Let $V \subseteq \mathbb{R}^3$ have smooth boundary $\partial V$. Suppose $f, g, \mathbf{J}$ are differentiable on $V \cup \partial V$, with $\nabla \cdot \mathbf{J} = 0$ in $V \cup \partial V$, then prove that
$$ \int_V (f\ \mathbf{J} \cdot \nabla g + g\ \mathbf{J} \cdot \nabla f)\ dV = 0. $$
My attempt: The integrand is $\nabla\cdot(fg\ \mathbf{J})$, so by Divergence Theorem, the integral above is equal to
$$ \int_{\partial V} fg\ \mathbf{J}\cdot d\mathbf{S}. $$
Furthermore, $\nabla \cdot \mathbf{J} = 0$, so $\mathbf{J} = \nabla \wedge \mathbf{A}$ for some $\mathbf{A}$, so the integral is equal to
$$ \int_{S}fg\ \nabla\wedge\mathbf{A}\cdot d\mathbf{S}. $$
How to show that this is $0$?