Prove the following relation:
$$\frac {1} {3} \int_S r \cdot da = V,$$
where V is the volume enclosed by the closed surface S.
Im not really sure where to even start with this problem. The most I can reason out is it's some kind of relation with the divergence theorem.
$$ \int_{\partial V} \vec {F}\cdot d\vec{S} =\int_V (\vec{\bigtriangledown} \cdot \vec{F})dV$$
I'm not exactly sure what the r represents, but I'm sure I could replace it as F, with F(r)=r. Beyond that I would try taking the dot product of r, but I'm not even sure if It's a vector, as there is no indication of that at all. Overall, just pretty confused on how to START the problem, I think I could get it if I just had a kickstart.
$\vec{r}$, which is the vector you're referring to above, represents the position vector, which in three space is $\vec{r} = (x,y,z)$. However that's cartesian coordinates, and it looks like you are working in spherical coordinates (see here and here) so that for an arbitrary vector field
$$\nabla \cdot F = \frac{1}{r^2} \frac{\partial (r^2 F_r)}{\partial r} + \frac{1}{r\sin \theta} \frac{\partial}{\partial \theta} (F_{\theta} \sin \theta) + \frac{1}{r\sin \theta}\frac{\partial F_{\phi}}{\partial \phi} $$
In your case, $F = \vec{r}$ is just the position vector. But in the notation of spherical coordinates, this would be
$$F = (F_r, F_{\theta}, F_{\phi}) = (|\vec{r}|, 0, 0) = (r, 0, 0)$$
where we usually denote $r$ as the magnitude of the position vector because it's always assumed to be positive. So like you have noted, apply the divergence theorem taking $S = \partial V$ to be the boundary. And recall that
$$\int_V dV = \text{Volume enclosed by S}$$
(Note: Don't be afraid of the complicated formula for $\nabla \cdot \vec{r}$ as two of those terms are automatically zero)