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Say $F=(F_1(x),F_2(x),F_3(x))$ and $x\in\mathbb{R}^3$.

The standard divergence theorem states: $\int\int F\cdot n dS=\int\int\int\nabla\cdot F dV$.

Then, if $G=(G_1(x),G_2(x),G_3(x))$, what can we say about $\int\int (F\cdot n)GdS$?

Or, if you prefer, you can think of $G$ as a scalar valued function of $x$.

Fozz
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  • I'm inclined to think that, if $G$ is 'sufficiently nice', then you could conclude that $\int\int(F\cdot n)GdS=\int\int\int\nabla\cdot FGdV$. But what sufficiently nice would be I am not sure, if $G$ is a scalar constant, then the result I gave follows by linearity of the (multiple) integral. – Justin Benfield Sep 14 '16 at 04:05
  • @JustinBenfield: What you've written holds only when $G$ is a constant function, whether we interpret the RHS as $(\nabla\cdot F)G$ or $\nabla\cdot(FG)$. – Ted Shifrin Sep 14 '16 at 05:49

1 Answers1

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Consider the case of a scalar function $\phi$ (you can apply this to each component of your $G$ if you wish). Then, by the Divergence Theorem, $$\iint (\phi F)\cdot n\,dS = \iiint \nabla\cdot(\phi F)\,dV = \iiint \big(\phi\,\nabla\cdot F+\nabla\phi\cdot F\big)\,dV.$$

Ted Shifrin
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