# You should use whichever method you think will be the

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Unformatted text preview: e the integral directly, or use the divergence theorem, or use Stokes’ theorem. You should use whichever method you think will be the easiest.) Solution: z The rim of S is the circle x2 + y 2 = a2 in the plane z = 0. This circle is also the rim of the disc x2 + y 2 ≤ a2 in the plane z = 0. Call the circle C , and the disc B . Then by Stokes’ theorem, n a y x ( × F) · n dS = S C F · dr = y dx + zx dy + y dz, C where C is taken in the anti-clockwise direction. Since the upwards pointing normal to B is k, we also have (by two applications of Stokes’ theorem) ( S × F) · n dS = C F · dr = ( B × F) · k dS. Note that this result also follows from Gauss’ theorem: On both S and B choose the normal outwards to the solid bounded by the closed surface S ∪ B . Then, by the divergence theorem, ( S ∪B × F) · n dS = 2 V ·( × F) dV = 0 . But ( × F) · n dS = S ∪B so that ( S ( × F)...
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## This document was uploaded on 04/08/2014.

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