor-
tion. These results were discovered by Archimedes.
171. Surfaces of Revolution.—In the preceding we
have regarded a sphere as generated by the revolution of a
circle around a diameter. In general, if any plane be sup-
posed to revolve around a fixed line situated in it, every point
in the plane will describe a circle, and any curve lying in the
plane will generate a surface.
Such a surface is called a surface of revolution; and the
fixed line, round which the revolution takes place, is called
the avis of revolution.
It is obvious that the section of a surface of revolution
made by any plane drawn perpendicular to its axis is a
circle.
If we suppose any solid of revolution to be cut by aseries
of planes perpendicular to its axis, the volume of the solid
intercepted between any two such sections may be regarded
as the limit of the sum of an indefinite number of thin cylin-
drical plates.
Now, if we suppose the generating curve to be referred to
rectangular axes, the axis of revolution being that of x, the
area of the circle generated by a point (2, y) 1s plainly equal
to wy’, and the cylindrical plate standing on it, whose thick-
ness is dz, is represented by wy’*dz.
Hence, the element of volume of the surface of revolution
is wy’ da, and the entire volume comprised between two sec-
tions, corresponding to the abscisse a and PB, is obviously
represented by the definite integral
B
7 | y’ da,
a
in which the value of y ia terms of x is to be got from the
equation of the generating curve.
The Sphere. 255
In like manner, the volume of the surface generated by
the revolution of a curve around the axis of y is represented
by wfa*dy, taken between suitable limits.
Again, we may regard the surface generated by any
element ds of the curve as being ultimately a portion of the
surface of a truncated cone, as in Art. 170; and hence the
surface generated by ds in a complete revolution round the
axis of x is represented by 2ryds; and accordingly the entire
surface generated is represented by
2m | yds
taken between proper limits.
‘We proceed to apply these formule to a few elementary
examples.
172. The Sphere.—Let 2’ + y’ = a’ be the equation of
the generating circle ; then, substituting a’ — x for y’, we get
for the volume
3
Ve= AG — 2) de = (ae - =) + const.
3
If we take o and a as limits, we get = for the volume of
3
the hemisphere; .*. the entire volume of the sphere is — ;
as in Art. 170.
To find the volume of a spherical cap, let # be the length
of the portion of the diameter cut off by the bounding plane,
and we get for the corresponding volume
@ h
7 (a? - x*) da = wh’ (« - 5}
a-h 3
¥
Again, to find the superficial area, we have
dy’\3 x a :
ds = (: + 5) ae-(1 + =) dx = go J. yds = adz.
Hence, the surface of the zone contained between two
parallel planes corresponding to the abscissa a, and a is
*
an adx = 27a (a, — %);
x
256 Volumes and Surfaces of Solids.
that is the product of the circumference of a great circle by
the breadth of the zone. This agrees with Art. 170.
173. Right Cone.—If a denote, as before, the angle
which the right line which generates a cone makes with its
axis of revolution, we get y = # tana, taking the vertex of the
cone as origin, and the axis of revolution as that of «; accord-
ingly, the element of volume is 7 tan’aa*dz. ;
Hence, if A denote the height of the cone, we get its
volume equal to
wh?
h
a tan’a | ve dx = = tan’a;
0
Le. g x area of its base, as in Art. 169.
3
Again, to find its surface, we have ds = sec adz ;
h
. 20 f yds = 2m tan a sec «| eda = wh tana seca ;
0
which agrees with the result already obtained.
EXAMPLEs.
1. The base of a cylinder is a circle whose area is equal to the surface of a
sphere of radius 5 ft.; being given that the volume of the cylinder is equal to
the sum of the volumes of two spheres of radii 9 ft. and 16 ft., find the height
of the cylinder. Ans. 64% ft.
2. A solid sector is cut out of a sphere of 10 ft. radius, by a cone the angle
of which is 120°; find the radius of the sphere whose solid contents are equal to
those of the sector. Ans. sh 2.
3. Two eones have a common base, the radius of which is 12 ft. ; the alti-
tude of one is 9 ft. ; and that of the other is 5 ft.; find the radius of a sphere
whose entire surface is equal to the sum of the areas of the cones.
Ans. 24/28 ft.
174. Paraboloid of Revolution.— Writing the equa-
tion of a parabola in the form y* = 2mzx, we get for the
volume of the solid generated by its revolution round the
axis of #
7
2am f adx = mmx + const. = S ye + const.
Surface of Spheroid. 257
Hence, the volume of the surface generated by the revo-
lution of the part of a parabola between its vertex and the
point (x, 71) is represented by . yi %, i.e. is equal to half the
volume of the ee cylinder.
Again, to find the surface of the paraboloid, we have
ieegl ial) a= = omina
yds=y\ 1+ 5) dy =~ (y+ m')bydy.
Hence, the surface of the paraboloid, between the same
limits as above, is represented by
20 "1 2 2 Bo 27 2 B\e .. pel
=|" EGS Va Se iE
175. Spheroids of Revolution.—If we suppose an
ellipse to revolve round its axis major, the surface generated
by the revolving curve is called a prolate spheroid. If it re-
volve round the axis minor the surface is called an oblate
spheroid.
The volume of a spheroid is easily obtained ; for, taking
2 2
— + a = 1 as the equation of the curve, we get, on substitut-
ae
ing 0° € -5) for y’,
2 2 2
V= ral - )de=To(a - =) + const.
a a 3
Hence the entire volume is ear In like manner, the vo-
lume of an oblate spheroid is obviously 4 oa’.
176. Surface of Spheroid.—In the case of a prolate
spheroid we have
btx?\3
ds = (: + wai) dx;
4 \h 2 3 2 i
“yds = (x =a) dz = G ao ea) do = (S - #) de.
e a
[17]
258 Volumes and Surfaces of Solids.
Henee, if CN = «,, CM = x, we get for S, the zone gene-
rated in a complete revo- Py
lution by the are PQ, “er a E,
Qi
2 (aq? 3
S-2n--| (= ss *) die.
& J xo c| MF IN JA
Now, if we take CD = <,
and construct an ellipse Pe
whose semiaxes are CD Fig. 39.
and CB, it is easily seen
(Art. 129) that the elementary area between two consecutive
2
wD 3 .
ordinates of this ellipse is a(S —2 | de. Hence it follows
q, f
that the area of the zone generated by the arc PQ is w times
the area of the portion P,Q,Q.P, of this ellipse.
/ _ Again, if AZ, be the tangent at the vertex of the original
ellipse, we see that the entire surface of the spheroid is 47
x the area BCAE, ; but this is seen, without difficulty, to be
Ds
aad? + 20 = sine. (1)
In like manner, we get forthe surface S generated by the
revolution of an ellipse round its minor axis
4 2 4
S= on.| ed = 20 IG + “ie r) dy
We bt \3
= 27 Fl(v+ 7) dy.
If this be integrated, as in Art. 151, we get, after some
obvious reductions,
' 2 2 2 4/2 4) 4
Sank @ey + OE ne log BE a
If this be taken between the limits o and 8, and doubled, we
get fo: the entire surface of the ellipsoid
I+eé
2 ac
and! + me tog ( ) (2)
ES ¢é
Ellipsoid of Revolution. 259
It is readily seen, as in the former case, that the surface ,
of any zone of this ellipsoid is 7 times the area of a corre- |
sponding portion of the hyperbola
ve eey?
a
bounded by lines drawn parallel to the axis of «.
The area of the surface generated by the revolution of a
hyperbola round either axis admits of a similar investigation.
EXamMPLes.
1, Find the volume of the surface generated by the revolution of a cycloid
round its base. :
Here, referring the cycloid to DA and WN L B
DB as co-ordinate axes, we have (see Diff. 7
Cale: ;*Art. 272) c
r=a(pt+sing), y=a(1 + cos¢),
where / PCL = 9. AN 0 D A’
Hence
dV=ryde = wa3(1 + cos dp 3 Fig. 40.
.. for the entire volume V, we get
Tr vr p
V= ana! | (1 + cos ¢)3dp = 16708 j cosé 2%
0 0
TT
= ana? : cos°@d0, making es 0.
0
f=
Hence V = sna’.
2. Find the whole surface generated in the same case.
Pes
Here San | ydr= grat | (x + 005 9) e085 do;
hence the entire surface is
64" a*
Zr ar i cos? * dp =
2 3
a a [17 a]
260 Volumes and Surfaces of Solids.
3. Find the volume and the surface of the solid generated by the revolution
of the tractrix round its axis.
(1). Here we have A
y? du =— (a — Ay dy ;
hence the volume generated by aan
the portion .4P is T
Bi 7
n{ Vege. ey
y
The volume generated by the
: Ss, 2a é
entire tractrix is — a3; i.e. half
the volume of the sphere whose Fig. 41.
radius is 0.4.
(2). The surface generated by 4P is
a
an vas = 27a | dy (see Ex. 2, Art. 154)
y
= 2na(a-y).
Hence the entire surface generated is 2ra?; i.e. half the surface of the sphere
of radius 0.4.
4. Find the volume, and also the surface, generated by the revolution of the
catenary around the axis of x.
(1). Here the volume of the solid gene- ,
rated by VP is represented by
2x 2a:
2: ee fF ES -=
=|" yae=™ | (e+e a+ 2) de
0 4 Jo
mittapm. =
=— 5 (e—< «)+ 2a}
4 (2
7a ~
=" (s+ az), o wx
where s = PP. Fig. 42,
(2). Again, since
we have
Surfaces of Revolution. 261
Consequently the surface generated by PV in a complete revolution is z
x the volume generated ; i.e. = r(ys + az). <
5. In the same curve to find the surface generated by its revolution round
the axis OV.
Here
2 =
Saar [edn sede +x| re “dx,
Again
ee z a = =
| verde = anet— a e* du = a (wet — aet + a).
0
Also the value of
0
Ee ere
| we adg
0
is obtained by changing the sign of a in the last result.
Hence
a ge pee we
| ze *dy=a®—are *— are 4;
0
x a x a
ete San faa? + an (« -«#) = a(a rea)
=2r(a* + xs — ay).
177. Annular Solids.—If a y <3
closed curve, which is symmetrical C IkK—
with respect to a right line, be made
to revolve round a parallel line, then é z -
P
the superficial area generated in a
complete revolution is equal to the
product of the length of the moving
curve into the circumference of the
circle whose radius is the distance § na x
between the parallel lines. ee
This is easily proved: for let euvs
APBP’ be any curve, symmetrical with respect to AB, and
suppose OX to be the axis of revolution; and draw PN, QU
two indefinitely near lines perpendicular to the axis. It is evi-
dent that PQ=P’Q’. Again, let PN=y, PN=y',PQ=P'Q
= ds, DN =}; then the sum of the elementary zones described
by PQ and P’@ in a complete revolution is represented by
an (y + y')ds = 4rbds.
262 Volumes and Surfaces of Solids
Consequently the surface generated by the entire curve is
2m 6S, where S denotes the whole length of the curve.
A similar theorem holds for the volume of the solid ge-
nerated : viz., the volume generated is equal to the product
of the area of the revolving curve into the circumference of
the saire circle as before.
For the volume of this solid is plainly represented by
[Wt -v de,
or by r[U-v)uty)ae = and | (y- yaa.
But the area of the curve is represented by
[@- nae:
consequently, denoting this area by A, and the volume by JV,
we have
V=27b x A.
In these results the axis of revolution is supposed not
to intersect the curve; if it does, the expression 27) x A
represents the difference between the volumes of the surfaces
generated by the portions of the curve lying at opposite sides
of the axis of revolution ; as is readily seen. A similar alte-
ration must be made in the former theorem in this case.
If a circle revolve round any external axis situated in its
plane, the surface generated is called a spherical ring. From
the preceding it follows that the entire surface of such a ring
is 47?ab; where ais the radius of the circle, and d the dis-
tance of its centre from the axis of revolution.
In like manner the volume of the ring is 27°a’b.
It would be easy to add other applications of these
theorems. Pace
178. Guldin’s* Theorems.—The results established in
the preceding Article are but particular cases of two general
* Guldin, Centrobaryica, seu de centro gravitatis trium specierum quantitatis
continue, 1635. Guldin arrived at his principle by induction from a small num-
ber of elementary cases, but his attempt at a general demonstration was an
eminent failure. See Montucla Hist. des Math., tom. ii. p. 34. Montucla has
shown, tom. ii. p. 92, that Guldin’s theorems can be established from geome-
trical considerations, without recourse to the Calculus.
Guildin’s Theorems. 263
propositions, usually called Guldin’s Theorems, but originally
enunciated by Pappus (see Walton’s Mechanical Problems,
p- 42, third Edition). They may be stated as follows :—
_ (1). If a plane curve revolve round any external axis,
situated in its plane, the area of the surface generated is equal
to the product of the perimeter of the revolving curve by the
length of the path described, during the revolution, by the centre
of gravity of that perimeter.
(2). Under the same circumstances, the volume of the solid
generated is equal to the product of the area of the generating
curve into the path described by the centre of gravity of the re-
volving area.
To prove the former, let s denote the whole length of the
curve, x, y, the co-ordinates of one of its points, z, ¥, those
of the centre of gravity of the curve; then, from the defi-
nition of these latter, we have
fyds
eee
*, 27ys = 20 | yds,
ive. the surface generated by revolution round the axis of vis
equal to the product of S, the length of the generating curve,
into 277, the path described by the centre of gravity.
To prove the second proposition ; let A denote the area
of the generating curve, and dA the element of area corre-
sponding to any point z, y. Also let x, 7 be the co-ordinates
of the centre of gravity of the area, then
y= a = Sydeay (substituting dw dy for dA) ;
*, amyA = 20 ffydedy =f y dz;
where the integral is supposed taken for every point round the
perimeter of the curve: but, from Art. 171, the integral at
the right-hand side represents the volume of the solid gene-
rated ; hence the proposition in question follows.
For example, the volume of the ring generated by the
revolution of an ellipse around any exterior line situated in
its plane is at once 2n°abe, where a and 6 are the semiaxes
264 Volumes and Surfaces of Solids.
of the ellipse, and ¢ is the distance of its centre from the axis
of revolution.
It may be noted that these results still hold if we suppose
the curve, instead of making a complete revolution, to turn
round the axis through any angle. For, let 6 be the circular
measure of the angle of rotaticn, and in the former case we
have
Oys = Of yds.
But Oy is the length of the path described by the centre
of gravity, and 0 yds is the area of the surface generated by
the curve; .*. &e.
In like manner the second proposition can be shown to
hold.
Again, Guldin’s theorems are still true if we suppose the
rotation to take place around a number of different axes in
succession ; in which case the centre of gravity, instead of
describing a single circle, would describe a number of arcs of
circles consecutively ; and the whole area of the surface ge-
nerated will still be measured by the product of the length of
the generating curve into the path of its centre of gravity ;
for this result holds for the part of the surface corresponding
to each axis of revolution separately, and therefore holds for
the sum.
Again, in the limit, when we suppose each separate rota-
tion indefinitely small, we deduce the following theorem. If
any plane curve move so that the path of its centre of gravity
is at each instant perpendicular to the moving plane, then the
surface generated by the curve is equal to the length of the
curve into the path described by its centre of gravity.
The corresponding theorem holds for the volume of the
surface generated.
These extensions of Guldin’s theorems were given by
Leibnitz (Act. Zrud. Lips., 1695).
179. Expression for Wolume of any Solid.—The
method given in Art. 171 of investigating the volume bounded
by a surface of revolution can be readily extended to a solid
bounded in any manner. For, if we suppose the volume
divided into slices by a system of parallel planes, the entire
volume may, as before, be regarded as the limit of the sum
Volume of Elliptic Paraboloid. 265
of a number of infinitely thin cylindrical plates. Thus,.if we
suppose a system of rectangular co-ordinate axes taken, and
the cutting planes drawn parallel to that of wy; then, if 4,
represent the area of the section made bya plane drawn at
the distance z from the origin, the entire volume is denoted
by
{ A.ds,
taken between proper limits.
The area A, is to be determined in each case as a function
of s from the conditions of the bounding surface.
For example, to find the volume of the portion of a cone
cut off by any plane; we take the origin at the vertex, and
the axis of s perpendicular to the cutting plane; then, if B
denote the area of the base, and / the height of the cone, it
is easily seen that we have
Bz?
A,:B=2: 2, or A, = Ty
h
Al ede =~ Bx h; asin Art. 169.
h’), 3 ;
If the cutting planes be parallel to that of yz, the volume
is denoted by { A,dv; where A, denotes the area of the sec-
tion at the distance x from the origin.
180. Volume of Elliptic Paraboloid.—Let it be
proposed to find the volume of the portion of the elliptic
paraboloid
x y?
— + =23,
P ¢
cut off by a plane drawn perpendicular to the axis of the sur-
face. Here, considering s as constant, the area of the ellipse
2 2
a <7 = 22, by Art. 128, is 2781/p9.
Hence, denoting by ¢ the distance of the bounding plane
from the vertex of the surface, we have
ce
V= on/74| sdz = 10 / pq.
0
266 Volumes and Surfaces of Solids.
This result admits of being exhibited in another form ; for if
B be the area of the elliptic section made by the bounding
plane, we have
B = 270,/ pq.
Hence V = 3 circumscribing cylinder, as in paraboloid of re-
volution.
181. The Ellipsoid.—Next, to find the volume of the
ellipsoid
a
The section of the surface at the distance s from the origin
is the ellipse
the area of this ellipse is
2 2
(1 5a, i.e. A, = ac ~5 ab
Hence, denoting the entire volume by V, we have
¢e 2
v= arab] (: -5) ae =4 abe.
é 3
0
182. Case of Oblique Axes.—It is sometimes more
convenient to refer the surface to a system of oblique axes.
In this case, if, as before, we take the cutting planes parallel
to that of zy, and if w be the angle the axis of s makes with
the plane of zy, the expression for the volume becomes
sin w { A,dz,
taken between proper limits, where .4, represents the area of
the section, as in the former case.
For example, let us seek the volume of the portion of an
ellipsoid cut off by any plane.
Case of Oblique Ames. 267
Suppose DED’E’ to represent the section made by the
plane, and 4BA’B’ the parallel central section. Take OA,
OB, the axes of this section as axes of
zand y respectively ; and the conju-
gate diameter OC as axis of z. g
Then the equation of the surface
is
2 2
2
a y g
A Ra ee
in nn
where OA =a’, OB=0', OC =¢.
It will now be convenient to transfer the origin to the
point C’, without altering the directions of the axes, when the
equation of the surface becomes
Ty
Fig. 44.
ec yp 22 3
The area A, of the section, by Art. 128, is
a
wa b (3 re =) ’ (3)
hence, denoting C’N by h, the volume cut off by the plane
DED is represented by
h 2
a 23 g
wav’ sinw ep pe dz,
0 \é c
he hh
wav sinw (7 = )
3e"
or
But, by a well-known theorem,* we have
ab’é sin w = abe, vi
where a, }, c, are the principal semiaxes of the surface.
Hence the expression for the volume V in question be-
comes a iB
Ve rabe( =) ; (4)
ce? 3¢
* Salmon’s Geometry of Three Dimensions, Art. 96.
268 Volumes and Surfaces of Solids.
¢
or, denoting a by &,
V = wabck? (: S -) (5)
This result shows that the volume cut off is constant for all
sections for which & has the same value. Again, since
=1-k, the locus of Nisa similar ellipsoid ; and we infer
that if a plane cut a constant volume from an ellipsoid, the locus
of the centre of the section is a similar and similarly situated
ellipsoid. :
183. Elliptic Paraboloid.—The corresponding results
for the elliptic paraboloid can be deduced from the preceding
by adopting the usual method of such derivation: viz., by
taking
@=pe, P= ¢e,
and afterwards making c infinite; observing that in this case
the ratio 7 becomes unity.
Making these substitutions in (4), it becomes
Ven /pqh (: - 5) or wh? ./pq, since c’ =o.
3¢
Hence, if a constant length be measured on any diameter
of an elliptic paraboloid and a conjugate’ plane drawn, then
the volume* of the segment cut from the paraboloid by the
plane is constant.
Again, the area of an elliptic section by (3) is
Wy 2h PW mabe (2h ih
rab'|— —--—)\, or———_(— - =}.
6 CP dsnw\ &
* Fora more direct investigation the student is referred to a memoir ‘‘ On
some Properties of the Paraboloid,’’ Quarterly Journal of Mathematics, June,
1874, by Professor Allman.
Elliptic Paraboloid. 269
' On making the same substitutions, this becomes for the
paraboloid
on / PH,
sinw
Now, if we suppose a cylinder to stand on this section,
the volume of the portion cut off by the parallel tangent
plane to the paraboloid is obtained by multiplying the area
of the section by 4 sinw; and, consequently, is
2m Vf py h’,
i. e. is double the corresponding volume of the paraboloid.
This is an extension of the theorem of Art. 180.
EXAMptes.
_ 1. Prove that the volume of the segment cut from a paraboloid by any plane
is $ths of that of the circumscribing cone standing on the section made by the
plane as base.
2. A cylinder intersects the plane of zy in an ellipse of semiaxes 0.4 = a,
OB = b, and the plane of xz in an ellipse of semiaxes 0A = a, OC =c; the
edges of the cylinder being parallel to BC; find the volume of the portion of the
cylinder bounded by the three co-ordinate planes. Ans. % abe,
3. The axes of two equal right cylinders intersect at right angles; find the
volume common to both. Ans. 4a, where ais the radius of either cylinder.
This surface is called a Groin.
184. Wolume by Double Integration.—In the ap-
plication of the preceding method of finding volumes the
area represented by .A,, instead of being immediately known,
requires in general a previous integration; so that the deter-
mination of the volume of a surface involves two successive
integrations, and consequently V is expressed by a double
integral.
Thus, as the area A, lies in a plane parallel to that of yz,
its value, as in Art. 126, may generally be represented by
f sdy, taken between proper limits. Hence V may be repre-
sented by
SL fsdy]de
or, adopting the usual notation, by
f[edydz,
taken between limits determined by the data of the question.
270 Volumes and Surfaces of Solids.
The value of z is supposed given by a relation s=/(a, y),
by means of the equation of the bounding surface; hence
fady =[f(a, y) dy.
In the determination of this integral we regard z as
constant (since all the points in the area have the same
value of x), and integrate with respect to y between its proper
limits.
Thus, if y, and y, denote the limiting values of y, the
definite integral
y,
i. J (a, y) dy
becomes a function of «: this function, when integrated
with respect to # between the proper limits, determines the
volume in question.
If x, and 2 denote the limits of x, V may be represented
by the double integral
TY.
| | Sle, y) dy de.
FJ Yo
We shall exemplify this by a figure, in which we suppose
the volume bounded by the plane of wy, by a cylinder
perpendicular to that plane, and
also by any surface.* Let a
RPR’Q represent the section of
the cylinder by the plane of zy;
and suppose PMNVQ to be the
section of the volume by a plane
parallel to yz at the distance x
from the origin. Let PL =%, 0
QL = y, then the area PUNQ re
Y
is represented by the integral
P
yy,
| ady. Fig. 45.
%
* The determination of a volume of any form is virtually contained in this.
For, if we suppose the surface circumscribed by a cylinder perpendicular to the
plane of zy, the required volume will become the difference between two
tylinders, bounded by the upper and lower portions of the surface, respectively.
See Bertrand, Cale. Int. § 447.
Volume by Double Integration. 271
The values* of y, and % in terms of z are obtained from
the equation of the curve RPR’Q.
Again, suppose P-M’ N’Q to represent the parallel section
at the infinitesimal distance dv from PUN Q, then the
elementary volume between PMNQ and P’'U’N ‘Q’ is repre-
sented by
y,
ae| sdy.
Yo
Now, if RT and RT’ be tangentsto the bounding curve,
drawn perpendicular to the axis of 2, and if OJ" =x, OT=m,
the entire volume is represented by
1 (Yr
| | a dy dx.
It should be observed that sdydx represents the volume
of the parallelepiped whose height is z, and whose base is the
infinitesimal rectangle having dz and dy as sides; and conse-
quently the volume may be regarded as the sum of all such
parallelepipeds corresponding to every point within the area
RPP.
It is also plain that we shall arrive at the same result
whether we integrate first with respect to w, and afterwards
with respect to y, or vice-versd ; i.e. whether we conceive the
volume divided into slices parallel to the plane of zz, or to
that of yz.
We shall illustrate the preceding by an example.
Suppose RPR’Q to be the circle
(@- a)? + (y - 6) = PR,
and the bounding surface the hyperbolic paraboloid
LY = C8;
* In our investigation we have assumed that the parallels intersect the
curve in but two points each; the general case is omitted, as the solution in
such cases can be rarely obtained, and also as the investigation is unsuited for
an elementary treatise. : St
z This ane the next example are taken from Cauchy's Applications Géomé-
triques du Caleul Infinitésimal, p. 109.
272 Volumes and Surfaces of Solids.
then we have
Yo= b - J F'- (@-@)', (w - a), y=bt+/R - (2-4)’,
and
(Yo a es
| ady = =|. ay dy = — = (yi - - Yo) = are J B= (e-ay.
Yo Yo
Again, a=a+R, n=a-R;
VAR Ree apie.
a
Now let 2x-a=F sin 6, and we get
Ve= el cos’ O(a + R sin 0) d0.
But f cos’ 6d0 =-, | cos’ 6 sin 0d0 = o,
2
a vege.
Again, if for the cylindrical surface which has for its
base the circle we substitute a system of four planes x = x,
w= X,y=Y,y = Y, we get
x (Y
Ve= | a dy da
J”, JY, c
= 7, (Xt 8) (¥* - ys)
LoYo t+ HY + Xy+ XV
4c
By + B,+ S34 By
4
= (X - a) (¥ — yo)
= (X- %)(Y - y)
Double Integration. 273
in which 2, 2, 23, %, are the ordinates of the four corner
points of the portion of the surface in question.
Again, from the well-known properties of the surface, in
order to construct the hyperbolic paraboloid it is sufficient
to trace the gauche quadrilateral whose summits are the
extremities of the ordinates 2, %, %:, ,; then a right line
moving on a pair of opposite sides of this quadrilateral, and
comprised in a plane parallel to the other pair, will generate
the paraboloid in question.
Hence we arrive at the following proposition :—
Having traced a gauche quadrilateral on the four lateral
faces of a right prism standing on a rectangular base, if a
right line move on two opposite sides of this quadrilateral
and be parallel to the planes of the faces which contain the
other two sides, then the volume cut from the prism by the
surface so generated is equal to the product of the area of
the rectangular base of the prism by one-fourth of the sum
of the edges of the prism between the vertices of the
rectangle and those of the quadrilateral.
185. Double Integration.—F rom the preceding Article
it is readily seen that the double integral
{lv (a, y) dy dx
can be represented geometrically by a volume ; and the deter-
mination of the double integral, when the limits are given, is
the same as the finding the volume of a solid with correspond-
ing limits. ; . ;
For instance, the example in the preceding page is equi-
valent to finding the value of the double integral
{| ay dudy
taken for all values of z and y subject to the condition
(w@- a)? + (y- 6)? - BR (a? — b7)3,
EXamp.ies.
1, A sphere is cut by a right cylinder, the radius of whose base is half that
of the sphere, and one of whose edges passes through the centre of the sphere ;
find the volume common to both surfaces.
3 B48
ae > a being the radius of the sphere.
ns.
2. If the base of the cylinder be the complete curve represented by the /
equation 7 = @ cos 6, where ” is any integer, find the volume of the solid be-
tween the surface of the sphere and the external surface of the cylinder.
187. It is readily seen, asin Art. 141, that the volume in-
cluded within the surface represented by the equation
(x g
zh i)-s
is abe x the volume of the surface
F(@, y, 8) = 0.
For, let = =a, =¥, = =’, and we shall have
sdxdy = abes' da’ dy’,
and ww. [ fedady = abe [fz da’ dy’ ;
which proves the theorem. ;
Hence, for example, the determination of the volume of
an ellipsoid is reduced to that of a sphere.
Again, if the point (2, y, s) move along a plane, the cor-
responding point (2’, 7’, 2’) will describe another plane. From
this property the expression for the volume of an ellipsoidal
cap (Art. 182) can be immediately deduced from that of a
spherical cap (Art. 170).
[18 a]
276 Volumes and Surfaces of Solids.
In like manner the volume included between a cone en-
veloping an ellipsoid and the surface of the ellipsoid is reducible to
the corresponding volume for a sphere. sa
188. Quadrature on the Sphere.—We next propose
to give a brief discussion of quadrature on a sphere, and
commence with the results on the subject usually given in
treatises on Spherical Trigonometry. In the first place,
since the area of a lune is to that of the entire sphere as the
angle of the lune to four right angles, the area of a lune of
angle A is represented by 24; where #& is the radius of
the sphere, and 4 is expressed in circular measure.
Again, the area of a spherical triangle ABC is expressed
by #?(4+B+C-—7); for, the sum of the three lunes
exceeds the hemisphere by twice the area of the triangle, as
is easily seen from a figure.
Hence, it readily follows that the area = of a spherical
polygon of m sides is represented by
=H {A+ B+ C+ &. - (n- 2)7};
A, B, C, &e., being the angles of the polygon.
This result admits of being expressed in terms of the
sides of the polar polygon; for, representing these sides by
a’, U', c’, &e., we have
A=7-a@, B=z7-V, &,
and consequently
B= BRilar- (a+ +e + &e)}.
Or, denoting the perimeter of the polar figure by S,
=+ RS = an PR’. (6)
This proof is perfectly general, and holds in the limit,
when the polygon becomes any curve; and, accordingly, the
area bounded by any closed spherical curve is connected with
the perimeter of its polar curve by the relation (6).
Again, the spherical area bounded by a lesser circle
(Art. 170) admits of a simple expression. If o denote the
circular radius of the circle, or the are from its pole to its
circumference, the area in question is represented by
2m R? (1 — cos p);
Quadrature on the Sphere. 277
for (see fig. Art. 170) we have
AN = AC - ON = R(1 - cosp).
This result also follows immediately as a simple case of
equation (6).
