Green s Theorem. Let be the boundry of the unit squre, y, oriented ounterlokwise, nd let F be the vetor field F, y e y +, 2 y. Find F d r. Solution. Let s write P, y e y + nd Q, y 2 y, so tht F P, Q. Let be the region, y. The boundry of, oriented orretly so tht penguin wlking long it keeps on his left, is the given urve. So, Green s Theorem sys tht F d r Q P y da 2 e y da. We ompute this by onverting it to n iterted integrl: F d r 2 e y da 2 e y d dy 2 e y e y dy y e y y 2 e y dy 2. Let be the oriented urve onsisting of line segments from, to 2, 3 to 2, bk to,, nd let F, y y 2, 2. Find F d r. Solution. Here is piture of the urve, long with the interior of the tringle, whih we ll ll : 3 2 2 The boundry of, oriented orretly so tht penguin wlking long it keeps onhis left side, is tht is, it s with the opposite orienttion. So, Green s Theorem sys tht F d r Q P y da, where F P, Q. We re looking for F d r, whih we know is the negtive of
F d r. Therefore, F d r Q P y da 2 2y da 2 3/2 2 2 2 2y dy d 2y y 2 y3/2 3 4 2 d 2 4 3 y d 2 3. Find the re of the region enlosed by the prmeterized urve rt t t 2, t t 3, t..4.2.2 Solution. Let be the region in question. We know from #2 on the worksheet Double Integrls tht the re of is da. Normlly, we would evlute this by onverting it to n iterted integrl; in this se, tht s quite hllenging, so we ll insted use Green s Theorem to evlute this integrl. If we n ome up with vetor field F, y P, y, Q, y suh tht Q P y, then Green s Theorem will sy tht da F d r, where is the boundry of the region, trveled ounterlokwise so tht penguin wlking long keeps on his left. One suh vetor field is F, y,. We re given prmeteriztion rt of the urve, nd this prmeteriztion does in ft trvel the 2
urve ounterlokwise. So, da F d r F rt r t dt, t t 2 2t, 3t 2 dt t t 2 3t 2 dt t t 2 3t 3 + 3t 4 dt 2 t2 3 t3 3 4 t4 + 3 5 t5 t 6 t 4. Let F, y P, y, Q, y be ny vetor field defined on the region in 2 shown in the piture, nd let nd 2 be the oriented urves shown in the piture. Wht does Green s Theorem sy bout F d r, F d r, nd Q P y da? 2 2 Solution. The boundry of onsists of two urves, nd 2. A penguin wlking long in the indited diretion would indeed keep on his left, but penguin wlking long 2 in the indited diretion would hve on his right. So, the boundry of is relly together with 2, whih mens Q P y da F d r F d r. 2 5. Let F, y P, y, Q, y 2 + y, y. You n hek tht P y Q. 2 2 + y 2 Wht is wrong with the following resoning? P y Q, so F is onservtive. Solution. F is not defined t the origin, so its domin is 2 eept the point,. This domin is not simply onneted, so we nnot onlude nything from the ft tht P y Q. b Let be ny simple losed urve in 2 tht does not enlose the origin, oriented ounterlokwise. This is not ompletely obvious, but there s n esy wy to tell t the end whether the prmeteriztion went the right wy we re looking for n re, so our finl nswer must be positive. 3
A simple urve is urve tht does not ross itself. Use Green s Theorem to eplin why F d r. Solution. Sine does not go round the origin, F is defined on the interior of. The only point where F is not defined is the origin, but tht s not in. Therefore, we n use Green s Theorem, whih sys F d r Q P y da. Sine Q P y, this sys tht F d r. Let be positive onstnt, nd let be the irle 2 + y 2 2, oriented ounterlokwise. Prmeterize hek your prmeteriztion by plugging it into the eqution 2 + y 2 2, nd use the definition of the line integrl to show tht F d r. Why doesn t the resoning from b work in this se? Solution. One possible prmeteriztion of is rt os t, sin t, t 2π. Then, F d r s we wnted. 