Line Integrals, Surface Integrals, and Integral Theorems

Transcription

1 Line Integals, uface Integals, and Integal Theoems onstuction of mathematical models of hysical henomena equies functional domains of geate comlexity than the eviously emloyed line segments and lane egions. This section makes ogess in meeting that need by eniching integal theoy with the intoduction of segments of cuves and otions of sufaces as domains. Thus, single integals as functions defined on cuve segments take on new meaning and ae then called line integals. tokes s theoem exhibits a stiking elation between the line integal of a function on a closed cuve and the double integal of the suface otion that is enclosed. The divegence theoem elates the tile integal of a function on a thee-dimensional egion of sace to its double integal on the bounding suface. The elegant language of vectos best descibes these concets; theefoe, it would be useful to eead the intoduction to hate 7, whee the imotance of vectos is emhasized. (The integal theoems also ae exessed in coodinate fom.) LINE INTEGRAL The objective of this section is to geometically view the domain of a vecto o scala function as a segment of a cuve. ince the cuve is defined on an inteval of eal numbes, it is ossible to efe the function to this imitive domain, but to do so would suess much geometic insight. A cuve,, inthee-dimensional sace may be eesented by aametic equations: o in vecto notation: whee (see Fig. 1-1). x ¼ f 1 tþ; y ¼ f 2 tþ; z ¼ f 3 tþ; b 1Þ x ¼ tþ tþ ¼xi yj zk 229 oyight 22, 1963 by The McGaw-Hill omanies, Inc. lick Hee fo Tems of Use. 2Þ

2 23 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP. 1 Fig. 1-1 Fo this discussion it is assumed that is continuously diffeentiable. While (as we ae doing) it is convenient to efe the Euclidean sace to a ectangula atesian coodinate system, it is not necessay. (Fo examle, cylindical and sheical coodinates sometimes ae moe useful.) In fact, one of the objectives of the vecto language is to fee us fom any aticula fame of efeence. Then, a vecto A½xtÞ; ytþ; ztþš o a scala,, isictued on the domain, which accoding to the aametic eesentation, is efeed to the eal numbe inteval b. The Integal A d 3Þ of a vecto field A defined on a cuve segment is called a line integal. eesentation The integand has the A 1 dx A 2 dy A 3 dz obtained by exanding the dot oduct. The scala and vecto integals X n tþ dt ¼ lim k ; k ; k Þt k n!1 k¼1 X n AtÞdt ¼ lim A k ; k ; k ÞtÞ k n!1 k¼1 4Þ 5Þ can be inteeted as line integals; howeve, they do not lay a majo ole [excet fo the fact that the scala integal (3) takes the fom (4)]. The following thee basic ways ae used to evaluate the line integal (3): 1. The aametic equations ae used to exess the integand though the aamete t. Then t2 A d ¼ A d t 1 dt dt 2. If the cuve is a lane cuve (fo examle, in the xy lane) and has one of the eesentations y ¼ f xþ o x ¼ gyþ, then the two integals that aise ae evaluated with esect to x o y, whicheve is moe convenient.

3 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM If the integand is a efect diffeential, then it may be evaluated though knowledge of the end oints (that is, without efeence to any aticula joining cuve). (ee the section on indeendence of ath on Page 232; also see Page 237.) These techniques ae futhe illustated below fo lane cuves and fo thee sace in the oblems. EALUATION OF LINE INTEGRAL FOR PLANE URE If the equation of a cuve in the lane z ¼ isgiven as y ¼ f xþ, the line integal (2) isevaluated by lacing y ¼ f xþ; dy ¼ f xþ dx in the integand to obtain the definite integal a2 a 1 Pfx; f xþg dx Qfx; f xþg f xþ dx which is then evaluated in the usual manne. imilaly, if is given as x ¼ gyþ, then dx ¼ g yþ dy and the line integal becomes b2 b 1 Pfg yþ; ygg yþ dy Qfg yþ; yg dy If is given in aametic fom x ¼ tþ; y ¼ tþ, the line integal becomes t2 t 1 PftÞ; tþg tþ dt QftÞ; tþg; tþ dt 7Þ 8Þ 9Þ whee t 1 and t 2 denote the values of t coesonding to oints A and B, esectively. ombinations of the above methods may be used in the evaluation. If the integand A d is a efect diffeential, d, then c;dþ A d ¼ d ¼ c; dþ a; bþ 6Þ a;bþ imila methods ae used fo evaluating line integals along sace cuves. PROPERTIE OF LINE INTEGRAL EXPREED FOR PLANE URE Line integals have oeties which ae analogous to those of odinay integals. 1: Px; yþ dx Qx; yþ dy ¼ Px; yþ dx Qx; yþ dy Fo examle: 2: a2 ;b 2 Þ Pdx Qdy¼ a1 ;b 1 Þ a 1 ;b 1 Þ a 2 ;b 2 Þ Pdx qdy Thus, evesal of the ath of integation changes the sign of the line integal. 3: a2 ;b 2 Þ Pdx Qdy¼ a3 ;b 3 Þ Pdx Qdy a2 ;b 2 Þ a 2 ;b 1 Þ a 1 ;b 1 Þ a 3 ;b 3 Þ Pdx Qdy whee a 3 ; b 3 Þ is anothe oint on. imila oeties hold fo line integals in sace.

4 232 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP. 1 IMPLE LOED URE, IMPLY AND MULTIPLY ONNETED REGION A simle closed cuve is a closed cuve which does not intesect itself anywhee. Mathematically, a cuve in the xy lane is defined by the aametic equations x ¼ tþ; y ¼ tþ whee and ae singlevalued and continuous in an inteval t t 2. If t 1 Þ¼t 2 Þ and t 1 Þ¼ t 2 Þ, the cuve is said to be closed. If uþ ¼vÞ and uþ ¼ vþ only when u ¼ v (excet in the secial case whee u ¼ t 1 and v ¼ t 2 ), the cuve is closed and does not intesect itself and so is a simle closed cuve. We shall also assume, unless othewise stated, that and ae iecewise diffeentiable in t t 2. If a lane egion has the oety that any closed cuve in it can be continuously shunk to a oint without leaving the egion, then the egion is called simly connected; othewise, it is called multily connected (see Fig. 1-2 and Page 118 of hate 6). As the aamete t vaies fom t 1 to t 2, the lane cuve is descibed in a cetain sense o diection. Fig. 1-2 Fo cuves in the xy lane, we abitaily descibe this diection as ositive o negative accoding as a eson tavesing the cuve in this diection with his head ointing in the ositive z diection has the egion enclosed by the cuve always towad his left o ight, esectively. If we look down uon a simle closed cuve in the xy lane, this amounts to saying that tavesal of the cuve in the counteclockwise diection is taken as ositive while tavesal in the clockwise diection is taken as negative. GREEN THEOREM IN THE PLANE This theoem is needed to ove tokes theoem (Page 237). Then it becomes a secial case of that theoem. Let P, be single-valued and continuous in a simly connected egion bounded by a simle closed Pdx Qdy¼ dx whee is used to emhasize that is closed and that it is descibed in the ositive diection. This theoem is also tue fo egions bounded by two o moe closed cuves (i.e., multily connected egions). ee Poblem 1.1. ONDITION FOR A LINE INTEGRAL TO BE INDEPENDENT OF THE PATH The line integal of a vecto field A is indeendent of ath if its value is the same egadless of the (allowable) ath fom initial to teminal oint. (Thus, the integal is evaluated fom knowledge of the coodinates of these two oints.) Fo examle, the integal of the vecto field A ¼ yi xj is indeendent of ath since x2 y 2 A d ¼ ydx xdy¼ dxyþ ¼x 2 y 2 x 1 y 1 x 1 y 1 Thus, the value of the integal is obtained without efeence to the cuve joining P 1 and P 2. This notion of the indeendence of ath of line integals of cetain vecto fields, imotant to theoy and alication, is chaacteized by the following thee theoems: Theoem 1. A necessay and sufficient condition that A d be indeendent of ath is that thee exists a scala function such that A ¼.

