Prove that $\iint_S \vec{r}\cdot \vec{n}\,\mathrm{d}S=3 V(A)$ 
Let $S$ be a closed regular surface and $\vec{n}$  a vector field of unit normal vectors to the surface. Prove that the flux of the vector field $\vec{F}=\vec{r}$ is equal to $3\, V(A),$ where $V(A)$ is the volume of the interior $A$ of $S$.

I am looking for a proof that $\displaystyle \iint_S \vec{r}\cdot \vec{n}\,\mathrm{d}S= 3 V(A)$, without use of Gauss divergence theorem. So far I can say that
 $$\iint_S \vec{r}\cdot \vec{n}\,\mathrm{d}S=\iint_D (\vec{r},\vec{r}_u,\vec{r}_v)\,\mathrm{d}u\,\mathrm{d}v,$$ where $(\vec{r},\vec{r}_u,\vec{r}_v)$ stands for the triple product of $\vec{r},\,\vec{r}_u,\,\vec{r}_v$.
Any ideas are welcomed.
Thanks in advance for the help.
 A: Too long for a comment:
 Just an idea for geometric intuition which might be useful to understand how to formalize the matter: The product $\frac{\vec{r}\cdot \hat{n}dS}{3}$ represents a volume of an infinitesimal pyramid whose base is a surface element $dS$ and the height is $\vec{r}\cdot \hat{n}$ the projection of $\vec{r}$ ortohgonal to the base. (In terms of using a parameterization, this volume can be thought of as a triple product of $\vec{r}, \vec{r}_u du, \vec{r}_v dv$. For simplicity, assume $\vec{0}\in A$ and that the domain is star-shaped. $V(A)$ is the sum of all these tiny pyramids.
In the general case, I would prove first for $A$ which is star-shaped in relation to $\vec{p}\neq \vec{0}$. This means a pyramid doesn't intersect twice the surface, and then for any $A$ partition it into star-shaped regions and sum the results.
I suppose that in order to truly formalize this, one must bound ($\vec{r},\vec{r}_u\Delta u,\vec{r}_v\Delta v)$ and show the error in relation to the volume of the "not really a pyramid", diminishes as $\Delta u,\Delta v \rightarrow 0$.