Again, the area bounded by the lesser circle and by two
arcs drawn to its pole is plainly represented by
FR’ a(t - cos p),
where a is the circular measure of the angle between the ares.
We can now find an expression for the area bounded by
any closed curve on a sphere; for
0
the position of any point P on the
surface can be expressed by means ve
of the arc OP drawn to a fixed 7
point, and of the angle POX y; y es
between this arc and a fixed arc
through O. These are called the
polar co-ordinates of the point, and
are analogous to ordinary polar
co-ordinates on a plane.
Now, let OP = p, and POX =w;
then any curve on the sphere may be supposed to be expressed
by a relation between p and w.
Again, suppose OQ to represent an infinitely near vector,
and draw PR perpendicular to OP; then, neglecting in
the limit the area PQR, the elementary area OPQ by the
preceding is represented by
R? (1 - cos p) dw.
Hence the area bounded by two vectors from O is
Fig. 46.
expressed by the integral al (1-cosp)dw, taken between
suitable limits.
If the curve be closed, the entire superficial area becomes
ar
al (1 — cos p) dw.
0
The value of cos p in terms of w is to be determined in
each case by means of the equation of the bounding curve.
278 Volumes and Surfaces of Solids.
ar
The integral R? | cose dw obviously represents the area
0
included between the closed curve and the great circle which
has O for its pole.
The ional of the curve can also be represented by a
definite integral ; for, regarding PRQ as ultimately a right-
angled triangle, we have in the limit,
PQ = PR + RQ: also PR =sinpdw.
Hence ds? = dr? + sin’p dw’,
' dp\*.
or ds = dw : sin’p + () ;
Ne
os =| dw J sin’p + () '
Again, it is manifest. from (6) that the determination of
the length of any spherical curve is reducible to finding the
area of its polar curve, and vice versd.
EXamMpLes.
I. Find the area of the portion of the surface of a sphere which is inter-
vepted by a right cylinder, one of whose edges passes through the centre of the
sphere, and the radius of whose base is half that of the sphere.
Here, the equation of the base may be written in the form r= RK sina,
£ being the radius of the sphere, and w being measured from the tangent to the
circular base.
Again, from the sphere we havery =F sinp; .-. p= is the equation of
the curve of intersection of the sphere and the cylinder; hence the area in
question is
wily
2n*| (I — cos w)dw = 2R? ( -x).
0
This being doubled gives the whole intercepted area = 27.R? —
This is the celebrated Florentine enigma, proposed by Whee Viviani as a
challenge to the Mathematicians of his time, in the following form :—<‘ Inter
venerabilia olim Greecize monumenta extat adhue, perpetuo quidem duraturum,
Templum augustissimum ichnographia circulari Alme Geometrie dicatum, quod
Testudine intus perfecte hemispherica operitur: sed in hac fenestrarum quatuor
sequales arez (circum ac supra basin hemisphere ipsius dispositarum) tali con-
figuratione, amplitudine, tantaque industria, ac ingenii acumine sunt exstructe,
Quadrature of Surfaces. 279
ut his detractis, superstes curva Testudinis superficies, pretioso opere musivo
ornata, Tetragonismi vere geometricisit capax.’’—Acta Eruditorum, Leipsic, 1692.
(See Montucla, Histoire des Mathématiques, tome ii., p. 94.
In general, if r =f(w) be the equation of the base of a cylinder, it is easily
seen that the equation of the curve of its intersection with the sphere may be
written in the form # sin p = f(w).
For example, let the diameter of the right cylinder be less than half that
of the sphere; then writing the equation of the base in the form r = a sin w,
where a is the diameter of the section, we get & sin p=asin a, orsinp =x sinw
(where « is < 1), as the equation of the curve of intersection of the sphere and
the cylinder.
Hence the intercepted area is denoted by
Tv wv
z 3
2R? i. (1 -f/i-2@ sin?w)dw = rR? - oe JI — sin? odw.
Hence the area in question depends on the rectification of an ellipse.
2. Find the area of the portion of the surface of the cylinder intercepted by
the sphere, in the preceding.
Here the area in question is easily seen to be represented by 2 f zds, where
ds denotes the element of the curve which forms the base, corresponding to the
edge z.
Now (1), when the diameter of the base is equal to the radius of the sphere,
we have
z=Rcosw, andds = Rdw;
wv
. area in question = 2R? \. cos wdw = 4R?; i.e. the square of the diameter of
‘ :
the sphere.
2. When the diameter is less than the radius of the sphere,
2 | sdo= 20 | / Ba aiate o = zak | 7 — x? sin?w dw; .°. &e.
189. Quadrature of Surfaces.—In seeking the area
of a portion of any surface we regard it as the limit of a
number of infinitely small elements, each of which is con-
sidered as a portion of a plane which is ultimately a tangent
plane to the surface. Now let dS denote such an element of
the superficial area, and do its projection on a fixed plane
which makes the angle @ with the plane of the element; then,
trom elementary geometry, we shall have
do = cos 0d8, or dS = sec Odo.
Hence S -| sec Odo,
taken between suitable limits.
280 Volumes and Surfaces of Solids.
The applications of this formula usually involve double
integration, and are generally very complicated ; there is,
however, one mode by which the determination of the area of
a portion of a surface can be reduced to a single integration,
and by whose aid its value can in some cases be found; viz.,
by supposing the surface divided into zones by a system* of
curves along each of which the angle @ between the tangent
plane and a fixed plane is constant; then, if dS denote the
superficial area of the zone between the two infinitely near
curves corresponding to the angles 9 and 0 + d0; and, if dA
be the projection of this area on the fixed plane, we shall
have dS = sec 0aA.
If we suppose the surface referred to a rectangular
system of axes, the fixed plane being that of wy; and
adopting the usual notation, if we take A, pu, v as the direction
angles of the normal at any point on the surface, we get
for dS, the area of the zone between the curves corresponding
to v and v + dy, the equation
dS = sec vd A,
where A denotes the area of the projection on the plane of
zy of the closed curve defined by the equation v = constant.
Now whenever we can express, the area A in terms of »
and constants, then the area of a portion of the surface,
bounded by two curves of the system in question, is reducible
to a single integration.
The most important applications of this method are
furnished by surfaces of the second degree, to which we
proceed to apply it, commencing with the paraboloid.
190. Quadrature of the Paraboloid.— Writing the
equation of the surface in the form
2 2
Pg
* This method has been employed in a more or less modified form by
M. Catalan, Liouville, tome iv., p. 323, by Mr. Jellett, Camb. and Dub. Math.
Journal, vol. i., as also bv other writers. The curves employed are called
parallel curves by M. Lebesgne, Liouville, tome xi., p. 332, and Curven isokliner
Normalen, by Dr. Schlémilch.
Quadrature of the Paraboloid. 281
the equation of the tangent plane at the point (2, y, 3) is
aX + yY_ gt+ Z,
p qg
where X, Y, Z are the co-ordinates of any point on the plane.
Comparing this with the equation
X cos + Y cosp + Z cos v = P,
x
we get co A can, eos m= —T cosy:
substituting in the identical equation
cos’ + cos’ + cos*y = 1,
a? 2
we get # + a tan’ y. (7)
Consequently the curve along which the tangent plane
makes the angle v with the tangent plane at the vertex is
projected on that plane into the ellipse
2 2
a + 2 = tan’y.
The area A of this ellipse is mpgtan’»y ; accordingly, we
have
aA = mrpgd (tan’ v) ;
.. aS = pq sec vd(tan?v) = mpg sec vd(sec?v) ;
hence the area of the paraboloidal cap bounded by the curve
v=ais
wpy [see vd(sec’v) = 3mpq (sec*a — 1).
Also the area of the belt* between the curves
v=aand v =a’ is 3rpq(secta’ — sec*a). (8)
* This form for the quadrature of a paraboloid is, I believe, due to Mr. Ji ellett:
see Camb. and Dub. Math. Journal, vol. i. p. 65. The proof given above is in
a great measure taken from Mr. Allman’s paper in the Quarterly Journal, already
referred to.
282 Volumes and Surfaces of Solids.
191. Quadrature of the Ellipsoid.—Proceeding in
like manner to the ellipsoid
ey Bs
a 8 @
2 2 2
= Ty
the equation of the tangent plane at the point (a, y, 3) is
Xe Vy , Ze
Cr eo
Hence, comparing with the equation
X cosA + Y cosp + Zcosv = P,
we get
x eux ey
cosh =~ cosy, C08 M = Fz COSY.
Hence, we have
cle y® :
cos’ y= | | + a = cos’A + cos’ = sin’ y ;
a \a
2 2 2
cia we oy zg
or, substituting 1 - aR for »
a y
a (« sin’ y + ¢ cos’ ») + AG sin’y + ¢ cos? ») sin’ v.
This shows that the projection on the plane of zy of
a curve along which v = constant is an ellipse.
Again the area A of this ellipse is
wa’ b* sin’ v
(a? sin’ y + c® cos’v)2 (0? sin’y + c? cos? vy)
and accordingly, the area dA of the elementary annulus
between two consecutive ellipses is
Fie
raed? a sin*y Hs
dv |(@ sin’ v + ¢’ cos’ v)3 (0? sin*y + ¢ cos*v)4
The corresponding elementary ellipsoidal zone dS is
represented by
rab? d sin’ y d
cos v dv (a? sin®y + ¢ cos*v)4 (0? sin?» + ¢ cos®y)af
Quadrature of the Ellipsoid. 283
Now, if S denote the superficial area* between two
curves corresponding to v = a and vy =a’, after one or two
reductions, it is easily seen that
S=nrahe (I+I’), (9)
a in v dy
where I= ‘ Bens
i (6° sin*y + ¢* cos’ y)! (a* sin? y +c’ cos* v)?
le e sin v dv
a(@ sin’ y + ¢? cos’ v)# (6? sin? y + ¢ cos*y)#
It is easily shown that the former of these integrals is
represented by an arc of an ellipse, and the latter by an are
of a hyperbola ; it being assumed that a > b>.
For, assuming @ - @ = ae, and 2 - @ = B é, and
making cos v = 2, we get
I COS a dx
4 col, (1 — e? a*)8(1 — ea)?
rr I fs di
7 ab cos a’ (1 rs ex") (1 = ea) B
Again, let ex = sin @ in the former integral, and ex = sin 0
in the latter, and we get
= | d0
~ ab? J (e — e? sin? 0)?
/2
rez j do
~ ab | (6? — & sin’ 0)?
Now, since e > e, the former integral represents an
are of an ellipse, and the latter an are of a hyperbola. (See
Ex. 19, p. 249).
* This form for the quadrature of an ellipsoid is given by Mr. Jellett in
the memoir already referred to. He has also shown that the ellipse and the
hyperbola in question are the focal conics of the reciprocal ellipsoid; a result
which can be easily arrived at from the forms of J and I’ given above.
For application to the hyperboloid, and further development of these results,
the student is referred to Mr. Jellett’s memoir.
284 Volumes and Surfaces of Solids.
192. Integration over a Closed Surface.— We shall
conclude this Chapter with the consideration of some general
formule in double integration relative to any closed surface.
We commence by adopting the same notation as in Art. 189,
where A, », v are taken as the angles which the exterior
normal at the element dS makes with the positive directions
of the axes of 2, y, 8, respectively. ;
Again, let each element of the surface be projected on
the plane of zy, and suppose* for simplicity that each z ordi-
nate meets the surface in but two points: then, if the indefi-
nitely small cylinder standing on any element dA in the
plane of wy intersects the surface in the two elementary por-
tions dS, and dS, (where dS, is the upper, and dS, the lower
element), and if v, and »; be the corresponding values of », it
is plain that 1 is an acute, and », an obtuse angle, and we
have
aA = COS v} d8, =-— COS v2 082.
Hence, if we take into account all the elements of the surface,
attending to the sign of cos v, we shall have
ffcos vdS =o.
In like manner we get
ffeosAdS = 0, and f{cosudS =0;
the integrals extending in each case over the whole of the
closed curve
These formule are comprised in the equation
{f(a cosA + B cosy + y cos v) dS = 0. (10)
Again, if z, and z, be the values of z corresponding to the
element dA, then, denoting by dV the element of volume
ous on dA and intercepted by the surface, we plainly
ave
aV = (& ~ &)dA = 2,08; cos v, + 22.48, cos v2,
* Itit easily seen that this and the following demonstrations are perfectly
general, inasmuch as each ordinate must meet a closed surface in an even number
of points, which may be considered in pairs.
Integration over a Closed Surface. 285
and the sum of all such elements, that is, the whole volume,
is evidently represented by
ffecosvdS.
Hence, denoting the whole volume by V, we have
V=ffecosAdS = fy cosudS = [f scosvd8;
the integrals, as before, being extended over the entire
surface.
Again, it is easily seen that we have
J[ecosvdS=0, ff[ycosvdS=0, [fwxcosudS =o,
ffy cosAdS=0, ffscosAdS=0, ffs cos udS =o.
For, as in the first case, it readily appears that the elements
are equal and opposite in pairs in each of these integrals.
These results are comprised in the equation
Sf (aw + By + yz) (a’ cosd + B’ cos + y’ cos v) dS
= (aa’ + BB+ yy’) Ve (11)
For a like reason, we have
flay cosudS=0, ffzxcospdS=0, ffys cosrAdS =o.
Also [fz cosvdS=0, ff2’ cosudS =o, &e.
Next, let us consider the integral
ffzz cos vd8.
This integral is equivalent to ff edV; consequently, if
2, y, 3, be the co-ordinates of the centre of gravity of the
enclosed volume V, we get [fz cosvdS =ffadV =2V; in
like manner f{ #2 cosAdS = 8V.
Again, the integral
ff # cos vdS
consists of elements of the form (z,” - #:°)¢A; but
(317 ey 2") dA = (8, + a) (& = 22) dA
= (a + &) a7.
286 Volumes and Surfaces of Solids.
But the z ordinate of the centre of gravity of dV is
2, + Be
2
plainly , and consequently
[| eos vas 2| [a7 = 237.
In like manner it can be shown that
[Ja cosAdS = 2zV, fy cospdS = 2yV.
Accordingly we have
Vi =43 fa cosrdS = {fy cos udS = ff az cosvd8,
Vy = ffye cosXdS =2) fy’ cosudS={f yz cosvd8,
Vz=ffee cosrdS = ffsy cosudS =1f fz? cos vd.
193. Expression for Volume of a Closed Surface.
—Next, if we suppose a cone described with its vertex
at the origin O, and standing on the elementary base dS,
its volume is represented (Art. 169) by 1pdS, where p is the
length of the perpendicular drawn from O to the tangent
plane at the point.
Also, if » be the distance of O from the point, and y the
angle which r makes with the internal normal, we have
p=Pr cosy.
Hence the elementary volume is equal to 17 cos ydS, and
it is easily seen that if we integrate over the entire surface,
the enclosed volume is represented by
LJfr cosyd8.
194. Again, if we suppose a sphere of unit radius described
with O as centre, and if dw represent the superficial portion
of this sphere intercepted by the elementary cone standing on
d&, then it is easily seen that cos ydS =7'dw ;
cos yd 8
2
. dw =
r
Now if O be inside the closed surface, and the integral
be extended over the entire surface, it is plain that f {dw =47,
being the surface of the sphere of radius unity ;
cos yd
ico
Expression for Volume of a Closed Surface. 287
Again, if O be outside the surface, the cone will cut the
surface in an even number of elements, for which the values
of cos y will bé alternately positive and negative, and, the
corresponding elements of the integral being equal but with
opposite signs, their sum is equal to zero, and we shall have
(Ae. 5
r*
If O be situated on the surface, it follows in like manner
that
aed dS = 27
Vid ;
Hence, we conclude that
[[?2as = 47, 27, or 0, (12)
according as the origin is inside, on, or outside the surface.
The multiple integrals introduced into this and the two
preceding Articles are principally due to Gauss.
The student will find some important applications of
this method in Bertrand’s Cale. Int., 8§ 437, 455, 456,
476, &e.
288 Examples.
ExamPies.
1. A sphere of 15 feet radius is cut by two parallel planes at distances of
3 and 7 feet from its centre; find the superficial area of the portion of the sur-
face included between the planes approximately. Ans. 376.9908 sq. feet.
2. Being given the slant height of a right cone, find the cosine of half its
vertical angle when its volume is a maximum. A I
NS. Tau
v3
3. Prove that the volume of a truncated cone of height / is represented by
h
= (BR? + Rr +72),
where R and ¢ are the radii of its two bases.
4. A cone is circumscribed to a sphere of radius R, the vertex of the cone
being at the distance D from the centre; find the ratio of the superficial area of
the cone to that of the sphere. Bes D- Rk
me
5. Two spheres, 4 and B, have for radii 9 feet and 4o feet ; the superficial
area of a third sphere Cis equal to the sum of the areas of A and B; calculate
the excess, in cubic feet, of the volume of C over the sum of the volumes of 4
and B. Ans. 17558.
6. If any arc of a plane curve revolve successively round two parallel axes,
show that the difference of the surfaces generated is equal to the product of the
length of the arc into the circumference of the circle described by any point on
either axis turning round the other.
If the axes of revolution lie at opposite sides of the curve, the sum of the
surfaces must be taken instead of the difference.
7. Find, in terms of the sides, the volume of the solid generated by the
complete revolution of a triangle round its side ce.
ies 8(s — a) = b)(s — e)
8. Apply Guldin’s theorem tu determine the distance, from the centre, of the
centre of gravity, (1) of a semicircular area; (2) of a semicircular arc.
Bae ges, aah:
30 ©
g. Ifa triangle revolve round any external axis, lying in its plane, find an
expression for the area of the surface generated in a complete revolution.
10. Prove that the volume cut from the surface
m= Ag? + By?
: F th part of the cylinder standing on
the plane section, and terminated by the plane of xy.
by any plane parallel to that of zy, is
Examples. 289
11. A cone is circumscribed to a sphere of 23 feet radius, the vertex of the
cone being 265 feet distant from the centre of the sphere ; find the ratio of the
superficial area of the cone to that of the sphere.
12, The axis of a right circular cylinder passes through the centre of a
sphere ; find the volume of the solid included between the concave surface of the
sphere and the convex surface of the cylinder.
3 ‘
Ans, * where ¢ is the length of the portion of any edge of the cylinder
intercepted by the sphere.
This question is the same as that of finding the volume of the solid generated
by the segment of a circle cut off by any chord, in a revolution round the
diameter parallel to the chord.
13. Find the volume of the solid generated by the revolution of an arc of a
(2a? + ¢*) sina Ps
ee ea
3
where a = radius, ¢ = distance of chord from centre, and cos a = <.
In this we suppose the arc less than a semicircle: the modification when it
is greater is easily seen.
circle round its chord. Ans. 20a |
14. If the ellipsoid of revolution,
ary?
P+ e+ = ak,
and the hyperboloid
2 72
a —F
Das Ge
e+e a Yee
be cut by two planes perpendicnlar to the axis of revolution, prove that the
zones intercepted on the two surfaces are of equal area.
15. Find the entire volume bounded by the positive sides of the three co-
ordinate planes, and
4 4 4
6) ~ (5) + (=) =I, Ans. ee
a b ce go
16. Find the volume of the surface generated by the revolution of an arc of
a parabola round its chord; the chord being perpendicular to the axis of the
curve.
Ans. Sei os where ¢ is the length of the chord, and 4 the intercept made
I
by it on the diameter of the parabola passing through the middle point of the
chord.
17. A sphere of radius r is cut by a plane at distance @ from the centre ; find
the difference of the volumes of the two cones having as a common base the
circle in which the plane cuts the sphere, and whose vertices are the opposite
the diameter perpendicular to the cutting plane.
mae Ans. 3nd (7? - d?).
[19]
290 Examples.
18. Find the area of a spherical triangle; and prove that if a curve traced
on a sphere have for its equation sin A = f(), A denoting latitude, and / longi-
tude, the area between the curve and the equator = J f (7) @/.
19. Show that the volume contained between the surface of a hyperboloid
of one sheet, its asymptotic cone, and two planes parallel to that of the real
axes, is proportional to the distance between those planes.
20. Find the entire volume of the surface
a\t y # z\é 4m abe
= Ze eo Ans. ——.
a) eG) eee eget
21. The vertex of a cone of the second degree is in the surface of a sphere,
and its internal axis is the diameter passing through its vertex ; find the volume
of the portion of the sphere intercepted within the cone.
22. Prove that the volume of the portion of a cylinder intercepted between
any two planes is equal to the product of the area of » perpendicular section
into the distance between the centres of gravity of the areas of the bounding
sections.
23 If A be the area of the section of any surface made by the plane of zy,
prove, ss in Art. 192, that
4 = ffoosvds,
the integral being extended through the portion of the surface which lies above
the plene of zy.
24. Tf a right cone stand on an ellipse, prove that its volume is represented
by
(04 .0A')3 sin? cosa;
where O is the vertex of the cone, A and 4’ the extremities of the major axis
of the ellipse, and a is the semi-angle of the cone.
25. In the same case prove that the superficial area of the cone is
"(04 + 04’) (0.4. 0A) sin a.
( 291 )
CHAPTER X.
INTEGRALS OF INERTIA.
195. Integrals of Inertia.—The following integrals are
of such frequent occurrence in mechanical investigations,
that it is proposed to give a brief discussion of them in this
Chapter.
If each element of the mass of any solid body be supposed
to be multiplied by the square of its distance from any fixed
right line, and the sum extended throughout every element
of the body, the quantity thus obtained is called the moment
of inertia of the body with respect to the fixed line or axis.
Hence, denoting the element of mass by dm, its distance
from the axis by p, and the moment of inertia by I, we have
T= Sp*dm. (1)
In like manner, if each element of mass of a body be
multiplied by the square of its distance from a plane, the
sum of such products is called the moment of inertia of the
body relative to the plane.
If the system be referred to rectangular axes of co-
ordinates, then the expression for the moment of inertia
relative to the axis of z is obviously represented by
= (2 + y*)dm.
Similarly, the moments of inertia relative to the axes of
xand y are represented by 3(y’ + s’)dm and B(x’ + s*) dm,
respectively.
Again, the quantities Sa2’°dm, Sy’dm, Xs’dm, are the
moments of inertia of the body with respect to the planes
of yz, wz, and ay, respectively. Also the quantities Saydm,
Zerdm, Syzdm, are called the products of inertia relative to
the same system of co-ordinate axes.
In like manner the moment of inertia of the body with
reference to a point is Sr’dm, where r denotes the distance of
the element dm from the point. Thus the moment of inertia
relative to the origin is 3 (a + y’ + 8°) dm.
[19 a]
292 Integrals of Inertia.
196. Moments of Inertia relative to Parallel
Axes, or Planes.—The following result is of fundamental
importance :—The moment of inertia of a body with respect to
any axis exceeds tts moment of inertia with respect to a parallel
axis drawn through its centre of gravity, by the product of the
mass of the body into the square of the distance between the
parallel axes.
For, let I be the moment of inertia relative to the axis
through the centre of gravity, J’ that for the parallel axis,
Mf the mass of the body, and a the distance between the axes.
Then, taking the centre of gravity as origin, the fixed
axis through it as the axis of s, and the plane through the
parallel axes for that of sv, we shall have
L=3(¢+yY)dm, I’ = 3{(@+ a)? + y*}dm.
Hence I’ - T= 2adtdn+¢Sdm =H,
since Sxdm = o as the centre of gravity is at the origin ;
eo D=T+ eM. (2)
Consequently, the moment of inertia of a body relative to
any axis can be found when that for the parallel axis through
its centre of gravity is known.
Also, the moments of inertia of a body are the same for
all parallel axes situated at the same distance from its centre
of gravity.
Again, it may be observed that of all parallel axes that
which passes through the centre of gravity of a body has the
least moment of inertia. ‘hs
It is also apparent that the same theorem holds if the —
moments of inertia be taken with respect to parallel planes,
instead of parallel axes.
A similar property also connects the moment of inertia
relative to any point with that relative to the centre of
gravity of the body.
In finding the moment of inertia of a body relative to
any axis, we usually suppose the body divided into a system
of indefinitely thin plates, or Jamine, by a system of planes
perpendicular to the axis; then, when the moment of inertia
is determined for a lamina, we seek by integration to find
that of the entire body.
Radius of Gyration. 2938
197. Radius of Gyration.—If % denote the distance
from an axis at which the entire mass of a body should be
concentrated that its moment of inertia relative to the axis
may remain unaltered, we shall have
Mh? = I= pdm. (3)
The length & is called the radius of gyration of the body
with respect to the fixed axis.
In homogeneous bodies, which shall be here treated of
principally, since the mass of any part varies directly as its
volume, the preceding equation may be written in the form
BV = sp'dV,
where dV denotes the element of volume, and V the entire
volume of the body.
Hence, in homogeneous bodies, the value of & is indepen-
dent of the density of the body, and depends only on its form.
We shall in our investigations represent the moment of
inertia in the form 7. Wee:
a >
and, it is plain that in its determination for homogeneous
bodies we may take the element of volume for the element of mass,
and the total volume of the body instead of its mass.
Also, in finding the moment of inertia of a lamina, since its
radius of gyrationis independent of the thickness of the lamina,
we may take the element of area instead of the element of
mass, and the total area of the lamina instead of its mass.
198. If A and B be the moments of inertia of an infi-
nitely thin plate, or lamina, with respect to two rectangular
axes OX, OY, lying in its plane, and if C be the moment of
inertia relative to OZ drawn perpendicular to the plane, we
have C=A+B. (4)
For, we have in this case 4 = 3y’dm, B= Sx2’dm, and
C= 3(x + y’) dn. .
Again, for every two rectangular axes in the plane of the
lamina, at any point, we have
Da?dm + Sy’ dm = const.
Hence, if one be a maximum, the other is aminimum, and
vice versd. ;
We shall, in all investigations concerning lamina, take C
for the moment of inertia relative to a line perpendicular to
the lamina.
294 Integrals of Inertia.
199. Uniform Rod, Rectangular Lamina.—We
commence with the simple case of a rod, the axis being perpen-
dicular to its length, and passing through either extremity.
Let x be the distance of any element dm of the rod from
the extremity; then, since the rod is uniform, dm is propor-
tional to dx, and we may assume dm = pdx: hence, the
moment of inertia J is represented by u Za’ dz, or by
i
| xv’ da,
0
where / is the length of the rod.
3 2
Hence a ae .
3 3
If the axis be drawn through the middle point of the rod,
perpendicular to its length, the moment of inertia is plainly
the same for each half of the rod, and we shall have in this case
2
pee,
12
Next, let us take a rectangular lamina, and suppose the
axis drawn through its centre, parallel to one of its sides
Here, it is evident that the lamina may be regarded as
made up of an infinite number of parallel rods of equal
length, perpendicular to the axis, each having the same
radius of gyration, and consequently the radius of gyration
of the lamina is the same as that of one of the rods.
Accordingly, we have, denoting the lengths of the sides
of the rectangle by 2a and 26, and the moments of inertia
round axes through the centre parallel to the sides, by 4 and
B, respectively,
I I
A=—-Me, B=-Me.
; 5 (5)
Hence also, by (4), the moment of inertia round an axis
through the centre of gravity and perpendicular to the plane
of the lamina, is
: (a? + &). (6)
By applying the principle of Art. 196 we can now find
its moments of inertia with respect to any right line either
lying in, or perpendicular to, the plane of the lamina.
Circular Plate, Cylinder. 295
200. Rectangular Parallelepiped.—Since a parallel-
epiped may be conceived as consisting of an infinite number
of lamine, each of which has the same radius of gyration
relative to an axis drawn perpendicular to their planes, it
follows that the radius of gyration of the parallelepiped is
the same as that of one of the lamineg.
Hence, if the length of the sides of the parallelepiped be
2a, 2b, and 2¢, respectively; and, if A, B, C be respectively
the moments of inertia relative to three axes drawn through
the centre of gravity, parallel to the edges of the parallel-
epiped, we have, by the last,
A=“ UP +e), B=-MC +e), C= M@+P). (1)
201. Cireular Plate, Cylinder.—If the axis be
drawn through its centre, perpendicular to the plane of a
circular ring of infinitely small breadth, since each point of
the ring may be regarded as at the same distance r from the
axis, its moment of inertia is r*dm, where dm represents its
mass.
Hence, considering each ring as an element of a circular
plate, and observing that dm = 2rrdr, we get for C, the
moment of inertia of the circular plate of radius a,
aw pat
ee,
a
C- ann | rdr = 7
0
Consequently, the moment of inertia of a ring whose
outer and inner radii are a and 4, respectively, with respect to
the same axis, is
a at — bt a +
ann rdr=wKE
b
2
Again, by (4), the moment of inertia of a circular plate
2
about any diameter is ? since the moments of inertia are
obviously the same respecting all diameters. ;
Tn like manner, the moment of inertia of a ring relative
to any diameter is
a+ 6
Mw ;
296 Integrals of Inertia.
Also, the moment of inertia of a right cylinder about its
axis of figure is
2
a
MU -,
2
a being the radius of the section of the cylinder.
Again, the moment of inertia relative to any edge of the
cylinder is 3 Md.
2
202. Right Cone.—To find the moment of inertia of a
right cone relative to its axis, we conceive it divided into an
infinite number of circular plates, whose centres lie along the
axis; and, denoting by # the distance of the centre of any
section from the vertex of the cone, and by a the semi-angle
of the cone, we have
Iz autan‘a eee _ Teoh
2 10
?
0
where / is the height of the cone, and 6 the radius of its base.
Hence, since by Art. 169 the volume of the cone is 5 om
we have
-3 up
r-5 uy. (8)
203. Elliptic Plate.—Next let us suppose the lamina
an ellipse, of semi-axes a and 0; and
let A and B be the moments of inertia
relative to these axes, respectively.
Describe a circle with the axis
minor for diameter, and suppose the
lamina divided into rods by sections
perpendicular to this axis. Let B’ be
the moment of inertia for the circle Fig. 47.
round its diameter.
Then, denoting by dB and dB’ the moments of inertia of
corresponding rods, we have
dB: dB = (np): (np’)* = (oa)? : (0b)? = a: 8;
“ Br B= @&: #.