5 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM 233 Theoem 2. Anecessay and sufficient condition that the line integal, is that A ¼. Theoem 3. A d be indeendent of ath If A ¼, then the line integal of A ove an allowable closed ath is, i.e., A d ¼. If is a lane cuve, then Theoem 3 follows immediately fom Geen s theoem, since in the lane case A educes EXAMPLE. Newton s second law fo foces is F ¼ dmvþ, whee m is the mass of an object and v is its velocity. dt When F has the eesentation F ¼, itissaid to be consevative. The evious theoems tell us that the integals of consevative fields of foce ae indeendent of ath. Futhemoe, showing that F ¼ is the efeed way of showing that F is consevative, since it involves diffeentiation, while demonstating that exists such that F ¼ equies integation. URFAE INTEGRAL Ou evious double integals have been elated to a vey secial suface, the lane. Now we conside othe sufaces, yet, the aoach is quite simila. ufaces can be viewed intinsically, i.e., as non-euclidean saces; howeve, we do not do that. Rathe, the suface is thought of as embedded in a thee-dimensional Euclidean sace and exessed though a two-aamete vecto eesentation: x ¼ v 1 ; v 2 Þ While the uose of the vecto eesentation is to be geneal (that is, inteetable though any allowable thee-sace coodinate system), it is convenient to initially think in tems of ectangula atesian coodinates; theefoe, assume ¼ xi yj zk and that thee is a aametic eesentation x ¼ v 1 ; v 2 Þ; y ¼ v 1 ; v 2 Þ; z ¼ v 1 ; v 2 Þ The functions ae assumed to be continuously diffeentiable. The aamete cuves v 2 ¼ const and v 1 ¼ const establish a coodinate system on the suface (just as y ¼ const, and x ¼ const fom such a system in the lane). The key to establishing the suface integal of a function is the diffeential element of suface aea. (Fo the lane that element is da ¼ dx; dy.) At any oint, P, ofthe suface dv 2 sans the tangent lane to the suface. In aticula, the diections of the coodinate cuves v 2 ¼ const and v 1 ¼ const ae designated by dx 1 1 and dx 2 dv 2, esectively (see Fig. 1-3). 2 The coss oduct dx 1 xdx 1 dv 2 2 is nomal to the tangent lane at P, and @v is the aea of a diffeential coodinate aallelogam. 2 11Þ

6 234 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP. 1 Fig. 1-3 (This is the usual geometic inteetation of the coss oduct abstacted to the diffeential level.) This stongly suggests the following definition: Definition. The diffeential element of suface aea is @v dv 1 dv 2 2 Fo a function v 1 ; v 2 Þ that is eveywhee integable on d ¼ v 1 ; v @v dv 1 dv 2 2 is the suface integal of the function : In geneal, the suface integal must be efeed to thee-sace coodinates to be evaluated. suface has the atesian eesentation z ¼ f x; yþ and the identifications ae made then and Theefoe, v 1 ¼ x; v 2 ¼ y; z ¼ f v 1 ; v 2 k ¼ # 2 12Þ 13Þ If the Thus, the suface integal of has the secial eesentation " ¼ x; y; zþ # dx dy 14Þ If the suface is given in the imlicit fom Fx; y; zþ ¼, then the gadient may be emloyed to obtain anothe eesentation. To establish it, ecall that at any suface oint P the gadient, F is eendicula (nomal) to the tangent lane (and hence to ). Theefoe, the following equality of the unit vectos holds (u to sign):

7 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM 235 F @ 2 15Þ [Now a conclusion of the theoy of imlicit functions is that fom Fx; y; zþ ¼ (and unde aoiate conditions) thee can be oduced an exlicit eesentation z ¼ f x; yþ of a otion of the suface. This is an existence statement. The theoem does not say that this eesentation can be exlicitly oduced.] With this fact in hand, we again let v 1 ¼ x; v 2 ¼ y; z ¼ f v 1 ; v 2 Þ. Then F ¼ F x i f y j F z k Taking the dot oduct of both sides of (15) yields F z jfj @v 2 The ambiguity of sign can be eliminated by taking the absolute value of both sides of the @v ¼ jfj 2 jf z j ¼ ½F xþ 2 F y Þ 2 F z Þ 2 Š 1=2 jf z j and the suface integal of takes the fom ½F x Þ 2 F y Þ 2 F z Þ 2 Š 1=2 dx dy jf z j The fomulas (14) and (16) also can be intoduced in the following nonvectoial manne. Let be a two-sided suface having ojection on the xy lane as in the adjoining Fig Assume that an equation fo is z ¼ f x; yþ, whee f is single-valued and continous fo all x and y in. Divide into n subegions of aea A ; ¼ 1; 2;...; n, and eect a vetical column on each of these subegions to intesect in an aea. 16Þ Fig. 1-4 Let x; y; zþ be single-valued and continuous at all oints of. Fom the sum X n ¼1 ; ; Þ 17Þ

8 236 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP. 1 whee ; ; Þ is some oint of. If the limit of this sum as n!1in such a way that each! exists, the esulting limit is called the suface integal of x; y; zþ ove and is designated by x; y; zþ d 18Þ ince ¼jsec j A aoximately, whee is the angle between the nomal line to and the ositive z-axis, the limit of the sum (17) can be witten x; y; zþj sec j da 19Þ The quantity j sec j is given by j sec j ¼ 1 jn kj ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi @y Then assuming that z ¼ f x; yþ has continuous (o sectionally continuous) deivatives in, (19) can be witten in ectangula fom as sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x; y; zþ 2 dx In case the equation fo is given as Fx; y; zþ ¼, (21) can also be witten qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F x Þ 2 F y Þ 2 F z Þ 2 x; y; zþ dx dy jf z j The esults (21) o(22) can be used to evaluate (18). In the above we have assumed that is such that any line aallel to the z-axis intesects in only one oint. In case is not of this tye, we can usually subdivide into sufaces 1 ; 2 ;...; which ae of this tye. Then the suface integal ove is defined as the sum of the suface integals ove 1 ; 2 ;... The esults stated hold when is ojected on to a egion on the xy lane. In some cases it is bette to oject on to the yz o xz lanes. Fo such cases (18) can be evaluated by aoiately modifying (21) and (22). 2Þ 22Þ THE DIERGENE THEOREM The divegence theoem establishes equality between tile integal (volume integal) of a function ove a egion of thee-dimensional sace and the double integal of the function ove the suface that bounds that egion. This elation is vey imotant in the exession of hysical theoy. (ee Fig. 1-5.) Divegence (o Gauss) Theoem Let A be a vecto field that is continuously diffeentiable on a closed-sace egion,, bound by a smooth suface,. Then Ad ¼ A n d 23Þ whee n is an outwadly dawn nomal. If n is exessed though diection cosines, i.e., n ¼ i cos j cos k cos, then (23) may be witten

