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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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