Gradient of a Harmonic Function I was asked the following vector calculus problem:

Let $D$ be the unit ball and let $S$ be the unit sphere in $\mathbb{R}^3$.  Suppose that $F:\mathbb{R}^3\rightarrow \mathbb{R}^3$ is a $C^1$ vector field on some open neighborhood of $D$ which satisfies:
$(i) \nabla\times F=0$
$(ii) \nabla\cdot F=0$
$(iii)$ On $S$, $F$ is orthogonal to the radial vector.
Prove that $F=0$ on all of $D$.

Conditions $(i)$ and $(ii)$ imply that $F=\nabla g$ for some $g:\mathbb{R}^3\rightarrow \mathbb{R}$ where $g$ must be harmonic as well.
I know one solution (see end), however my initial instinct was to try to use the max/min property of harmonic functions, and I couldn't get it to work.  Since the gradient is always orthogonal to the sphere, there must be a point on the sphere where it is $0$.  (Hairy ball)  If that was a local max or min in $\mathbb{R}^3$ we would be done, by taking a small neighborhood around it.  If it is a saddle point this doesn't work.  (We know that it must be a local max/min on $S$ since it is harmonic)
My question is:  Is there any way to modify this approach, and solve the problem?
Thanks!
Other Solution: Here is one solution that first uses the fact that the radial vector is orthogonal, and then applies Gauss's Divergence theorem to the function $gF$. ($\nabla g=F$) That is $$0=\iint_S (gF\cdot n)dS=\iiint_D \nabla\cdot (gF)dV=\iiint_D \|F\|^2dV,$$ and since the integrand on the right hand side is non-negative, continuous and integrates to give zero, it must be zero.

Your second proof is a good one and entirely appropriate within vector calculus. The first attempt has a gap (I think). 
A third proof relies on Hopf's Lemma (commonly taught in graduate level classes on partial differential equations) which implies here that if a function $u$ satisfying $\Delta u \le 0$ in $D$ attains a strict minimum at $z \in \partial D = S$, then the outer normal derivative at that point satisfies $\nu \cdot \nabla u(z)  < 0$. Applying this with $g = u$, it follows that $g$ must attain its minimum in the interior of $D$  and hence must be constant. 