9 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM d A 1 cos A 2 cos A 3 cos Þ d The ectangula atesian comonent fom of (23) d ¼ 1 dy dz A 2 dz dx A 3 dx dyþ 24Þ 25Þ EXAMPLE. If B is the magnetic field vecto, then one of Maxwell s equations of electomagnetic theoy is B ¼. When this equation is substituted into the left membe of (23), the ight membe tells us that the magnetic flux though a closed suface containing a magnetic field is zeo. A simle inteetation of this fact esults by thinking of a magnet enclosed in a ball. All magnetic lines of foce that flow out of the ball must etun (so that the total flux is zeo). Thus, the lines of foce flow fom one ole to the othe, and thee is no disesion. TOKE THEOREM tokes theoem establishes the equality of the double integal of a vecto field ove a otion of a suface and the line integal of the field ove a simle closed cuve bounding the suface otion. (ee Fig. 1-6.) uose a closed cuve,, bounds a smooth suface otion,. If the comonent functions of x ¼ v 1 ; v 2 Þ have continuous mixed atial deivatives, then fo a vecto field A with continuous atial deivatives on A d ¼ n Ad 26Þ whee n ¼ cos i cos j cos k with ;, and eesenting the angles made by the outwad nomal n and i; j, and k, esectively. Then the comonent fom of (26) is A 1 dx A 2 dy A 3 dzþ 3 If A ¼, tokes theoem tells us that cos A d ¼. This is Theoem 3 on Page 237. d 27Þ

10 238 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP. 1 Fig. 1-6 olved Poblems LINE INTEGRAL 1.1. Evaluate 1;2Þ ;1Þ x 2 yþ dx y 2 xþ dy along (a) astaight line fom ; 1Þ to 1; 2Þ, (b) staight lines fom ; 1Þ to 1; 1Þ and then fom 1; 1Þ to 1; 2Þ, (c) the aabola x ¼ t, y ¼ t 2 1. (a) (b) An equation fo the line joining ; 1Þ and 1; 2Þ in the xy lane is y ¼ x 1. Then dy ¼ dx and the line integal equals 1 fx 2 x 1Þg dx fx 1Þ 2 xg dx ¼ x¼ 1 2x 2 2xÞ dx ¼ 5=3 Along the staight line fom ; 1Þ to 1; 1Þ, y ¼ 1; dy ¼ and the line integal equals 1 x 2 1Þ dx 1 xþþ ¼ 1 x¼ x 2 1Þ dx ¼ 2=3 Along the staight line fom 1; 1Þ to 1; 2Þ, x ¼ 1; dx ¼ and the line integal equals 2 1 yþþy 2 1Þ dy ¼ 2 y¼1 1 y 2 1Þ dy ¼ 1=3 (c) Then the equied value ¼ 2=31=3 ¼ 8:3. ince t ¼ at; 1Þ and t ¼ 1at1; 2Þ, the line integal equals 1 ft 2 t 2 1Þg dt ft 2 1Þ 2 tg 2tdt¼ t¼ 1 2t 5 4t 3 2t 2 2t 1Þ dt ¼ If A ¼3x 2 6yzÞi 2y3xzÞj 1 4xyz 2 Þk, evaluate A d fom ; ; Þ to 1; 1; 1Þ along the following aths : aþ x ¼ t; y ¼ t 2 ; z ¼ t 3 bþ The staight lines fom ; ; Þ to ; ; 1Þ, then to ; 1; 1Þ, and then to 1; 1; 1Þ cþ The staight line joining ; ; Þ and 1; 1; 1Þ A d ¼ f3x 2 6yzÞi 2y3xzÞj 1 4xyz 2 Þkgdxi dyj dzkþ ¼ 3x 2 6yzÞ dx 2y3xzÞ dy 1 4xyz 2 Þ dz (a) If x ¼ t; y ¼ t 2 ; z ¼ t 3, oints ; ; Þ and 1; 1; 1Þ coesond to t ¼ and t ¼ 1, esectively. Then

11 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM A d ¼ ¼ t¼ 1 t¼ f3t 2 6t 2 Þt 3 Þg dt f2t 2 3tÞt 3 Þg dt 2 Þf1 4tÞt 2 Þt 3 Þ 2 g dt 3 Þ 3t 2 6t 5 Þ dt 4t 3 6t 5 Þ dt 3t 2 12t 11 Þ dt ¼ 2 Anothe method: Along, A ¼3t 2 6t 5 Þi 2t 2 3t 4 Þj 1 4t 9 Þk and ¼ xi yj zk ¼ ti t 2 j t 3 k, d ¼i2tj3t 2 kþ dt. Then 1 A d ¼ 3t 2 6t 5 Þ dt 4t 3 6t 5 Þ dt 3t 2 12t 11 Þ dt ¼ 2 (b) Along the staight line fom ; ; Þ to ; 1; 1Þ, x ¼ ; y ¼ ; dx ¼ ; dy ¼, while z vaies fom to 1. Then the integal ove this at of the ath is 1 z¼ f3þ 2 6ÞzÞg f2þ3þzþg f1 4ÞÞz 2 Þg dz ¼ 1 z¼ dz ¼ 1 Along the staight line fom ; ; 1Þ to ; 1; 1Þ, x ¼ ; z ¼ 1; dx ¼ ; dz ¼, while y vaies fom to 1. Then the integal ove this at of the ath is 1 y¼ f3þ 2 6 yþ1þg f2y 3Þ1Þg dy f1 4Þ yþ1þ 2 g ¼ 1 y¼ 2ydy¼ 1 Along the staight line fom ; 1; 1Þ to 1; 1; 1Þ, y ¼ 1; z ¼ 1; dy ¼ ; dz ¼, while x vaies fom to 1. Then the integal ove this at of the ath is 1 x¼ Adding, f3x 2 61Þ1Þg dx f21þ3x1þg f1 4x1Þ1Þ 2 g ¼ A d ¼ ¼ 3: 1 x¼ 3x 2 6Þ dx ¼ 5 (c) The staight line joining ; ; Þ and 1; 1; 1Þ is given in aametic fom by x ¼ t; y ¼ t; z ¼ t. Then 1 A d ¼ 3t 2 6t 2 Þ dt 2t3t 2 Þ dt 1 4t 4 Þ dt ¼ 6=5 t¼ 1.3. Find the wok done in moving a aticle once aound an ellise in the xy lane, if the ellise has cente at the oigin with semi-majo and semi-mino axes 4 and 3, esectively, as indicated in Fig. 1-7, and if the foce field is given by F ¼3x 4y 2zÞi 4x 2y 3z 2 Þj 2xz 4y 2 z 3 Þk In the lane z ¼ ; F ¼3x 4yÞi4x2yÞj 4y 2 k and d ¼ dxi dyj so that the wok done is F d ¼ f3x 4yÞi 4x2yÞj 4y 2 kgdxi dyjþ ¼ 3x 4yÞ dx 4x2yÞdy hoose the aametic equations of the ellise as x ¼ 4cost, y ¼ 3 sin t, whee t vaies fom to 2 (see Fig. 1-7). Then the line integal equals 2 t¼ 2 ¼ f34costþ 43 sin tþgf 4 sin tg dt f44costþ23 sin tþgf3costg dt t¼ 48 3 sin t cos tþ dt ¼48t 15 sin 2 tþj 2 ¼ 96 y t Fig. 1-7 = xi + yj = 4 cos t i + 3 sin t j x

