Math 32B Discussion Session Week 8 Notes February 28 and March 2, f(b) f(a) = f (t)dt (1)


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1 Green s Theorem Mth 3B isussion Session Week 8 Notes Februry 8 nd Mrh, 7 Very shortly fter you lerned how to integrte singlevrible funtions, you lerned the Fundmentl Theorem of lulus the wy most integrtion tully gets done. The version you probbly found yourself using most frequently sys tht f(b) f() = f (t)dt () for suffiiently nie funtions f. Our primry use for this eqution is tht it llows us to evlute definite integrls. If we find n ntiderivtive for the funtion we re trying to integrte, we n evlute this ntiderivtive t the endpoints of our intervl nd ompute the differene. In theory, we ould lso use () to ompute this differene. Given funtion f, perhps we wnt to know how muh f(x) hnges s x vries from to b. Tht is, we wnt to know how muh elevtion the grph of f gins between nd b: Eqution () sys tht this vlue is given by the sum of the instntneous hnges in f. Even though f is sometimes inresing nd sometimes deresing, the totl hnge in f between nd b is given by summing these instntneous hnges. The Fundmentl Theorem of lulus hs generl form tht we now wnt to mimi. It sys tht the behvior of our funtion f on the boundry of the intervl [, b] tht is, t the endpoints nd b is relted to the behvior of the derivtive of f ross the intervl. Let s rell how we defined the line integrl of vetor field F long urve. We imgined thin wire submersed in some fluid, nd on this wire ws bed whih the fluid would push long the wire. We defined the line integrl F dr to be the totl work done by the fluid s it pushed the bed long the wire, nd determined tht this must equl F Tds, where T is positively oriented unit tngent vetor field to the urve.
2 Figure : The urve =. Let s restrit our ttention now to losed urves. In simply onneted domin, losed urve n be thought of s the boundry of some region. Tht is, the losed urve is to some region s the endpoints re to n intervl. If F is onservtive vetor field, we know tht the work done by F round is zero: F dr =. But wht bout when F is not onservtive? Will there lwys be nie wy to ompute F dr? n we ompre the behvior of F long to the behvior of some derivtive of F long region for whih is the boundry? This is problem we d like to ddress tody. To perform nïve nlysis, we ll restrit our ttention even further to losed urves of speifi type. Let s onsider retngle = {(x, y) : x b, y d} whih lies in the domin of vetor field F on R, nd we tke our urve to be the boundry of this retngle, written =. A piture of our setup n be seen in Figure As with the line integrl long, we would like to ompute the work done by F long the urve, oriented ounterlokwise, but this time we ll tke different pproh. onsider the work done long the bottom line of our retngle the work done from (, ) to (b, ). At eh point long this line, our positivelyoriented unit tngent vetor T is the vetor,, so the quntity F T is given by F T = F, F, = F. We find the work done long the bottom line by integrting with respet to x (this is the vrible tht hnges s we move long this line): work done long bottom line = F (x, )dx.
3 Next, onsider the work done long the top line of the retngle the line from (, d) to (b, d). This time our positivelyoriented tngent vetors re pointing to the left, so we hve T =,, mening tht F T = F. So work done long top line = F (x, d)dx. This mens tht the totl work done long the top nd bottom lines is given by work done long top nd bottom lines = (F (x, ) F (x, d))dx. Now beuse we re looking t retngle whih lies in the domin of F, we n use the Fundmentl Theorem of lulus to rewrite the quntity inside of this integrl, for eh x b: F (x, ) F (x, d) = Substituting this in, we see tht d F d (x, y)dy = work done long top nd bottom lines = F (x, y)dy. F (x, y)dydx. () A similr nlysis works for the left nd right lines. On the right, T =,, so F T = F, while on the left T =, nd F T = F. Then work long left nd right lines = (F (b, y) F (, y))dy = F (x, y)dxdy. (3) x Of ourse we find the totl work done long by summing the vlues in () nd (3). Using Fubini s theorem to interhnge the order of integrtion, we see tht the totl work is given by Tht is, F (x, y)dydx + F dr = F (x, y)dxdy = x ( F x F ) dydx. ( F x F ) da. (4) We ll the result in (4) Green s Theorem, nd it n be used to simplify lot of lultions. Notie tht the integrnd here is preisely the zoordinte of url(f), where we onsider F to be vetor field in three spe. So in words Green s theorem is sying tht the work done long is given by integrting the url of F over. Figure ttempts to mke this seem geometrilly resonble. If F is spinning in the ounterlokwise diretion ner, this will generte positive work long. We hve helpful relity hek here: if F is onservtive vetor field, then the line integrl of F bout the losed loop must be zero. But we lso know tht onservtive vetor fields hve zero url, so Green s theorem gives zero, s desired. 3
