Calculate CurlF and then use Stokes’ theorem to compute the flux of CurlF through the given surface as a line integral. F = < y, x, x2 + y2 > , the upper hemisphere; x2 + y2 + z2 = 1, z ≥ 0.
Answer; curlF =
Now expanding the terms, the first component of curl of F is, (curlF)x = ∂yFz − ∂zFy and the 2nd term here has to be zero, as there is no z dependence. Thus the first term is 2y.
Similarly the y component of curl of F, is: (curlF)y = ∂xFz − ∂zFx = 2x.
But the 3rd term of the curl, (curlF)z = ∂xFy − ∂yFx is zero, since x-partial of 2nd component, ∂xFy, and y-partial of the 1st component of F, ∂yFx, are both 1.
So, CurlF = < 2y, 2x, 0 > .
Now Stoke’s Theorem is the following; ∫S curlF.dS = ∮F.dl.
The line integral of F is over the contour of the circle on the z = 0 plane, as S on the surface integral on LHS is the upper hemisphere.
So, the dot product of F and dl is this: (yi + xj).(dxi + dyj) = ydx + xdy.
Consequently the integral is ∮ydx + xdy where the limits of integration vary from (0, 1).
So RHS = ∫01dx√1 - x2 + ∫01dy√1 − y2 = 2∫01dx√1 − x2.
We replace x = sinθ so that dx = − cosθdθ.
We then have RHS = − 2∫0 π/2dθcos2θ = − 2[1/2(θ + sinθcosθ)]0 π/2 = − π/2.
So the flux of the CurlF through the given surface is − π/2.
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