Sphere. 9297
1p
°
’
But B’, by Art. 201, is
sca poe ea
406 4
Similarly, Mee = i.
Hence the moment C round a line through the centre of
the ellipse, perpendicular to its plane, is
= (@ + b*). (9)
It is plain, as before, that the expression for the moment
of inertia of an elliptical cylinder relative to its axis is of the
same form.
204. Sphere.—If we suppose a sphere divided into an
infinite number of concentric spherical shells, the moment of
inertia of each shell is plainly the same for all diameters;
and accordingly, representing the mass of any element of a
shell by dm, and by 2, y, s any point on it, we have
Sedm = Sydm = 2 dm.
But S(e+y?+2)dm = =r'dm;
S(v’+y)dm = 5 br.
Hence, (a) the moment of inertia of a shell whose radius
is r with respect to any diameter is - mr*, where m repre-
sents the mass of the shell. ;
Again, (0) for a solid sphere of radius R, since the volume
of an indefinitely thin shell of radius r is 4rr’dr, we get
R
ar dv = an | ips a2 VR,
0 3 5
When this is substituted, the moment of inertia of a solid
homogeneous sphere relative to any diameter is found to be
= UR. (10)
298 Integrals of Inertia.
205. Ellipsoid.—Let the equation of an ellipsoid be
we
ah
*
ae
cogs
and suppose A, B, C to be the moments of inertia relative to
the axes a, 0, c, respectively ; then
=pd(e?+y)dV =p ii} (a? + y*) dady da.
x f , & /
Now, let woes ay, so,
and we get
C = pabe iil (aa? + By”) de'dy'ds’,
where the integrals are extended to all points within the
sphere
ety +e ad,
But, by the last example we have
{J x” da! dy! dx | {[ysaerayae = Re T3
0s 4 a uabe (a? + 6°) es 6). (11)
15 5
In like manner,
A sO @ +e), B= fe +a’).
5 5
It should be remarked that the moments of inertia of the
ellipsoid with respect to its three principal planes are
uM a’, at B, ze, respectively..
5
Moments of Inertia of a Lamina. 299
206. Moments of Inertia of a Lamina.—Suppose
that any plane lamina is referred to two rectangular axes
drawn through any origin O, and that a is the angle which
any right line through 0, lying in the plane, makes with the
axis of x; then, if J be the moment of inertia of the lamina
relative to this line, we have
[= Zp'dm = X(y cosa — x sin a)*dm
= cos'a Dy'dm + sin’a Sa*dm — 2 sinacosaSaydm
- =a cosa + bsin’a - 2h sina cosa; (12)
where a and 6 represent the moments of inertia relative to
the axes of # and y, respectively ; and his the product of
inertia relative to the same axes.
Again, supposing X and ¥ to be the co-ordinates of a point
taken on the same line at a distance R from the origin, we
a bs
get cosa = Re sa=z; and, consequently,
IR? = aX’? + bY* - 2h XV.
Accordingly, if an ellipse be constructed whose equation is
aX?’ + bY? — 2hXY = const, (13)
we have :
IF? = const. ;
and, consequently, the moment of inertia relative to any line
drawn through the origin varies inversely as the square of
the corresponding radius vector of this ellipse.
The form and position of this ellipse are evidently inde-
pendent of the particular axes assumed ; but its equation is
more simple if the axes, major and minor, of the ellipse had
been assumed as the axes of co-ordinates. Again, since in
this case the coefficient of XY disappears from the equa-
tion of the curve, we see that there exists at every point in
a body one pair of rectangular axes for which the quantity
hor Saydm = o.
This pair of axes is called the principal ares at the
point ; and the corresponding moments of inertia are called the
principal moments of inertia of the lamina, relative to the point.
300 Integrals of Inertia.
Again, if A and B represent the principal moments of
inertia, equation (12) becomes
I= A cos’a + B sin’a. (14)
Hence, for a lamina, the moment of inertia relative to
any axis through a point can be found when the principal
moments relative to the point are determined.
The equation of the ellipse (13) becomes, when referred
to the principal axes,
AX? + BY’ = const.
207. Momental Ellipse.—Since the moments of inertia
for all axes are determined when those relative to the centre
of gravity are known, it is sufficient to consider the case
where the origin is at the centre of gravity. With reference
to this case, the ellipse
AX? + BY’ = const. (15)
is called the momental ellipse of the lamina.
Again, if two different distributions of matter in the
same plane have a common centre of gravity, and have the
same principal axes and principal moments of inertia, at
that point, they have the same moments of inertia relative to
all axes.
This is an immediate consequence of (14). Hence it is
easily seen that the moments of inertia for any lamina are
the same as for the system of four equal masses, each —,
placed on the two central principal axes, at the four dis-
tances + a and + 8, from the centre of gravity, where a and J
are determined by the equations
= = B == Me.
Again, if two systems of the same total mass, in a plane,
have a common centre of gravity, and have equal moments
of inertia relative to any three axes, through their common
centre of gravity, they have the same moments of inertia for
all axes.
Momental Ellipse. 301
This follows immediately since an ellipse is determined
when its centre and three points on its circumference are
iven.
: Again, it may be observed that the boundary of an
elliptical lamina may be regarded as the momental ellipse of
the lamina.
For, if I be the moment of inertia relative to any
diameter making the angle a with the axis major, we have
I= A cos’a + B sin’a.
But, by Art. 203,
-28, Barat;
M :
o2= a: (6 cos’a + a’ sin’a)
M ,,,/cosa_ sin’a
Fan(s =)
M a®b*
fe
Hence the moment of inertia varies inversely as the square
of the semi-diameter 7; and, consequently, the ellipse may be
regarded as its own momental ellipse.
208. Products of Inertia of Lamina.—Suppose the
lamina referred to its principal axes at a point O; and let p
and q be the distances of any element dm. from two axes,
which make the angles a and 6 with the axis of z; then we
have
Spqdm = By cosa — # sin a)(y cos 3 — # sin B) dm
= cosa cos 3 Sy*dm + sin a sin B adm
—sin(a + B) Zaydm
= A cosa cos 3 + B sina sin ,
since A= Zy'dm, B= adm, and Sxydm =o.
Hence, if-Spqdm =o, we have
A cosa cos + B sina sinB = 0,
302 Integrals of Inertia.
and accordingly the axes are a pair of conjugate diameters
of the momental ellipse
AX? + BY® = const.
Hence, if two lamina in the same plane have for any point
two pairs of axes for which Spgdm=o and Sp'/dm' =o,
they have the same principal axes at the point. This follows
from the easily established property, that if two ellipses have
two pairs of conjugate diameters in common, they must be
similar and coaxal.
209. Triangular Lamina and Prism.—Suppose a
triangular lamina, whose sides are a, 0, ¢, to be divided into
a system of rods parallel to a side a; A
and let A represent the moment of eg
inertia relative to a line parallel to x
the side a, and drawn through the
opposite vertex; also let p be the
perpendicular of the triangle on
the side a, and z the distance of an a&
elementary rod from the vertex; then
we have, since the mass dm of the
Fig. 48.
elementary rod may be represented by pu a de,
A = dx’ dm = pda de
a (? apy M
=pu—| ede=p— =—p*.
e Al ee ee
In like manner, let B and C be the moments of inertia
relative to lines drawn through the other vertices parallel to
6 and c; and let g, r be the corresponding perpendiculars of
the triangle, and we have
eG at
Bee, Oa,
Again, if 4), Bo Co, represent the moments of inertia
Triangular Lamina and Prism, ~ 3803
relative to three parallels to the sides, drawn through the
centre of gravity of the lamina, we have, by (2),
I I I
Ay = = Mp’, B= Ue, Co = = Mr’. (16)
Also, if 4, B, C,, be the moments of inertia relative
to the sides a, 0, ¢, respectively, it follows, in like manner,
from (2), that
A, = = Mp’, B, = = Mi, C= 2 Mr. (17)
Again, it is readily seen that the values of A, Ay, A,, &e.,
are the same as if the whole mass If were divided into three
equal masses, placed respectively at the middle points of the
sides of the lamina.
Consequently, by Art. 207, the moments of inertia of the
triangular lamina relative to all axes are the same as for
three masses, each x placed at the middle points of the
sides of the triangle.
Hence, if I be the moment of inertia of a triangular
lamina with respect to the perpendicular to its plane drawn
through its centre of gravity, we have
Te oll + D4 0), (18)
This expression also holds for the moment of inertia of a
right triangular prism with respect to dts axis.*
In like manner the moments of inertia of the triangular
lamina relative to the three perpendiculars to its plane,
drawn through its vertices, are
E 2 2_ @ cues 5) iu(ess-S);
iMac a pee ee a 3
‘and the same expressions hold for a triangular prism relative
to its edges.
* By the axis of a prism is understood the right line drawn through its
centre of gravity parallel to its edges.
804 Integrals of Inertia.
210. Momental Ellipse of a Triangle.—It can be
shown without difficulty that the ellipse which touches at the
middle points of the sides
may be taken for the mo-
mental ellipse of the triangle.
For, let 2, y, s be the Z y
middle points of the sides,
and it is easily seen that o
is the centre of this ellipse ; 5 a
also, if L, L, I, be the
moments of inertia of the
lamina relative to the lines aw, by, cz, respectively, it can be
readily shown from (17), that we have
Fig. 49.
I I I
(ax)? * (by)? (cs)*
I I I
~ (oa)? * (oy)** (oa)
Accordingly, by Art. 207, the ellipse zyz may be taken for
the momental ellipse of the lamina.
211. Wetrahedron.—Ii a solid tetrahedron be supposed
divided into thin laminew parallel to one of its faces, and if
A, B, C, D represent its moments of inertia with regard
to the four planes drawn respectively through its vertices
parallel to its faces; then, denoting the areas of the corre-
sponding faces by a, b, c, d, and the corresponding perpen-
diculars of the tetrahedron by p, g, r, 8, respectively, it is
easily seen, as in Art. 209, that we shall have
GL? bts
2 P
A = 32°dm = pdaa— da = 15 x! dee
P P'do
ap _ 3
=p— =— Mp’.
Me 5 5 2p
In like manner we have
B-iuy, c=3ur, D=-3 M8.
5 5 5
Solid Ring. 3805
Again, if Ay, Bo. Co, Dy be the corresponding moments of
inertia relative to the parallel planes drawn through the
centre of gravity of the tetrahedron, we have, by (2),
3 3 3
Ay= 55 Up’, By = 5 My’, Oo= so Ur", Dy = 2 Me. (19)
Also, if A,, B,, C,, D, be the moments of inertia relative
to the four faces of the tetrahedron, we have
I I 1
A,= To ’: B= To’: C, = 7 Mr’, D,= * Us’. (20)
212. Solid Ring.*—If a plane closed curve, which is
symmetrical with respect to an axis 4B, be made to revolve
round a parallel axis, lying in
its plane, but not intersecting the
curve, to prove that the moment
of inertia I of the generated solid, a
taken with respect to the axis of
revolution, is represented by
M (h? + 3h),
where UM is the mass of the solid, > NT $
h the distance between the parallel
axes, and & the radius of gyration
of the generating area relative to its axis.
For, if the axis of revolution be taken as the axis of z,
and, if y, Y be the distances of any point P within the
generating area from AB, and from OX, respectively ; and,
if dA be the corresponding element of the area, then the
volume of the elementary ring generated by dA is 27 YdA,
and its mass 27u YdA; hence the moment of inertia of this
elementary ring, relative to the axis of X, is 2muY°*dA.
Accordingly, we have
I= arp V*daA = arpd(ht+y)dAa
= 27 d(h + 3h’y + 3hy + ¥) dA.
Fig. 50.
* The theorems of this Article were given by Professor Townsend in the
Quarterly Journal of Mathematics, "(ac ]
20
306 Integrals of Inertia.
Moreover, since the curve is symmetrical with respect to
the axis AB, it is easily seen that we have
SydA=0, SydA=o.
Also, by definition, Sy’dA = Ak’.
Hence T= 2arpha(h’ + 3k).
Again, by Art.177, M=a2rpuha;
o T= UW (h + 3%). (21)
This leads immediately to some important cases.
Thus, for example, the moment of inertia of a circular
ring, of radius a, round its axis is
m1 +30)
4
Again, if a square of side a revolve round any line in its
plane, situated at the distance A from its centre, we have
I=U(i +@).
There is no difficulty in adding other examples.
213. General Expression for Products of Inertia.
—We shall conclude this Chapter with a short discussion of
the general case of the moments and products of inertia, for
any body, or system.
Let us suppose the system referred to three rectangular
planes, and let p, g, 7 represent the respective distances of
any element dm from the three planes
2 cosa + y¥ cos + 2 cosy = 0,
@ cosa’ + y cos 3’ + 2 cosy’ =0,
a cosa” + y cos [3” + 8 cosy” =0.
Then
Spqdm=3 (xcosa+ycos+scosy) (xcosa’+ycos[3’+zcosy’) dm
= cosa cosa’ Za dm + cos B cos’ Syd + cos y cosy’ Ss"dm
+ (cos a cos 9’ + cos B cosa’) Sxydm
+ (cosy cos a’ + cosa cos y’) Sardm
+ (cos 8 cosy’ + cos y cos 3’) Sysdm ;
and we get similar expressions for Zprdm and Sgrdm.
Principal Axes. 307
Now, suppose that we take
sa’dm=a, Bydm=b, Sdn =c,
Sysdm=f, Bazdm=g9, Beydm=h;
then the preceding equation may be written
=pqdm = cosa(a cos a’ + h cos 23’ + g cos y’)
+ cos 8 (2 cos a’ + b cos 3’ + f cos 7’)
+ cos y (g cos a’ + f cos 3’ + ¢ cos y’) ; (22)
along with similar expressions for Zrpdm and Sgrdm.
214. Principal Axes.—Next, let us suppose that the
planes are so assumed as to satisfy the equations
=pgdm=0, Irpdm=o, Sqrdm=o;
then it is easily seen* that these planes are a system of con-
jugate diametral planes in the ellipsoid represented by the
equation
aX? + bY? + cZ? + 2fVZ + 2gZX + 2hXY = const. (23)
Hence it follows that at any point there exists one system of
rectangular planes for which the corresponding products of
inertia, for any body, vanish: viz., the principal planes of the
preceding ellipsoid.t
These three planes are called the principal planes of the
body relative to the point, and the right lines in which they
intersect are called the principal aves for the point.
Again, every two solids have for every point at least one
common system of planes for which Spqgdm = 0, Srpdm =o,
Sqrdm = 0, Spd dm = 0, Sr pdm’ =0, Sq rdw = 0;
where the unaccented letters refer to the elements of one
solid, and the accented to those of the other.
This is obvious from the property that every two con-
centric ellipsoids have one common system of diametral planes.
*, Salmon’s Geometry of Three Dimensions, Art. 72. e .
+ The exceptional cases when the ellipsoid is of revolution, or is a sphere,
will be considered subsequently.
[20a]
308 Integrals of Inertia.
Again, if two solids have for any point more than one
system of planes for which the foregoing six products of
inertia vanish, they must have the same principal planes at
the point. This follows since the two ellipsoids in that case
must be similar and coaxal.
215. Principal Moments of Inertia.— Let us now
suppose the co-ordinate planes to be the principal planes of
the body for the origin, then the moment of inertia relative
to the plane
; #COSa+Y COS +8 COSY =O
is
Sp’dm = B(x cosa + y cos B + 8 cos y)’dm
= cos’a Sa" dm + cos’B Xy’dm + cos*y Bs’dm, (24)
since in this case we have
Seydm=0, Seedm=o0, Sysdm =o.
Again, let I be the moment of inertia of the body relative
to the line through the origin whose direction angles are
a, 3, y; then we have
I+ 3p’dm = irdm = 3(e + y+ )dm;
*, [= cosa d(y’ + 2°) dm + cos’B B(s° + a”) dm
+ cos’ y B (2 + y*)dm;
or I= A cos’a + B cos’B + C cos’y, (25)
where A, B, C are the moments of inertia of the body
relative to its three principal axes.
A, B, Care called the three principal moments of inertia
of the body relative to the origin.
If the centre of gravity be taken as the origin, the
corresponding values of A, B, C are called the principal
moments of inertia of the body.
We suppose, in general, that 4 is the greatest, and C the
least of the three principal moments.
It follows from (25) that the moment of inertia of a body
relative to any line passing through a given point is known,
whenever the angles which the line makes with the principal
axes are known, as also the moments of inertia relative to
these axes.
Momental Ellipsoid. 309
216. Ellipsoid of Gyration.—Suppose, as before, the
solid referred to its three principal axes at any point, and let
a, b, ¢ be the corresponding radii of gyration, i.e. let
A=Me, B=M, C=MNe,
and J = Mk’; then equation (25) becomes
K? = @ cos’ a + b’ cos’ + c*cos’y. (26)
Now, if we suppose an ellipsoid described having the
principal axes for the directions, and a, 6, ¢ for the lengths
of its corresponding semi-axes; then (26) shows that. the
radius of gyration of the body, relative to the perpendicular
from the origin on any tangent plane to this ellipsoid, is
equal in length to this perpendicular. (Salmon’s Geometry
of Three Dimensions, Art. 89.)
The foregoing ellipsoid is called the ellipsoid of gyration
relative tothe point. It should, however, be observed that
by the ellipsoid of gyration of a body is meant the ellipsoid
in the particular case where the origin is at the centre of
gravity of the body.
217. Momental Ellipsoid.—lIi X, Y, Z be the co-
ordinates of a point R taken on the right line through the
origin O, whose direction angles are a, /3, y, we have
X=ORcosa, Y=ORcosP, Z=OR cosy.
Substituting the values of cos a, cos 8, cos y, deduced
from these equations, in (25), it becomes
I. OR’ = AX’ + BY’ + C2.
Suppose, now, that the point £ lies on the ellipsoid
AX’? + BY’ + CZ = const., (27)
and we get I. OR’ =X, denoting the constant by d ;
A
“Ls OR* (28)
Hence the moment of inertia relative to any axis, drawn
through the origin, varies inversely as the square of the cor-
responding diameter of the ellipsoid (27).
310 Integrals of Inertia.
From this property the ellipsoid is called the momental
ellipsoid at the point. :
_ When the origin is taken at the centre of gravity of the
body, this ellipsoid is called the central ellipsoid of the body.
If two of the principal moments of inertia relative to any
point be equal, the momental ellipsoid becomes one of re-
volution, and in this case all diameters perpendicular to its
axis of revolution are principal axes relative to the point.
If the three principal moments at any point be equal, the
ellipsoid becomes a sphere, and the moments of inertia for all
axes drawn through the point are equal. Every such axis is
@ principal axis at the point. :
For example, it is plain that the three principal moments
for the centre of a cube are equal, and, consequently, its
moments of inertia for all axes, through its centre, are equal.
218, Equimomental Cone.—Again, since
cos*a + cos’B + cos’y = I,
equation (25) may be written in the form
(A - I) cos’a + (B-T) cos*?B + (C- I) cosy = 0;
hence the equation
(4-1) X?+(B-I)V*?+(C-I)Z=0 (29)
represents a cone such that the moment of inertia is the same
for each of its edges. Such a cone is called an eguimomental
cone of the body.
Again, the three axes of any equimomental cone, for any
solid, are the principal axes of the solid relative to the vertex
of the cone.
When I= B, the cone breaks up into two planes; viz.,
the cyclic sections of the momental ellipsoid.
For a more complete discussion of the general theory of
moments of inertia and principal axes, the student is referred
to Routh’s Rigid Dynamics, chapters 1. and u1.; as also to
Professor Townsend’s papers in the Camb. and Dub. Math.
Journal, 1846, 1847.
Examples. 811
EXAMPLES.
_ Find the expressions for the moments of inertia in the following, the bodies
being supposed homogeneous in all cases :—
1, A parallelogram, of sides a, 4, and angle @, with respect to its sides.
MN
Ans. — # sin’ 9, # a’ sin? 0.
3 3
z. A rod, of length @, with respect to an axis perpendicular to the rod and
at a distance d from its middle point.
2
Ans. M (5+2") :
3. An equilateral triangle, of side a, relative to a line in its plane at the
distance d from its centre of gravity..
2
Ans. IL 5 + #)
24
4. A right-angled triangle, of hypothenuse ¢, relative to a perpendicular to
its plane passing through the right angle.
c
Ans. M e
5. A hollow circular cylinder, relative to its axis.
rer?
Ans. UM
, where 7 and 7’ are the radii of the bounding circles.
6. A truncated cone with reference to its axis.
3M 05 — B6
co BaeP where 8 and 0’ are the radii of its bases.
Ans.
4. Aright cone with respect to an axis drawn through its vertex perpen-
dicular to its axis.
2
Ans. ae e+ =) , where / denotes the altitude of the cone,
and 6 the radius of its base.
8. An ellipsoid with respect to a diameter making angles a, 8, y with its
axes.
Ans. S(# sin? + 4? sin? B + c* sinty) 5
g. Area bounded by two rectangles having a common centre, and whose
sides are respectively ‘parallel, with respect to an axis through their centre
perpendicular to the plane.
M (a? + B)ab—-(a'? 4b) ab’
Ans. ——————_,;——_—
12 ab — a'b
312 Examples.
10. A square, of side a, relative to any line in its plane, passing through its
centre.
2
Ans. Mo.
12
11. A regular polygon, or prism, with respect to its axis.
Ans. . R?+42r?), where R and r are the radii of the
circles circumscribed, and inscribed to the polygon.
12. Prove that a parallelogram and its maximum inscribed ellipse have the
same principal axes at their common centre of figure.
13. Prove that the moments and products of inertia of any triangular
lamina, of mass U, are the same as for three masses, each er placed at the
three vertices of the triangle, combined with a mass 3 placed at its centre of
gravity. .
14. Prove that the moments and products of inertia of any tetrahedron are
the same as for four masses, each - placed at the vertices of the tetrahedron,
combined with a mass si placed at its centre of gravity.
15. Lf a system of equimomental axes, for any solid, all lie in a principal
plane passing through its centre of gravity, prove that they envelop a conic,
having that point for centre, and the principal axes in the plane for axes.
16. Prove also that the ellipses obtained by varying the magnitude of the
moment of inertia form a confocal system.
17. Prove that the sum of the moments of inertia of a body relative to any
three rectangular axes drawn through the same point is constant.
18. Prove that a principal axis belonging to the centre of gravity of a body
is also a principal axis with respect to every point on its length.
19. Prove that the envelope of a plane for which the moment of inertia of
a body is constant is an ellipsoid, confocal with the ellipsoid of gyration of the
body.
20. If a system of equimomental planes pass through a point, prove that
they envelop a cone of the second degree.
21. For different values of the constant moment the several enveloped cones
are confocal P
22, The common axes of this system of cones are the three principal axes of
the body for the point ?
23. The three principal axes at any point are the normals to the three sur-
faces confocal to the ellipsoid of gyration, which pass through the point.
(M. Binet, Jour. de l’ Ee. Poly. 1813.)
re 81s
CHAPTER XI.
MULTIPLE INTEGRALS,
219. Double Integration.—In the preceding Chapters we
have considered several cases of double and triple integra-
tion in the determination of volumes and other problems
connected with surfaces. We now proceed to a short treat-
ment of the general problem of Multiple Integration, com-
mencing with double integrals.
The general form of a double integral may be written
| [ re nacay
in which we suppose the integration first taken with respect
to y, regarding x as constant. In this case, Y, yo, the limits
of y, are, in general, functions of «; and the limits of # are
constants.
For example, let us take the integral
a?
ais
U0 -| | ay" drdy,
Out
in which / is supposed greater than m.
as
re 1 /a™
Here | yy" dy = 7 (= - ”);
x
1 (2 am 2qitm
2-1 m \ de =
=—| a —_-a™ |dr= :
therefore U h I ( 5a ) are
Tt should be observed that in many cases the variables are
to be taken so as to include all values limited by a certain
condition, which can be expressed by an inequality : for in-
stance, to find
U= [fe y”"' dxdy,
extended to all positive values of # and y subject to the con-
dition e+ yu and y 24 dedy dz,
taken between the same limits, has for its value
P (2) T(m) T(n)
l+m4n-1
r(d@+m+n) ane me
Dirichlet’s Theorem. 319
Accordingly, the value of the multiple integral
Sf[Fle+y +3) ay v dady dz,
extended to all positive values of the variables, subject to the
condition
et+yt+es; also D represents ae
336 Multiple Integrals.
Here, by Leibnitz’ Theorem (Diff. Cale., Art. 48), the
general term in the development of
Do ((@ ~ a)" (w ~ 6)")
is of the form
(m+n) (m+n—- Ls ..(m+n-—P7r+1) D-r(e— a)". Dr(e- 8).
Moreover, as this is evanescent so long as 7’ is less than n,
we can assume 7 = 7 + p, and the preceding may be written
| m +n
[n+ p|m—P
D"? (¢—a)™. D™*? (a - b)™,
Ee eel ,
or, [n+p |m—p [p [m—n—p (w - a)? (w— b)™-"-®,
Accordingly, the expression
(aa)? (w= 8)" D™™ {(@— a)" (w-B)")
P= m+n m m
po |[n - | m—p = aaa (x—b)"*. (39)
Again, the general term in
; Do {(@ ~ a)" (@ — by")
may be written
|m—n
ae
DP (a= G) Dag b)™
|m—n | m | m
= =~ (x — a)? (w — 6)”,
|p |m-n-p|n+p |m—-p
Comparing this with (39), the theorem in (37) follows im-
mediately.
Application to Spherical Harmonics. 3837
Again, if we substitute m for 2, s for n, and make / = 1,
a = — 1, this result can be written in Rodrigues’ form, viz.,
mM+s
(u? 3 1)§ Ds (uw? me 1)” ee ne Divs (w? a 1)". (40)
Hence, since
(ut — 3)* De" (y4— 1)
T,,() =
2” | m
>
we get
s
ac ete at) ee eee”
Ln
| -8 2™| m
Consequently, multiplying the two expressions,
(Tn)? ie fone D8 (u?— ie D"* (uy? - Ly.
Therefore,
+1
| (Zin)? de =
-1
mt+s I
[aa 2)
Again, integrating by parts, and observing that the term
outside the sign of integration vanishes for eitler limit,
we get
| D™ 8 (y?- 1)" D™ (y?- 1)" du.
+1
| D5 (4? = rye De? an oe du
<1
+1
Si | pms (Ww eh 1)” Drs (wv? = 1)"du3
ei
hence, by successive integration by parts, we get finally
le pms (uw? a ep (Ww? mis 1)"du
eas [eve ~ 1)"}?du.
~ [22]
338 Muitiple Integrals.
Consequently
FL | m +g +1 D” (uw? - ry
(8))\2 dy = (— 1)8 eS he,
[cenntdne TS | (ee)
| m -s
wee : [m+ I
Si tyes (Pa) aa= f4) [m= 8 2m+1
(41)
Hence, from (36),
2 | m —8 im
a= (-1)* —=—— 0s s¢’ T,,; (42)
| m +8
and the complete expression for Lm can be immediately
written down. (Compare Diff. Cailc., p. 428.)
234. Expansion of a Function in Spherical Har-
monics.—We next proceed to prove that any function
F(u, ¢), which is finite and continuous, can be expanded
in a series of spherical harmonics, 7. e. that
F(m, o)=Yot Vit Yot+..-+¥n+&e. (43)
For if we assume this result, multiply both sides of the
equation by Zp, and integrate, we get, from (26),
‘am (+1 am (+1
| \ PM ?) Lindy dp = ic Y,Lndudp
pele Jo Js
ae :
anti Pay ONES
Again, writing y’, ¢’ for u, ¢ in (43), we have
(us @) = Yo + Yi + Yo'+.. 24+ Yn’ + &o.
+1
=F Bene 1) ||" Aso) Ladd. (44)
Expansion of a Function in Spherical Harmonics. 389
We shall verify this result by proving that /(u’, ¢’) is the
limit of the expression at the right hand side of (44) when
n is increased indefinitely.
For, suppose h =o then since, by hypothesis, 7’ is less
than r, equation (28) may be written
I
a ne ee (45)
where
d = cos POP’ = pp + /1 — 2/1 — pn? 008 (p - 9).
If we differentiate (45) with respect toh, and multiply by
2h, we get
2h(A-h)
(1— 2hd + h?)*
Adding to (45), we have
= 2AL,+ 4 L,+ ..4+ 2nh"L,+...
-/
as oat = Ly + 3hLy + 51?Ly+...+(2n+ 1)h" Ln ++. (46)
Hence
ae ii = (1 —h?) fu, o) aS
& (amt ayh [| 70. ?) E,d8 =| he (47)
where the integrals are supposed to be extended over the
surface of a unit sphere, of which dS is an element.
Hence we infer that
> (2n+ 1) {[ 70 ¢) Lidudg
is the limiting value of
{| O =f F(u, $) 48 oes when A = 1.
(1 - 24d + Wy?
Again, when 1 - / is indefinitely small, the coefficient of
[22a]
340 Multiple Integrals.
every element in the latter integral is indefinitely small except
those for which (1 — 2h + h*)4, or
PP’, is indefinitely small, ¢.e. for P
which the point P is taken in- A
definitely near to the point 4 on \
the sphere. Consequently the
integral has ultimately the same
value as if it were only taken
over a very small portion of the
surface around the point 4; but
throughout this portion we may
assume /(u, ¢) =f lu’, ?)s namely,
its value at the point 4. Hence
the limiting value of
Fig. 53.
(=F) flu, GES _ ae gn (f_(- aS -
{| (1 — 2hd + h*)8 “SW, 9)|| 7 “oma ae when / = 1.
Again, since X = cos ACP, we may write dd dg, for AS,
where ¢, is the angle the plane ACP makes with a fixed
plane drawn through C’A, and we have
| aS 7 amr c+l ddd,
J (1 Saunt |. I, (1 — 2hA + hP)®
ae ie ee
-(1-2hA +h) 1 - dF
Accordingly, for all values of 2,
when taken over the surface of a unit sphere; and we con-
clude that
ae :
47 mal
(an +0) | "[" Puy) Ladd = Stu’, 9% (48)
thus verifying equation (44).
