B i so that f n ds s zx y f k z 1 1 since

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Unformatted text preview: · n dS = ( j y ( × F) · (− k) dS, k × F = ∂/∂x ∂/∂y ∂/∂z = (1 − x) i + (z − 1) k, Now S ( B × F) · ( k) dS. B i so that ( × F) · n dS + S zx y × F) · k = z − 1 = −1 (since z = 0 on B ). Therefore × F) · n dS = ( B × F) · n dS = B (−1)dS = −(area of the disc B ) = −πa2 . Alternately, evaluating the line integral above, we have C given by x = a cos θ, y = a sin θ, z = 0, θ : 0 → 2π, and ( S × F) · n dS = y dx + zx dy + y dz C 2π = (a sin θ)(−a sin θ) dθ y dx = C 0 2π = − a2 0 sin2 θ dθ = −πa2 . x2 y 3 dx + dy + z dz , where C is x2 + y 2 = a2 , z = 0, taken once, in an 4. Evaluate C anti-clockwise direction when viewed from above. Solution: The evaluation of this integral is a little easier using Stokes’ theorem, rather than integrating directly. N...
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This document was uploaded on 04/08/2014.

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