Stokes' Theorem Examples
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1 Stokes' Theorem Eamples Stokes' Theorem relates surface integrals and line integrals. STOKES' TEOREM Let F be a vector field. Let be an oriented surface, and let be the boundar curve of, oriented using the right-hand rule. Then: (( curl afb. A œ * F. s EXAMPLE Let be the curve defined b the parametric equations B œ! C œ cos> D œ sin>! Ÿ > Ÿ Use Stokes' Theorem to evaluate ( B /.B B cosc.c C.D. &D SOLUTION The parametric equations above describe a circle of radius on the CD-plane: z Let be the disc whose boundar is the given circle. B Stokes' Theorem: ( ˆ &D (( ˆ &D B / i B cosc j C k. s œ curl B / i B cosc j C k. A e compute the curl: curlˆ B / &D ˆ &D i B cosc j C k œ œ &B / C `Î`B `Î`C `Î`D i j cos k â &D B / B cosc C â So we have to evaluate (( ˆ &D i &B / j cosc k. A.
2 This integral can be evaluated geometricall. The vector. A for the disc points in the positive B direction. (Stokes' theorem uses the right-hand rule: if ou curl the fingers of our right hand in the direction of, then our thumb points in the direction of.a.) So: Therefore: ˆ &D i &B / j cosc k. A œ.e (( ˆ &D i &B / j cosc k. A œ ((.E œ athe area of the discb œ è EXAMPLE 2 Let be the surface D œ B a BbC a C b for! Ÿ B Ÿ and! Ÿ C Ÿ. Evaluate the integral (( B k. A, where. A is the upward-pointing normal vector. SOLUTION If we wish to use Stokes' theorem, we must epress Bk as the curl of some vector field F. The formula for the curl is: curl F œ `J `J `J `J `Î`B `Î`C `Î`D D `JC B D `JC B œ Œ ß ß J J J â B C D â So what we need is: `JD `JC `JB `JD `JC `JB œ! œ! œ B C B D It is not hard to guess that J œ B, with J œ J œ!. Indeed: curlœ B j œ `Î`B `Î`C `Î`D œ B k â! B! â B Stokes' Theorem, we conclude that (( B k. A œ * B.C, where is the boundar curve of the surface. So what is the boundar of? ell, the equation D œ B a BbC a Cb specifies a surface whose D-coordinate varies with horizontal position:
3 z The allowed values of B and C are determined b the inequalities! Ÿ B Ÿ and! Ÿ C Ÿ, which describe a square on the BC-plane: The sides of this square are along the lines B œ!, C œ!, B œ, and C œ. Looking at the equation, we see that D œ! for these values of B and C, so the boundar of the surface is just the boundar of the square on the BC-plane: e can evaluate * geometricall: B.C * B.C œ ( B.C ( B.C ( B.C ( B.C top bottom left right The top and bottom sides are horizontal, so.c œ!. Furthermore, B œ! along the left edge, and B œ
4 along the right edge, so: * B.C œ!!! (.C œ right è EXAMPLE 3 Let be the upper hemisphere of the unit sphere B C D œ. Use Stokes' theorem to evaluate (( ˆ C C B / i B / j. A, where. A is the upward-pointing normal vector. SOLUTION If we wish to use Stokes' theorem, we must epress B / C i B / C j as the curl of some vector field F. The formula for the curl is: curl F œ `J `J `J `J `Î`B `Î`C `Î`D D `JC B D `JC B œ Œ ß ß J J J â B C D â So what we need is: `JD `JC C `JB `JD C `JC `JB œ B / œ B / œ! C If we guess that J and J are zero, it is not too hard to figure out that J œ B /. Indeed: C B D C curlˆ B / k œ `Î`B `Î`C `Î`D œ â C!! B / â C C B / i B / j B Stokes' Theorem, we conclude that (( ˆ C C C B / i B / j. A œ * B /.D, where is the boundar curve of the surface. Since is the upper hemisphere of the unit sphere, is just the unit circle on the BC-plane. B the right-hand rule, is oriented counterclockwise. (It would be clockwise if we had started with the downward-pointing. A.) Then * B /.D is zero, since D is not changing over the course of the circle. C e conclude that (( ˆ C C B / i B / j. A œ!. è
5 EXAMPLE 4 Let be the surface defined b D œ B C for D Ÿ %. Use Stokes' Theorem to evaluate (( ˆ BD i D k. A, where. A is the upward-pointing normal vector. SOLUTION If we wish to use Stokes' theorem, we must epress BD i D j as the curl of some vector field F. The formula for the curl is: curl F œ `J `J `J `J `Î`B `Î`C `Î`D D `JC B D `JC B œ Œ ß ß J J J â B C D â So what we need is: `JD `JC `JB `JD `JC `JB œ BD œ! œ D If we guess that J and J are zero, it is not hard to figure out that J œ BD. Indeed: B D C curlˆ BD j œ `Î`B `Î`C `Î`D œ BD i D â! BD! â B Stokes' Theorem, we conclude that (( ˆ BD i D k. A œ * BD.C, where is the boundar curve of the surface. So what is the boundar of? ell, the equation for the surface can be epressed as D œ <, where D Ÿ %. This appears as a parabola on the <D-plane: z (2, 4) z C k r Therefore, the surface is a bowl-shaped paraboloid. Its boundar curve is a counterclockwise circle on the plane D œ % with radius, with parameterization: B œ cos> C œ sin>! Ÿ > Ÿ D œ %
6 (The circle is counterclockwise b the right-hand rule.) Thus: * BD.C œ ( a cos> ba% b a cos>.> b! œ &'( cos >.>! œ &'( a cos> b.>! œ ) > sin>! (double-angle formula) œ &' è
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