This is the well-known general formula of Laplace; from
which we infer that every finite continuous function of » and ¢
Expansion of a Function in Spherical Harmonics. 841
can be expressed in a series of Spherical Harmonics. There
is no difficulty in showing that the series is unique: i.e. that a
given function can only be expanded in one way in a series
of Spherical Harmonics.
235. It may be observed that the determination of the
value in spherical harmonics of a given function of § and ¢
is usually best obtained by means of the corresponding solid
harmonic functions. We shall illustrate this by an example.
To transform wu = cos 6 sin®@ sin’ cos ¢.
Here r*u = wyz*; and we readily see that we may suppose
w= Y,+ V¥,. (49)
This gives rys =PV,+ Vy (50)
where VP, and V,, are solid harmonics.
Operating with v* on both sides, we get
2ry = V(r? V.) = 2.7V23
hence V, = Lay, and therefore Y,=1,/1 — py’ cos ¢. Also
from (50),
V, ele
= ay|2? — —},
4 mt z
+ Kus pot ~ pcos {(1 - p’) sin’ - 4).
ae ‘ COS @ — COS 3
Again, since cos ¢ sin’¢ = —~—_—+,
4
we readily get
4 _ ont
Fy = BVI Og — pF) cos p — BIE) 008 39,
4 4
: : wfi- pe
Hence cos # sin’ sin’ cos ¢ = a cos p
~— uw ~ 2)?
if mye (2 — 2) cos ¢ a cos 3¢.
It is readily seen that a function cannot be exhibited in a
Sinite series of spherical harmonics unless the corresponding
expression in 2, y, s is rational, or becomes rational when
multiplied by +.
342 Examples.
ExampLes.
1. If U=acosu+osinucosv+c sine sine,
am par A +1
prove that | | S (UV) sin u du dv =2r | f (Aw) de,
0 Jo =
where A=Vei+R4 0.
Let g2=cosv, y=sinwcosy, z=sinwsiny;
then (x, y, 2) are the coordinates of a point on a sphere of unit radius, with
centre at the origin. .
Also let a= Aa, b= AB, c= Ay; then u, 8, y is also a point on the same
sphere, and
acosu+bsinu cosy +csine sinv =A cos 8,
where @ is the arc joining the point a, 8, y to 2, y,2. Again, the element of
the surface of the sphere at the latter point may be represented by sinu du dv,
or by sin 6 40 d@, indifferently. Consequently,
f (acosu + bsin w cosy + ¢sin u sin v) sin u du dv = f (A cos 6) sin 6 dd dp.
Integrating each of these over the entire surface, we get
7 on . an (3 4 7 ‘
| | F(T) sin w dud =| | F(A cos 6) sin9 do dp =2n [ FS (4 cos @) sin 6 40.
o Jo 0 Jo 0
2. Hence deduce the following :
7 pan 2math
| | F(T) sinw cos udu do = =2"l f (Aw) wd,
o Jo -1
27e
“ i f (Aw) wa.
a
| | I(T) sin?u cos v du dv =
0 Jo
These are deduced from (1) by differentiation under the sign of integration.
3. Show that the integral
T=Sfflery) ayn dedy,
supposed extended to all positive values subject to the condition 7 +y fe [Fe y) da dy,
where k
344 Examples.
7. Prove that
aes
(’ { 2 (m? cos?6 +n? cos) dd dp
Jo Jo vr — m= sin? @) (I —n'sin® )
T
=i)
when m+ n=.
This is an immediate consequence of (9), Art. 222.
8. Show that Legendre’s Theorem connecting complete elliptic integrals
with complementary moduli follows immediately from the preceding example.
v wT
z do 2
Let F(m) = [ Vicaane Elm™= {, VI — m? sin? do,
then the equation in Ex. 7 is immediately transformed into
F(m) E(n) + E(m) F(n) — F(n) F(m) =4.
g. Prove that the area of a surface in polar coordinates is represented by
of, arty ar
snzg [92 4 Se gh
{[fsin Q (: + a) + ag rd@ dp,
taken between suitable limits.
10. Find the value of
ar
| Ln do. Ans. 20 Py P'n.
0
11, Adopting the notation of Art. 232, prove the relation
D{ut*! Ds Pn D® Pn} = (Tonle)? + (mm = 8)(m + 6 + 1){ Tn)?
where w= yw? — I.
Here Dus Ds Py Ds Py) = ust) (Ds) Py)?
+ U8 D8 Pm (WD? Pm + 2u($ + 1) D* Py).
Also, Art. 335, Diff. Calc., since P,» satisfies the equation
D(uDPm) = m(m + 1) Pn,
we have Du DPm) = m(m + 1) Dt Pm;
hence uD? Pm + 24 (8 + 1) D* Py = (m —s)(m +641) Dt Pine
The result in question follows immediately.
Examples. 845
12. Hence determine the value of the definite integral
+1
| (Loul*)? du.
-1
Multiplying the result in Ex. 11 by du, and integrating between the limits
+ 1 and — 1, we get
iy (Tmt)? du = — (m+s+1)(m—s) |
41
i L
(Zm'5))? du.
Tlence, substituting s — 1 for s,
+1 +1
| (Tia)? du = — (m+ 8) (m+ 1-8) | (Zl)? da
a1
+1
=(m + 8)(m+s—1)(m+1—s)(m—s) | (Lal)? de
-1
= &.
But when s = 0, Tix‘8) becomes Py; hence, by equation (35), Art. 232,
tees
.
+1
())?du = (—
iE (Tn) du. ( 1)8 2m +1 [m a5
Compare Art. 233.
13. Express cos”@ sin?@ sin @ cos@ in Surface Harmonics.
Proceeding as in Art. 235, we easily get
cos?@ sin?@ sin @ cosq = Py (I — w?)sin 2p
+ (x — a) (et — 9) sin 29.
( 346 )
CHAPTER XII.
ON MEAN VALUE AND PROBABILITY.
236. A very remarkable application of the Integral Calculus
is that to the solution of questions on Mean or Average
Values and Probability. In this Chapter we will consider a
few of the less difficult questions on these subjects, which
will serve to give at least some idea of the methods em-
ployed. We will suppose the student to be already ac-
quainted with the general fundamental principles of the
theory of Probability.
Mean Values.
237. By the Mean Value of n quantities is meant their
arithmetical mean, i.e. the n™ part of their sum.
To estimate the mean value of a continuously varying
magnitude, we take a series of » of its values, at very close
intervals, » being a large number, and find the mean of these
values. The larger n is taken, and consequently the smaller
the intervals, the nearer is this to the required mean value.
This mean value, however, depends on the law accord-
ing to which we suppose the » representative values to be
selected, and will be different for different suppositions.
Thus, for instance, if a body fall from rest till it attains the
velocity v, and it be asked—What is its mean velocity
during the fall? If we take the mean of the velocities at
successive equal infinitesimal intervals of time, the answer
will be 40; but if we consider the velocities at equal intervals
of space, it will bev. The former is the most natural sup-
position in this case, because it is the answer to the question
—What is the velocity with which the body would move,
uniformly, over the same space in the same time P—a question
which implies the former supposition. We might frame a
similar question, of a less simple kind, to which the second
value above would be the answer.
Case of One Independent Variable. 347
Again, if we wish to determine the mean value of the
ordinate of a semicircle, we might take the mean of a series
of ordinates equidistant from each other; or through equi-
distant points of the circumference; or such that the areas
between each pair shall be equal: in each case the mean
value will be different.
Thus we see that the Mean Value of any continuously
varying magnitude is not a definite term, as might be sup-
posed at first sight, but depends on the law assumed as to its
successive values.
238. Case of One Independent Wariable.— We
will therefore suppose any variable magnitude y to be ex-
pressed as a function ¢(x) of some quantity x on which it
depends, and its mean value taken as w proceeds by equal
infinitesimal increments 4 from the value a to the value 0.
Let » be the number of values, then xh =b-—a. The mean
value is
p(a)+g(ath)+o(at 2h) +}
But (Art. 90),
A} ola) +9(a+ A) +9 (a+ 2h) +. | -| o (x) de.
a
Hence the mean value is
M- ra ee. (1)
EXAMPLES.
1. To find the mean value of the ordinate of a semicircle, supposing the
series taken equidistant.
Lit jae eo
u=+|,ve- et
viz., the length of an arc of 45°.
2. In the same case, let us suppose the ordinates drawn through equidistant
points on the circumference.
eae [s sin 0 d0 = a ry; the ordinate of the centre of gravity of the arc.
wo w
348 On Mean Value and Probability.
3. Determine the mean horizontal range of a projectile ix vacuo for different
angles of elevation from 45° — @ to 45°+ 6; given the initial velocity V.
If a be the angle of elevation, the range is
2
Ris = sin 2a.
o
TrP?, aida Fy
Hence M= a \ oF sin 2ada, between the limits 45° + 0
V2 si
therefore wa oe
g 20
2
The mean value for all elevations, from 0° to 90°, is : *.
Tv
4. A number x is divided at random into two parts; to find the mean value
of their product.
M=" "e(n #) da => n2
=:/’ 6 7
5. To find the mean distance of two points taken at random on the circum-
ference of a circle.
Here we may evidently take one of the points, A, as fixed, and the other, B,
to range over the whole circumference; since by altering the position of 4 we
should only have the same series of values repeated: let @ be the angle between
AB and the diameter through 4: as we need only consider one of the two semi-
circles,
Tr
2
==| ar cos do =,
wT So T
6. To find the mean values of the reciprocals of all numbers from % to 2”,
when x is large; that is, to find the mean value of the quantities
that is, the mean value of the function =, as # increases by equal increments
from 1 to 2; therefore
u=[ Os Pig 2)
1 ne n
7. To find the mean values of the two roots of the quadratic
w-axr+b=0,
she roots being known to be real, but ) being unknown, except that itis positive.
Case of One Independent Variable. 34S
In this case 5 is equally likely to have any value from 0 to ~ ; hence, for the
greater root, a, :
a?
lop
u =r | add
&" Jo
eh
Rh
a
|. 2(2— 4) da; since b=a(a—a);
therefore M=a.
“Ain
The mean value of the smaller root is :
The mean squares of the two roots are 7 a, ~at, These might be deduced
from the former results, since
M (x?) — aM (x) + M(b) =0.
8. Find the mean (positive) abscissa of all points included between the axis
of x and the curve
x2
¢
Ans. —->+
y=ae ?,
Tv
The mean square of the abscissa is }c.
239. If M be the mean of m quantities, and MW’ the mean
of »’ others of the same kind, and if « be the mean of the
whole m + m’ quantities, we have evidently
mM + mM’
(
m +m’ ?)
B \
Thus if it be required to find the mean distance of one
extremity of the diameter of a semicircle from a point taken
at random anywhere on the whole periphery of the semicircle ;
since the mean value when it falls on the diameter is 7, and
the mean value when it falls on the arc is =, we have
r
27 Pea
wT 6r
°
OS or+ar | 24m
350 On Mean Value and Probability.
240. Case of Two or More Independent Variables.
—If s = ¢(«) be any function of two independent variables,
and 2, y be taken to vary by constant infinitesimal increments
h, k, between given limits of any kind, the mean value of the
function 2 will be
[fedxdy
both integrals being taken between the given limits.
The easiest way of seeing this is to suppose w, y, s the
coordinates of a point ; and to conceive the boundary, repre-
senting the limits, traced on the plane of zy, and then ruled
by lines parallel to w, y at intervals k, h apart. We have
thus a reticulation of infinitesimal rectangles hk; and if at
each angle an ordinate s be drawn to the surface s = (2, y),
as the number of ordinates will be the same as that of rect-
angles, we shall have
volume ff sdxrdy = sum of ordinates x hk;
also the plane area {{dzdy = number of ordinates x hk;
so that dividing the sum of the ordinates by their number,
the above expression results.
It may be shown, in like manner, that for three or more
independent variables a similar expression holds.
It is evident that the above expression, viewed geometri-
cally, gives the mean value of any function of the coordinates
of a series of points uniformly distributed over a given plane
area.
Exampres,
1. Suppose a straight line @ divided at random at two points; to find the
average value of the product of the three segments.
Let the distance of the two points X, Y, from one end of the line, be
called z, y. Consider first the cases when x > y; the sum of the products for
these is half the whole sum; hence
wee "(ioe - (ea) drdy = has
a jo
2, A number a is divided into three parts; to find the mean value of one
part.
Case of Two or More Independent Variables. 35]
Let 7, y, a—a% — y be the parts;
a a-x
\ | xdady
_ Jo Jo
I
“la pane “ee
\ \ dx dy
0 Jo
This value might be deduced, without performing the integrations, by consider-
ing that the expression is the abscissa of the centre of gravity of the triangle
OAB; OA, OB being lengths taken on two rectangular axes, each = a.
Of course the result in this case requires no calculation; as the sum of the
mean values of the three parts must be = @; and the three means must be equal.
The mean square of a part is z
3. A number a is divided at random into three parts: to find the mean value
of the deast of the three parts ; also of the greatest, and of the mean.
Let z, y, a— 2 —y, be the greatest, mean, and least parts. The mean value
‘ dudy Boo B
f th t = [ededy ‘i
of the greatest is (fdedy the limits of both |
integrations being given by
E>yr>a-xu—y>o. M Vv
If x, y be the coordinates of a point, referred
to the axes 04, OB, taking O04 = OB = a, the
above limits restrict the point to the triangle 4V
(AM being drawn to bisect OB); and the above
value of M is the abscissa of the centre of gravity of © A
I i ‘
this triangle; i.e. 3 of the sum of the abscissas of its
angles; hence
The ordinate of the same centre of gravity, viz.,
1/1 tN Bey
3 c etary ae
is the mean value of the mean part; hence the mean values of the three parts
required are respectively
ae +9
3” 1” 9°
4. To find the mean square of the distance of a point within a given square
(side = 2a), from the centre of the square.
DE pea erage a
unl [ern # 3
852 On Mean Value and Probability.
It is obvious that the mean square of the distance of all points on any plane
area from any fixed point in the plane is the square of the radius of gyration of
the area round that point.
5. To find the mean distance of a point on the circumference ofa circle from
all points inside the circle. :
Taking the origin on the circumference, and the diameter for the axis, if dS
be any element of the area, we have
wT
es are
wa® = wa? J_™
241. Many problems on Mean Values, as well as on
Probability, may be solved by particular artifices, which, if
attempted by direct calculation, lead to difficult multiple
integrals which could hardly be dealt with.
EXAMPLES.
1. To find the mean distance between two points within a given circle.
If MW be the required mean, the sum of the whole number of cases is repre-
sented by
(xr)? M.
Now let us consider what is the differential of this, that is, the sum of the new
cases introduced by giving r the increment dr. If Mo be the mean distance of
a point on the cirewmference from a point within the circle, the new cases intro-
duced, by taking one of the two points 4 on the infinitesimal annulus 2rrdr,
are
ar* My. 2nrrdr:
doubling this, for the cases where the point B is taken in the annulus, we get
ad. { (mr)? M} = qn? Mordr,
Now M= ie (Ex. 5, Art. 240);
OT
therefore wrt M = 1 ba if dr;
9 0
therefore M= ae
457
2. To find the mean square of the distance between two points taken on any
plane area 0.
ne a8, dS’ be any two elements of the area, A their mutual distance, and
we have
M= = [f[farasas’.
Case of Two or More Independent Variables. 353
Now, fixing the element dS, the integral of A?d8” is the inerti
! moment of
of the area 2 round dS; so that if K be the radius of gyration af the ae aun
a8, :
M= = fj Kas:
let r = distance of dS from the centre of it i
ee oe gravity G of the area, & the radius of
Krea=r+k;
therefore M=k 4 = Sf 2a8 = 2k;
ina eee ba oo the square of the radius of gyration of the area
242. The mean distance of a point P within a given area
from a fixed straight line (allie: dacs not meet the area) is
evidently the distance of the centre of gravity G of the area
from the line. Thus, if 4, B are two fixed points on a line
outside the area, the mean value of the area of the triangle
APB is the triangle AGB.
From this it will follow, that if X, Y, Z are three points
taken at random in three given spaces on a plane (such that
they cannot all be cut by any one straight line), the mean
value of the area of the triangle X YZ isthe triangle GG’G”,
determined by the three centres of gravity of the spaces.
EXxamp.e.
1, A point P is taken at random within
a triangle ABC, and joined with the three
angles. To find the mean value of the
greatest of the three triangles into which
the whole is divided.
Let @ be the centre of gravity; then if H kK
the greatest triangle stands on 4B, P is
restricted to the figure CHGK, and_ the
mean value of 4PB is the same as if P
were restricted tothe triangle GCH ; hence
c
we have to find the area of the triangle
whose vertex is the centre of gravity of A B
GOK, and base AB, Fig. 55.
I I
therefore Mas (A0B+ AKB+ AGB) =~ (: +o+ ;) ABC;
I ;
hence the mean value is = of the whole triangle.
The mean values of the least and mean triangles are respectively ; and 3
of the whole.
This question can readily be shown to be reducible to Question 3, Art. 240.
[28]
354 On Mean Value and Probability.
243. If M be the mean value of any quantity depending
on the positions of two points (e.g. their distance) which are
taken, one in a space 4, the other in a space B (external to
A); and if I’ be the mean value when both points are taken
indiscriminately in the whole space 4 + B; M4, Wy, the
mean values when both points are taken in A, or both in B,
respectively ; then
(4+ BPM = 2ABM + AM, + BM. (4)
If the space A = B,
4M’ = 2M+M,+ M,;
if, also, I, = If,
2M’ =M+N,:
thus if If be the mean distance of a point within a semi-
circle from a point in the opposite semicircle, Jf, that of two
points in one semicircle, we have (Art. 241),
M + I, 1= aoe r.
457
To determine UM or I, is rather difficult, though their
sum is thus found. The value of Jf is pee r
1357
EXaMPLets.
1. Two points, X, Y are taken at random within a triangle. What is the
mean area M of the triangle XYC, formed by joining them with one of the
angles of the triangle ?
Bisect the triangle by the line CD; let 24 be the mean value when both
points fall in the triangle ACD; Mp the value when one falls in ACD and the
other in BCD; then
2M = Mi + Me.
I
But Ih = su ; and Mz = GG'C, where G, G' are the centres of gravity of
ACD, BCD, this being a case of the theorem in Art. 242; hence
2 4
M,=-ABC, and M=— ABC.
9 27
2. To find the mean area of the triangle formed by joining an angle of a
square with two points anywhere within it.
Case of Two or More Independent Variables. 355
By a similar method this is found to be
I
— of the whole square.
3. What is the mean area of the triangle formed by joining th t
points with the centre of the square ? : ae aes eens
We may take one of the points X always in the square O04; take the whole
square as unity; then if M be the mean, the sum D
of all the cases is c
I I I I
gt pth tt git Gls Z
IM, Mz, Mz being the mean areas when the second Ne
point Yis taken respectively in 0.4, OB, and OC. ft oo o
But Mz; = Mh, for to any point Yin VC there cor- xX 4
responds one Y’ in OA, which gives the area Y¥
OXY' = OXY;
I I A B
th = - ye .
erefore Mu > M+ 5 Me Fig. 56.
13 «OC I 5
B =~, - oc-—: = — .*
ut mM ae! mM TAL hence If ae of the whole square
244. If two spaces 4 + C, B+ C have a common part C,
and M be any mean value relating to two points, one in 4 + C,
the other in B+ C; and if the whole space 4 + B+ C= W,
and IM, be the mean value when both points are taken indis-
criminately in W; WM, when taken in A, &e., then
2(4+C)(B+C) M=W'? Uy + CM - AM, - BM, (5)
as is easily seen by dividing the whole number W® of cases
into the different classes of cases which compose it.
* In such questions as the above, relating to areas determined by points
taken at random in a triangle or parallelogram, we may consider the triangle as
equilateral, and the parallelogram asasquare. This will appear from orthogonal
projection ; or by deforming the triangle into a second triangle on the same
base and between the same parallels, when it is easy to see that to one or more
random points in the furmer there correspond a like set in the latter, determining
the same areas. This second triangle may be made to have a side equal to a
side of an equilateral triangle of the same area; and then be deformed in like
manner into the equilateral triangle itself. Likewise a parallelogram may be
deformed into a square.
[23 a]
856 On Mean Value and Probability.
Example.
Two segments, 4B, CD, of a straight line have a common part CB; to
find the mean distance of two points taken, one in AB, the other in CD.
I
24B.CD.M=AD*. 342 ae OB. CB AC?.* AO- BD*.- BD,
since the mean distance of two points in any line is ; of the line;
AD + CB — AC? — DB
therefore N= ja
245. The consideration of probability may often be made
to assist in determining mean values. Thus, if a given
space S is included within a given space A, the chance of a
point P, taken at random on 4, falling on S is
aS
p= A
But if the space S be variable, and If (S) be its mean value,
_H(S)
Se ie
For, if we suppose S to have equally probable values
8,, S8., S;...., the chance of any one S;, being taken, and of
P falling on S,, is
_ I 8, 2
eA
now the whole probability p =p, +p. + pa+...3 which leads
at once to the above expression.
The chance of ¢wo points falling on S is
p= “a (7)
In such a ease, if the probability be known, the mean value
follows, and vice versd. Thus, we might find the mean value
of the distance of two points X, Y, taken at random in a line,
Case of Two or More Independent Variables. 357
by the consideration that if a third point Z be taken at random
in the line, the chance of it falling between X and Yis-; as
one of the three must be the middle one. Hence the mean
distance is 5 of the whole line.
2a"
j ; n+1)(n +2)
where a=wholeline. For if p is the probability that n more
points taken at random shall fall between X and ¥,
M (XY)" = ap.
Now, the chance that out of the n + 2 points X shall be
Sl he ee
one of the extreme points is =
n+2
Again, the mean n‘* power of the distance is
; and if it is so, the chance
that Y shall be the other extreme point is aa
n+i
EXamMP es.
1. From a point X, taken anywhere
in a triangle, parallels are drawn to two
of the sides. Find the mean value of
the triangle UX).
If a second point X’ be taken at
random within ABC, the chance of
its falling in XUV is the same as the
chance of X falling in the correspond-
ing triangle X’U’ V’; that is, of X’
falling in the parallelogram XC. Hence
the mean value of UXV=mean value 4
of XC. But the mean value of (UXV 4
+XC) is 2 ABC; as the whole triangle Fig. 57.
can be divided into three such parts by drawing through X a parallel to 4B.*
Thus
M(UXV) =; ABC.
The mean value of UV is 1 4B. For UV is the same fraction of 4B that the
altitude of X is of that of C: see Art. 242.
* The triangle may be considered equilateral: see note, Art. 243.
358 On Mean Value and Probability.
Cor. Hence, if p be the perpendicular from X on AB, h the altitude of
the triangle ABC, we get
I
2) — 72
U(p*)= 6% ;
If the area ABC be taken as unity, we have, since UXV: AXB=AXB: ABC,
(AXB)? = UX.
Thus the mean square of the triangle 4XB is z If two other points Y, Z are
taken at random in the triangle, the chance of both falling on 4XB is thus the
same as that of a single point falling on UXV; i.e. z Hence we may easily
infer the following theorem :—
If three points X, Y, Z are taken at random in a triangle, it is an even
chance that ¥, Z both fall on one of the triangles 3 Cc
AXB, AXC, BXC.
2. Ina parallelogram ABCD a point X is taken at
random in the triangle ABC, and another, Y, in 4DC.
Find the chance that X is higher than Y.
Draw XH horizontal: the chance is
mean area of AHK = ADC.
But AHK= XUV, and the mean area of XUV'is ; ACB H}-~-7fe
(Ex. 1); hence the chance is ~
xX
2 A U VW B
3. If O be a point taken at random on a triangle, and
lines be drawn through it from the angles, to find the Fig. 58.
mean value of the triangle DEF. (Mr. Mruuer.)
It will be sufficient to find the mean area of the triangle 4 EF, and subtract
three times its value from 4BC. If we put a, B, y for the triangles BOC,
AOC, AOB, it is easy to prove that
epi Pe
A
- ABC.
(a + B)(a + ) Ze :
If we put the whole area ABC =1, and if E <7
dS be the element of the area at 0, } LR
Bxis UY
M (AEF) = {J Sey ale
42) = )) aE a) i &
he integration extending over the whole triangle. Fig. 59.
But if », g are the perpendiculars from O on the sides }, c, it may be easily
shown that the element of the area is
s _apdq_ 4
“sind besin A
aB dy = 2dBdy.
Case of Two or More Independent Variables. 3859
Thus the mean value of AFF becomes
Fi 1(1-8 2BydBdy
He) | Gene
Again, by Art. 95, the definite integral
1 2B
= —1—loe — ad,
{,@ — log 8) 48
1 Blog B rd
I, I-£ Pee ee
therefore M=-1-2(1-F) = 2-3.
Hence the mean value of the triangle DEF is
10 — 7,
that of ABC being unity.
It is remarkable that the same value, 10 — 7, has been found by Col. Clarke
to be the mean area of a triangle formed by the jintersections of three lines,
drawn from .4, B, C to points taken at random in a, b, ¢ respectively.
4. To find the average area of all triangles having a given perimeter (2s).
By this is meant that the given perimeter is divided at random in every seats
way into three parts, a, b, ¢, and only those cases are taken in which a, 4, ¢ can
form a triangle ; then the mean value of
SS ed
= V58(s — a)(s — b)(s—¢) A x ¥ B
Fig. 60.
has to be found.
Take AB = 2s, let X, Y be the two points of division, dX =z, AY=y:
these are subject to the conditions
Z<8, Y>s Y-Ux>0, a>y>0, and ay < . We have
accordingly to integrate for y from a to 0, when x is between o and 7 and from
a to o, when z is between 2 and a; thus
alo
a QQ? a a
dedy =| ade + | 2 iis BE Toph
Sfdady ; ie ata
:
H Be + log 2
ence 2p ata g 2.
3.. Two points are taken at random in a given line a; to find the chance
that their distance asunder shall exceed a given value c.
It is easy to see that the distances of two such points from one end of the
line are the coordinates of a point taken at random B L D
in a square whose side is a. Thus to every case
of partition of the line corresponds a point in the va
square—such points being uniformly distributed over
its surface.
Thus, if in the above question , y stand for the
distances of the two points, from one end of the line,
y being greater than x, we have to find the chance H
of y—« exceeding c. The point P whose co-
ordinates are #, y, in the square OD (side = a),
may take all possible positions in the triangle OBD, A
if no condition is imposed on it. Butify—z>e,
then if we measure OH = ¢, the favourable cases Fig. 63.
Probabilities. 3865
occur only when P is in the triangle BHI; hence the probability required
BHI a-c\*
P= OBD ( ) :
a
In fact this is only performing the integrations in the expression
ay-c
ti dxdy
_ ve 40
~Tapy.
["["aeay
ovo
4. Two points being taken at random in a line a, to find the chance that no
one of the three segments shall exceed a given p ZEKE N D
length c.
The segments being as before, 2, y—%, a-y
PH=2, PK=a-y, Pl=y-«. There will y
be two cases :-— M
I as P
(1). If e> 54s take OV=BV =DZ=BN=c;
I
then it is easy to see that the only favourable Vv J
cases are when P falls in the hexagon UZNMJV;
OBD-—3.UBZ a—c\?
y= See -3 (“= oO A
OBD a Fig. 64.
(2). Ife< aa; take OU = BV =¢, as before; then the only favourable
cases are when P falls in the triangle RST; B KZ D
RST 3c —a\? ae
thereft SS
erefore pe OBD ( z ) fat
I i R a
since RST = > RT?, and RI= VI'+ RH-VH . ae
= 2¢—(a—c).
Such cases of discontinuity in the functions
expressing probabilities frequently present them-
selves. ‘The functions are connected by very
remarkable laws. Thus, in the present question, 0 A
if p=f(c), p2 = F(c), we have Fig. 65.
So) —f(a- 0) = Fao).
5. A floor is ruled with equidistant parallel lines; a rod, shorter than the
distance between each pair, being thrown at random on the floor, to find the
chance of its falling on one of the lines (Buffon’s problem).
Let x be the distance of the centre of the rod from the nearest line, @ the
inclination of the rod to a perpendicular to the parallels, 2a the common distance
of the parallels, 2¢ the length of rod; then as all values of « and 6 between their
366 On Mean Value and Probability.
extreme limits are equally probable, the whole number of cases will be repre-
sented by
TT
a2
[| eaede = we.
ove
. xz
Now, if the rod crosses one of the lines, we must have ¢ > con? that the
favourable cases will be measured by
T
2 ccnBs@
| de | du = 20,
Jo
2
Thus the probability required is = = °
This question is remarkable as having been the first proposed on the subject
now called Local Probability. It has been proposed, as a matter of curiosity,
to determine the value of m from this result, by making a large number of trials
with a rod of length 2a: the difficulty, however, here consists in ensuring that
the rod shall fall really at random. ‘The circumstances under which it is thrown
may be more favourable to certain positions of the rod than others. Though we
may be unable to take account ¢@ priori of the causes of such a tendency, it will
be found to reveal itself through the medium of repeated trials.
249. Sometimes a result depends upon a variable (or
variables) all the values of which are not equally probable, but
are such that the probability of a certain value for a variable
depends, according to some law, on the magnitude of that
value itself (and also, perhaps, on the values of other variables).
Thus a point may be taken in a straight line so that all
positions are not equally probable, but the probability of the
distance from one end having the value , being proportional
to x itself. This would be in fact supposing the series of
points in question as ranged along the line with a density
proportional to x; as, e.g., if they were the projections, on
the line, of points taken at random in the space between the
line and another line drawn through one of its extremities.
To give an example :—
Two points are taken in a line a, with probabilities vary-
ing as the distance from one end 4; to find the chance of
their distance exceeding a length c.
Let x, y, be the distances from A, and suppose y > #.
Probabilities. 367
Here the probability of a point falling between # and 2 + dz
is not proportional to dz, but to «dx; and the result will be
ra Yy-C
| vay| w dee ; aa
feet (3
| yay | eds a2 a
0
The mean values of the three divisions of the line, in the
same case, will be found to be
8
—A, a a, ra.
