Suppose the boundary of $\Omega\subset \mathbb{R^3}$ is a smooth surface $S$, prove $$v(\Omega)=\frac{1}{3}\iint _S\textbf{r} \cdot d\textbf{S}. $$
When $S$ under the general sphere coordinates $(x,y,z)=(\arcsin\phi \cos\theta ,br\sin\phi \sin\theta, cr\cos\phi)$, $r=r(\phi ,\theta)$, where $ (\phi, \theta)\in D$, then prove $$v(\omega)=\frac{abc}{3}\iint_Dr^3(\phi,\theta)\sin\phi d\phi d\theta.$$
Actually I have considered the case which is the sphere, then it is obvious it is correct. But in general, I think we consider the Gauss's Theorem.