12 24 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP. 1 In tavesing we have chosen the counteclockwise diection indicated in Fig We call this the ositive diection, o say that has been tavesed in the ositive sense. If wee tanvesed in the clockwise (negative) diection, the value of the integal would be Evaluate yds along the cuve given by y ¼ 2 ffiffiffi x fom x ¼ 3tox¼24. ince ds ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx 2 dy 2 ¼ 1 y Þ 2 dx ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1=x dx, wehave 24 yds¼ 2 ffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 24 ffiffiffiffiffiffiffiffiffiffiffi x 1 1=x dx ¼ 2 x 1 dx ¼ x 1Þ3=2 ¼ GREEN THEOREM IN THE PLANE 1.5. Pove Geen s theoem in the lane if is a closed cuve which has the oety that any staight line aallel to the coodinate axes cuts in at most two oints. Let the equations of the cuves AEB and AFB (see adjoining Fig. 1-8) be y ¼ Y 1 xþ and y ¼ Y 2 xþ, esectively. If is the egion bounded by, b Y2 dx dy x¼a y¼y 1 dy dx b b ¼ Px; yþj Y 2xÞ y¼y 1 xþ dx ¼ Y 2 Þ Px; Y 1 ÞŠ dx x¼a a½px; b a ¼ Px; Y 1 Þ dx Px; Y 2 Þ dx ¼ Pdx Then Then a b Pdx¼ dx imilaly let the equations of cuves EAF and EBF be x ¼ X 1 yþ and x ¼ X 2 yþ f X2 f dx dy y¼c x¼x 1 dx dy ¼ ½QX 2 ; yþ QX 1 ; yþš dy c c f ¼ QX 1 ; yþ dy QX 2 ; yþ dy ¼ Qdy f c Adding (1) and (2), Qdy¼ dx dx dy Fig. 1-8 Then 1.6. eify Geen s theoem in the lane fo 2xy x 2 Þ dx xy 2 Þ dy whee is the closed cuve of the egion bounded by y ¼ x 2 and y 2 ¼ x.

13 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM 241 The lane cuves y ¼ x 2 and y 2 ¼ x intesect at ; Þ and 1; 1Þ. is as shown in Fig Along y ¼ x 2,the line integal equals The ositive diection in tavesing 1 f2xþx 2 Þ x 2 g dx fx x 2 Þ 2 g dx 2 Þ¼ x¼ Along y 2 ¼ x the line integal equals f2 y 2 ÞyÞ y 2 Þ 2 g d y 2 Þfy 2 y 2 g dy ¼ y¼ x 3 x 2 2x 5 Þ dx ¼ 7=6 4y 4 2y 5 2y 2 Þ dy ¼ 17=15 Then the equied line integal ¼ 7=6 17=15 @ dx @x x 2xy x2 Þ dx dy ¼ 1 2xÞ dx dy ¼ ¼ ¼ 1 x¼ 1 y 2xyÞj Hence, Geen s theoem is veified. ffiffi 1 x dx y¼x 2 x¼ ffiffi x y¼x 2 1 2xÞ dy dx x 1=2 2x 3=2 x 2 2x 3 Þ dx ¼ 1=3 Fig Extend the oof of Geen s theoem in the lane given in Poblem 1.5 to the cuves fo which lines aallel to the coodinate axes may cut in moe than two oints. onside a closed cuve such as shown in the adjoining Fig. 1-1, in which lines aallel to the axes may meet in moe than two oints. By constucting line T the egion is divided into two egions 1 and 2, which ae of the tye consideed in Poblem 1.5 and fo which Geen s theoem alies, i.e., 1Þ 2Þ TU T Pdx Qdy¼ 1 Pdx dx dy; dx dy Fig. 1-1 Adding the left-hand sides of (1) and (2), we have, omitting the integand Pdx Qdy in each case, ¼ ¼ ¼ TUT TU T T TU T T TU T TUT using the fact that ¼. T T Adding the ight-hand sides of (1) and (2), omitting the integand, ¼ egions 1 and 2. Then Pdx Qdy¼ dx dy and the theoem whee consists of A egion such as consideed hee and in Poblem 1.5, fo which any closed cuve lying in can be continuously shunk to a oint without leaving, iscalled a simly connected egion. A egion which is not

14 242 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP. 1 simly connected is called multily connected. We have shown hee that Geen s theoem in the lane alies to simly connected egions bounded by closed cuves. In Poblem 1.1 the theoem is extended to multily connected egions. Fo moe comlicated simly connected egions, it may be necessay to constuct moe lines, such as T, toestablish the theoem how that the aea bounded by a simle closed cuve is given by 1 2 xdy ydx. In Geen s theoem, ut P ¼ y; Q ¼ x. xdy xþ yþ dx dy ¼ 2 dx dy ¼ 2A whee A is the equied aea. Thus, A ¼ 1 2 xdy ydx Find the aea of the ellise x ¼ a cos ; y ¼ b sin. 2 Aea ¼ 1 2 xdy ydx¼ 1 2 a cos Þb cos Þ d bsin Þ a sin Þ d ¼ abcos 2 sin 2 Þ d ¼ ab d ¼ ab 1.1. how that Geen s theoem in the lane is also valid fo a multily connected egion such as shown in Fig The shaded egion, shown in the figue, is multily connected since not evey closed cuve lying in can be shunk to a oint without leaving, asisobseved by consideing a cuve suounding DEFGD, fo examle. The bounday of, which consists of the exteio bounday AHJKLA and the inteio bounday DEFGD, istobetavesed in the ositive diection, so that a eson taveling in this diection always has the egion on his left. It is seen that the ositive diections ae those indicated in the adjoining figue. In ode to establish the theoem, constuct a line, such as AD, called a coss-cut, connecting the exteio and inteio boundaies. The egion bounded by ADEFGDALKJHA is simly connected, and so Geen s theoem is valid. Then ADEFGDALKJHA dx dy But the integal on the left, leaving out the integand, is equal to ¼ AD DEFGD DA ALKJHA DEFGD ALKJHA since ¼. Thus, if 1 is the cuve ALKJHA, 2 is the cuve DEFGD and is the bounday of AD DA consisting of 1 and 2 (tavesed in the ositive diections), then ¼ and so 1 dx dy 2 Fig. 1-11

15 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM 243 INDEPENDENE OF THE PATH Let Px; yþ and Qx; yþ be continuous and have continuous fist atial deivatives at each oint of a simly connected egion. Pove that a necessay and sufficient condition that Pdx Qdy¼ aound evey closed ath in is identically in. ufficiency. Then by Geen Pdx Qdy¼ dx whee is the egion bounded by. Necessity. uose Pdx Qdy¼ aound evey closed ath in and at some oint of. In > attheoint x ; y Þ. By ae continuous in, sothat thee must be some egion containing x ; y Þ as an inteio oint >. If is the bounday of, then by Geen Pdx Qdy¼ dx contadicting the hyothesis that Pdx Qdy¼ foall closed cuves cannot be ositive. imilaly, we can cannot be negative, and it follows that it must be identically zeo, identically in Let P and Q be defined as in Poblem Pove that a necessay and sufficient condition that B A Pdx Qdy be indeendent of the ath in joining oints A and B is identically in. ufficiency. thenbypoblem 1.11, Pdx Qdy¼ A D 1 2 E Fig B ADBEA (see Fig. 1-12). Fom this, omitting fo bevity the integand Pdx Qdy, wehave ¼ ; ¼ ¼ and so ¼ 1 2 ADB BEA ADB BEA AEB i.e., the integal is indeendent of the ath. Necessity. If the integal is indeendent of the ath, then fo all aths 1 and 2 in we have ¼ ; ¼ and ¼ 2 1 ADB AEB ADBEA Fom this it follows that the line integal aound any closed ath in is zeo, and hence by Poblem 1.11