15 15 5
The above value of p is also the value of the chance, that
the difference of the altitudes of two points within a triangle
shall exceed a given fraction “ of the altitude of the triangle.
EXxaMPLes.
1. Two points being taken on the sides 0.4, OB, of a square a’, the chance
of their distance being less than a given value 6 is easily seen without calcula-
2
tion to be _ provided 5 < a; as it is the chance of a point taken at random in
a
the square falling within a quadrant of a given circle. Suppose now that two
points are taken on OA, and two on OB, and that we take X, Y, the two points
furthest from O on each side, to find the chance that their distance XY is less
than a given length 4; (b y—--.
BS SY S ’ y ei
ea dy dt 1, Bee ae
Hence o={f., “go B23
2, An urn contains a large number of black and white balls, the proportion
of each being unknown: if on drawing m + balls, m are found white and
n black, to find the probability that the ratio of the numbers of each colour lies
between given limits. : Vv
The qtennon will not be altered if -
we suppose all the balls ranged in a line
AB, the white ones on the left, the
black on the right, the point X where
they meet being unknown, and all posi-
tions for it in AB being d priori equally a XK 3B
probable; then m+ points being taken Fig. 69.
at random in 4B, mare found to fall on ee ie
AX,non XB. That is, all we know of X is, that it is the (m + 1)* in order,
[24 a]
872 On Mean Value and Probability.
beginning from 4, of m+ +1 points falling at randomon 4B. If AX =z,
AB = 1, the number of cases for X between x and x + dz is measured by
[™ +n
pdr, an (1 "ass x)" dx.*
[mee
Hence the probability that the ratio of the white balls in the urn to the
whole number lies between any two given limits a, 8—that is, that the distance
from 4 of the point X lies between a and 6—is
(i am (1 — 2x) dx
a
p=———————«
I
| a” (I — 2)" de
0
The curve of frequency for the point X will be one whose equation is
y= am (l— a).
The maximum ordinate KV occurs at a point K dividing AB in the ratio
m:n. This is of course what we should expect: the ratio of the numbers of
black and white balls is more likely to be that of the numbers drawn of each
than any other. The value for py above is simply the area of the above curve
between the values a, B, of x, divided by the whole area.
Let us suppose, for instance, that 3 white and 2 black balls have been
drawn ; to find the chance that the proportion of white balls is between = ana 4
of the whole—that is, that it differs by less than + : from > its most natural
value. Here
4
° _ p)2
[pee ada 2256 18
p= = — = —, nearly.
a.
| #2 (1 —2)*dx S° 75
0
The above results will apply to any event that must turn out in one of
two ways which are mutually exclusive, this being the whole of our 4 priori
knowledge with regard to it—the ratio of the black or white balls to the
whole number, meaning the real probability of either event, as would be
manifested by an infinite number of trials. We will give one more example of
the same kind.
3. An event has happened m times and failed times in m+ trials. To
find the probability that, on » + q further trials, it shall happen y times and
tail ¢ times.
* For a specified set of m points, out of the m + m, falling on AX, the
m+n
[m [2
number is #™ (1 — 2)"dx ; the number of such sets is
Curve of Frequency. 373
That is, that y+ more points being taken at random in AB, p shall fall in
AX, and gin BX. The whole number of cases is as before:
[m+n u m+nph
—— (anyre| a" (1 —a)ndx = - | am (I~ a)dx.
L™|2 0 (Io
When any particular set of » points, out of the p 4 dditi i i
9 + q additional trials, falls in
AX, the number of favourable cases is 4 .
inal
L™[”
Tut the number of different sets of p points is eeraa es pale ah
1.2.3...p.1.2.3...¢
Hence the probability is, putting as before |p for1.2.3...9,
gintp (I fas aja dz.
0
1
[et | amp (I—a)™odz
0
Le lg : f. qm (1 as x)" dx
By means of the known values of these definite integrals (p. 117), we find
A=
[p+9 |[m+p[n+¢ [m+n+t
“Tela [mn [mint pegar
For instance, the chance that in one further trial the event shall happen is
m+
mtn+2"
sections by the m+” + I points in it, including X. Now, if one more trial 18
made, i.e. one more point taken at random, it is equally likely to fall in any
section ; and m + 1 sections out of the entire number are favourable.
This is easily verified, as the line 4 B has been divided into m+n+2
4. Trace the curve of frequency of the ratio 3 aand b being numbers taken
at random within the limits + 1. Y
If we measure the values of
the ratio as abscissas along an _
axis OX, and make OA =1; B B
OA’ =—-1, AB=A'B'=13; ,
then the line whose jana e c Cc
are proportional to the fre- ;
anveney will be, for values of A oO A x
; comprised between the limits Fig. 70.
+ 1, the straight line BB’; but, for values beyond these limits, will consist of
the arcs BC, B'C’ of the curve a’y = 1.
It is thus an even chance that the ratio ; lies itself between the limits + 1:
this would also appear by a construction such as that given in the next Article.
374 On Mean Value and Probability.
251. Errors of Observation.—One of the practically
most important, as well as the most difficult, departments
of the theory of Probability is that which treats of Errors of
Observation. We will give here an example of the simplest
description.
Two magnitudes 4 and B are measured ; each measure-
ment being subject to an error, of excess or defect, which
may amount to + a, all values between these limits being
supposed equally probable.* To determine the probability
that the error in the sum, 4 + B, of the two magnitudes,
shall lie within given limits; also its mean value.
Thus the angular distance of two objects A, C is some-
times found by measuring the angle between A and B, an
intermediate object ; and afterwards that between B and C,
and adding the two angles. If each measurement is liable
to an error + 5’, all values being equally probable, to find the
probability of the error of the result falling within assigned
limits: its extreme limits being of course + 10’.
The question is more easily comprehended by means of
a geometrical construction than by B K
integration. J
Take AB = 2a; then all the values :
of the first error aie the distances - NS
from O of points P taken at random , YW Ns
in AB; positive when in OB; ee
negative when in OA. Make also I
A’B’ = 2a; the values of the second
error are given by points in A’B’.
Take any values, OP = x for the first,
OP’ = «' for the second: these values
taken as co-ordinates determine a point V corresponding to
one case of the compound error # +2’; and such points V
will be uniformly distributed over the square HK. The value
of the compound error « corresponding to the point V is
e=a2+a' = OS,
if VS be drawn at 45° to the axes. Now all values of the
Fig. 71.
* This supposition must not be taken to be practically correct. The Theory
of Errors shows that the probability of an error of magnitude z is proportional
to e427,
Errors of Observation. 375
errors x, 2 which give « + 2 the same, give the same value
for <; hence all points on the line JZ correspond to com-
pound errors of amount OS. Take Ss = de; the number of
compound errors between « and < + de is the number of
points between JJ and a parallel to it throughs. Now the
area of this infinitesimal strip is evidently
(2a — «) de.
Hence the probability of the error being between « and
e+ dé is
(2a - «)de
2
4a
This holds for negative values of ¢, provided we only consider
their arithmetical magnitude.
Thus the frequency of an error of magnitude « = OS is
proportional to JJ, the intercept of a line through S sloping
at 45°. The probability of the error « falling between any
two given limits OS, OS’ is found by measuring these
lengths (with their proper signs) from O, along AB, and
dividing, by the area of the wnole square, the area intercepted
on the square by parallels through S and 8’, sloping at 45°.
Thus the chance of the error falling between the limits
+a (those of the two component errors) is 2
The mean value of the error, strictly speaking, is 0; but it
is evident that for this purpose we ought to consider negative
errors as positive; and consequently take the mean of the
arithmetical values of all the errors, which is the same as the
mean of the positive errors only; hence the mean error
required is
M(s)=+
a.
wid
The most probable value, such that it is an even chance that
the error exceeds it (since the triangle JJ must be of the
whole square, for that value of OS), is
ta(2—/2) = +.586a.
376 On Mean Value and Probability.
Let it be now proposed to find the probability of a given
error in the sum of A and B, assuming, according to the
modern theory of errors, that the probability of an error
between w and # + dz in either is
= ee e “de:
CVn
I
es
dition that the differential, being integrated from o to — a,
must give unity, as the error must lie between these limits.*
Referring to the above construction, the number of values
of the first error between # and x + dx being proportional to
2 dx,
and the number of values of the second error between 2 and
a +da being proportional to
ae dx’,
the corresponding number of values of the compound error is
proportional to
the coefficient
being determined by the necessary con-
eet a2
e ” dada.
Hence the number of points, corresponding each to a case
of the compound error, in any element dS of the plane at a
distance r from the origin, is measured by
e “dS :
which shows that the points have the same density along any
* It is of course absurd to consider infinite values for an error: but the
a?
curve y=e © tends so rapidly to coincide with its asymptote, the axis of 2,
that the cases where # has any large values are so trifling in number, that it is
indifferent whether we include them or not.
Errors of Observation. 377
circle whose centre is 0. Now the probability of this com-
pound error being between « and ¢ + dé is proportional to the
number of points between JZand the consecutive line; making,
as before, OS = «, Ss = de. But this number is the same
as when the strip JJ is turned round O through an angle of
45°) because the points lie in concentric circles of equal den-
sity. Hence the number is proportional to
a f. ee jel
a 2 ts if 2
as the perpendicular from 0 on JI is —=.
2
Thus the probability of a compound error between « and
« + de is proportional to
e2
e °°" de,
and as this, when integrated between the limits + 0, must
give the probability 1, the value of p is
1 -s
e *? de.
Cif 27
It thus follows the same law as the two component errors,
¢./2 taking the place of ¢.
252. Various artifices have been employed for the solution
of different interesting questions on Probability, which would
be found extremely tedious, or impracticable, if attempted
by direct integration. Forexample: _
‘'wo points are taken at random within B
a sphere of radius 7; to find the chance that
their distance is less than a given value ec.
Let F = number of favourable cases, "9 5
W = whole number; then
F 4 ‘)
p=, We (3 mr).
ve Fig. 72.
Let us consider the differential dF, or
the additional favourable cases introduced by giving r the
increment dr, ¢ remaining unchanged.
378 On Mean Value and Probability.
If one of the points A is taken anywhere (at P) in the
infinitesimal shell between the two spheres, then drawing a
sphere with centre P, radius c, all positions of the second
point, B, in the lens ED common to the two spheres, are
favourable; let Z = volume FD, then the number of favour-
able cases when A is in the shell is
arrdr. L:
doubling this, for the cases when B is in the same shell,
dF = 8977" Ldr.
Now it may be easily proved, from the value for the volume
of a segment of a sphere, that
gt ge Fe.
3 4e°
2 I
hence F= 87° (3 er = cfr + e);
C being an unknown constant; ie. involving c, but not r;
F & 9g & gC
th = a
erefore p ae pe a tS oe
9
Now the probability = 1 if r = =e :
Cc 2
therefore 1 =8- Cm OR oe cane! Bs
ots x 64 53 oo ba?
hence iS a a
If the two points be taken within a circle, instead of a
sphere, it may be proved by a similar process that
2 2 os
¢ 2 ec : c I ec C Zl
paSei(1-$ sin — —- —.-(24— oo,
r 7 r 2r aw r iad r
Errors of Observation. 3879
It is a remarkable fact, pointed out by Mr. 8S. Roberts,
that if we draw the chord ED, the probability in the case of
the circle is,
ite 2.segment HQD + segment HPD |
area of circle HHD :
and also, in the case of the sphere,
_2: volume EQD + volume EPD
volume of sphereKHD —~
These results evidently suggest that there must be some
manner of viewing the question which would conduct to
them in a direct way.
Exampres.
1. Three points being taken at random within a sphere, to find the chance
that the triangle which they determine shall be acute-angled.
As the probability is independent of the radius of the sphere, it is easy to
see that we may take the farthest from the centre of the three points as fixed on
the surface of the sphere. For if p be the probability of an acute-angled triangle
in this case, p will also be the probability of an acute-angled triangle for each
position of the farthest point, as it travels over the whole volume of the sphere.
Hence p will be the probability when no restriction is put on any of the points.
Take then A, one of the points on the surface of the sphere ; two others, B, C,
heing taken at random within it, and let us find the
chance of ABC being obtuse-angled : to do this, we
will find separately the chance of the angles 4, B, C
leing obtuse: the events being mutually exclusive,
the probability required will be the sum of these
three.
(1). To find the chance that A is obtuse, let us fix
B; then, drawing the plane 4V perpendicular to 4B,
the chance required is
volume of segment A4HV
volume of sphere * Fig. 73
Let r be the radius of sphere, p = 4B, 9 =/OAB; then the volume of tle
segment AHV is
dar (1 — cos 8)? (2 + G08 8) 5
therefore when B is fixed the chance is
2(1 — cos 6)? (2 + cos 6).
380 On Mean Value and Probability.
Now let B move over the whole volume of the sphere, and we have for Pa,
the probability that 4 is obtuse,
2 jf fee — cos 6)? (2 + cos 6)dV
Mae
sphere
Tv
2
3 2r cos@ e
= | | (2 — 3.cos @ + cos?@) p*sin 640 dp.
73 JoJo
Hence Pa =>
(2). To find the chance, Pz, that Bis obtuse. Fix B as before; then the
chance that B is acute is
segment MHN
sphere
Now, volume MHWN = 3m7% (2+ I — cos e)’ (2 + cos 0 -£) ; so that the
chance is
_ {2 — 30080 + costo + 3 2 (1 pot 49 © wees ae
4 r re 3)"
Hence the whole probability (1 — Pz) that B is acute is
T
3 Zp2rcosg + p ‘ ? p3 5
a\,, {2-3 cos 6 + cos? @ + 3 7 (E008 4)+3 rm) cos 6 — et p* sin 6 dO dp.
3 Z ¥ I
Performing the integrations, we find Pg = Zz.
The probability for Cis, of course, the same as for B; hence the whole pro-
bability of an obtuse-angled triangle is
Pers Pettey eee,
70 70° 70 70
Hence, the chance of an acute-angled triangle is 3.
For three points within a circle the chance of an acute-angled triangle is
4 1
mm 8
2. Two points, 4, B, are taken at random inatriangle. If two other points,
C, D, are also taken at random in the triangle, find the chance that they shall lie
on opposite sides of the line 4B.
Errors of Observation. 381
The sides of the triangle 4 BC produced divide the whole triangle into seven.
spaces. Of these, the mean value of
those marked (a) is the same, viz., the
mean value of ABC; or, ps of the
whole triangle, as we have shown in
Art. 245; the mean value of those
marked (8) being % of the triangle.
This is easily seen: for instance,
if the whole area = 1, the mean value
of the space PBQ gives the chance
that if the fourth point D be taken
at random, B shall fall within the
triangle 4DC: now the mean value
of 4 BC gives the chance that D shall
fall within ABC; but these two
chances are equal. Fig. 74.
Hence we see that if 4, B, C be
taken at random, the mean value of that portion of the whole triangle which
lies on the same side of 4B as C does is 44 of the whole; that of the opposite
Portion is +.
Hence the chance of and D falling on opposite sides of AB is 75.
253. Random Straight Lines.—If an infinite number
of straight lines be drawn at random in a plane, there will
be as many parallel to any one given direction as to any other,
all directions being equally probable; also those having any
given direction will be disposed with equal frequency all
over the plane. Hence, if a line be determined by the co-
ordinates, p, w, the perpendicular on it from a fixed origin O,
and the inclination of that perpendicular to a fixed axis; and
if p, w be made to vary by equal infinitesimal increments,
the series of lines so given will represent the entire series of
random straight lines. Thus the number of lines for which
p falls between p and p + dp, and w between w and w + dw,
will be measured by dpdw, and the integral
f J dpdw,
between any limits, measures the number of lines within those
limits.
It is easy to show from this that the number of random
lines which meet any closed convex contour of length L is
measured by L. . :
For, taking O inside the contour, and integrating first
for p, from o to p, the perpendicular on the tangent to the
contour, we have {pdw: taking this through four right angles
382 On Mean Value and Probability.
for w, we have by Legendre’s theorem (p. 232), W being the
measure of the number of lines,
2
v= | wise Ee
0
Thus if a random line meet a given contour, of length Z,
the chance of its also meeting another convex contour, of
length /, internal to the former, is
fe
Bee
If the given contour be not convex, or not closed, NV will
evidently be the length of an endless string, drawn tight
around the contour.
EXAMPLes.
1. Ifa random line meet a closed convex contour, of length L, the chance
of it meeting another such contour, external to the former, is
Xx=¥7
ts
where X is the length of an endless band
enveloping both contours, and crossing
between them, and Y that of a band also
enveloping both, but not crossing.
This may be shown by means of
Legendre’s integral above ; or as fol-
lows :—
Call, for shortness, N(A) the number
of lines meeting an area 4; N (A, 4’) Fig. 75.
the number which meet both 4 and 4’; then
N(SROQPH) + N(S'Q'OR'P’H') = N(SROQPH + S'Q'OR' P’H’)
+ N(SROQPH, S'QOR'P'H’,
since in the first member each line meeting doth areas is counted twice. But
the number of lines meeting the non-convex figure consisting of OQPHSR and
OQS'H'P'R’ is measured by the band Y, and the number meeting both these
areas is identical with that of those meeting the given areas 2, 0’; hence
X=Y+ Na, 0’).
Thus the number meeting both the given areas is measured by X— Y. Hence
the theorem follows.
Random Straight Lines. 383
2. Two random chords cross a given convex boundary, of length Z and area
Q; to find the chance thut their intersection falls inside the boundary.
Consider the first chord in any position; let @ be its length ; considering it
as a closed area, the chance of the second chord meeting it is
2c.
TZ
and the whole chance of its co-ordinates falling in dp, dw, and of the second
chord meeting it in that position, is
20 dp dw 2
But the whole chance is the sum of these chances for all its positions ;
2
therefore prob. = BR | | Cdp dw.
Now, fora given value of w, the value of [ Cdp is evidently the area 0; then
taking w from 7 to 0,
required probability = =
The mean value of a chord drawn at random across the boundary is
{{Cdpdw 72
M =~ =,
Sfdpde T
3. A straightband of breadth ¢ being traced on a floor, and a circle of radius
ry thrown on it at random, to find the mean area of the band which is covered by
the circle. (The cases are omitted where the circle falls outside the band.)*
If 8 be the space covered, the chance of a random point on the circle falling
on the band is
_ (8)
are”
This is the same as if the circle were fixed, and the band thrown on it at
random. Now let A be a position of the f
random point : the favourable cases are when
HK, the bisector of the band, meets a circle,
centre 4, radius $c; and the whole number
are when HX meets a circle, centre O, radius
r+ 4c; hence (Art. 245) the probability is
2m.de
ar(rthe) arte’
p=
This is constant for all positions of 4 ;
hence, equating these two values of p, the
* Or the floor may be supposed painted with parallel bands, at a distance
asunder equal to the diameter ; so that the circle must fall on one.
384 On Mean Value and Probability.
mean area required is
e
2ar+e
The mean value of the part of the cirewmference which falls on the band is
(8) =
ar,
of the whole circumference.
4 ¢
the same fraction
2ar+e
If any convex area Q, of perimeter L, be thrown on the band, instead of a
circle, the mean area covered is
(5)
“T+ *
254. Application to Evaluation of Definite Inte-
grals.—The consideration of probability sometimes may be
applied to determine the values of Definite Integrals. For
instance, if n+ 1 points are taken at random in a line, /, and
we consider the chance that one of them, X, shall be the last,
beginning from the end A of the line, the number of favour-
able cases, when X is in the element dz, is, if AX =x,
measured by
x” de.
i
Hence | ae
p= eters
but the chance must be- - ~ 3 we thus have an independent
proof that
2 (nt
| "dx = ,
. n+t
when # is an integer.
Again, ifm +n+1 points are taken, to find the chance
that X shall be the (m+ 1) in order; the number of favour-
able cases when X falls in dv and a particular set of m points
fall to the left of X, is
2” (1 —«)"dw; taking 7=1;
hence the whole number of favourable cases is
[m+ np
{ a” (1 — a)"da;
[m L#Jo
Evaluation of Definite Integrals. 385
this is the required probability, since 7" = 1, But the
value is
penny 2 OY point is equally likely to fall in
the (m + 1)" place: we thus deduce the definite integral
1 min
[ie (1 fied a ;
0 | m +nN+1
when m, are integers. (See Art. 92.)
255. To investigate the probability that the inclination
of the line joining any two points in a
given convex area Q shall lie within
given limits.
We give here a method of reducing
this question to calculation, for the sake
of an integral to which it leads, and q
which is not easy to deduce otherwise.
First, let one of the points, A, be
fixed; draw through it a chord PQ = C, Fig. 77.
at an inclination @ to some fixed line; b
put AP =r, AQ=7’; then the number of cases for which
the direction of the line joining A and B lies between 0 and
6 + d0 is measured by
P
4(r° +r”) db.
Now, let A range over the space between PQ and a
parallel chord distant dp from it, the number of cases for
which A lies in this space, and the direction of AB
from @ to 0 + dO, is (first considering A to lie in the
element drdp)
4 dpio| (40) dr= 40% dp dd.
0
Let p be the perpendicular on the chord C from any
origin O, and let w be the inclination of p to the prime vector
(we may put dw for d@), then C' will be a given function of p, w ;
and integrating first for w constant, the whole number of cases
for which w falls between given limits w’, w”, is
| de forays
the integral { O°dp being taken for all positions of C between
[25]
386 On Mean Value and Probability.
two tangents to the boundary parallel to PQ. The question
is thus reduced to the evaluation of this integral; which,
of course, is generally difficult enough; we may, however,
deduce from it a remarkable result; for if the integral
Lf[ Cdp dw
be extended to all possible positions of C, it gives the whole
number of pairs of positions of the points A, B which lie
inside the area; but this number is Q’; hence
[J C%dp dw = 3.0%,
the integration extending to all possible positions of the
chord C’; its length being a given function of its co-ordinates
Py w.
"Cor. Hence if L, Q, be the perimeter and area of any
closed convex contour, the mean value of the cube of a chord
2
drawn across it at random is .
It follows that if a line cross such a contour at random,
the chance that three other lines, also drawn at random, shall
2
meet the first inside the contour, is 24 ies
Some other cases of definite integrals deduced from the
theory of Probability are given in a paper in the Philo-
sophical Transactions for 1868, pp. 181-199. See also Pro-
ceedings London Math. Soe., vol. viii.
Several examples on Mean Values and Probability are
annexed; some of them, as also some of the questions which
have been explained in this chapter, are taken from the
papers on the subject in the Educational Times, by the Editor,
Mr. Miller, as also by Professor Sylvester, Mr. Woolhouse,
Col. Clarke, Messrs. Watson, Savage, and others. Some few
are rather difficult; but want of space has prevented our
giving the solutions in the text.
We may refer to Mr. Todhunter’s valuable History of
Probability for an account of the more profound and dif_i-
cult questions treated by the great writers on the theory of
Probability.
Examples. 387
Examp.ies.
1. A chord is drawn joining two points taken at random on a circle: find the
mean area of the lesser of the two segments into which it divides the circle.
2. Find the mean latitude of all places north of the Equator.
Ans. 32° .704.
3. Find the mean square of the velocity of a projectile in vacuo, taken at all
instants of its flight till it regains the velocity of projection.
Ans. V* cos’a+4V*sin’a: where Vis the initial velocity, and ais the
angle of projection.
4. If x and y are two variables, each of which may take independently any
value between two given limits (different for each), show that the mean value
of the product xy is equal to the product of the mean values of # and y.
5. If X, ¥ are points taken at random in a triangle ABC, what is the
chance that the quadrilateral ABXY is convex?
Ans. 2
3
For, it is easy to see that of the three quadrilaterals ABXY, ACXY, BCXY,
one must be convex, and two re-entrant.
6. Find the mean area of the quadrilateral formed by four points taken at
random on the circumference of a circle.
Ans. 2 (area of circle).
T
7. Aclass list at an examination is drawn up in alphabetical order; the num-
ber ofnamesbeing”. Ifa name be selected at random, find the chance that the
candidate shall not be more than m places from his place in the order of merit.
2m+1 m(m+1)
Ang SS ae
nv
chance after the selection has been made : this may easily be found.)
(N.B.—This is not, of course, the value of the
8. A traveller starts from a point on a straight river and travels a certain
distance in a random direction. Having quite lost his way, he starts again at
random the next morning, and travels the same distance as before. Find the
chance of his reaching the river again in the second day’s journey.
Ans. -.
4
. Two lengths, 3, 8’, are laid down at random in a line a, greater than
either: find the chance that they shall not have a common Oe erene snen ¢.
; a—b-b'+e)
Ans. ee
[25 a]
388 Examples.
1o. A person in firing 10 shots at a mark has hit 5 times, and missed 5. Find
the chance that in the next 10 shots he shall hit 5 times, and miss 5.
rae = 756 . Ifthe first 10 shots had not been fired, so that
19.17.13 4199 ade I
nothing was known as to his skill, the chance would be sae if he
had been found to hit the mark half the number of times out of a
large number, the chance would be 3.
11. lf a line 7 be divided at random into 4 parts, the mean square of one
of the parts is = ?: but if the line be divided at random into 2 parts, and
each part again divided into 2 parts, then the mean square of one of the 4 parts
is +22,
9
12. Three points are taken at random in aline?. Find the mean distance
of the intermediate point from the middle of the line.
3
Ans. 16 l.
13. A certain city is situated on a river. The probability that a specified
inhabitant 4 lives on the right bank of the river is, of course, 3, in the absence
of any further information. But if we have found that an inhabitant B lives on
the right bank, find the probability that 4 does so also.
2 :
Ans. —. (N.B.—It is here assumed that every possible partition of the
number of inhabitants into 2 parts, by the river, is equally probable
a@ priori.)
14." If 4, B, CO, D, are four given points ix directum, and 2 points are taken
at random in 4D, andione is taken in BC: find the chance that it shall fall
between the former two.
I I
Aves ps2 BO(AB+ CD) + 24B.0D| ;
15. Ifz=x-+y, where x may have any value from 0 to a, and y any value
ak to 6: find the probability that zis less than an assigned value; (suppose
e>b, poe
pee
It =
G3) te>a, ps =
If we denote the functions expressing the probability in the three
a by fi (a, 4, ¢), fo(a, 2, ¢), fa (a, 6, c), we shall find the re-
ation
Si (4, b, e) + fa (4, 4, ¢) =fa(a, , 0) +fo(b, a, ¢).
Examples. 389
16. In the cubic equation
a+ put q=o,
p and g may have any values between the limits + 1. Find the chance that the
three roots are real.
2 iS
Ans. — ,
ns 45 J 3
17. Two observations are taken of the same magnitude, and the mean of the
results is taken as the true value. If the error of each observation is assumed
to lie within the limits + a, and all its values to be equally probable, show that
it is an even chance that the error in the result lies between the limits + 0.293 a.
18. A point is taken at random in each of two given plane areas. Show
that the mean square of the distance between the two points is
+24 A;
where A is the distance between the centres of gravity of the areas; and 4, k’
are the radii of gyration of each area round its centre of gravity.
19. Show that the mean square of the area of the triangle formed by joining
any three points taken in any given plane area is 3 h? k®; where h, & are the
radii of gyration of the area round the two principal axes of rotation in its plane.
If one of the points is fixed at the centre of gravity, the value is 3A? 4°.
(Mr. Woornovsz.)
20. A line is divided at random into 3 parts. Find the chance—(1) that they
will form a triangle ; (2) an acute angled-triangle.
Ans. (1). pr =}
(2). po = 3 log 2-2.
21. A line is divided into » parts. Find the chance that they cannot form a
polygon.
Ans. Sant
22. If two stars are taken at random in the northern hemisphere, find the
chance that their distance exceeds 90°. :
Ans. -.
wT
23. The vertices of a spherical triangle are points taken at random on a
sphere. Find'the chance—(1) that all its angles are acute ; (2) that all are obtuse.
I BAT
I
Ans. (1). an & (2). a=
I . ‘ :
24. Show that the mean value of -, where p is the distance of two points
: . 16
taken at random within a circle, is a
390 Examples.
2s. Two equal lines of length a include an angle @: find the chance that if
two points P, Q are taken at random, one on each line, their distance PQ shall
be less than a.
wT w 30-7
2 7. yp =? + 2 c08 0.
Ans. (1). Mica te oe Sang
T wr—@
(2). When 6>-3 Pao
Here the functions are connected by the relation F(0) +F (w — 8) =f (0) +f (w— 8).
26. The density of a city population varies inversely as the distance from a
central point. Find the chance that two inhabitants chosen at random within a
radius 7 from the centre shall not live further than a distance 7 from each other.
T
It 2 V3\ , § ode 3 6d0 |
are a (:-3) oleae ane
1a
wl
whence p = 0.7771. This result is easily obtained by employing
the values given in Question 25.
27. Four points are taken at random within a circle or an ellipse. Show
that the chance that they form a re-entrant quadrilateral is 2.
6
28. Find the mean distance of two points within a sphere. Azs. . r.
29. Three points 4, B, C are taken within a circle, whose centreisO. Find
the chance that the quadrilateral 4BCO is re-entrant.
r.4
Ans. Z + at”
30. Find the chance that the distance of two points within a square shall
not exceed a side of the square.
Ans, p=% — 2.
31. In the same case, find the chance that the distance shall not exceed an
assigned value ¢ ; a being the side of the square.
Ans. (1). When ea; p=4—sin-l-—@#— 42 — yVert-a2t-2 ——-—
2) aC ¢ es a a 2a4 3
32. Three points are taken at random on a sphere ; show that the chance that
in the spherical triangle some one angle shall exceed the sum of the other two
- 1 4 . ee
is >. Also the chance that its area shall exceed that of a great circle is e
33- Ifa line be divided at random into 4 parts, show that it is an even chance
that one of the parts is greater than half the line.
Examples. 391
34. The mean distance of a point within a triangle from the vertex C is
T(a+b (a—b)(a®— 3%) 2 at+b+e
(ee) eww Doers)
3 2 2¢ ¢ atb-e
where / is the altitude of the triangle. (See Ex. 6, Art. 242.)
