Proof of $\iint \limits_{\delta V} f \overrightarrow{dA} = \iiint \limits_V \nabla f \, dV$

Question

In my aerodynamics class, we often use the identity:

$\iint \limits_{\delta V} f \overrightarrow{dA} = \iiint \limits_V \nabla f \, dV$

for a closed surface (can't seem to get \oiint to work) and scalar f. It's supposed to be a simple corollary of the divergence theorem, but I can't seem to find a proof to convince me anywhere.

Can anyone offer a link or a quick outline of the main proof idea?

Apply divergence theorem to $f\vec{\eta}$ where $\vec{\eta}$ is an arbitrary constant vector. — achille hui, Oct 17 '14 at 16:57

score 3 · Accepted Answer · answered Oct 17 '14 at 17:04

The divergence theorem states that $$\iint_{\delta V} F \cdot n dA = \iiint_V \mathrm{div} F \, dV$$ whenever $F$ is a smooth vector field on $V$ and $n$ is the outward normal unit vector. If $f : V \to \mathbb R$ let $F = (f,0,0)$ so that $$\iint_{\delta V} f n_1 dA = \iiint_V \frac{\partial f}{\partial x_1} \, dV$$ where $n_1$ is the first component of the outward normal unit vector. Repeat with $ F = (0,f,0)$ and $F = (0,0,f)$ to get $$\iint_{\delta V} f n_i dA = \iiint_V \frac{\partial f}{\partial x_i} \, dV$$for $i=1,2,3$. In vector form this is exactly $$\iint_{\delta V} f n dA = \iiint_V \nabla f \, dV.$$

Proof of $\iint \limits_{\delta V} f \overrightarrow{dA} = \iiint \limits_V \nabla f \, dV$

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