16 244 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP Let P and Q be as in Poblem (a) Pove that a necessay and sufficient condition that Pdx Qdy be an exact diffeential of a function x; yþ is (b) how that in such case oints. (a) Necessity. B A Pdx Qdy¼ B A d ¼ BÞ AÞ whee A and B ae any two If Pdx Qdy¼ dx dy, anexact diffeential, then ¼ P, ¼. Thus, by diffeentiating (1) and (2) with esect to y and x, since we ae assuming continuity of the atial deivatives. ufficiency. By Poblem 1.12, then Pdx Qdy is indeendent of the ath joining two oints. In aticula, let the two oints be a; bþ and x; yþ and define x; yþ ¼ Then x x; yþ x; yþ ¼ ¼ x;yþ a;bþ xx;y a;bþ xx;yþ x;yþ Pdx Qdy Pdx Qdy Pdx Qdy x;yþ a;bþ Pdx Qdy ince the last integal is indeendent of the ath joining x; yþ and x x; yþ, wecan choose the ath to be a staight line joining these oints (see Fig. 1-13) so that dy ¼. Then by the mean value theoem fo integals, x x; yþ x; yþ ¼ 1 x x xx;yþ x;yþ Pdx¼ Px x; yþ <<1 Taking the limit as x!, we ¼ P. imilaly we can show ¼ Q. Thus it follows that dx dy y (x, y) (x + Dx, y) (a, b) Fig x

17 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM 245 (b) Let A ¼x 1 ; y 1 Þ; B ¼x 2 ; y 2 Þ. Fom at (a), x; yþ ¼ x;yþ a;bþ Then omitting the integand Pdx Qdy,wehave B A ¼ x2 ;y 2 Þ x2 ;y 2 Þ x1 ;y 1 Þ ¼ x 1 ;y 1 Þ a;bþ a;bþ Pdx Qdy ¼ x 2 ; y 2 Þ x 1 ; y 1 Þ¼BÞ AÞ (a) Pove that 3;4Þ 1;2Þ 6xy 2 y 3 Þ dx 6x 2 y 3xy 2 Þ dy is indeendent of the ath joining 1; 2Þ and 3; 4Þ. (b) Evaluate the integal in (a). (a) P ¼ 6xy 2 y 3 ; Q ¼ 6x 2 y 3xy 2. ¼ 12xy 3y 2 and by Poblem 1.12 the line integal is indeendent of the ath. (b) Method 1: ince the line integal is indeendent of the ath, choose any ath joining 1; 2Þ and 3; 4Þ, fo examle that consisting of lines fom 1; 2Þ to 3; 2Þ [along which y ¼ 2; dy ¼ ] and then 3; 2Þ to 3; 4Þ [along which x ¼ 3; dx ¼ ]. Then the equied integal equals Evaluate 3 x¼1 24x 8Þ dx 4 y¼2 54y 9y 2 Þ dy ¼ ¼ 236 Method @x ; we must ¼ 6xy2 y 3 ¼ 6x2 y 3xy 2 : Fom (1), ¼ 3x 2 y 2 xy 3 f yþ. Fom (2), ¼ 3x 2 y 2 xy 3 gxþ. The only way in which these two exessions fo ae equal is if f yþ ¼gxÞ ¼c, aconstant. Hence ¼ 3x 2 y 2 xy 3 c. Then by Poblem 1.13, 3;4Þ 1;2Þ 6xy 2 y 3 Þ dx 6x 2 y 3xy 2 Þ dy ¼ 3;4Þ 1;2Þ d3x 2 y 2 xy 3 cþ ¼ 3x 2 y 2 xy 3 cj 3;4Þ 1;2Þ ¼ 236 Note that in this evaluation the abitay constant c can be omitted. ee also Poblem 6.16, Page 131. We could also have noted by insection that 6xy 2 y 3 Þ dx 6x 2 y 3xy 2 Þ dy ¼6xy 2 dx 6x 2 ydyþ y 3 dx 3xy 2 dyþ fom which it is clea that ¼ 3x 2 y 2 xy 3 c. x 2=3 y 2=3 ¼ a 2=3 : ¼ d3x 2 y 2 Þ dxy 3 Þ¼d3x 2 y 2 xy 3 Þ x 2 y cos x 2xy sin x y 2 e x Þ dx x 2 sin x 2ye x Þ dy aound the hyocycloid P ¼ x 2 y cos x 2xy sin x y 2 e x ; Q ¼ x 2 sin x 2ye x ¼ x 2 cos x 2x sin x 2ye x by Poblem 1.11 the line integal aound any closed ath, in aticula x 2=3 y 2=3 ¼ a 2=3 is zeo. URFAE INTEGRAL If is the angle between the nomal line to any oint x; y; zþ of a suface and the ositive z-axis, ove that

18 246 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP. 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F j sec j ¼ 1 z 2 x z 2 x 2 Fy 2 Fz 2 y ¼ jf z j accoding as the equation fo is z ¼ f x; yþ o Fx; y; zþ ¼. If the equation of is Fx; y; zþ ¼, a nomal to at x; y; zþ is F ¼ F x i F y j F z k. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F k ¼jFjjkj cos o F z ¼ Fx 2 Fy 2 Fz 2 cos Then fom which j sec j ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Fx 2 Fy 2 Fz 2 jf z j as equied. In case the equation is z ¼ f x; yþ, we qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi can wite Fx; y; zþ ¼z f x; yþ ¼, fom which F x ¼ z x ; F y z y ; F z ¼ 1 and we find j sec j ¼ 1 z 2 x z 2 y Evaluate Ux; y; zþ d whee is the suface of the aaboloid z ¼ 2 x 2 y 2 Þ above the xy lane and Ux; y; zþ is equal to (a) 1, (b) x 2 y 2, (c) 3z. Give a hysical inteetation in each case. (ee Fig ) The equied integal is equal to qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ux; y; zþ 1 z 2 x z 2 y dx dy whee is the ojection of on the xy lane given by x 2 y 2 ¼ 2; z ¼. ince z x ¼ 2x; z y ¼ 2y, (1) can be witten qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ux; y; zþ 1 4x 2 4y 2 dx dy 2Þ 1Þ (a) If Ux; y; zþ ¼1, (2) becomes qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 4x 2 4y 2 dx dy Fig To evaluate this, tansfom to ola coodinates ; Þ. Then the integal becomes 2 ¼ ffiffi 2 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d d ¼ ¼ ffiffi Þ 3=2 2 d ¼ 13 3 ¼ (b) Physically this could eesent the suface aea of, othemass of assuming unit density. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi If Ux; y; zþ ¼x 2 y 2,(2) becomes x 2 y 2 Þ 1 4x 2 4y 2 dx dy o in ola coodinates 2 ffiffi 2 3 ¼ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d d ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi whee the integation with esect to is accomlished by the substitution ¼ u. Physically this could eesent the moment of inetia of about the z-axis assuming unit density, o the mass of assuming a density ¼ x 2 y 2.