_ 35. The mean value of the distance between any two points in an equilateral
triangle is
M
3 (: I )
=a(-+-lo :
oe ae
This question may be solved by proving that If = 3 Mo, where Mp is the
mean distance of an angle of the triangle from any point within it. For, let
My = na}, where p» is constant, and A = area of the triangle. Take now any
element dS of the triangle, draw from it parallels to the sides to meet the base ;
let & be the area of the equilateral triangle so formed: the sum of the whole
number of cases will be equal to
6S fd. ud? . dS = Ma,
if dS is made to range over the whole triangle: if we call the whole triangle
unity, and put dS = 2dadp as in Ex. 3, Art. 245, 5 = a?, and the integral
becomes Toh= M. The result then follows from (34).
36. From a tower of height 4, particles are projected in all directions in
space, with a velocity due to a fall through a height A. Show that the mean
value of the range is
eset
er | fin# de. (Prof. Wolstenholme.)
0
. each of which takes indepen-
37. If there be x quantities a, d,¢,d...
... &c., (the number
dently a given series of values a, @2, 3)... « 51, ba, bs, .
of values is different for each), if we put
Ba=atb+e+dt+....,
and for shortness denote ‘‘the mean value of x’? by Mx, prove that
Ms2= Ma+ Mb+Me+....=3 Ma,
M (3a)? = 3 (Ma)? — 3 (Ma) + 3M (a).
38. Two points are taken at random in a triangle; find the mean area of the
triangular portion that the line joining them cuts off from the whole triangle.
Ans, ; of the whole triangle.
( 392 )
CHAPTER XIII.
ON FOURIER’S THEOREM.
256, Expansion in Trigonometrical Series.—In many
physical investigations it is of importance to express a
function f(x) in a series of sines and cosines of multiples of
z. We propose to investigate the form of such expression,
and the conditions under which it is possible.
Let us commence by assuming that f(x), between the
limits + w and - z, is capable of being represented by a
series of the required form: thus suppose
Ff () =do+ G COS H+ A,COS 27 4+...4+ A, COSNL +...
+b, sing+6,sin2¢+...+0,sinne+... (1)
Here, since this relation is supposed to hold for all values
of w between + 7, we get, as in Art. 33, on multiplying by
cos nz and integrating,
An = “| J (2) cos nadx -2 |" 40) cos nv dv. (2)
Also On = ie J (v) sin no do,
and A= =|" S(v) do
Substituting in (1) it becomes
fe)= 2" Fv) dot — a2, cos ne |” cos nv f(v) dv
+> ye sin nal sin nv f(v) dv
n=l 7
= =|" 70) dv+2 [: cosn (v-2x) f(v) dv. (3)
7 wT
Fourier’s Theorem. 393
_ It should be noticed that when /(x) is an even function of
x its development in general consists only of cosines ; if ¢(z)
be an odd function, its development contains sines only.
257. We proceed to give an ‘a posteriori verification of
equation (3), and to examine the conditions under which it
holds good.
The right-hand side of this equation may be written
*|70) dv ($+ cos 0+ cos 20+... +008 20 + &e.),
T jer
where 6 =v-«.
But, by Trigonometry, we readily get
‘ 1
$+ cosO+ cos 20+...+c08 6 Spay .
2 sin $0
Hence, to verify (3), it remains to prove that
fle) (1 = h?) f (0) dO
Oa cos (0 - 3) ra (13)
Thus, when #3 is outside the limits,
3[ 70) a0+ 5" ['70) cos (6-) dd=0. (14)
In particular we have
{7 (0) 40+" [7o) cosn(0+f)d0=0, (15)
i
v0
where {3 is positive.
Again, if 8 ties between the limits of the integral in (12),
we need only consider the portion of the integral arising from
values of @ which are indefinitely near to 6. Accordingly,
if f(6) be continuous,
=|. (1-h?) f (0) dd: -/(B) on (1 —h?) dO
o 1-2hc0s (0-8) +h? | a 1-2hc0s (0-3) +h"
h=l
Fourier’s Theorem. 397
Again, whatever be the value of h,
i: (1-H) dO | fe (1-H) de
o 1-2hc0s (0-3)+h? Jig 1-2hcosst i?
a (1h) dz
> 1—2heose+h® 7" (Art. 18).
Hence
lim.f?™ (1 -h*) f(0)d0
ah i 2hoos (0-By aw 277 8). (16)
Consequently by (12),
F(8) = [710 40+ 5 ["A0) c08n(6-) dB (17)
This is usually called Fourier’s theorem.
Also, by aid of (15),
md (B) = =e sin nls (0) ain x6 dO.
259. We shall next investigate the limit when a = 0 of
the integral
‘a (b
| | ¢(t) cos ux cos ut du dt
0 Ja
ll
ol
sin a (a - Zt) (> sin a (x + #)
ie =e o(t) dt + eae ea ¢(2) dt
0 at B42 gi
af = o(e +2) as+3| == g(e—) de.
a a4
Now, by (7), the latter integral vanishes when a = © and vis
positive; and by (10), when a lies between a and 4, the former
integral becomes 7 (2).
Also, when 2 does not lie between a and 4, the former in-
tegral vanishes, and we have
ob
| | g(t} cos ux cos ut du dt =o. (18)
0
a
398 Fourier’s Theorem.
When « lies between a and 8,
{ if p(t) cos ux cos ut du dt = < ¢ (#).
Hence, if be positive, we have
j ko cos ux cos ut du dt = ; ¢ (2).
Likewise it is easily seen that
| I (t) sin ue sin utdudt=~ 9 (c),
when 2 is positive.
We readily see that
| [ (4) cos wx cos utdudt
0 =
eo
Daw
= | | o(t) sin ux sin ut dudt =m (a).
Also 2p (x) -[ i ¢(t) cos u (¢- a) du dt,
the form in which the theorem was originally
Fourier.
(19)
(20)
(21)
(22)
(23)
given by
Examples. 399
EXampPtes.
1, When has any value between / and — 7, prove that
if T Gre (? nm (v— 2)
9 (2) = tls $ (») dv + > >. {_,00) cos —T—* dv.
2. For all values of z between 0 and /, prove that
1 neo pl
af o(0) +> \ (v) cos fh ye
0 ner 0° Z
3. For all values of z between 0 and /, prove that
2qar? | nx (i nny
(2) = 7 a sin —- \, ¢(v) sin oF dv.
4. Prove that, for all values of « between +m and - 7,
da =sing—fsin 27+4s8in3¢—tsinqga+...
5. For all values of « between 7 and — = prove that
2 2
pe # (ane 34 a sin 54 he: )
9 25
6. Prove that
metwz—eoe sing 2s8in2¢ 3s8in3r
2em—Ee%m gt+t +4 a@+9g
Here (Art. 21),
Mm COS 111
a +m”
a .
| (ea® — e-at) sin nade =— (eam — ear)
7
7. Find a function of # which has the value ¢ when # lies between 0 and a,
and the value zero when lies between a and JU.
a mux 1 , 27a 2x
Ans. ol)= 5 += (sin cos 7 + 5 sin ~~ eos > ~
+ Ein 5 cos 34...)
t “Tv ‘a. amv cl na
Here i, cos —- o(v) dv =e| cos 5 dv = ap iD
0
400 Examples.
‘3 t
8. Find a function of « which is equal to 4x when « lies between 0 and 2
z
and is & (2? — x) when « lies between = and 2.
kl 8ki ft
os 27” I cee ys I cos ST 4 &e.)
nS. 7Z = ae ex 8 >— + & 00s 7 To? a .
Here
z
Q 2 1 nTrV
| (0) cos“ du= [* kv co" dv+ |, (2-0) cos do.
0 “0
2
72
This = — = when » is of the form 4m + 2; and is zero for other values
of n. me
g. If o(x) = - when z lies between 0 and a, and ¢ (x)= . when = lies
between a and r—a, and $(x) = a when « varies from 7 — ato 7, prove that
I
9 (2) ==
Tt ‘ Tse ,
(sin sin a + — sin 3z sin 3a +— sin $4 sinSat+.. ‘) .
a\ D) 25
10. When z lies between + 7, prove the relations
, 2. sing 2sn2¢ 3sin 3x
sin mz = — sin mr - =a ee
7 I-—m? 2%-—m? 3%—m?
Bi I mCosx mcos2% mcos 3a
cos ma =~ sin mm {— + ——, —-—7—_- + ——— _.,..}-
7 m lm 2% — m' 37 — m?
11, Hence prove the relation
‘ 2, 2u 4 2u &
cot u =-+— 5—, + =—_ +t...
uo wom wt — qn?
12, Find a function which shall be unity for all values of # between + 1,
and zero for all other values of x.
F(#)= { dp [ F(&) cos pé cos wx dt = an dp ie cos wé cos wardt
0
2° cospzx sin uw
= - |) ee im.
TSO B&
This result can be verified independently.
13. Find a function which shall be equal to cos x for all values of x between
0 and zm, and to — cos for values between — 7 and o.
Here we easily find
" J (%) cosnzdz =0,
-7T
and we get
cosx = 4 i sin 2 +
wtl.3
sin 4% + 2 sin 6% + }
go ge aes
( 401 )
CHAPTER XIV.
ON LINE AND SURFACE INTEGRALS.
260. We have already considered in Arts. 226 and 227 a
general theorem, commonly called Green’s theorem, that
connects volume with surface integration. We now propose
to consider the analogous theorem that connects integration
taken over a portion of any surface with integration along
the curve or curves that bound that portion of the surface—
that is, which connect what are styled susface integrals with
line integrals. All these relations can be shown to be based
on the following elementary theorem concerning integration
in a plane.
261. Integration over a Plane Area.—If we suppose
P and Q to be real functions of x and y, that are finite and
continuous for all points within a certain plane region, then
we shall have
{| (2 = =) dady = | (Pde + Qdy), (1)
where the double integral is taken for all points within the
region, and the single integral is taken round the boundary
of the region. This can be immediately deduced from the
theorem given in Art. 226, by supposing the volume in
question to be a portion of a cylinder intercepted between
two planes drawn perpendicular to its edges, the edges being
supposed parallel to the axis of s.
The theorem can be also proved independently as
follows :—
Taking the positive directions of the axes as in the
accompanying figure, we define the positive direction along
the boundary to be that for which the bounded surface is on
the right hand; then, if we have to exclude any portion,
e.g. a space without the outer boundary in the accompanying
[26]
402 On Line and Surface Integrals.
figure, the positive direction of the boundary is that indicated
by the arrows in the figure. In all the simple integrals the
integration must be effected in the positive direction thus
defined.
‘ 2 uQ :
First, to integrate le dadywith |y
respect to 2, let us divide the
region into elements by parallels
to the axis of z. Select any one
of these parallels, and, reading
from left to right, denote the
values of Q, where the line crosses
the boundary at its entrances into
No
Ss MY
es an
y
the region, by Q,, Q,, &c., and at [O Xx
its exits by Q’, Q”, &e. ; Fig. 78.
dQ ; ”
then {> dt=-Q, + Q- Q.+ Q" &e.,
and accordingly
dQ 4 "r
(JF aay =—| cs | ty - | y+ [Ody ~ be
Now in each of these integrals y passes through all its values
from the least to the greatest, therefore dy is always to be
taken positively. Again, observing that in the figure the
directions for the outer and inner boundaries must be taken
as opposite, and denoting by dy, dy2, &c., and by dy’, dy”,
&e., the projections on the axis of y of the arcs of the
boundary cut off by the consecutive parallels as above,
we have
dy =— dy, =- dy,=+ dy’ =+ dy” = &e.;
thus
[ess fae anr fastest fa
where the integral is taken along the entire boundary in the
positive direction.
In like manner, dividing the region into elements
parallel to the axis of y, and denoting the values of P at the
Stokes’ Theorem. 403
entrances, proceeding from below upwards, by P,, P., &.,
and at the exits by P’, P”, &c., we have
(z dady = - | Pade + | Par - | Pye + &e.,
where dz is positive. Hence, as before, taking account of
the positive direction of the boundary, we have
dx =+ day = + dit, = - dx’ =— du”, &e.,
and consequently
(Fe dady = ~| Pade -| Pde’ ~| Pate, — &. = -|Pae,
the integral being taken in the positive direction along the
entire boundary. Accordingly, we have
(ae 2 ely =| (Pac + Qdy)
I
de dy
-|(2%+ am) ee (2)
taken around the entire boundary.
We have assumed that there were no points within the
region at which P or Q are discontinuous. If there were
such, we should have to surround them with closed lines, as
small as we please, and thus exclude them, by introducing
these lines as parts of the boundary of the region.
262. Stokes’ Theorem.*—Suppose u, v, w to be con-
tinuous functions of 2, y, 2, the coordinates of a point; and
let dS be any element of a surface, and /, m, n the direction
cosines of its outward drawn normal, then we shall have
dw dv dues dw dv du
r= {as (z = 7)*™ (ae - z) + (- 3)
dy dz
d.
= [wae + ody + wads) = {(w + oF. +05) ds, (3)
* This theorem was given by Professor Stokes at the Smith Prize Examination
for 1854, and is of extensive application, both in the theory of ‘‘vortex motion,”’
as also in electricity and magnetism.
[26 a]
404 On Line and Surface Integrals.
the former, or surface integral, being taken over any portion
of the surface S, and the latter, or Jine-infegral, round the
boundary, or boundaries of the surface.
Here, from the equation of the surface, we may regard z
as being a function of # and y at all points on the surface;
hence, if p = a. g= 3 we have, by elementary geometry,
I —P ~Y
Tape Sipe "Jape |
and dS =./1+p'+¢ dudy (see Art. 224).
Hence we get
dv du dv dw dw du
r-(( ~ ay TD ye “Pa, Ta ~ 9 Fe dey. (5)
Again, if the total differential of » with respect to x be
represented by is (v), we have:
d dv dv win a du — du
an) Soe ea and likewise a = a qo3
also,
Va ~? ay 1 ag Pg @)- GO) ~ ge:
i dp _ ay
since oa
Hence (5) becomes
r-{{{- (v + wg) - = (w+ up) dedy,
Accordingly, from (1), we get
r-({\z (v + wg) - 5 ( + wp)| dae dy
Z | (wae + ody + wds), (6)
since dz = pdx + qdy.
This establishes the theorem in question.
Example in Solid Harmonies. 405
_263. Example in Solid Marmonies.-—Let V be a
solid harmonic (Art. 230), i.e. let it satisfy the equation
CV. ay a
dc * ap * ag = (7)
then with the same notation we shall have
d* ad d\ dV aV aV
where the integrals are, as before, respectively, taken over
any portion of a surface, and along its boundary.
For, substituting
eee f Cy
ae Pe dy? or aw ’
the double integral becomes
{| jo ey EU gt OV og
dadx ”” dady ” de® ne)
{ Fg AT BV OT 82
-\ll gee 1 sdy * de a .
Wel) +a) aw
ave dV -| adVde dVdy
“(7 aa ~ dy) = ra = He Be ds.
264. Lemma on Solid Angles.—If dw represents the
elementary solid angle subtended at any external point
by an element dS situated at a point P on a surface; then,
‘sin Art. 193, we deduce immediately the expression
q
cos y dS
= mo ’
dw
where r= PP’, and y is the angle made by r with the
normal to the surface.
406 On Line and Surface Integrals.
Again, it is obvious that the direction cosines of PP’
are, respectively,
and it follows that
I(a-—a’)+m(y-y/)+n (s- 2’)
cos y =
Y r
Hence
L(a-x)+m(y-y)+n (8-21) 1g
dw = 3
Again, since
P=(e—2')? + (y-y)*+ (8-8),
d(i\_a-«@ d(1\_y-y a(x _e 8. (0)
(2) xg lt) a ale)a ae :
1-20.40). 26)-e6
Hence we get
d d d\/t
dy = - 48 (14 +m 24 nZ\(2)
i 8 tS ae)
Consequently, if Q be the solid angle subtended at P’
by the boundary of any portion of a surface, we have
d ad d\/t
o=(tw=-[fas(eF +m + nz)
ad d a\ [1
where the integral is taken for each element of the surface
within the boundary.
5 (12)
Differential Equations relative to Solid Angles. 407
265. Differential Equations relative to Solid
Angles.—If we differentiate (12) with respect to x’, we get,
since /, m, are independent of the coordinates of P,
ay = [lesz da’ * ma+ a) (;)
= {jes(eZrme +n 5) 2 (2)
(aZQ)-0£0) os
by (8), since - satisfies equation (7).
Again, by (10),
jon ()--[oal)--@)/t
and likewise
[ax a ft\__(a\(&
dy\r)— \dy [>
d fi d (dz
and [as a(t)--wle
in which the integrals
de, (ay,
| r? Jr? Jr
are supposed taken round the entire circuit that bounds the
solid angle.
If we now put
F-{®, e-(% a-[%, (14)
r r r
and suppose F, G', H to denote /ine-integrals taken round the
408 On Line and Surface Integrals.
circuit, or boundary, we get, from (13), and the analogous
equations,
dq dH dG
de dy ds’
dQ _aF aH \. (15)
dy dx dx
dQ adG dF
moe
266. Neumann’s Theorem.—If we suppose the point
ays’ to be taken on any surface S’, and if dS’ be the element
of the surface at the point; also, if J’m’n’ be the direction
cosines of the normal at P’, we get
(Be am dQ +e) ds’
da’ dy’
dH dG dF dH dG dF .
“WG - Be 5 7) (F- 7 dy wh aS
Hence, by (3), we have
d d d
-{ wack
fase + w wt" H)2
(ln 2 oY 2 a ®\ ae
-\(7 abs He de, (16)
where the former integral is taken
over any portion of 8’, and the latter
round the boundary.
If now we substitute for F, G, H, Fig. 79.
and Q their values as given in (12) and (14), the preceding
equation becomes
od d ,d\fd a@ a\fx
Jasas (ro +m Daw Z)(i etma tne |()
= dada dydy’ dsdsz <
“Ee ye Eo) aed, (17)
Neumann’s Theorem. 409
where the former integral is taken for all elements on both
the surface regions, and the latter along the boundaries of
the regions.
The latter integral may be written
je dsds’,
where ¢< is the angle between the directions of the tangents
to ds and ds’.
The foregoing, when interpreted in the theory of
magnetism, leads to Neumann’s theorem connecting the
energy of electric currents with that of magnetic shells.
See Clerk Maxwell’s “ Electricity and Magnetism,” vol. 2,
§ 637.
We shall conclude with the consideration of two or three
surface integrals taken over
a sphere, which are of great 2
importance in the theory of
attraction.
267. We commence with O
the determination of the
single integral | ee where
p
dS is an element of a
spherical surface at'any point P, and p is othe. distance of P
from any fixed point O.
Let LPCO=90, CP =a, CO=/;
then we have ;
pr=a’+f? - 2af cos 8;
therefore
pde = af sin dda. (18)
Again, as in Art. 230, we may write
dS = a’ sin 0d0 do,
where ¢ is the angle that the plane OCP makes with a fixed
plane passing through OC.
410 On Line and Surface Integrals.
Accordingly
{> =a [[™ 0d0 dp
p p
Again, since p is independent of ¢,
3 (= Odd dps | sin 9d0_ | dp, by (17),
p , f
Hence, (1°), when 0 is inside the sphere, since the limits for
pare a+f and a—f, we have
[2=4na, (19)
p
or the expression is constant in this case.
(2°) When the point is outside the sphere, we get
2
| a8 _4me | surface of sphere : (20)
p Ff co’
These show that a homogeneous sphere attracts an external
mass as if the whole mass of the sphere was concentrated at its
centre. Also, that a homogeneous spherical shell exerts no
attraction at an internal point (Williamson and Tarleton’s
Dynamics, Art. 126).
More generally we have
as at dp
or Fla
Hence, when 0 is inside the surface,
aS an a I ) “ae oe ;
SE \@+F) -(a-f) |, (21)
and, when O is outside,
| dS e2ra I
p” n—2f (Pay (assy (sah
Neumann’s Theorem. 411
Also, if ¢(o) be any function of p, we have
Y a a
[8 610) 20 2)” 9 (0), (22)
It may be observed that we can take the point from which
pe is measured either inside or outside the surface according
to pleasure; for if O and O’ are inverse points, 7.¢. if CO:
CB =CB: CO’, the triangles CO’P and CPO are similar,
and therefore PO: PO’=CO:CP. Consequently the ratio
PO: PO’is the same for all points on the sphere.
268. More generally, to find the integral
dS
V-| opzoR,
taken over the surface of a sphere,
where P, is a point on the surface,
and O and O, are any two fixed
points.
We may take one of these
points inside, and the other out-
side the surface, since, as shown
B
above, the ratio of the distances of
two inverse points from any point
on the surface of the sphere is
constant. 7
Produce OP, to meet the surface =
again in P,, and take O, on 00, 2
such that 00; . OO, = OP,. OP, Fie. 81
=f? ~a*, where f= CO, as before. a
Now, let OP, =r, O.P, =p, OP, =, P20.=p,; and
let dS’ be the indefinitely small element at P, intercepted by
the cone whose vertex is at O, and which passes through the
element dS; then, it is immediately seen that we have
a (23)
= ’
ry?
y-(S (23 is.
=| lino fx a np
therefore
412 On Line and Surface Integrals.
Again, since the triangles P,0,0, and O,P,0 are equi-
angular, we have
OP,: P,0,= 00,: O,P: 5;
therefore
pni= OO0,sp.. (24)
I 1 (dS’
Ve ag 66:
Accordingly
Hence, since O, is inside the sphere, we see by (19), that
47a
“F=@ 00; (25)
The preceding is a modification of Sir W. Thomson’s proof
of his well-known theorem on the distribution of electricity
on spherical conductors (Cambridge and Dublin Mathematical
Journal, 1848; also Thomson and Tait’s Natural Philosophy,
vol. 11., § 474).
More generally still, if m+n = 4, we see by the preceding
that, taking O inside,
r-| d8 =| ads’
vm p” ym ry? p”
_ @- fry fds’
- ae [om (26)
This can be immediately expressed by aid of (21).
( 413 )
CHAPTER XV.
CALCULUS OF VARIATIONS.
269. In this Chapter it is proposed to give a short account
of the elementary principles of the Calculus of Variations,
especially in connexion with the theory of maximum and
minimum integrals.
The origin of the Calculus of Variations may be traced
to John Bernoulli’s celebrated problem, published in the
Acta Eruditorum of Leipsic, a.p. 1696, under the following
form, Datis in plano verticali duobus punctis A et B, assignare
mobili M viam AMB per quam gravitate sua descendens, et
movert incipiens a puncto A, brevissimo tempore perveniat ad
punctum B. This problem introduced considerations en-
tirely different from those hitherto involved in the discussion
of curves, for in its treatment it is necessary to conceive a
curve as changing its form in a continuous manner, that is,
as undergoing what is styled deformation. This change of
form can be treated analytically as follows :—Suppose y = f(x)
to represent the equation of a curve, and let us write
y =f (#) + ab(2), (1)
where a is an infinitesimal quantity, and (x) any function of
x, subject only to the condition of being finite for all values
of x within the limits of the problem. Then, equation (1)
represents a new curve indefinitely close to the curve y=/(z) ;
and by varying the form of (7) we may regard (1) as
representative of any curve indefinitely near to the original.
270. Definition of Variation dy.—Here, ay(z) is the
difference between the y ordinates of the two curves for the
same value of «. ‘This indefinitely small difference is called
the variation of y, and is denoted by dy.
414 Calculus of Variations.
If the ordinate of the second curve be denoted by ym, we
may write
dy =~ -y = ap(z). (2)
Then dy may be regarded as the change in y arising solely
from a change in the relation that connects y with 2 while x
remains unaltered.
More generally, if « be any function of # and y, we may
write
du=w — 4, (3)
where w, is the value that « assumes when y becomes y + dy.
Again, when y becomes y + dy ey becomes at + arey
gain, » oan ie ae
Hence we see that
dty a
8 aa a (dy). (4)
This equation may also be written in the form
oD"y = Dby, (5)
where D stands for the symbol of differentiation -
More generally, if y be a function of any number of
independent variables 2, 72,...,, then dy represents any
indefinitely small change in y arising solely from a change in
the form of the function while , m, &c., are unchanged.
Thus the variable y may receive two essentially distinct kinds
of increment—one arising from a change in one or more of
the variables, the other arising solely from a change in the
relation which connects y with these variables. The former
increments are those contemplated and treated of in the
ordinary calculus; the latter are those principally considered
in the calculus of variations.
We shall follow Strauch, Jellett, Moigno, and the
principal modern writers on the subject, by restricting, in
general, the symbol 6 to the latter species of increment.
Total Variations. 415.
271. Total Variations.—In many cases, especially for
the limiting values of the variables, we have to take account
of both kinds of increment.
_ Thus, if y =f(m, a, 23), and if Ay denote the total
increment of y, we have
= dy dy dy
Ay= by + a An Sg eT Azs, (6)
where Az,, Av, Az; denote indefinitely small increments in
the variables x,, x, 7, respectively.
In the case of a single independent variable this gives
du
Au = 8u+ Fm Ag, (7)
272. Wariation of a Function.—We shall adopt
2
Newton’s notation and write ¥ for - y for os
ad” ‘ a
a and proceed to consider the variation of the general
... y™ for
expression V = f (2, Y5 Y% ¥, ...y™), in which the form of the
Junction f is given, while that of y in terms of x is indeterminate.
Here, considering w as unchanged, we have
dV dV aV
es Se . ee (2),
OV: a Ytay y+... + ayo 8Y
Now let
dV Pp dV dV P,, (3)
then we have
SV = Pdy + P,Ddy + P,D'sy+...+P,D"%y. (9)
It may be observed that in all cases in finding the
variation of a function we neglect terms of the second and
higher degrees in the increments. ‘
416 Calculus of Variations.
273. Wariation of a Definite Integral.—We shall
next consider the variation of the definite integral
7-{" Var, (10)
where V is of the form stated in the preceding Article and
does not contain either of the limits, %, 2.
Here, when the limits are unchanged, we evidently have
su=[" ares. (11)
%
And, when the limits undergo variation,
7
av-(" &Vdx “+ ViAa, =; Vi Axo. (12)
0
If now we suppose that the expression [J’]} represents the
difference between the limiting values of any function F,
i. e. the difference between its values when we substitute
=, Y=Y1, and =A, Y =Yo, respectively, we may write the
last equation in the form
au-|"3Pdes [Waa]. (13)
We shall suppose in general that y, 4, #, &c. are con-
tinuous and finite, and that dy, dy, &c. are indefinitely
small, between the limits of integration.
Again, as any relation between 2 and y can be represented
by a curve, we can always give a geometrical meaning to
the definite integral U, and we may speak of the limiting
values of z and y as the coordinates of the Himiting points.
274. Case of V= f(x, y, y).— We proceed to transform
(12), commencing with the case in which V is a function of
2, y, and y solely.
Here,
av~|"(Pay+ Piz ay des [Packs (14)
zo 7
d dP.
but [Pig b= Pade [ay dz;
Variation of a Definite Integral. 417
therefore
AU= { ‘( - =) oyda + | Paby + Vax) :
Again, if Ay, and Ay be the total variations of the
limiting coordinates, y, and y., we have, by (7),
Ayn= OY +7 An, AY = oYo + GYoAd. (15)
Accordingly, substituting in the above, it becomes
P
AU= [, (2 -=) Sigs te Py) Ae Pray |. (16)
0
Next, if we suppose the limiting point (m, y:) to be
restricted to lie on a fixed curve (y=/1(%) suppose) ; we shall
have
An =fi' (a) Ax).
If the other limiting point be likewise restricted to lie
on the curve y=/(z), we shall have
AYo = fo (t») Ato.
In such cases equation (16) becomes
au-|"(P - =) Syd + | V+ Pf’ (0)- a} ac] .
0
275. Case of V=/ (2, y, y, /).—lf V be a function of
2, Y, Y, and ¥, we have
Ge [(eey + P, Diy + P, D’sy) de + [ Vas |
% 0
Hence, since
@P, d
P,Diby - 8y = 4 (P, Diy - by
we have
[-P.D'sy -|
X Zs
a)
ds
_
da + | P.Day ~ dy
[a7]
=|:
ds
418 Calculus of Variations.
Hence,
A v-("(P = os + =) oy da + | 74* |,
+ fey (P.- Z)+ Psi (18)
If we now suppose
P- = + oe (P), P,- nt (P,);
we may write our equation in the following simple form :—
‘x T
av-| : (P) dyde + {Vac + (P,) dy + P.dy,- (19)
Zo 0
Again, as before, we have
oy =An- pany, oy = An -YAXy,
OYo = AY ae YoAXoy oY pt AY — YoAXo.
Substituting in (19), it becomes
AU= Le Syda+ [( V-(P,)y -Pi) as|
1
0
+ [ (Bray + Pasi |: (20)
It is often convenient to write this in the abbreviated form
AUah, ~In+ |." Mbyte, (21)
where ae
M=P- “ + ae
and LZ, and Z, represent the terms at the upper and lower
limits, respectively.
276. Maxima and Minima. — The most important
applications of the Calculus of Variations have reference to
Maxima and Minima. 419
the determination of the form of one or more unknown func-
tions contained in a definite integral, in such a manner
that the integral shall have a maximum or minimum value.
For instance, to determine the form of the function y
. . 71 . . .
which renders the integral v-| Vde a maximum or a mini-
zo
mum.
Here, when we substitute y+aw instead of y in V, where
a is an infinitesimal, and also vary the limits, let us suppose
that U becomes
2
a
U0+a0,4+ = U,4+ &e.
I.
Then, as in the Differential Calculus, if U be a maximum
or a minimum, the expression
a
oi U,+ &e.
must have the same sign for all variations that are con-
sistent with the conditions of the problem.
Now, since dy, or aif, is, in general, restricted solely by the
condition that it should be very small, we see that we can
generally change the sign ofa without violating the conditions
of the problem. Hence, as in the Differential Calculus, Art.
138, we see that U cannot be either a maximum or a
minimum unless
UT, =0.
