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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

2 Answers 2

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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 out2012-06-26
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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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    True ... there is not $I_1$ if it it not closed at the bottom. So you cannot apply Divergence Theorem.2012-06-27