19 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM 247 (c) If Ux; y; zþ ¼3z, (2) becomes qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3z 1 4x 2 4y 2 dx dy ¼ 3f2 x 2 y 2 Þg 1 4x 2 4y 2 dx dy o in ola coodinates, 2 ¼ ffiffi 2 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 32 2 Þ d d ¼ Physically this could eesent the mass of assuming a density ¼ 3z, othee times the fist moment of about the xy lane Find the suface aea of a hemishee of adius a cut off by a cylinde having this adius as diamete. Equations fo the hemishee and cylinde (see Fig. 1-15) ae ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi given esectively by x 2 y 2 z 2 ¼ a 2 (o z ¼ a 2 x 2 y 2 Þ and x a=2þ 2 y 2 ¼ a 2 =4(ox 2 y 2 ¼ ax). ince x z x ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi and y z y ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 x 2 y 2 a 2 x 2 y 2 we have qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Requied suface aea ¼ 2 1 z 2 x z 2 a y dx dy ¼ 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx dy a 2 x 2 y 2 Fig Two methods of evaluation ae ossible. Method 1: Using ola coodinates. ince x 2 y 2 ¼ ax in ola coodinates is ¼ a cos, theintegal becomes =2 2 ¼ a cos ¼ a =2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d d ¼ 2a ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 2 a 2 2 ¼ a cos ¼ d =2 ¼ 2a 2 1 sin Þ d ¼ 2Þa 2 Method 2: The integal is equal to a ffiffiffiffiffiffiffiffiffiffi ax x 2 2 x¼ y¼ a ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 x 2 y 2 a dy dx ¼ 2a ¼ 2a sin 1 y ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ax x 2 ffiffiffiffiffiffiffiffiffiffiffi x sin 1 dx a x x¼ a ffiffiffiffiffiffiffiffiffiffi ax x 2 y¼ dx Letting x ¼ a tan 2,thisintegal becomes =4 4a 2 tan sec 2 d ¼ 4a tan2 j =4 1 2 ¼ 2a 2 tan 2 j =4 =4 =4 n ¼ 2a 2 =4 tan Þj =4 tan 2 d sec 2 1Þ d o ¼ 2Þa 2 Note that the above integals ae actually imoe and should be teated by aoiate limiting ocedues (see Poblem 5.74, hate 5, and also hate 12).

20 248 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP Find the centoid of the suface in Poblem qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi zd z 1 4x 2 4y 2 dx dy By symmety, x ¼ y ¼ and z ¼ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d 1 4x 2 4y 2 dx dy The numeato and denominato can be obtained fom the esults of Poblems 1.17(c) and 1.17(a), esectively, and we thus have z ¼ 37=1 13=3 ¼ Evaluate A n d, whee A ¼ xyi x 2 j xzþk, is that otion of the lane 2x 2y z ¼ 6 included in the fist octant, and n is a unit nomal to. (ee Fig ) A nomal to is 2x 2y z 6Þ ¼ 2i 2i 2j k 2i 2j k 2j k, and so n ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼. Then A n ¼fxyi x 2 2i 2j k j xzþkg 3 ¼ 2xy 2x2 x zþ 3 ¼ 2xy 2x2 x 6 2x 2yÞ 3 ¼ 2xy 2x2 x 2y 6 3 The equied suface integal is theefoe Fig. 1-16! 2xy 2x 2! x 2y 6 2xy 2x 2 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2y 6 d ¼ 1 z 2 x z 2 y dx dy 3 3! 2xy 2x 2 x 2y 6 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ dx dy 3 ¼ ¼ 3 3 x x¼ y¼ 3 x¼ 2xy 2x 2 x 2y 6Þ dy dx xy 2 2x 2 y xy y 2 6yÞj 3 x dx ¼ 27= In dealing with suface integals we have esticted ouselves to sufaces which ae two-sided. Give an examle of a suface which is not two-sided. Take a sti of ae such as ABD as shown in the adjoining Fig Twist the sti so that oints A and B fall on D and, esectively, as in the adjoining figue. If n is the ositive nomal at oint P of the suface, we find that as n moves aound the suface, it eveses its oiginal diection when it eaches P again. If we tied to colo only one side of the suface, we would find the whole thing coloed. This suface, called a Mo bius sti, Fig. 1-17

21 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM 249 is an examle of a one-sided suface. This is sometimes called a nonoientable suface. A two-sided suface is oientable. THE DIERGENE THEOREM Pove the divegence theoem. (ee Fig ) Fig Let be a closed suface which is such that any line aallel to the coodinate axes cuts in at most two oints. Assume the equations of the lowe and ue otions, 1 and 2,tobez ¼ f 1 x; yþ and z ¼ f 2 x; yþ, esectively. Denote the ojection of the suface on the xy lane by. d f2 x;yþ dz dy dx ¼ z¼f 1 dz dy dx f 2 ¼ A 3 x; y; zþ dy dx ¼ ½A 3 x; y; f 2 Þ A 3 x; y; f 1 ÞŠ dy dx z¼f 1 Fo the ue otion 2, dy dx ¼ cos 2 d 2 ¼ k n 2 d 2 since the nomal n 2 to 2 makes an acute angle 2 with k. Fo the lowe otion 1, dy dx ¼ cos 1 d 1 ¼ k n 1 d 1 since the nomal n 1 to 1 makes an obtuse angle 1 with k. Then A 3 x; y; f 2 Þ dy dx ¼ A 3 k n 2 d 2 2 A 3 x; y; f 1 Þ dy dx ¼ A 3 k n 1 d 1 1 and A 3 x; y; f 2 Þ dy dx A 3 x; y; f 1 Þ dy dx ¼ A 3 k n 2 d 2 2 ¼ A 3 k n d 1 A 3 k n 1 d 1

22 25 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP. 1 so that d ¼ A 3 k n d imilaly, by ojecting on the othe coodinate 2Þ d ¼ A 1 i n d d ¼ A 2 j n d o Adding (1), (2), and d Ad ¼ A 1 i A 2 j A 3 kþnd A n d The theoem can be extended to sufaces which ae such that lines aallel to the coodinate axes meet them in moe than two oints. To establish this extension, subdivide the egion bounded by into subegions whose sufaces do satisfy this condition. The ocedue is analogous to that used in Geen s theoem fo the lane eify the divegence theoem fo A ¼2x zþix 2 yj xz 2 k taken ove the egion bounded by x ¼ ; x ¼ 1; y ¼ ; y ¼ 1; z ¼ ; z ¼ 1. We fist evaluate A n d whee is the suface of the cube in Fig Face DEFG: n ¼ i; x ¼ 1. Then DEFG A n d ¼ ¼ f2 zþi j z 2 kgi dy dz 2 zþ dy dz ¼ 3=2 Face ABO: n ¼ i; x ¼. Then ABO A n d ¼ ¼ ziþ iþ dy dz zdydz¼ 1=2 Face ABEF: n ¼ j; y ¼ 1. Then 1 1 A n d ¼ f2x zþi x 2 j xz 2 kgjdx dz ¼ 1 1 ABEF x 2 dx dz ¼ 1=3 Fig Face OGD: n ¼ j; y ¼. Then A n d ¼ 1 1 OGD f2x zþi xz 2 kg jþ dx dz ¼

