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This 2 page Class Notes was uploaded by Ms. Shyann Nader on Saturday October 3, 2015. The Class Notes belongs to MATH 211 at Bucknell University taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/218082/math-211-bucknell-university in Mathematics (M) at Bucknell University.
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Date Created: 10/03/15
Math 211 Stokes Theorem Day 51 Stokes Theorem Let S be an oriented piecewise smooth surface with a boundary of a union of closed curves with the boundary orientation If F is a vector eld with continuous partials then jigsFdsScurlFdS The following example veri es Stokes7 Theorem for the vector eld F lty27m722gt on the surface 5 described by z 2 y2 and 2 g 1 with the outward normal There are two parts to this veri cation We will compute if F ds and curlF dS showing 38 5 that they are equal which veri es the theorem First we consider curlF dS The surface S can be parametrized by 5 ltIgtuv u 11142 U2gt with u 6 01 and v 6 7V 17 u2x17 742 take note of the bounds on 1 Next we compute the normal vector lt1 10214 ltIgtv 012ugt 50111 Pu X by lt72u72v1gt At the origin this normal vector points straight up ltIgt0 0 0 and 50 0 lt0 0 1 which is an inward normal So we set nu U 75uv 214 21171 Note that this points down at the origin and so is an outward pointing normal The last computation we make before the integral is curlF curlF V x F 0017 2y 1 2 Finally we approach the integral scurlFdS curlFltIgtuvnuvdA DltO0172vgtlt2u2v71gtdA 1 w17u2 0017 2Ugt 214 21 71gtdvdu 0 7w17u2 1 w17u2 2v 7 1dvdu 0 7w17u2 1 1 72V17 uzdudv 72 V17 uzdudv 0 0 this integral is asking for the area of a half of the unit circle which is 7r2 727r2 77139 Now for the line integral Notice that 85 is a circle which is given by z 1 and 2 y2 1 The boundary orientation requires that the circle or its projection on the zy plane be oriented in a clockwise direction This can be parametrized by Ct ltcost7sint1gt with t E 0 27139 We will also need the derivative of this parametrization ct lt7 sint 7 cost 0 We now compute the integral 27r fl F d3 Fct and as 0 27r ltsin2tcost1gt lt7 sint 7 cost 0 0 27r 7sin3t 7 cos2tdt 77r 0 This last step is some work with single variable calculus In conclusion we see that in fact the integrals that Stokes7 Theorem claims are equal are indeed equal
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