Let $S=\{z=2(1-x^2-y^2),\ z\geqslant0\}$. How to calculate flux of $rot\ A$ outside $A$ through $S$ for $A=(y,z,x)$ ? I know the definition, but don't know how to actually calculate what's mentioned in this exercise...
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Use Stokes' Theorem. The flux of $\nabla \times A$ through the surface equals the line integral of $A$ around the boundary which is the unit circle $C$ in the x-y plane
$$\int_{S}\nabla \times A \cdot nd\sigma = \oint_{C} A \cdot dl.$$ The only non-zero term in the line integral is
$$\oint_{C} A \cdot dl= \oint_{C}ydx= \int_{0}^{2\pi}\sin(\theta)d\cos(\theta)=-\int_{0}^{2\pi}\sin^2(\theta)d\theta=-\pi.$$
Depending on how you define flux "from outside through S" the answer is $\pi$ or $-\pi$.
RRL
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Oh, sorry I meant flux outside $A$. Umm, I believe now I translated it correctly. – Jules Jun 04 '14 at 18:21
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In that case the flux is through the surface from above, the unit normal points inward and the line integral would be clockwise with value $+\pi$. – RRL Jun 04 '14 at 18:24