23 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM 251 Face BDE: n ¼ k; z ¼ 1. Then 1 1 A n d ¼ f2x 1Þi x 2 yj xkgkdx dy ¼ BDE Face AFGO: n ¼ k; z ¼. Then A n d ¼ Adding, AFGO 1 1 A n d ¼ ¼ 11 6 : Ad ¼ the divegence theoem is veified in this case Evaluate n d, whee is a closed suface. By the divegence theoem, n d ¼ d whee is the volume enclosed by. 1 1 f2xi x 2 yjg kþ dx dy ¼ ince 2 x 2 2xzÞ dx dy dz ¼ 11 k @z d ¼ 3 d ¼ xdxdy 1= Evaluate xz 2 dy dz x 2 y z 3 Þ dz dx 2xy y 2 zþ dx dy, whee is the entie suface of the hemisheical egion bounded by z ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 x 2 y 2 and z ¼ (a) bythe divegence theoem (Geen s theoem in sace), (b) diectly. (a) ince dy dz ¼ d cos ; dz dx ¼ d cos ; dx dy ¼ d cos, the integal can be witten fxz 2 cos x 2 y z 3 Þ cos 2xy y 2 zþ cos g d ¼ A n d whee A ¼ xz 2 i x 2 y z 3 Þj 2xy y 2 zþk and n ¼ cos i cos j cos k, theoutwad dawn unit nomal. Then by the divegence theoem the integal equals x2 y z 2xy y2 zþ d ¼ x 2 y 2 z 2 Þ d whee is the egion bounded by the hemishee and the xy lane. By use of sheical coodinates, as in Poblem 9.19, hate 9, this integal is equal to 4 =2 =2 ¼ ¼ ¼ 2 2 sin d d d ¼ 2a5 5

24 252 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP. 1 (b) If 1 is the convex suface of the hemisheical egion and 2 is the base z ¼ Þ, then a ffiffiffiffiffiffiffiffiffi xz 2 a 2 y 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a ffiffiffiffiffiffiffiffiffi dy dz ¼ z 2 a 2 y 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 y 2 z 2 dz dy z 2 a 2 y 2 z 2 dz dy y¼ a z¼ y¼ a z¼ 1 a ffiffiffiffiffiffiffiffiffi x 2 y z 3 a 2 x 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Þ dz dx ¼ fx 2 a 2 x 2 z 2 z 3 g dz dx x¼ a x¼ 1 a ffiffiffiffiffiffiffiffiffi a 2 x 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f x 2 a 2 x 2 z 2 z 3 g dz dx x¼ a z¼ a ffiffiffiffiffiffiffiffiffi 2xy y 2 a 2 x 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi zþ dx dy ¼ ffiffiffiffiffiffiffiffiffi f2xy y 2 a 2 x 2 y 2 g dy dx x¼ a y¼ a 2 x 2 1 xz 2 dy dz ¼ ; x 2 y z 3 Þ dz dx ¼ ; 2 2 a ffiffiffiffiffiffiffiffiffi 2xy y 2 zþ dx dy ¼ f2xy y 2 a 2 x 2 Þg dx dy ¼ ffiffiffiffiffiffiffiffiffi 2xy dy dx ¼ x¼ a y¼ a 2 x By addition of the above, we obtain a ffiffiffiffiffiffiffiffiffi a 2 y 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 z 2 a 2 y 2 z 2 y¼ x¼ a dz dy 4 x¼ a 4 x¼ ffiffiffiffiffiffiffiffiffi a 2 x 2 z¼ ffiffiffiffiffiffiffiffiffi a 2 x 2 y¼ x 2 y 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 x 2 z 2 dz dx qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 x 2 y 2 dy dx ince by symmety all these integals ae equal, the esult is, on using ola coodinates, a ffiffiffiffiffiffiffiffiffi a 2 x 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi =2 a 12 y 2 a 2 x 2 y 2 dy dx ¼ 12 2 sin 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 2 d d ¼ 2a5 x¼ y¼ ¼ ¼ 5 TOKE THEOREM Pove tokes theoem. Let be a suface which is such that its ojections on the xy, yz, and xz lanes ae egions bounded by simle closed cuves, as indicated in Fig Assume to have eesentation z ¼ f x; yþ o x ¼ gy; zþ o y ¼ hx; zþ, whee f ; g; h ae single-valued, continuous, and diffeentiable functions. We must show that AÞnd ¼ ½ A 1 i A 2 j A 3 kþš n d ¼ A d whee is the bounday of. onside fist ½ A 1 iþš n d: i @ ince A @z A 1 k;

25 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM 253 Fig. 1-2 ½ A 1 iþš n d n n k d 1Þ If z ¼ f x; yþ is taken as the equation of, then the osition vecto to any oint of is ¼ xi yj zk ¼ xi yj f x; yþk k ¼ But is a vecto tangent to and thus eendicula n, sothat ubstitute in (1) to ¼ n n k ¼ o n @y n n n k d n n k d o ½ A 1 iþš n d 1 n 2Þ Now on, A 1 x; y; zþ ¼A 1 ½x; y; f x; yþš ¼ Fx; yþ; 1 and (2) ½ A 1 iþš n d n k d Then ½ A 1 iþš n d dx whee is the ojection of on the xy lane. By Geen s theoem fo the lane, the last integal equals Fdxwhee is the bounday of. ince at each oint x; yþ of the value of F is the same as the value of A 1 at each oint x; y; zþ of, and since dx is the same fo both cuves, we must have Fdx¼ A 1 dx

26 254 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP. 1 o ½ A 1 iþš n d ¼ A 1 dx imilaly, by ojections on the othe coodinate lanes, ½ A 2 jþš n d ¼ A 2 dy; ½ A 3 kþš n d ¼ A 3 dz Thus, by addition, AÞnd ¼ A d The theoem is also valid fo sufaces which may not satisfy the estictions imosed above. Fo assume that can be subdivided into sufaces 1 ; 2 ;...; k with boundaies 1 ; 2 ;...; k which do satisfy the estictions. Then tokes theoem holds fo each such suface. Adding these suface integals, the total suface integal ove is obtained. Adding the coesonding line integals ove 1 ; 2 ;...; k,theline integal ove is obtained eify toke s theoem fo A ¼ 3yi xzj yz 2 k, whee is the suface of the aaboloid 2z ¼ x 2 y 2 bounded by z ¼ 2 and is its bounday. ee Fig The bounday of is a cicle with equations x 2 y 2 ¼ 4; z ¼ 2 and aametic equations x ¼ 2cost; y ¼ 2 sin t; z ¼ 2, t < 2. Then A d ¼ 3ydx xz dy yz 2 dz Also, ¼ ¼ sin tþ 2 sin tþ dt 2costÞ2Þ2costÞ dt 12 sin 2 t 8cos 2 tþ dt ¼ 2 i @ A z 2 xþi z3þk 3y xz yz 2 Fig and Then n ¼ x2 y 2 2zÞ jx 2 y 2 2zÞj ¼ xi yj k ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : x 2 y 2 1 dx dy AÞnd ¼ AÞn jn kj ¼ xz 2 x 2 z 3Þ dx dy 8! 9 < ¼ x x2 y 2 2 x 2 x2 y 2 = 3 dx dy : 2 2 ; In ola coodinates this becomes 2 2 ¼ ¼ f cos Þ 4 =2Þ 2 cos 2 2 =2 3g d d ¼ 2

