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If $\vec{u}=4y\hat{i}+x\hat{j}+2z\hat{k}$, calculate the double integral

$$\iint(\nabla \times \vec{u})\cdot d\vec{s}$$ over the hemisphere given by,

$$x^{2}+y^{2}+z^{2}=a^{2}, \quad z\geq 0.$$

I approached it like this, $d\vec{s}$ can be resolved as $ds\vec{n}$ where $\vec{n}$ is the normal vector to the differential surface. Which translates the integral into the surface integral in Divergence Theorem of Gauss, which implies the volume integral will be Div of Curl of u, but this Div(Curl u) is zero. I dont think this question is this trivial

Help appreciated Soham

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up vote 2 down vote accepted

Divergence Theorem of Gauss (-Ostrogradsky) applied to integrals over closed surfaces: those that don't have any edge. You have one half of a sphere, so the equator makes an edge of your surface.

Try the Stokes' theorem instead: it will reduce the surface integral to a line integral over the equator.

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Much thanks. I missed that it is not a closed surface, my bad.Thanks for pointing me out – Soham Jun 26 '12 at 19:32

The curl is constant $ \nabla \times (4y\hat{i}+x\hat{j}+2z\hat{k}) = - 3$, Hence the surface integral reduces to $$I_1 = \iint_{circle} -3 \hat k \cdot (-\hat n) ds = 3 \pi a^2 $$ $$I_2 = \iint_{hemisphere} (\nabla \times u) . \hat n ds \implies \text{ Stoke's theorem } \implies \\ \iint_{circle} -3 \hat k \cdot \hat n ds = -3 \pi a^2$$ Adding $I_1 + I_2 = 0$, If it was a closed surface, you would get same result by Divergence theorem.

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Hi X, thanks for the effort taken. Much appreciated – Soham Jun 26 '12 at 19:32
you are welcome!! – Santosh Linkha Jun 26 '12 at 19:54
Hi X, one question, as I was thinking about this question, I realised that the surface so happen is just a hollow hemisphere, and as such is "open" from the bottom => there is no surface to integrate it per se in I1. Please do correct me. – Soham Jun 27 '12 at 10:28
True ... there is not $I_1$ if it it not closed at the bottom. So you cannot apply Divergence Theorem. – Santosh Linkha Jun 27 '12 at 11:31

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