Divergence Theorem with a vector field. I have been stuck on the following question for quite a while and my professor have not been helpful at all. I'm not sure how to make delta(f) a scalar so I can apply the theorem. Any useful hints are appreciated!

 A: The idea is this that the question is actually a lot more straightforward than you might be thinking. We start with the divergence theorem
$$ \int \int \int \nabla \cdot \vec{F} \mathrm{d}V = \int \int \vec{F} \cdot \hat{n}\mathrm{d} S$$
We substitute in the vector field $\vec{F}=\vec{c}f$ for constant vector $\vec{c}$ and scalar function $f$.
$$ \int \int \int \nabla \cdot (\vec{c}f) \mathrm{d}V = \int \int (\vec{c}f) \cdot \hat{n}\mathrm{d} S$$
Now apply the product rule to the left hand side
$$ \int \int \int \nabla \cdot (\vec{c}f) \mathrm{d}V =\int \int \int (\nabla \cdot \vec{c})f+ \vec{c}\cdot (\nabla f) \mathrm{d}V$$
The first term is zero since it is the divergence of a constant vector, so we are left with
$$\int \int \int \vec{c}\cdot (\nabla f) \mathrm{d}V - \int \int (\vec{c}f) \cdot \hat{n} \mathrm{d} S =0$$
(It is easier to see if we bring the surface integral to the left hand side) This is rearranged slightly to yield
$$ \vec{c}\cdot \left( \int \int \int  \nabla f \mathrm{d}V - \int \int f \hat{n} \mathrm{d} S\right) =0$$
Since $\vec{c}$ is an arbitrary nonzero constant vector the term in the parentheses must be zero from which the required identity is obtained
