# Proving $\oint\vec{F}(\vec{G}\cdot \hat{n}) d\sigma=\iiint[\vec{F}(\nabla \cdot \vec{G})+(\vec{G}\cdot\nabla)\vec{F}]dV$

Given $C^1$ vector fields $\vec{F}, \vec{G}$, show that: $$\unicode{x222F}_\Sigma\vec{F}(\vec{G}\cdot \hat{n}) d\sigma=\iiint_\Omega [\vec{F}(\nabla \cdot \vec{G})+(\vec{G}\cdot\nabla)\vec{F}]dV$$

I know that I need to start with the components of $\vec{F}(\vec{G}\cdot \hat{n})$, and use the divergence theorem, but I'm not sure where to start.

Note that

$$\hat x_i\cdot \oint_\Sigma \vec F(\hat n\cdot \vec G)\,d\sigma=\oint_\Sigma F_i(\hat n\cdot \vec G)\,d\sigma$$

Next, we use the product rule identity $\nabla \cdot(F_i\vec G)= F_i \nabla\cdot \vec G+\vec G\cdot \nabla F_i$ along with the Divergence Theorem to write

\begin{align} \oint_\Sigma F_i(\hat n\cdot \vec G)\,d\sigma&=\int_\Omega \left( F_i \nabla\cdot \vec G+\vec G\cdot \nabla F_i\right)\,dV\\\\ &=\hat x_i \cdot \int_\Omega \left( \vec F \nabla\cdot \vec G+\vec (G\cdot \nabla) \vec F\right)\,dV \end{align}

Inasmuch as this is true for all $i$, we arrive at the coveted equality

$$\oint_\Sigma \vec F(\hat n\cdot \vec G)\,d\sigma=\int_\Omega \left( \vec F \nabla\cdot \vec G+\vec (G\cdot \nabla) \vec F\right)\,dV$$

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