fe Peas of ser ste res? Rae s eer et eee Math : QA 308% Wye 184] 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. 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