Then k f 3x2 y 2 k c y now choose some surface which

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Unformatted text preview: ote that C is in the plane z = 0, so that x2 y 3 dx + dy + z dz = C x2 y 3 dx + dy , C and in this case Stokes’ theorem is the same as Green’s theorem. z Let F = x2 y 3 i + j + z k . Then k × F = − 3x2 y 2 k . C y Now choose some surface which has C as its rim. The simplest such surface is the disc x x 2 + y 2 ≤ a2 , z = 0 . Call this disc S , and note that, since C is taken anti-clockwise, we choose the normal pointing up from S to use in Stokes’ theorem. That is, we take n = k, 3 and so ( × F) · n = −3x2 y 2 . We therefore have x2 y 3 dx + dy + z dz = C C F · dr = ( × F) · n dS S −3x2 y 2 dS = S a 2π −3r2 cos2 θ r2 sin2 θ rdr dθ = 0 0 1 = − a6 2 1 = − a6 2 1 = − a6 8 2π cos2 θ sin2 θ dθ 0 2π sin2 2θ dθ 4 2π 6 1 − cos 4θ dθ = −πa 8 . 2 0 0 5. By using a suitable integration theorem, or otherwise,...
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