F / x everywhere in some domain containing R. Then, + ). (10.4.1)

Transcription

1 0.4 Green's theorem in the plne Double integrls over plne region my be trnsforme into line integrls over the bounry of the region n onversely. This is of prtil interest beuse it my simplify the evlution of n integrl. The trnsformtion n be one by the following theorem. Theorem : Green's theorem in the plne (Trnsformtion between ouble integrls n line integrls) Let be lose boune region (see Se. 0.3) in the xy -plne whose bounry onsists of finitely mny smooth urves (see Se. 0.). Let F( x, y) n F ( x, y) be funtions tht re ontinuous n hve ontinuous prtil erivtives F / y n F / x everywhere in some omin ontining. Then, ç - = ( + ). (0.4.) æ F F ö ç x y F x F y ç è x y ø Here we integrte long the entire bounry of in suh sense tht is on the left s we vne in the iretion of integrtion (see Fig. 3). Setting F FiF j n using eqn (9.9.), the vetoril form of eqn (0.4.) is, (url F) kx y = F r. (0.4.*) The proof follows fter Exmple. Exmple : Verifition of Green's theorem in the plne Before proving Green's theorem in the plne, we get use to the theorem by verifying it for F( y 7 y) i(xy x) j n the irle x y. Solution: In eqn (0.4.) on the left we get, æ F F ö - x y = [( y + ) -( y - 7)] x y = 9 x y 9 ç è x y ø =, sine the irulr isk hs n re of. June 5, /6

2 We now show tht the line integrl in eqn (0.4.) on the right gives the sme vlue of 9. We must orient ounterlokwise, sy, r() t osti sintj. Then r() t sintiostj, n on, F y 7y sin t7sin t, F xyxos tsin t os t. Hene the line integrl in eqn (0.4.) beomes, ( + ) = ( + ) F x F y F x F y t, [(sin t 7sin t)( sin t) (ostsin t os t)(os t)] t, 0 = ( sin t 7sin t os tsin t os t) t, 0 = = = 9, verifying Green's theorem. Proof: Green's theorem in the plne is first prove for speil region tht n be represente in both forms, n xb, u( x) y v( x), (see Fig. 33), y, p( y) x q( y), (see Fig. 34). Using eqn (0.3.3), we obtin for the seon term on the left sie of eqn (0.4.) tken without the minus sign (see Fig. 33), F bé v( x) F ù x y = y x y ê u( x) y ú ë û. (0.4.) Integrte the inner integrl, v ( x ) F ( ) (, ) y = y = F x y v x = F ( ) [ x, v ( x )]- F [ x, u ( x )]. u( x) y y= u x By inserting the bove expression into eqn (0.4.) we fin, June 5, /6

3 y F b b x y = F x v x x - F x u x x [, ( )] [, ( )], =- F[ x, v( x)] x- F[ x, u( x)] x. b b ** * Sine y vx ( ) represents the urve (Fig. 33) n y ux ( ) represents, the lst two integrls my be written s line integrls over n 33); therefore, ** * (oriente s in Fig. y F x y =- ** F x y x - * F x y x =- (, ) (, ), F( x, y) x. (0.4.3) This proves eqn (0.4.) in Green's theorem if F 0. The result remins vli if hs portions prllel to the y -xis (suh s n in Fig. 35). Inee, the integrls over these portions re zero beuse in eqn (0.4.3) on the right we integrte with respet to x. Hene these integrls my be e to the * ** integrls over n to obtin the integrl over the whole bounry in eqn (0.4.3). We now tret the first term in eqn (0.4.) on the left in the sme wy. Inste of eqn (0.3.3) we use eqn (0.3.4), n the seon representtion of the speil region (see Fig. 34). Then, F é F x ë x q( y) x y = êp( y) ù x y, úû = F [ q( y), y] y - F [ p( y), y] y, = F [ q( y), y] y + F [ p( y), y] y, = F ( x, y) y. Together with eqn (0.4.3) this gives eqn (0.4.) n proves Green's theorem for speil regions. June 5, /6

4 We now prove the theorem for region tht itself is not speil region but n be subivie into finitely mny speil regions (Fig. 36). In this se we pply the theorem to eh subregion n then the results; the left-hn members up to the integrl over while the right-hn members up to the line integrl over plus integrls over the urves introue for subiviing. The simple key observtion now is tht eh of the ltter integrls ours twie, tken one in eh iretion. Hene they nel eh other, leving us with the line integrl over Some pplitions of Green's theorem Exmple : Are of plne region s line integrl over the bounry In eqn (0.4.) we first hoose F 0, F x n then F y, F 0. This gives, n x y = x y x y =- y x, respetively. The ouble integrl is the re A of. By ition we hve, A= xy-yx, (0.4.4) ( ) where we integrte s inite in Green's theorem. This interesting formul expresses the re of in terms of line integrl over the bounry. It is use, for instne, in the theory of ertin plnimeters (mehnil instruments for mesuring re). For n ellipse x / y / b or x os t, y bsin t we get xsin t, ybost; thus from eqn (0.4.4) we obtin the fmilir formul for the re of the region boune by n ellipse, A= ( xy - yx ) t= éb t b t ùt 0 ê - - ú = 0 ë û b os ( sin ). Exmple 3: Are of plne region in polr oorintes Let r n be polr oorintes efine by x ros, y rsin. Then, x os r r sin, y sinr r os, n eqn (0.4.4) beomes formul tht is well known from lulus, nmely, A =. (0.4.5) r As n pplition of eqn (0.4.5), onsier the rioi r ( os ), where 0 (Fig. 37). Here, 3 A= - = ( os ). 0 June 5, /6

5 Exmple 4: Trnsformtion of ouble integrl of the Lplin of funtion into line integrl of its norml erivtive Use Green's theorem for eriving bsi integrl formul involving the Lplin. Tke funtion wxy (, ) tht is ontinuous n hs ontinuous first n seon prtil erivtives in omin of the xy -plne ontining region of the type inite in Green's theorem. We set F w/ y n F w/ x. Then F / y n F / x re ontinuous in, n in eqn (0.4.) on the left we obtin, x y x y F F w w w, (0.4.6) the Lplin of w. Furthermore, using those expressions for F n F, we get in (0.4.) on the right, x y w x w y ( + ) = æ + ö = ç æ - + ö ç, (0.4.7) F x F y F F s s èç s s ø çè y s x s ø where s is the r length of, n is oriente s shown in Fig. 38. The integrn of the lst integrl my be written s the ot prout, w w y x w y w x (gr w) n i j i j. (0.4.8) x y s s x s y s The vetor n is unit norml vetor to, beuse the vetor r() s r/ s x / s iy / s j is the unit tngent vetor of, n r n 0, so tht n is perpeniulr to r. Also, n is irete to the exterior of beuse in Fig. 38 the positive x -omponent x / s of r is the negtive y -omponent of n, n similrly t other points. From this n eqn (9.7.4) we see tht the left sie of eqn (0.4.8) is the erivtive of w in the iretion of the outwr norml of. This erivtive is lle the norml erivtive of w n is enote by w/ n; tht is, June 5, /6

6 w/ n(gr w) n. Beuse of eqns (0.4.6)-(0.4.8), Green's theorem gives the esire formul relting the Lplin to the norml erivtive, = wxy w s. n (0.4.9) For instne, w x y stisfies Lple's eqution w 0. Hene, its norml erivtive integrte over lose urve must give 0. This n be verifie iretly by integrtion, sy for the squre 0 x, 0 y. Green's theorem will be the essentil tool in the proof of very importnt integrl theorem, nmely, Stokes's theorem in Se June 5, /6