# Error In Divergence Theorem

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?

• Could you show your attempt so that we can figure out where you went wrong? 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.