I'm TAing a vector calculus course and the professor has asked the students to prove that $\nabla \cdot (\nabla \times \vec{F}) = 0$. I'm meant to teach this problem in recitation tomorrow and I think the problem is incorrect. The specific problem statement is:
Problem. Let $C$ be a simple, closed curve, $S_1, S_2$ two surfaces whose boundary is $C$ and $\vec{F}$ a vector field that is defined and differentiable throughout a simply connected region containing $C, S_1$, and $S_2$. Use Stokes' theorem and the divergence theorem to show that $\nabla \cdot (\nabla \times F)$ is zero.
This is obviously super easy to do if one uses Euclidean coordinates (for example, on page 3). And since we're dealing with curl, I think it's safe to assume that the domain is $\mathbb{R}^3$, so that Euclidean coordinates are certainly available. But proof by computation in coordinates does not require the divergence theorem or Stokes' theorem, so I think the professor has a coordinate-free approach in mind.
Specifically, I think he's thinking that the shared boundary $C$ will be oriented in opposite directions when considered as the boundary of $S_1$ versus $S_2$. Therefore, one can compute
\begin{align*}
\iiint_V \nabla \cdot (\nabla \times \vec{F}) dV &= \iint_{S_1 \cup S_2} (\nabla \times \vec{F}) \cdot d\vec{A} \\
&= \iint_{S_1} (\nabla \times \vec{F}) \cdot d\vec{A}_1 + \iint_{S_2} (\nabla \times \vec{F}) \cdot d\vec{A}_2 \\
&= \oint_{C^+} \vec{F} \cdot d\vec{r} + \oint_{C^-} \vec{F} \cdot d\vec{r} \\
&= 0,
\end{align*}
where $V$ is the region enclosed by $S_1 \cup S_2$, where $d\vec{A}_1$ and $d\vec{A}_2$ are outward normal to $S_1$ and $S_2$ respectively, and where $C^+$ indicates $C$ oriented counterclockwise while $C^-$ indicates clockwise orientation. However, that computation only shows that the integral of $\nabla \cdot (\nabla \times \vec{F})$ is zero, not that the function itself is zero. Is there some way to show from here that the function itself is zero? Another approach that I am missing? Should I just tell the students to do the computation in coordinates and ignore the divergence/Stokes stuff?
Thanks in advance!
Best Answer
The problem is, as @Mark S. commented a while ago, very badly written. I am going to reiterate the suggestions of that comment as an answer, because this is an important technique.
Yes, surely the professor intends to assume $\vec F$ is $C^2$. The gist of the argument is a common one in math and physics. If $\nabla\cdot(\nabla\times\vec F)$ is not everywhere zero, then at some point $P$ it is nonzero, hence — say — positive. By continuity, it is positive on a small ball around $P$. Now you get a contradiction by taking your little surfaces $S_1$ and $S_2$ within that ball.