4 Figure : ounterlokwise url leds to positive work long. Note. It s importnt here tht lies entirely within the domin of F otherwise, we might not be ble to use the FT. This subtlety is relted to the requirement tht vetor field s domin be simply onneted in order for the url test for onservtive vetor fields to work. If were not simply onneted, we ould not lwys pply the FT, nd the sitution would beome muh hirier. Note. espite the ft tht we ve only given n explntion for Green s theorem in the se tht is retngle, the eqution ontinues to hold s long s is region in the domin of F whose boundry onsists of finite number of simple, losed urves, nd we orient these urves so tht is lwys to the left. Exmple. (Setion 8., Exerise 8) ompute (ln x+y)dx x dy, where is the retngle with verties (, ), (3, ), (, 4), nd (3, 4), oriented lokwise. (Solution) In our symboli nottion, we re being sked to ompute F dr, where F = ln x + y, x. Sine we don t like integrting terms suh s ln x, this is very diffiult line integrl to ompute priori. Green s theorem simplifies it quite bit though, sine F = x nd F =. Sine is the boundry of the retngle = {(x, y) : x 3, y 4}, we hve 3 4 (ln x + y)dx x dy = ( x )da = ( x )dydx = 3 3 (x + )dx = 3 [ x + x ] 3 = 3. 4
5 So we see tht Green s theorem is nie tool for turning diffiult line integrls into double integrls. We n lso use Green s theorem to go the other diretion, turning double integrl into line integrl. In prtiulr, it s osionlly helpful to use Green s theorem to ompute the re of region by rewriting the double integrl A s line integrl long. The trik is to find vetor field F so tht F F =. Three suh x vetor fields re given by F(x, y) =, x, F(x, y) = y,, F(x, y) = y, x, mening tht re n be written s Are = da = xdy = ydx = xdy ydx. Exmple. Use one of these vetor fields to ompute the re of the ellipse x where, b >. + y b =, (Solution) We ll use the expression xdy to ompute the re, but first we ll need prmetriztion of. The stndrd prmetriztion of the boundry of n ellipse is given by r(t) = ( os t, b sin t), where t vries from to π. Then the re is π A = xdy =, os t sin t, b os t dt = π = b b os tdt = b [t + ] π sin t π ( + os t)dt = π b = πb. The first nontrivil step of the bove omputtion n be thought of in this wy: the expression xdy in line integrl mens tht we hve F(x, y) =, x, so F(r(t)) =, os t. Notie tht if we tried to ompute the re s double integrl we would hve something like A = x / dydx b x / in retngulr oordintes, nd the polr oordintes setup isn t nie either. In either se, our line integrl is muh simpler. 5
6 Flux Rell tht the divergene of vetor field is mesure of the extent to whih the vetor field is expnding or ontrting t given point. In week 5 we derived formul for the divergene of vetor field F = F, F by drwing smll retngle round given point nd mesuring the degree to whih mtter ws leving or entering this retngle. This sounds eerily similr to flux omputtion, nd we n ply gme similr to wht we did bove for work to expliitly hrterize this reltionship. onsider the retngle is Figure, with ll the sme ssumptions s before. If we d like to ompute the flux through, we n ompute this linebyline, s we did with work. Along the bottom line our outwrdpointing unit norml vetor is given by,, so flux through bottom line = nd long the top we hve n =,, so F (x, )dx Together this gives flux through top line = F (x, d)dx. flux through top nd bottom lines = Similrly, nd so flux through right line = flux through left line = flux through left nd right lines = Altogether, Tht is, flux out of the retngle = F n ds = (F (x, d) F (x, ))dx = F (b, y)dy F (, y)dy, (F (b, y) F (, y))dy = F x + F dydx. div(f)da. F dydx. F x dxdy. Just s with Green s theorem, this eqution lso nie interprettion: the totl flux out of is given by integrting the divergene of F ross. This is n unsurprising reltionship, espeilly if we go bk nd look t some of the figures from week 5. 6
7 Exmple. (Setion 8., Exerise 38) ompute the flux of fluid flow with veloity vetor field F(x, y) = os y, sin y ross the boundry of the retngle x, y π. (Solution) Sine the flux is given by div(f) = x (os y) + (sin y) = os y, F n ds = π/ os ydxdy = [sin y] π/ =. Note tht omputing the flux s line integrl would require us to prmetrize the retngle x, y π. This prmetriztion would lmost ertinly be defined in four piees, so we d hve to ompute four different norml vetors nd then evlute four different integrls. This pproh is muh lener. 7
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