I am working through some problems and have a doubt about how to proceed with the following, as I find the notation somewhat confusing:
Show that $$\iiint_V \nabla f \, dV = \unicode{x222f}_S \;f \,\hat{n} \; dS$$
It suggests using the result $\nabla \cdot (f \, \vec{c}\,) = \nabla f \cdot \vec{c} \, + \, f\nabla \cdot \vec{c}$.
What I find confusing is the initial triple integral, as it is written: they don't specify the nature of $f$, but I'm assuming it is a scalar field, and so $\nabla f$ would be the gradient of $f$; however, in that case, what exactly is meant by the triple integral over the volume? Am I to understand it as $\iiint_V \nabla f \cdot d\vec{V}$?
Here is my attempt, assuming the above interpretation, and assuming that $d\vec{V}$ would have some sort of analogous definition to $d\vec{S}$ in surface integrals of vector-functions (i.e. $d\vec{S} = \hat{n}\,dS$):
$$ \iiint_V \nabla f \cdot d\vec{V} = \iiint_V \bigl(\nabla f \cdot \hat{n} \bigr) dV = \iiint_V \bigl( \nabla \cdot (f\, \hat{n}) - f \, \overbrace{\nabla \cdot \hat{n}}^{=\,0} \bigr) \, dV = \iiint_V \nabla \cdot (f \, \hat{n}) \, dV $$
At which point, we can apply the divergence theorem to show:
$$ \iiint_V \nabla \cdot (f \, \hat{n}) \, dV = \unicode{x222f}_S (f \, \hat{n})\cdot d\vec{S}$$
I am not sufficiently comfortable with the ideas here, nor the notation, to be confident in this interpretation and would appreciate any guidance as to the appropriate interpretation of the question.
EDIT: $f$ apparently is an arbitrary, smooth scalar field and the identity is meant to be understood as a vector identity. This doesn't clarify much for me, but in case it is helpful...