Again, for a maximum JU, must be negative for all values
of dy that are compatible with the conditions of the
problem, and for a minimum J, must be positive for all
values of dy under similar conditions.
In many cases we can see from the nature of the problem
that it necessarily admits of a maximum, or of a minimum,
value ; in such cases when we have obtained the solution by
aid of equation U,=0, we may dispense with the labour of
investigating the second condition.
[27 a |
420 Calculus of Variations.
Again, it is easily seen that
A U= a 0,
and accordingly for a maximum or a minimum value of U
we must have AU'=o0, or
Iy~In+ |" Mdyde= o. (22)
%
Now, without restricting the value of dy, this equation
cannot be satisfied unless we have
LT, - In=0, and | ‘ USydx =0.
%
For if Z, — Z, be not zero, we must have
al ve) Mae=[/" aayde= Lo ~ Ly
0 %
This would require that the integral of an arbitrary function
can be expressed in terms of the limiting values of the
variables solely, but this is manifestly in general impossible ;
we have therefore
mu
D,- In =0, and | 'MSy de =0. (23)
%
Again, since the value of any definite integral depends
on the form of the function to be integrated, it is plainly
impossible in general to determine the value of the definite
integral in (23) without fixing the form of the function
represented by oy.
Accordingly, since the form of éy is by hypothesis
perfectly arbitrary, we infer that in general it is impossible
to satisfy the equation
i Moy dx =o,
0
unless by making
H=0. (24)
Case of Geometrical Restrictions. 421
In any particular case the form of y in terms of 2, i.e.
the equation of the curve, is determined by the integration of
the differential equation M=o: and also the arbitrary con-
stants introduced in this integration are in general determined.
by aid of the equation Z;-L,=0. Again, this equation
cannot be always zero unless the coefficient of each of the
independent variations be separately zero.
It can be shown that the equations thus obtained are in
general sufficient for the determination of the above-
mentioned arbitrary constants; this will appear more fully
when we come to discuss our applications.
Whenever the solution thus arrived at does not satisfy
the criterion respecting the function U2, such solution is not
either a maximum or a minimum, and is called a stationary
solution.
277. Case of Geometrical Restrictions.— We have
here supposed that there is no restriction on dy, so that for
any value of x the increments + dy and - dy are equally
compatible with the conditions of the problem. The
reasoning consequently will not apply if the conditions
render this impossible. For instance, if a curve be restricted
to lie within a given boundary, then for all points on the
boundary the displacements must be inwards, and the
opposite displacements are impossible. In this case it is
easily seen that the curve satisfying a required maximum
or minimum condition consists partly of portions of the
boundary and partly of portions of a curve satisfying the
equation M =o.
We shall now show that the integration of the equation
M =o is much simplified for particular cases of the form of
the function V.
278. V a Function of z and ¢ solely.—In this case
1 2
the equation I= 0 becomes “ =o, and accordingly we
have P, = const. = ¢. (25)
More particularly, if V be a function of solely, then
P, is a function of y solely, and we get y = const., or
=ce+¢; hence the line joining the limiting points 1s the
solution of the problem in this case.
422 Calculus of Variations.
279. V a Function of y and 7 solely.—In this case
we have
dP,
P _ ae =O.
Also
aP. ad,
“+9 Pi== (YP);
ad 5 o ‘:
ae V=yP+9Pi=9 7
accordingly we have in this case
V=c+yP,, (26)
where c is a constant.
280. V a Funetion of z, 7, and 7 solely.—Here
the equation =o becomes
aPy_ OP s_
dx da
Hence we get
dP,
(P,) = Pi- ar = const. (27)
281. V a Function of y, 7 and ¥ solely.—In this
case we have
dP, @&P.
P27 aa = (28)
1 d : 7 es
ae ap 7 9P + yPit yPs
.aP, .@P, . 2%
Uae oe TPP eRe by (28) ;
therefore
d ds. OP,
aa = ay UP) - 9 ae tyPe
Again
d/.a. a ie
€ iP, .) P, a yP».
da\9 de 4?) 9 ae
Case of two Dependent Variables. 423
Consequently we get
d 3 Qf. GPs. \.
ie" gg OP) ~ 3,(0 GS -aP)s
therefore Vuac+yPi-y ei + ¥Po;
ae
or V=ct+y (Pi) + GP. (29)
In particular if V be a function of y and # solely, this
becomes, by (27),
Vac+cy+ GP. (30)
282. Case where JV contains the Limits.—In the
equation
v-| f Vda,
%
if V contain explicitly the limiting values of one or more
of the quantities 2, y, y, &c., the expression for AU, whenever
such limiting values are not fixed, contains terms additional
to those given in Art. 275.
For instance, suppose
V=f (a, Y y 00 My Yiy Y)s
then SV will contain additional terms arising from the
changes in z,, y, and 71; thus
d ( pee gs av av
da," dx, ayy dg
and the additional terms in AU are
"1 /AdV adVe.adV =
Sr ig ge — \d. — d
aa). (Se va dy, ~~ 7) ae x dy, -
. (1dV
+ An, fe ti dz. (31)
We shall illustrate the method of dealing with such
additional terms subsequently. :
283. Case of Two Dependent Wariables.— It is
easy to extend the preceding method to the solution of
424 Calculus of Variations.
problems of maxima and minima when there are two or
more dependent variables.
Thus let us consider the variation of the expression
v-| ; Vaz,
%
where V is a given function of 2, y, z, y, 2, ¥, #, and y, s are
both undetermined functions of z.
As before let
dV dV dV
Page SO gage Sage
and suppose
dV dV dV
Q=7) Qa a = ae
then
AU- | rax| ‘ [ (P3y + P,Ddy + P,D'8y) de
+ i (Q8s+ Q,D8s + Q,D%Ss) de.
Proceeding as in Art. 275 we readily find
xy TC
au=|. (P) dydz +| (Q) Ssduv + | 7ae|
Xo 0
- AB
+ | (Py) y+ Pady+ (Q) 854 Qs |
0
where (P), (P:) have the same meaning as in Art. 275, and
(Q), (Q:) are the corresponding expressions relative to the
variable z.
Again, as before, this expression admits of being trans-
formed into
(Q)8s dex
ay
x,
av=|" (P) ayaes |
% ( P= ipo P18 Qi) ae |
0
1
0
i | (Paay + PAy + (Q:) As+ Q.aé| (32)
1
0
LIsoperimetrical Problems. 425
This equation may be written for convenience in the
form
AU= [z), +{? (Mody + Nz) dx. (33)
284. Application to Maxima and Minima.—The
. . 7 . .
determination of y and s when | Vde is a Maximum or a
y
. . . 9
minimum leads, as in Art. 276, to the equations
M=0, N=o, (34)
along with the equation
[Z],=0 (35)
at the limits.
In the latter equation the coefficient of each independent
limiting variation must be zero as before; and the equa-
tions thus obtained enable us, in general, to determine the
arbitrary constants which appear in the solution of the
equations
P
dP, &P, dQ PQ _
ie ae a ae (36)
When VY is of particular forms it is easily seen that
results similar to those given for a single variable still hold
good.
For instance, when V does not contain # explicitly,
we see by the method adopted in Art. 281, that we shall
have
V=c+ (Pry + (Q)e+ Poy + Qi. (37)
Again, if V does not contain either 2, y, or explicitly,
this becomes, as in (30),
Vecscy+ c's + Proj t+ Q8. (38)
The foregoing results can be readily extended to the case of
three or more dependent variables.
285. Relative Maxima and Minima. Isoperi-
metrical Problems.—lIn the discussion of the curve which
possesses a maximum or a minimum property if we limit the
investigation to all curves of a given length, or that satisfy
426 Calculus of Variations.
some other condition, we get a new class of problems, called
problems of relative maxima and minima. These questions
originated in the isoperimetrical problems of James Bernoulli.
For example, let it be proposed to determine the form of
that renders U = . Vde a maximum or a minimum, and
y¥
that also satisfies the relation U’ -| V'd« = constant, where
to
Vand V’ are given functions of 2, y, 7, &e.
Here it is obvious that if Ube a maximum or a minimum
so also is U + aU’, where a is any arbitrary constant.
Accordingly the problem reduces to the determination of
the maximum or minimum value of
[2 (7447?) aes (39)
regarding a as a constant whose value is to be determined by
aid of the given value of U’.
286. Equations of Condition.— Another class of
problems closely allied with the preceding is that in which
the variables x, y,7,#, &c., are connected by a relation W=0;
in this case we may plainly write
rs ie (V+) ar, (40)
where X is any indeterminate function of wz.
A very important case of this principle arises whenever
we take the arc of the curve for the independent variable.
287. Case of Are being Independent Wariable.—
For instance, in looking for the maximum or minimum
solution of the integral
Sy
U -| uds,
59
dz
ds
the relation 4°+y?-1=0. In this case we seek for the
maximum or minimum solution of the expression
v= [lus @ere—9)| ds=|"'Vas, (41)
5
. dy... :
let — = &, = =y, and we have at each point of the curve
Case where w does not contain s Explicitly. 427
where V = » + 3A (# +9 — 1) and X is undetermined:
the coefficient is written in the form 3A for convenience.
288. Case where 1», does not Contain 5 Explicitly.—
For example, let « be a function of x and y solely; then,
since — = 3 (a+ 4-1)— = 0, we get, by equation (37),
ds
V=Pywet+ QYy+¢;3
ds
hence
ma=A+t+e.
This may be written
A=ut+a,
where a is an arbitrary constant, to be subsequently
determined.
Again, the equation [Z}{=0 at the limits becomes in thi
case
[mAs + A (ede + ydy) |} = 0,
or, since ox, = Axm—a%As,, &e.,
[(u-A)Ash+[A(@Ax + YAy) i = 03
therefore
— aA (8, - 8) + [A (@Avt yAy) ]}=0.
Hence we obtain
aA (81-8) =0, and [A (Ar +yAy)]}=0. (42)
Now, whenever the length of the curve is given the
former of these equations vanishes identically, but when the
length is undetermined we must have a = o.
Accordingly, in the latter case we have
A= pm, (43)
while in isoperimetrical problems
A=p+4, (44)
in which the arbitrary constant is to be determined from the.
given length of the curve. :
In the former oase, substituting mu for X in (41), we see
428 Calculus of Variations.
that, whenever » does not contain s explicitly, we may
write
eo
‘S.
=i a (+ y+ 1) ds. (45)
0
Hence, if u be a function solely of one of the variables,
y suppose, the maximum or minimum solution is given by
aid of the equation
Mae = C. (46)
Examples of this will be found further on.
289. In general, for a maximum or a minimum we
have, from (45),
d d, . a ad, ,
de” tthe) (47)
but
du. du dy
ds da” tay dy
Hence we get
Pe . du d
pena (g Ea a by = é(sf-9F) (48)
Again, let @ be the angle that ds, the element of the
curve, makes with the positive direction of the axis of x, and
we have
&=Cos p, Y=sin gd;
therefore
dp. dp
TG yas,”
Accordingly, either of the equations in (48) becomes
dp _ du du
aS « [Tein pF 005 9 |
Hence, adopting the usual convention as to the sign of p
the radius of curvature (Differential Calculus, Art. 226), we
may write the last equation in the form
i 2 du _ au
ra i(#s in ¢ Gyo 6): (49)
Case where V = ja + pat + poy. 429
Further integration of this expression is impossible without
previously fixing the form of wu.
Again, at the limits we have
[u (Ax + yAy)], = 0.
This, when the limiting points are independent, leads to the
equations
[@Ae+yAy]=0, and [#Ar+yAy]o=0, (50)
provided yu does not become zero at either limit.
When the limiting points are restricted to fixed curves,
equations (50) show that the curve for which | ‘yds isa maxi-
x
mum or a minimum must cut the limiting curves ortho-
gonally,
It may be observed that if the proposed integral had
Zo,
x,
boon | i. equation (49) would become
0
5 -1(Hsing - cos 6),
and consequently we see that the two curves contained under
the equation p’ = / (+, Y; z) are such that, if one renders { uds
a maximum or a minimum, the other possesses the same
property with regard to |
290. Case where V =p + it + poy.—Next let us consider
the case where V is of the form w+ mé+ wy, in which p, jm,
and pi. are given functions of # and y solely.
Here, as before, we assume
=f a tmd + ns + AE +=} ae (51)
50
Then, as in the previous case, we get
V=Pye+ Qy +6.
This leads immediately to the relation
paAte.
430 Calculus of Variations.
Again, the equation [Z]} =o at the limits reduces to
[(ut mat + poy) As]} + [unde + pody]t + [A (ede + dy) ]) =0.
Substituting, as before, for [87]j and [Sy]j, this ex-
pression immediately reduces to
cA (8 — 80) +[mAat woAy + (u-e)(#Art yAy) =o.
Hence, as in the preceding case, c= 0 except for isoperime-
trical problems, also we have
[mAv+ mAy+t mu (de +ydy) |i=0. (52)
In this case we have X = pw, and equation (51) may be
written v-| th (P+ P+1)+ wet wy ds. (53)
i)
If u, wm, and pw. be functions of one variable (y suppose)
solely, the differential equation for determining the curve
assumes the simple form
pe +p. =e. (54)
291. In general we get by (34) the equations
du du | dar
ie de ae a ee
du « duy @ dpe 1 ad e dp
dy "ay *4 ay = gy (Hd) + 3
Hence, as in Art. 280, we readily get
(55)
I (du. du. du, dw
po da” dy” * dy te (56)
We shall illustrate the results arrived at in the preceding
articles by the consideration of a few elementary problems
of maxima and minima.
292. Lines of Shortest Length.—In the case of plane
curves the length of the curve between any two points is
represented by
{ ‘ J t+e7 de.
*
Hence, since Vis a function of 7 solely, we have, by Art.
278, =const. ; consequently, as is obvious geometrically, the
curve of shortest length between two points is a straight line.
Lines of Shortest Length. 431
If the limiting point ay be restricted to the curve
¥=fo(z), we have Ayo=fo(%)A%, and the limiting equation
(17) gives
(1+ Ufo() )o Aa = 0,
or 1+y o(%o) =0. (5 7)
This shows that the right line cuts the boundary orthogonally.
Hence the problem reduces to the drawing a normal, or
normals, from the point 2 y to the bounding curve.
It is easily seen that if the point ay. is between the
curve and the corresponding centre of curvature, the distance
is a minimum, and leads to a real minimum solution. If
the point lies beyond the centre of curvature the normal in
question furnishes a stationary solution, but not a minimum
solution.
If each of the limiting points lies on a given curve the
solution is a line normal to both the curves.
Let us consider whether P, PB
oe ae »
such a solution is a true mini-
mum or only a stationary Pe el
solution. {
1°. Let the curves be con- Ce =
vex to each other along the 2
common normal P;,P,, in this Cy
case P,P, is a minimum. Fig. 82.
Tf the curves be concave relative to P,P., the distance
P,P, is not a true minimum, and consequently our solution
is but a stationary solution.
If the curves lie as in the figure, then it is easily seen
that P,P, is a true minimum only when C,, the centre of
curvature corresponding to P,, lies beyond C,, the centre of
curvature corresponding to P:.
More generally, we have for the length of any curve in
space
ey en)
u-| Sit Pre de.
ae
0
Hence, for a maximum or a minimum, we have, as in
Art. 278, 7 =const., ¢=const., and accordingly the curve is
the right line that joins the limiting points.
432 Calculus of Variations.
If we suppose the limiting point 2 y.%, restricted to lie on
the surface
u=f (x,y, 8) =0;
then we have
du du du
ae a gas =0. 8
[ae ar Ay a az| Oo (58)
Again, by (32), the term in dU that correspond to this.
limit is
Ax YoAYo + SAL
ae * ° 3°
AUF E+E Slt Yor t de?
This gives
Ax+ YAYo a So Aso =0;
hence, from (58) we see that we must have at the limiting
point
du itdu idu (59)
G aa ee oe
This shows that the right line is normal to the surface u=0
at the point ayo%o.
Hence when one of the limiting points is fixed, and the
other lies on a given surface, the problem reduces to drawing
normals to the surface from the given point.
It is easily seen from elementary geometry that for a
true minimum solution the point must be nearer to the
surface than either of the two corresponding principal centres
of curvature.
293. Brachystochrone. — We shall next consider
Bernoulli’s problem (Art. 269) of the line of quickest descent
under the action of gravity.
Let us take the axis of 2 vertically downwards, and that
of y horizontal. And suppose the particle to start from the
point 2. with the velocity due to the height 4; then, if v
be the velocity at any point we shall have
P= 2g (e+h—%X),
also, if ¢ be the time of motion, we have v = 7
The Brachystochrone. 433
Therefore
au J i+97 dx
a J 2g Ve +h— 2X9
Hence, neglecting the constant »/29, we may write
Be feat
aoe (60)
Zyf/eUt+h—xXy
Here, since V does not contain y explicitly, and 2 is con-
stant, we have, by (25),
¥
SSS SSS EU, 61
Sit P Sath-a | en
Now, let ¢ be the angle that the tangent to the curve at
the point zy makes with the axis of a, then
y=tang; .. Jath—w =,
or et+h-am=a sin’¢, (62)
writing @ instead of 53
hence dx = 2a sin } cos pd ;
also
dy = tan ¢ dx = 2a sin’¢ do ;
therefore
y= 2aJsin’¢ dp = ag - at + const. (63)
Hence (Diff. Cal., Art. 272), we see that the curve of
quickest descent is a cycloid.
The construction of the curve in any case depends on the
limiting conditions. Thus, if the particle starts from rest,
we have h = 0; again, if the upper limiting point be fixed,
taking the origin at the point we have a= 0, y= 0, and
equations (62) and (63) become
2=asin’?¢, y=a(6-
sin 29
a
These represent a cycloid, wie the origin at a cusp.
[28
434 Caleulus of Variations.
In general, if we suppose the lower point x,y, to lie on the
curve y=/(x), then the term outside the sign of integration
corresponding to this limit gives
V+P,(f(2)-9) =o. (64)
But, by (61), we have
P,=0, vac htth),
4 ’
y
hence (64) becomes
1+tan gif’ (m) =0.
This shows that the cycloid cuts the limiting curve at right
angles.
294. Whe Arc taken as Independent Variable.—It
will be instructive to illustrate the method of Article 288
by applying it to the problem of the brachystochrone.
Here we have
I
be
SoeLhaa,
and equation (45) gives, neglecting the constant multiplier,
3 (@+y +1)
ae,
ft +h— x
Again, as this does not contain y explicitly, we have, as
in equation (46),
“%
y ‘
—-=— = const.=¢c, where y= -
a J t+h—x
* Tho student must be careful not to confound the symbol ¥ in this in-
vestigation with the same symbol in the previous article. In fact ¥ in all
cases represents the fluxion of y relative to the independent variable.
The Brachystochrone. 435
If ¢ be the angle that the tangent makes with the axis
of x, we get, as before,
2+h-a=asin’¢,
and also
L=COKP; .. ds= ae = 2a sin od. (65)
This gives s=- 2a cos ¢ + const.
Hence (Diff. Cal., Art. 276), we see immediately that the
curve is a cycloid.
Again, equation (50) for the limit 2,y: gives
[eAr+ yAy], = 0, (66)
which shows that the cycloid intersects the limiting curve
orthogonally.
If we now suppose the point 2 to be variable, we see,
by Art. 282, that we must introduce in AU the additional
term g
ad
tac | :
So (« + h = o)*
Again, by (65), .
51 ds _ (fi 2 0) _ 2 z .
Sa [sot cot |
accordingly the additional term introduced is
or [cot go — cot 6:| .
Also, the term — po (¢Aa + YAy)o, see (42), becomes
ane (cos ¢oA% + Sin poAYo),
Ah
or = Fi (cot poAay + Ayo) ; a
adding this to the former term, we get
I
— —= (Am cot ¢: + Ayo) = 0.
io [28 a]
436 Calculus of Variations.
If we compare this with (66), we see that the tangents to
the limiting curves at the upper and lower limiting points
are parallel. [See Moigno, Calcul des Variations, § 113.]
For example, to find the curve
of quickest descent between a given
curve AB anda straight line CD, ¢ A
situated in the same vertical plane, B
we draw tothe curve a tangent AT’
parallel to the line CD. Then, sup- T
posing the particle to start from rest, D
the cycloid satisfying the condition
must start from a cusp at A, and Fig. 83.
cut CD orthogonally.
295. To find a curve of given length whose extremities lie
on a given curve, and such that the area comprised between
the two curves shall be a maximum.
Let y =/(x) be the equation of the given curve, then
taking the arc as independent variable, we may, as in Art.
290, write
U=slee@+y +1) + (y— f(x) a} ds.
Hence, by (34), we get
L=c¥3 2. cy=eta, (67)
and it readily follows that the curve is a circle of radius e.
Again, since at the limits we have y = /(2), the limiting
equation becomes, by (50),
c(#Ae +yAy)} =o, or, [# +9f'(z)}i=0.
This shows that the curve cuts the given curve y=/(2) at
right angles at each limiting point.
296. Minimum Surface of Revolution.—To find a
curve such that the surface generated by its revolution
around a given line shall be a minimum.
Take the given line as the axis of z, then the surface in
question is represented by
31
| yds ;
s0
Minimum Surface of Revolution. 437
hence, neglecting a constant multiplier, we may write, as in
Art. 287;
v{" (y+ Bd (# 4g = 1)} ds.
0
Again, by (45), this may be written in the form
U= 3{"y (+9 +1) ds.
80
Consequently, as in (46), the curve is determined by the
equation
yt =a, OY y=a sec ¢. (68)
From this we infer, by (3), Art. 131, that the curve is a
catenary: a result which can be readily seen, since (68)
leads immediately to the equation
dy ——
aT Ve Has
or x=alog (y+ Jy a) + const.
If now the origin be taken as in fig. 7, p. 184, we get
4s 2 72
v= alog V2 VF -*,
and hence
ale
ote Ht. ©
yn (e+6 ) = 4 cosh = (69)
Consequently, when the extreme points are given, the
problem reduces to drawing a catenary passing through
these points, and having the given line for its axis. Fora
general investigation of the possibility of this construction
the student is referred to T'odhunter’s Researches on the Calculus
of Variations, § 62. :
438 Caleulus of Variations.
We limit the investigation here to the particular case
where the limiting points
are equidistant from the
fixed line.
We shall first show
that all catenaries re-
presented by the equa-
tion
x’ Cc. H x
Fig. 84. /
where a is a variable parameter, have two common tangents,
which pass through the origin C.
For, let CN be a tangent to the curve, and let
NCO =a, CH=2, NH=y;
then, as in Art. 131, we have
HL =a, NL =are ON=s;
also from the figure
y x
—= COseC a, — = 8€C a;
a a
hence, by equation (69),
2 cosec a =e 8a 4 e—seca, and 2 cota =eseca—e- seca;
therefore
eseca=cosecat+cota=
1+cosa {ime
sina I- cosa’
Ii we make u = sec a, this gives
ou ft I ce.
u-iI
Properties of the Catenary. 439
The value of u can readily be found approximately from this
equation ; for let «=1+2, then (70) transforms into
see =24+83
but
e= 2.718..., .. & = 7.4, approximately ;
hence we get
e(r4 ant ass 4 t+ be.) = ee
3 3737
From this we find without difficulty
Z=4,q9.p; .. seca=4, g.p.
Hence, by a table of natural sines, we find a = 33° 30, q. p.
The equation tan a = 7 fumishes the maximum value of - for
the catenary.
Accordingly, if y = y,, and if d be the distance between
the limiting points, we may write 2z,=d, and we see that
de, a aoe ‘ :
whenever — =— > tan a it is impossible to describe a
Yo
catenary through the limiting points so as to have the fixed
line for its axis.
When ~? < tan a, it can be readily shown that two, and
: 7 , Co
but two, catenaries can be so drawn ; for let = tan B, and
0
“= =, then equation (69) gives
2 ut
u
zu cotBaetseta2(14 0 + ot
&e.). (71)
This equation cannot have more than two real roots, and
the preceding analysis shows that it has no real root when
B>a.
440 Calculus of Variations.
Again, the roots are equal when 8 =a; for the condition
for equal roots gives 2 cot (3 =e"—e, and we see that in that
case w is given by equation (70), and we must have B =a.
Next, we have already
seen in Ex. 4, p. 261, that if
She the surface generated by
the revolution round its axis
ofany portion OV, measured
from the vertex ofacatenary,
we have
S =m (ys + ax),
where x and y are the
coordinates of WN, and
8=ON=NL.
Again, if = be the
surface of the cone generated F P
by PN in a complete revo- ‘
lution, we have ee 0s
Z=a (PN. HN) =7y (NL+ PL) =rys+ary.PL;
but, by similar triangles, we have
NH.PL=PH.HL=a.PH;
therefore
=n (ys+a.PH).
S-S=na. CP.
If Pi be a second tangent drawn from P to the curve, we
have, in like manner,
S’- 3’=- ra0P,
where S’ and 3’ are the surface areas generated by OM and
by PH, respectively ; hence we see that
S+S’= 34+’, (72)
Consequently, if from any point P on the axis of a
catenary, tangents, PIf and PN, be drawn to the curve, the
surface generated by the arc IN in its revolution around the
axis is equal to the surface generated by the broken line
UPN in the same revolution.
Accordingly
Properties of the Catenary. 44]
_ More generally, if the tangent at IZ meets the axis in a
point P’, we have, adopting a similar notation,
S’- 2 =-naCP,
and we get, in general,
S+ 8’ -(35+ 3’) =+7aPP’. (73)
In this equation the upper or lower sign is taken according
as the tangents intersect below or above the fixed axis.
In fig. 84 it can be shown without difficulty that the
upper curve A’A gives a true minimum, while the lower
curve corresponds only to a stationary solution. (Art. 276.)
297. Lo find the curve for which the area between the
curve, its evolute, and the extreme radii of curvature shall be a
mininein
Here we have
v-|" pds -|" ae dz.
80 Xo y
Hence, since V does not contain either # or y, we have,
by (30),
Vectcy+ Py;
Pj ay V;
therefore we may write
Vac, + cy, (74)
where c, and ¢, are arbitrary constants.
2* This equation may be written in the form
but in this case
5a ls ar
On 1+# r+¥
= ¢,C0S p + CG SID, (75)
where @, as before, represents the angle that the tangent to
the curve makes with the axis of 2; therefore
i =, COS @ + Cz SiN
a 1 ote, Q.
This gives ;
$= C, SIN p — Cz COS P+ C35
and accordingly the curve is a cycloid.
442 . Calculus of Variations.
Again:
Qa 8, if the limiting points be fixed, but not the tangents
at these points, equation [L}}=0 becomes
[P.dy]} =o.
(P2)o = O, and (P.)1 =O.
Hence p =o at both limits, and consequently the extreme
points are cusps on the cycloid; therefore the curve in this
ease is a complete cycloid.
(2°). Suppose the extreme points lie on given curves.
This gives
Here, the equation [Z]} = o becomes, by (20),
[2¢,.Ae+ 2¢,Ay+ P.Ay}i=o.
Hence, since Ay is arbitrary at the limits, we must have
P,=0 at each limit, and therefore the extreme points are
cusps, as in the former case.
Also we must have
(Av+c,Ay=o
at each limit; this shows that the curve touches each of the
bounding curves; and also that the line joining the limiting
points is a maximum or a minimum distance between the
limiting curves.
(3°) If either limiting point be completely indeterminate,
we must have
¢,=0, and ¢,=0,
and the equation reduces to p = 0 at all points, an impossible
equation which shows that there is no solution in this case.
298. Ksoperimetrical Problem.— Again, in the same
problem, if the length of the curve be given we should have
g-[ {or +aJfit+y" de.
Proceeding, as before, we readily get
Cy a
PY S497 ” ae
+ b, where b=—3a;
General Transformation. 443
therefore i =¢, CSP +e sing + J;
or 8 = 0 81nd — ¢, cosg + bP +d. (76)
This is the intrinsic equation to the curve, and we see, as be-
fore, that it becomes a cycloid when b = o, i.e. when the
length of the curve is not given.
Again the equations
dy
—=sin ae 0
ds % me
give
— =sin ¢(¢c,cos¢ + ¢,8in ¢ + b)
d
4 ‘ (77)
Z Cos p(¢, Cos ¢ + c,8in d + 6)
dp (C1 COS + C28IN
From these we can express x and y in terms of ¢, and thus
determine the equation to the curve by elimination.
299. General Transformation.—In general if V be
a function of 2, y, ¥, 7, ...y™, we have
AU -| (P3y + P,D8y +... P,D"dy) de + [Va]
Now let D, denote the symbol £ when operating on dy solely,
and let D, denote the symbol when operating on any of the
functions P, P,,... Pn, then we may write
PnDdy - (- 1)" 8yD" Pm = (Di - (- 1)"D2”) (Pndy)
= (D, “b D;) (Dy = Dy? D, Shea) 5 ie a (- 1)™"D"*) (Pudy)
ad ack aP. m m— ged m1 a Pm e
=5, (Pad Oa: D *oy tenet ( 1) dam oy
Hence we get
Ly d™ Py
-~ dP, 1
+ | Paddy 2 ws Dvtdy+...(-1)" SF av)
444 Calculus of Variations.
Applying this to the different terms in the value of A U given
above, we get
A v-| (P) 8ydex + [VAx}
oy
2
1
a [(P) Sy +(P2) 89+... + P,8y],
oe P, @P. d” P,
dP, 2 Pa ” Pr
aha o et aa eet da
dP, @&P. dP,
(Qj=Pis 3, tae tet act
and so on.
We may write this, as before, in the form
av-|" (P) Syde + (LP; (78)
and it is plain, as in Art. 276, that for a maximum or
minimum value of the definite integral, we must have
(P) =0, and [L}=0. (79)
As before, the coefficient of each independent variation in
[LZ]! must be zero.
300. Criterion for Maxima and Minima. — In
Art. 276 we have seen that further conditions are requisite
in order that the solution of any problem, obtained from the
equation AU =o, should be a real maximum or minimum
solution. These conditions were investigated by Legendre,
Lagrange, and other eminent mathematicians, but the
complete solution of the problem was first supplied by Jacobi.