27 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM Pove that a necessay and sufficient condition that A ¼ identically. ufficiency. uose A ¼. Then by tokes theoem A d ¼ AÞnd ¼ A d ¼ fo evey closed cuve is that Necessity. uose A d ¼ aound evey closed ath, and assume A 6¼ at some oint P. assuming A is continuous, thee will be a egion with P as an inteio oint, whee A 6¼. Let be asuface contained in this egion whose nomal n at each oint has the same diection as A, i.e., A ¼ n whee is a ositive constant. Let be the bounday of. Then by tokes theoem A d ¼ AÞnd ¼ n n d > which contadicts the hyothesis that A d ¼ and shows that A ¼. It follows that A ¼ is also a necessay and sufficient condition fo a line integal indeendent of the ath joining oints P 1 and P 2. P2 Then P 1 A d to be Pove that a necessay and sufficient condition that A ¼ is that A ¼. ufficiency. If A ¼, thena ¼ ¼ by Poblem 7.8, ha. 7, Page 179. Necessity. If A ¼, thenbypoblem 1.28, A d ¼ aound evey closed ath and of the ath joining two oints which we take as a; b; cþ and x; y; zþ. Let us define A d is indeendent Then x; y; zþ ¼ x;y;zþ A d ¼ x;y;zþ a;b;cþ a;b;cþ A 1 dx A 2 dy A 3 dz x x; y; zþ x; y; zþ ¼ xx;y;zþ x;y;zþ A 1 dx A 2 dy A 3 dz ince the last integal is indeendent of the ath joining x; y; zþ and x x; y; zþ, wecan choose the ath to be a staight line joining these oints so that dy and dz ae zeo. Then x x; y; zþ x; y; zþ ¼ 1 x x xx;y;zþ x;y;zþ whee we have alied the law of the mean fo integals. Taking the limit of both sides as x! ¼ A 1. imilaly, we can show ¼ A 2 ¼ A 3 : Thus, A ¼ A 1 i A 2 j A 3 k ¼: A 1 dx ¼ A 1 x x; y; zþ << (a) Pove that a necessay and sufficient condition that A 1 dx A 2 dy A 3 dz ¼ d, anexact diffeential, is that A ¼ whee A ¼ A 1 i A 2 j A 3 k. (b) how that in such case, x2 ;y 2 ;z 2 Þ x 1 ;y 1 ;z 1 Þ A 1 dx A 2 dy A 3 dz ¼ x2 ;y 2 ;z 2 Þ x 1 ;y 1 ;z 1 Þ d ¼ x 2 ; y 2 ; z 2 Þ x 1 ; y 1 ; z 1 Þ

28 256 LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM [HAP. 1 (a) Necessity. If A 1 dx A 2 dy A 3 dz ¼ @ dx dy ¼ A ¼ A ¼ A 3 Then by diffeentiating we have, assuming continuity of the atial deivatives, which is ecisely the condition A Anothe method: If A 1 dx A 2 dy A 3 dz ¼ d, then A ¼ A 1 i A 2 j A 3 k ¼ fom which A ¼ ¼. ufficiency. If A ¼, thenbypoblem 1.29, A ¼ and A 1 dx A 2 dy A 3 dz ¼ A d @ dx dy dz @z (b) Fom at (a), x; y; zþ ¼ x;y;zþ a;b;cþ A 1 dx A 2 dy A 3 dz. Then omitting the integand A 1 dx A 2 dy A 3 dz, wehave x2 ;y 2 ;z 2 Þ x 1 ;y 1 ;z 1 Þ ¼ x2 ;y 2 ;z 2 Þ x1 ;y 1 ;z 1 Þ a;b;cþ a;b;cþ ¼ x 2 ; y 2 ; z 2 Þ x 1 ; y 1 ; z 1 Þ (a) Pove that F ¼2xz 3 6yÞi 6x 2yzÞj 3x 2 z 2 y 2 Þk is a consevative foce field. (b) Evaluate F d whee is any ath fom 1; 1; 1Þ to 2; 1; 1Þ. (c) Give a hysical inteetation of the esults. (a) Afoce field F is consevative if the line integal F d is indeendent of the ath joining any two oints. A necessay and sufficient condition that F be consevative is that F ¼. i @ ince hee @z ¼ ; F is consevative 2xz 3 6y 6x 2yz 3x 2 z 2 y 2 (b) Method 1: By Poblem 1.3, F d ¼2xz 3 6yÞ dx 6x 2yzÞ dy 3x 2 z 2 y 2 Þ dz is an exact diffeential d, whee is such ¼ 2xz3 6y Fom these we obtain, ¼ 6x ¼ 3x2 z 2 y 2 ¼ x 2 z 3 6xy f 1 y; zþ ¼ 6xy y 2 z f 2 x; zþ ¼ x 2 z 3 y 2 z f 3 x; yþ These ae consistent if f 1 y; zþ ¼ y 2 z c; f 2 x; zþ ¼x 2 z 3 c; f 3 x; yþ ¼6xy c, in which case ¼ x 2 z 3 6xy y 2 z c. Thus, by Poblem 1.3, 2;1; 1Þ 1; 1;1Þ F d ¼ x 2 z 3 6xy y 2 z cj 2;1; 1Þ 1; 1;1Þ ¼ 15

29 HAP. 1] LINE INTEGRAL, URFAE INTEGRAL, AND INTEGRAL THEOREM 257 Altenatively, we may notice by insection that F d ¼2xz 3 dx 3x 2 z 2 dzþ6ydx 6xdyÞ 2yz dy y 2 dzþ ¼ dx 2 z 3 Þd6xyÞ dy 2 zþ¼dx 2 z 3 6xy y 2 z cþ fom which is detemined. Method 2: ince the integal is indeendent of the ath, we can choose any ath to evaluate it; in aticula, we can choose the ath consisting of staight lines fom 1; 1; 1Þ to 2; 1; 1Þ, thento 2; 1; 1Þ and then to 2; 1; 1Þ. The esult is 2 2x 6Þ dx yÞ dy 1 x¼1 y¼ 1 z¼1 12z 2 1Þ dz ¼ 15 (c) whee the fist integal is obtained fom the line integal by lacing y ¼ 1; z ¼ 1; dy ¼ ; dz ¼ ; the second integal by lacing x ¼ 2; z ¼ 1; dx ¼ ; dz ¼ ; and the thid integal by lacing x ¼ 2; y ¼ 1; dx ¼ ; dy ¼. Physically F d eesents the wok done in moving an object fom 1; 1; 1Þ to 2; 1; 1Þ along. In a consevative foce field the wok done is indeendent of the ath joining these oints. MIELLANEOU PROBLEM (a) If x ¼ f u; vþ; y ¼ gu; vþ defines a tansfomation which mas a egion of the xy lane into a egion of the uv lane, ove yþ dx dy vþ du dv (b) Inteet geometically the esult in (a). (a) If (assumed to be a simle closed cuve) is the bounday of, thenbypoblem 1.8, dx dy ¼ 1 xdy ydx 2 Unde the given tansfomation the integal on the ight of (1) becomes @v dv ¼ y y whee is the maing of in the uv lane (we suose the maing to be such that is a simle closed cuve also). By Geen s theoem if is the egion in the uv lane bounded by,theight side of (2) equals x y du du yþ vþ du dv whee we have inseted absolute value signs so as to ensue that the esult is non-negative as is dx dy In geneal, we can show (see Poblem 1.83) yþ Fx; yþ dx dy ¼ Ff f u; vþ; gu; vþ du dv 3Þ 1Þ 2Þ