If $\vec{F}=y\vec{i}+(x-2xz)\vec{j}-xy\vec{k}$, Evaluate $$\iint_S (\nabla \times \vec{F})\cdot n \vec{ dS}$$ where $S$ is the surface of the sphere of radius $a$ with center at origin above the $xy$ plane and $n$ is the unit normal vector to $S$.

When I Applied Gauss Divergence Theorem, $$\iiint_V \nabla\cdot (\nabla \times \vec{F}) dv$$ should be equal to zero so the surface integral must be zero but it's not the answer Why?

  • $\begingroup$ Could you show your attempt so that we can figure out where you went wrong? $\endgroup$
    – abhi01nat
    Aug 21 '20 at 8:33

The divergence theorem is about a three-dimensional body $B$, its boundary surface $\partial B$, and a vector field ${\bf v}$ with domain $\Omega\supset B$. The theorem says that $$\int_{\partial B}{\bf v}\cdot{\bf n}\>dS=\int_B{\rm div}\,{\bf v}({\bf x})\> dV\ .\tag{1}$$ In your case $B$ is a half ball, and ${\bf v}={\rm curl}(\vec F)$. It follows that ${\rm div}\,{\bf v}({\bf x})\equiv 0$ in $B$. But $\partial B$ is not just the spherical part of $\partial B$. The boundary of $B$ also contains a plane circular disc in the $(x,y)$-plane. It follows that the right hand side of $(1)$ consists of two terms, one of which you have already computed. Computing the flow of ${\bf v}$ through the disc is simpler, since ${\bf n}\equiv(0,0,-1)$ and $dS=dA$, where $dA$ is the standard area element in the $(x,y)$-plane.


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