_it is easily seen, as in the extension of Taylor’s theorem,
that U, is derived from U, by the same process that VU, is
derived from JU.
Again, assuming that the limiting values of 2, y, y, &e.,
are fixed, we have, by (21),
sU- i Mdyde,
%
Application to V= f (a, y, 7). 445
and consequently
ens so -(" SUSyde, (80)
where we also have U=o.
Accordingly, as in Art. 276, the conditions for a real
maximum or minimum in general depend on the value of the
definite integral
if SMByde.
0
301. Application to V= f(z, y, y).—It would be beyond
the limits proposed in this chapter to enter into a general
discussion of the foregoing problem ; we shall merely consider
the case where V is a function of z, y, and ¥ solely.
In that case we have
ope". spp 2 ap
dx dx
Now, observing that
dP = a and writing s for dy,
dy
dP dP,. d/fdP, adP,.
we have ge da s- EG)
_.|¢P £(2)]-a(G*)
-+|F-5 a da \ ay
If we now suppose that w is any solution of the differential
equation 62f=o0, we have
dP da (=) d/dPy |
Ujo—-sl alloca e=Oo
a dx \ dy | za dy ) :
and, in general, we may write
I d (aP, . ad (dP, .
sui [ss (F 4) ue (G4)
ta (elves lo Pal ag Ce
_ wel a ue -ué) |= cma” @ a(n)
ihe aP,
writing Q, for aw
446 Calculus of Variations.
Henge
sof soMdx
-- alee G)]
ae [ 0: = (2) + { uw? Qe E (2) ae
-PeleG)le-LabaG)!®
since dy or s vanishes at both limits, by hypothesis.
Hence we see that, provided the other factor does not
vanish, the distinctive character of a maximum or a minimum
2.
depends on the function Q, or a .
If Q.dz is positive for all values between the limits of
integration, then U hasa minimum value; if Q.dx be negative,
UV has a maximum value. Again, if Quda changes its sign
between the limits of integration there is neither a maximum
nor a minimum value for U; for in that case we can dispose
of the arbitrary function dy so that either the positive or the
negative part of the integral shall be the greater, at our
pleasure. ~
It is easy to apply this criterion to the examples
previously considered: this is left for the student. For
further investigations, as well as for the treatment of double
and multiple integrals, the student is referred to Jellett’s
Calculus of Variations, or to that of Moigno and Lindeléf.
Examples. 447
EXAMpLes.
: i Prove that the curve of given length that encloses a maximum area is a
circle.
_ 2. Prove that for any system of coplanar forces the curve of quickest descent
is such that at each point the pressure on the curve due to the forces is equal to
that due to the motion.
Here we have, by mechanical considerations,
4 mv =f (Xdx + Yay),
also
"di
v={ : at a minimum,
*0
or, by (45), ‘
v=3{ : = (+ 9+ 1) ds;
0
hence, by (49), ;
we Xsing— Y cos 9,
P
which proves the theorem in question.
3. Find the differential equation of the curve, such that the surface generated
by its revolution round a given line shall be constant, and the contained volume
shall be a maximum.
Here, by (53), Art. 290, we get
&.
v=|) (Patzay(#+y7+1)} ds;
0
hence the differential equation of the curve, by (54), is
ayety*=c.
4. Hence show that if the surface is closed the curve is a circle.
It is readily seen that in this case we must have ¢ =o.
5. Show that in general the curve is the roulette described by the focus of
an ellipse or hyperbola rolling on the given right line. [Dz Launay, Journal de
Liouvitle, tome vi. ]
6. Find the differential equations of a curve of shortest length on a given
surface. : en «(DEED
Let «= 0 be the equation of the surface, then we have #?7+9?+'=1, and
fds a minimum; consequently we may write
3
o=("{(#+r+#+1) + uu} ds;
0
hence we get
du, ae ay au :
Mae Pay Y, Bo
448 Calculus of Variations.
and the differential equations of the curve are
# yy F
du au du’
dz dy dz
From these it follows that the osculating plane to the curve at each point
passes through the normal at the point. Such curves are called geodesics.
7. Show that the differential equation of the plane curve which makes
JSo(p)ds a maximum or a minimum, p being the radius of curvature, is given
by
po’ (p) = ant by +e.
8. In the preceding, if the extreme tangents to the curve are not given,
show that p? p'(p) must vanish at each of the limiting points.
g. Prove that a sphere is the only closed surface of revolution which con-
tains a maximum volume under a given surface.
10. Let it be required to draw between two points a curve for which fu ds
is a maximum or a minimum, where uw is a function of « and y solely. Show
that if it be possible to draw more than one curve satisfying the condition
3fuds=o,
and if these curves be arranged according to the angles which their initial tan-
gents make with a given line, then no two consecutive curves will give real
maxima or minima.
Miscellaneous Examples. 449
MiscEntannous Exampres.
1, Find the value of (= qe :
tT+eNe+sd
2 Find the area of the inverse ofah
p eer yperbola, the centre being the
See ee fe ae area of the inverse of an ellipse, aie fe pane
Y an arithmetic m i i
pen ain } eo ean between the areas of the circles described
dz |qz_ 72
wale ge
Ans. an ee see
e-—B b ONe — B
3. Find the integral of
4. Prove that
Xx X=24
[rofl@) 4e=(~4) $(8) log (==),
where é lies between X and x.
_ 5. In a spiral of Archimedes, if P, Q and P’, Q' be the points of section
with any two branches of the curve made by a line passing through its pole,
prove that the area bounded by the right line and by the two branches is half
the area of the ellipse whose semiaxes are PP’ and P’Q.
6. If @ be the sagitta of a circular segment whose base is 4, prove that the
area of the segment is, approximately,
7. If an ellipse roll upon a right line, show that the differential equation of
the locus of its focus is
dy a ee See,
(y? + 3?) a= WV (2ay + y? + 8) (2ay — y? — 0%):
8. A circle rolls from one end to the other of a curved line equal in length
to the circumference of the circle, and then rolls back again on the other side of
the curve; prove that, if the curvature of the curve be throughout less than
that of the circle, the area contained within the closed curve traced out by the
point of the circle which was first in contact with the fixed curve is six times
the area of the circle. (Camb. Math. Tripos, 1871.)
g. In the same case show that the entire length of the path described is
eight times the diameter of the circle.
10. Prove that the area of the locus formed by the points of intersection of
normals to an ellipse, which cut at right angles, is 7 (a— 4)?.
[29]
450 Miscellaneous Examples.
11. Prove that the area between two focal radii of a parabola and the curve
is half the area between the curve, the corresponding perpendiculars on the
directrix, and the directrix.
12. Evaluate the following integrals :—
dz
ey ———— (I+a)4dx
| ims | vise —T ae, | CaS
2 R .
13. If R= (22+ ax)?+ bx, and u=log tact VR find the relation
ve+ax—VR
between the integrals | 2 {=
n the integr: = =
et wee 4 ve lve
at 8 j # 43
Ans. | Ve 3)vR %
14. If a curve be such that the area between any portion and a fixed right
line is proportional to the corresponding length of the curve, show that it isa
catenary.
15. Prove that the volume of a rectangular parallelepiped is to that of its
circumscribed ellipsoid as 2: x V3.
dé \ B de
———— _ |" ———_, where sin B = x sina.
VI—«?sin?@ Jo Vi? — sin2@
16. Prove that 1
0
17. If any number of triangles be inscribed in one ellipse and circumscribed
to another ellipse, concentric and similar, prove that these triangles have all the
game area.
18, Show that the value of the integral { ay may be exhibited by the
a vy -I
following geometrical construction. Let the curve whose equation is
Zz m
rm? cos
m+
w=!
roll on the axis of x; take the points (x, 1) (#2, yz) on the roulette described
by the pole, such that yi=a, y2=0; then
=22- 21. (JELLETT.)
i dy
a Vym—1
19. If s be the length of the arc of a spherical curve measured to any point
P, and ¢ be the intercept on the great circle touching at P, between the point of
contact and the foot of the perpendicular from the pole, prove that
s—t=Jsin pdw.
The proof is similar to that of the corresponding theorem ix plano. See Art. 1 58.
Miscellaneous Examples. 451
20. Prove that the volume of a polyhedron, having for bases any two
polygons situated in parallel planes, and for lateral faces trapeziums, is ex-
pressed by the formula
ye
G (B+ B+ 4B");
where H is the distance between the parallel planes, B and B’ the areas of the
polygonal bases, and B” the area of the section equidistant from the two bases.
21. If S be the length of a loop of the curve 7™ = a" cos 0, and A the area
of a loop of the curve 7?” =a?” cos 2n@, prove that
3
TH
AxS=—-
2%
22. Find approximately the area, and also the length, of a loop of the curve
$= a3 cos *. (See Diff. Calc., Art. 268.)
Ans. area =a x 0.56616; length=a x 2.72638.
23. Show from Art. 134 that if a parabola roll on a right line, the locus of
its focus is a catenary.
24. If A be the area of any oval, B that of its pedal with respect to any
internal origin 0, and C' that of the locus of the point on the perpendicular
whose distance from 0 is equal to the distance of the point of contact from 0;
prove that 4, B, C are in arithmetical progression.
25. The arc of a curve is connected with the abscissa by the equation s*= kz;
find the curve.
26. If the coordinates of a point on a curve be given by the equations.
g=csin 20(I1+ cos 20), y=c cos 20 (I — cos 28),
prove that the length of its arc, measured from its origin, is fe sin 36.
27. Show how to find the sum of every element of the periphery of an ellipse
divided by any odd power (2r + 1) of the semi-diameter conjugate to that which
passes through the element, and give the result in the case of the fifth power.
(W. Rozerts.)
wT
4 (2 + 20)r-
Ans. (as (a? cos?6 + 5B? sin?6)r-! de.
(a? +8")
058
This gives when r = 2.
28. A sphere intersects a right cylinder; prove that the entire surface of the
cylinder teelnded within the sphere is equal to the product of the diameter of
the cylinder into the perimeter of an ellipse, whose axes are equal to the greatest
and least intercepts made by the re on is edges of the cylinder.
29a
452 Miscellaneous Examples.
29. Show that the equations of the involute of a circle are of the form
z=acosptapsing, y=asinp —ag cos¢,
and prove that the length of the arc of this involute, measured from ¢=0, is
one-half of the arc of a circle which would be described by a radius equal to the
arc of its evolute moving through the angle 9.
30. Show that the area of the cassinoid
74 — 2077? cos 26 + at = 4
is expressed by aid of an elliptic are when >a; and by a hyperbolic arc
when a> 6.
31. A string AB, with its end J fixed, lies in contact with a plane convex
curve; the string is unwound, and B is made to move about A till the strin
is again wound on the curve, the final position of B being B’ ; prove that for
variations of the position of 4 the arc traced out by B will be a maximum ora
minimum, when the tangents at B and B’ are equally inclined to the tangent at
4; and will be the former or the latter, according as the curvature at 4 is
greater or less than half the sum of the curvatures at B and B’.—(Camb. Math.
Tripos, 1871.)
32. Find the value of | ee Ans Jee oe
0 £ B
33. Find the length, and also the area, of the pedal of a cissoid, the vertex
being origin.
8a mat
Ans. — log (2 + V3) — a; —.
G g (2+ V3) — 4 a4
34. Prove that the length of an arc of the lemniscate r?= a? cos 20 is repre-
sented by the integral
“f__4@
v2) Vi-}am'g
35. Integrate the equation
cos 6 (cos # — sin a sin @) dé + cos ¢ (cos @ — sin a sin 6) dp =0.
If the arbitrary constant be determined by the condition that the equation must
be satisfied by the values (0, a) of (8, $), show that the equation is satisfied by
putting 6+ o¢=a.
36. Each element of the surface of an ellipsoid is divided by the area of the
parallel central section of the surface ; find the sum of all the elementar i
extended through the entire ellipsoid. aa
37. Hence, show that
Tv
=e.
2
[ { (u? — v?) dudy
aVe?— ht VRB BVA py? VI
0
Miscellaneous Examples. 453
This depends on the expression for an element of the surface of an ellipsoid in
terms of elliptic coordinates. See Salmon’s Geometry of Three Dimensions,
Art. 411. This proof is due to Chasles (Liowville, tome iii. p. 10.)
38. Hence prove the relation
F(m) E(n)+F(n) E(m)- F(x) F(m) =5,
where
z Tv
2 dg 2
F(m) =| —_——, E(m)=| VI — m? sin?9 0,
o VI — m?sin’6 0
and m?+ n?=1.
Let v=hsin 0, and w= VA? sin’ + #cos*¢, in the preceding example,
and it becomes
T =
i, f fi? sing + k? cos? — h? sin?@
0 VA? sin? + 4? cost Vi? — 1? sin20
wT oT Tv Tv
do dp
0
_ [? (? VA? sin’ + 2? cos’ 2 pe Vik WP sind
o Jo §«©69VA®— he? sin2a o Jo Vi? sin? + 4? cos’
Tv wT
i fe. kdodp
o Jo Vk? sin? + k2 cos? Vk? — h? sin?@
This furnishes the required result on making 4 = mk.
The preceding formula, which is due to Legendre, gives a general relation
between complete elliptic functions of the first and second species, with com-
plementary moduli.
39. Ifthree curves be described on the surface of an ellipsoid, along the first
of which the perpendicular to the tangent plane makes the constant angle y with
the axis of z, along the second 6 with the axis of y, and along the third a with the
tana tanB tany.
axis of x, and if the angles be connected by the relations ey ee
then, if As, 2, 41 be the included portions of the ellipsoid surface, prove that
A3— Ag A, — A2 4,-A,_
oe ge
40. Show that the results given in Arts. 161 and 162 hold good for
spherical conics, where the tangents are arcs of great circles on the sphere.
oO. (JELLETT.)
41. Prove that
a dz Re e dx
i {(e—#)(b-2)(e—2)} 7 [rat (@—2)(b=2)(c—2)}¥
where @, 3, ¢ are in the order of magnitude.
454 Miscellaneous Examples.
42. If w be an imaginary cube root of unity, show that, if
(@- wo?) +0723 dy (@ — w*) dx ;
= , then = a 3
1 0? (w— w) 2? (1—y)® (1+ wy")? (1-2) 2 (1 + w2)
(PRrorEsson CAYLEY.)
43. Prove that the value of
" cosbesinaz , . 7 7
————- dziso, -, or-,
0 x 4 e
according as ) is >, =, or < a.
we .
sin bx sin ax
44, Prove that | a
0
numbers @ and 0b.
Z dz => multiplied by the lesser of the
45. Ife be the eccentricity of an ellipse whose semiaxis major is unity, and
£ the length of its quadrant, prove that
i aus zh (W. Rozerts.)
o(1—#) Vea) avi
46. If S represent the length of a quadrant of the curve r™=a™ cos mé,
and S, the quadrant of its first pedal, prove that
m+
§.8,=
na’,
Here (Ex. 3, Art. 156), we have
. r(;
gal =)
2m r (=)
2m
_ Also, since the first pedal( Dif’. Calc., Art. 268) is derived by substituting =
instead of m, m+t
>
a r(r+)
2m
I
cag Ceo Naa) anaes
qm? r (r+ I ) 2m
2m
Miscellaneous Examples. 455
47. In general, if Sn be the quadrant of the n* pedal of the curve in the
last, prove that
mn+1
Sn Sn = 3a,
Here it is readily seen that the nth pedal is got by substituting _~
mn +
instead of m in the equation of the proposed; .°. &c.
(W. Rozerrs, Liouville, 1845, p. 177.)
48. Ifan endless string, longer than the circumference of an ellipse, be passed
round the ellipse and kept stretched by a moving pencil, prove that the pencil
will trace out a confocal ellipse.
49. If two confocal ellipses be such that a polygon can be inscribed in one
and circumscribed to the other, prove that an indefinite number of such polygons
can be described, and that they all have the same perimeter.
(Cuastes, Comp. Rend. 1843.)
50. To two ares of a hyperbola, whose difference is rectifiable, correspond
equal arcs of the lemniscate which is the pedal of the hyperbola. (Zbid.)
51. Prove that the tangents drawn at the extremities of two arcs of a conic,
whose difference is rectifiable, form a quadrilateral whose sides all touch the
same circle. (JZbid.)
52. In the curve
2
a af ya = as,
prove that any tangent divides the portion of the curve between two cusps into
arcs which are to each other as the segments of the portion of the tangent
intercepted by the axes.
53. If two tangents to a cycloid cut at a constant angle, prove that their
sum bears a constant ratio to the arc of the curve between them.
54. If AB, ab, be quadrants of two concentric circles, their radii coinciding,
show that if an arc 4d of an involute of a circle be drawn to touch the circles
at 4, 5, the arc 4d is an arithmetical mean between the arcs 4B and ab.
55. If ds represent an infinitely small superficial element of area at a point
Outside any closed plane curve, and ¢, ¢ the lengths of the tangents from the
point to the curve, and @ the angle of intersection of these tangents, prove that
in 6d: ‘i .
a 5 taken for all points exterior to
t
the sum of the elements represented by in
the curve, is 27°. (Pror. Crorton, Phil. Trans., 1868.)
56. Show that, for all systems of rectangular axes drawn through a given
point in a given plane area,
{JJ -y)dady}? + 4{ fay dx dy},
taken over the whole of the area, is constant; and that for a triangle, the point
being its centre of gravity, this constant value is
(By A)? (at + 04 + cf — Bc? — ca? — a*b?).
(Mr. J. J. WaLkEn.)
456 Miscellaneous Examples.
57. If ab=a'd’, prove that
f c 9 (ax + by)— 6 (48D) ayy,
0 Jo vy
=log (%) tog (5) (o()- 91),
ovided the limits @ (0) and ¢ (co) are both definite.
_ on (Mr. Extiorr, Proceedings, Lond. Math. Soc., 1876.)
58. If 8 denote the surface, and V the volume, of the cone standing on the
focal ellipse of an ellipsoid, and having its vertex at an umbilic, prove that
S=na(B—2)%, = 4ne(b?—c2),
where @, 4, ¢ are the principal semiaxes of the ellipsoid.
59. Prove that, if » be positive and less than unity,
it ada T I
i. (a? + x) log (1+ ol pane et 3 (1)
and
d dz I
i, (xP + xP) log (1— 2) np ten om (2)
where (1) may be deduced from (2) by putting 2? for x.
(Pror. WoLSTENHOLME.)
60. If u, v be the elliptic coordinates of a point in a plane, prove that the
area of any portion of the plane is represented by
| | (u? — v?) dudy
V(@—2) (@—)
taken between proper limits.
61. Prove that the differential equation, in elliptic coordinates, of any tan-
gent to the ellipse u= ju, is
du " dy 26
V (atm 08) (Wa?) V (= 9) (a? = 9)
62. Hence show that the preceding differential equation in w and y admits
of an algebraic integral.
63. Prove that the differential equation of the involute of the ellipse uz =p is
agit mato
we wy wey
q inl dv=0.
(a HIN Gp
Miscellaneous Examples. 457
_ 64. Show that, for a homogeneous solid parallelepiped of any form and
‘dimensions, the three principal axes at the centre of gravity coincide in direc-
tion with those of the solid inscribed ellipsoid which touches at the six centres
of gravity of its six faces; and that, for each of the three coincident axes, and
‘therefore for every axis passing through their common centre of gravity, the
moment of inertia of the parallelepiped is to that of the ellipsoid in the same
‘constant ratio, viz. that of 10 to x. (TowNnsEND.)
65. Show that the volumes of any tetrahedron, and of the inscribed ellipsoid
which touches at the centres of gravity of its four faces, have the same principal
axes at their common centre of gravity; and that their moments of inertia for
all planes through that point have the same constant ratio (viz. 18/3 : 7).
(Ibid.)
66. A quantity M of matter is distributed over the surface of a sphere of
tadius a, so that the surface density varies inversely as the cube of the distance
from a given internal point S, distant 5 from the centre; prove that the sum of
the principal moments of inertia of M at § is equal to 2 (a? - 8).
(Camb. Math. Tripos, 1876.)
-1
67. If (t-2az4+a%) 7=14+0X,+atX,...+a"Xat+..., prove that
+1 +1
| XnXmdz = 0, \ X,2dz = .
1 -1 2n+1I1
68. A closed central curve revolves round an arbitrary external axis in its
plane. Prove that the moments of inertia I and J, with respect to the axis of
revolution and to the perpendicular plane passing through the centre of inertia
of the solid generated by the revolving area, are given respectively by the
‘expressions
uu
T2nW@iw, 750 (# =) :
where m represents the mass of the solid, a the distance of the centre of the
generating area from the axis of revolution, # and & the radii of gyration of
the area with respect to the parallel and perpendicular axes through its centre,
and J the arm length of its product of inertia with respect to the same axes.
(TownsEnD, Quarterly Journal of Mathematics, 1879.)
6. oa ie (@ —2)"* f(z) dz, find the value of i : die $6.
70. Prove that the superficial area of an ellipsoid is represented by
2 (1 — ee 2") dx
o V(r — ea*)(1 — eta
where a- BP=ate, P—ce= eb,
2c? + onab |
(Jetierr, Hermathena, 1883.)
+1. Find the mean distance of two points on opposite sides of a square
whose side is unity.
An 2=V? oer 4/2).
458 Miscellaneous Examples.
72. A cube being cut at random by a plane, what is the chance that the
section is a hexagon ? (Cou. CLARKE.)
-aAVz <1
Ana, Va cot Vg — V2 00t V2 oye,
47
73. Three points are taken at random, one on each of three faces of a tetra-
hedron : what is the chance that the plane passing through them cuts iD a
face P ua.
Ans. 3,
4
74. Two stars are taken at random from a catalogue: what is the chance
that one or both shall always be visible to an observer in a given ee i a
Oe
I a.
| | (C3 — 3a7C'+ 205) dp dw,
go. Show, by means of Landen’s transformation in elliptic functions or
-otherwise, that
v do = {- dp
0 (a? cos?@ +b? sin?a)2 J 9 (a? cos’ + 2? sin?o)2’
where a and 4; are respectively the arithmetic and the geometric means
‘between @ and b.
Point out the value of this result in the calculation of the numerical value
-of the definite integral. (Camb. Math. Tripos, part ii., 1889.)
INDEX.
——j—_
ALLMAN, on properties of paraboloid,
268, 281.
Amsler’s planimeter, 214,
Annular solids, 261.
Approximate methods of finding
areas, 211.
Archimedes, on solids, 254.
spiral of, 194.
area of, 194.
rectification of, 227.
Areas of plane curves, 176.
Ball, on Amsler’s planimeter, 215.
Bernoulli’s series, by integration by
parts, 128.
Binet, on principal axes, 312.
Brachystochrone, 432.
Buffon’s problem, 365.
Calculus of Variations, 413.
Cardioid, area of, 192.
rectification of, 227, 238.
Cartesian oval, rectification of, 239.
Catenary, equation to, 183.
rectification of, 223.
surface of revolution by, 260.
Cauchy, on exceptional cases in defi-
nite integrals, 128.
on principal and general values of
a definite integral, 132.
on singular definite integrals, 134.
on hyperbolic paraboloid, 271.
Chasles, on rectification of ellipse, 234,
248.
on Legendre’s formula, 153.
Clerk Maxwell, 409.
Cone, right, 256. e
Crofton, on mean value and probabi-
lity, 346-391, 455, 458.
Cycloid, 189.
Definite integrals, 30, 115.
exceptional cases, 128.
infinite limits, 131, 135.
principal and general values, 132.
singular, 134.
differentiation of, 148, 147.
deduced by differentiation, 144.
integration under the sign J, 148.
double, 149, 313.
Definition of variation dy, 413.
Descartes, rectification of oval of, 239.
Differentiation under the sign of inte-
gration, 107.
Dirichlet’s theorem, 316.
Elliott, extension of Holditch’s theo-
rem, 209.
on Frullani’s theorem, 157.
Ellipse, are of, 226.
Ellipsoid, 266.
quadrature of, 282.
of gyration, 309, 312.
momental, 309.
central, 310.
Elliptic integrals, 29, 173, 226, 232,
235, 243, 279.
coordinates, 249.
Epitrochoid, rectification of, 237.
Equimomental cone, 310.
Errors of observation, 376.
Euler, 102,
theorem on parabolic sector, 198.
Eulerian integrals, 117, 124, 159.
definition of—
T'(n) and B(m, n), 124, 160.
T'(m) I (n)
=e
B (m, n) T (m+ ni)’ 161.
Tv
MOTE Shs 162.
462
Eulerian integrals—
I 2 n—-I
valueofr (=) r(=) vod (=)
n n n
164.
table of log (I'm), 169.
Fagnani’s theorem, 229.
Volium of Descartes, 192, 218.
Fourier’s theorem, 392.
Frequency, curve of, 368.
Frullani, theorem of, 155, 456.
Gamma functions, 124, 159.
‘Gauss, on integration over a closed
surface, 287.
Genucchi, rectification of Cartesian
oval, 240, 242.
Graves, on rectification of ellipse, 234.
Green’s theorem, 327.
Groin, 269.
‘Gudermann, 183.
Guldin’s theorems, 262, 263, 288.
Gyration, radius of, 293.
Helix, rectification of, 244.
Hirst, on pedals, 202.
Holditch, theorem of, 206.
Hyperbola, rectification of, 233.
Landen’s theorems on, 232.
Hyperbolic sines and cosines, 182.
Hypotrochoid: see epitrochoid.
Inertia, integrals of, 291.
moments of, 291.
products of, 291, 306.
principal axes of, 307.
momenta! ellipsoid of, 309.
Integrals, definitions of, 1, 114.
elementary, 2.
double, 149, 313.
of inertia, 291.
transformation of multiple, 320.
Integration, different methods of, 20.
by parts, 20.
xm dx
-, 58.
w—t
of
by successive reduction, 63.
by differentiation, 71, 144.
of binomial differentials, 75.
by rationalization, 92, 97.
by pone under sign f,
1
by infinite series, 110.
Index.
Integration, regarded as summation,
31, ‘
double, 269, 313.
change of order in, 314.
over a closed surface, 284.
Jacobians, 323, 326. ates
Jellett, on quadrature of ellipsoid, 288,
457.
Kempe, theorem on moving area, 210.
Lagrange’s series, remainder in, 158.
Lambert, theorem on elliptic area, 196.
Landen, theorem on hyperbolic arc,
on difference between asymptote
and arc of hyperbola, 233.
Laplace’s theorem on spherical har-
monics, 338.
Legendre, on Eulerian integrals, 160.
formula on rectification, 228. ‘
relation between complete elliptic
functions, 453.
Leibnitz, on Guldin’s theorems, 264.
Lemniscate, area of, 191.
rectification of, 452.
Leudesdorf, 157, 210, 220.
Limagcon, area of, 192.
rectification of, 237.
Limits of integration, 33, 115.
Line and surface integrals, 401.
Maxima and Minima, 419.
Mean Value and probability, 346.
Mean Value, definition of, 346.
for one independent variable,
347,
two or more independent va-
riables, 350.
Method of quadratures, 178.
Miller, 358.
Momental ellipse, 300.
of a triangle, 304.
Moments of inertia, 291.
relative to parallel axes, 292.
uniform rod, 294.
parallelepiped, cylinder, 295.
cone, 296.
sphere, 297.
ellipsoid, 298.
prism, 302.
tetrahedron, 304.
solid ring, 305.
Index.
M'‘Cullagh, on rectification of ellipse
and hyperbola, 236.
Neil, on semi-cubical parabola, 224,
249.
Newton, method of finding areas, 177.
by approximation, 213.
on tractrix, 219.
Observation, errors of, 374.
Panton, on rectification of Cartesian
oval, 240.
Paraboloid, of revolution, 256.
elliptic, 265, 268.
Partial fractions, 42.
Pedal, area of, 199.
of ellipse, 190.
Steiner’s theorem on area of, 201.
Raabe, on, 202.
Hirst, on, 202.
Roberts, on, 455.
Planimeter, Amsler’s, 214.
Popoff, on remainder in Lagrange’s
series, 159.
Probability, used to find mean values,
356.
Probabilities, 349.
Products of inertia, 301, 306.
Quadrature, plane, 176.
on the sphere, 276.
of surfaces, 279.
paraboloid, 280.
ellipsoid, 282.
Raabe, theorem on pedal areas, 202.
Radius of gyration, 293.
Random straight lines, 381.
Rectification of, plane curves, 222.
parabola, 223.
catenary, 233.
semi-cubical parabola, 224.
of evolutes, 224.
arc of ellipse, 226.
hyperbola, 231.
epitrochoid, 237.
roulettes, 238.
Cartesian oval, 239, 247.
twisted curves, 243. re
Recurring biquadratic under radical
sign, 101.
463
Reduction, integration by, 63.
by differentiation, 71, 80.
Roberts, W., on Cartesian oval, 240.
on pedals, 455.
Roulette, quadrature of, 205.
rectification of, 238.
Simpson’s rules for areas, 213.
Sphere, surface and volume of, 252.
quadrature on, 276.
Spherical harmonics, 332.
Spheroid, surface of, 257, 258.
Spiral, hyperbolic, 191.
of Archimedes, 194, 227, 380.
logarithmic, 227.
Steiner, theorem on pedal areas, 201.
on areas of roulettes, 203.
on rectification of roulettes, 238.
Stokes’ theorem on line integrals, 403.
Surface of, solids, 250.
cone, 251.
sphere, 252.
revolution, 254.
spheroid, prolate, 257.
oblate, 258.
annular solid, 261.
Taylor's theorem, obtained by inte-
gration by parts, 126.
remainder as a definite integral,
127.
Thomson, 412.
Townsend on moments of inertia of a
ring, 305.
on moments of inertia in general,
310.
Tractrix, area of, 219.
length of, 2265.
Van Huraet, on rectification, 249.
Variation of a definite integral, 416.
Viviani, Florentine enigma, 278.
Volumes of solids, 250, 264, 286.
Wallis, value for m, 122.
Weddle, on areas by approximation,
213.
Woolhouse, on Holditch’s theorem,
206.
Zolotareff, on remainder in Lagrange’s
series, 158.
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