2π 2π 2π, F rt r t dt os t os t2 + sin t, sin t sin t, os t dt 2 os t2 + sin t 2 dt We nnot use the resoning from b sine F is not defined in the whole interior of in prtiulr, it s not defined t the origin, whih is inside. d Let be ny simple losed urve in 2 tht does enlose the origin, oriented ounterlokwise. Eplin why F d r. Hint: Use nd #4. Solution. No mtter wht looks like, we n drw gint irle 2 +y 2 2 round the origin tht enloses ll of. Let s orient this gint irle ounterlokwise nd ll it, nd let s hve be the region between nd : Notie tht F is defined on ll of beuse it is defined everywhere eept the origin, nd 4
doesn t inlude the origin. So, #4 tells us tht Q P y da We showed in tht F d r, so this simplifies to Q P y da F d r F d r. F d r. Sine Q P y inside of, the double integrl is relly double integrl of, so it s equl to. Therefore, we onlude tht F d r s well. e Is it vlid to onlude from the bove resoning tht, if F, y P, y, Q, y is vetor field defined everywhere eept the origin nd P y Q, then F is onservtive? Solution. No! The lultion in only pplied to this prtiulr vetor field F, y P, y, Q, y 2 + y, y. 2 2 + y 2 There re vetor fields tht re defined everywhere eept the origin nd stisfy P y Q but re still not onservtive; the vetor field in #4b of the worksheet The Fundmentl Theorem for Line Integrls; Grdient Vetor Fields is n emple. 6. In this problem, you ll prove Green s Theorem in the se where the region is retngle. Let F, y P, y, Q, y be vetor field on the retngle [, b] [, d] in 2. Show tht [Q, y P y, y] da d [Qb, y Q, y] dy b [P, d P, ] d. Solution. Let s first brek the given double integrl into differene of two double integrls: [Q, y P y, y] da Q, y da P y, y da. Now, we ll onvert the double integrls on the right side to iterted integrls. This is esy, sine the region is just retngle. However, we re going to do the two iterted integrls in different orders: it mkes sense to first integrte Q with respet to sine it s derivtive with respet to nd to first integrte P y with respet to y: [Q, y P y, y] da d b Q, y d dy b d P y, y dy d. When we integrte Q with respet to, we just get Q; similrly, when we integrte P y with respet to y, we just get P : d b [Q, y P y, y] da Q, y dy P, y d d whih is etly wht we were sked to show. b [Qb, y Q, y] dy b yd y [P, d P, ] d, 5
b Let be the boundry of, trversed ounterlokwise. d [Qb, y Q, y] dy b [P, d P, ] d. Show tht F d r is lso equl to Solution. Here is piture of : y d b As we n see, it s omposed of 4 piees, nd we ll prmeterize eh seprtely. The bottom piee hs y, so only vries, nd we n prmeterize it using r t t, with t b. The right piee hs b, so only y vries, nd we n prmeterize it using r 2 t b, t, t d. The top piee hs y d, so only vries, nd we d like to prmeterize it using r 3 t t, d. The slight problem with this is tht it goes the wrong diretion: s t inreses, t, d goes to the right. This is tully not problem, s long s we ount for it lter. So, we ll go hed nd use r 3 t t, d with t b. Similrly, for the left piee, we ll use r 4 t, t, t d. Here s digrm showing the vrious things we ve prmeterized: y d r 3 t r 4 t r 2 t r t b As we n see from the two digrms, F d r F d r + r t r 2t Plugging the four prmeteriztions into this, b F d r F d r F d r. r 3t r 4t F d r is equl to d b d F t,, dt + F b, t, dt F t, d, dt F, t, dt. Writing F, y P, y, Q, y, we n simplify this to b d b F d r P t, dt + Qb, t dt P t, d dt d Q, t dt. This is etly wht we were supposed to show, whih is more obvious if we renme t to be in the first nd third integrls, renme t to be y in the seond nd fourth integrls, nd rerrnge 6
the terms: F d r b d d P, d + Qb, y dy d d Qb, y dy Q, y dy [Qb, y Q, y] dy b b b P, d d P, d d + d b [P, d P, ] d Q, y dy P, d 7