Let $U$ be an open set of $\mathbb{R}^3$, $f:U \to \mathbb{R}$ a $c^2$ function, $B \subset U$ an solid sphere (closed), $\mathcal{S}$ the spherical surface that bounds $B$, and $N$ the outward normal unit normal vector field to $\mathcal{S}$.
By divergence's Theorem, it could be shown that: $$\iint_{S}f\nabla{f}=\iiint_{B}||\nabla{f}||^2 dxdydz$$ if f is harmonic $(\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2}=0)$.
But now, if we consider $G:U \to \mathbb{R}^3$ a $c^1$ vector field such that $curl(G)=0,div(G)=0$ for $(x,y,z) \in B$ and $\langle G,N \rangle=0$ for $(x,y,z) \in \mathcal{S}$, this should lead to $G_{|B}=0$, but I can't find a good argument. If $curl(G)=0$, then $G$ is closed in B, so that $G$ is a gradient field in $int(B)$, but could this be continuously extended to $B$ so that by the result shown above, we could conclude that as f is harmonic and $\langle f\nabla{f},N \rangle=\langle fG,N \rangle=0$ for $||\nabla{f}||^2=||G||^2=0$ for all $(x,y,z) \in B$?