Show that $\int r\cdot n ds$ equals three time the volume of $\omega$.

Let $\Omega$ be an open region in $\mathbb{R}^3$ with surface $∂\Omega$ on every point $P$ of which the unit outward pointing normal $n = n(P)$ is well defined and smoothly varying. Let $r = (x, y, z)$. Show that

$$\int_{∂\Omega} r\cdot n \;ds$$

equals three time the volume of $\Omega$.

What I do know is that the divergence theorem states for $\Omega$ a solid space in $\mathbb{R}^3$ where $∂\Omega$ is bounded we have $$\int\int\int div F\; dv = \int\int_{∂\Omega}f\cdot n\; ds$$ where n is the normal vector and f is also a vector. However, I have no idea where this gets with this application. Also, what are the integration bounds?
• I think the problem's giving $\;\Omega\;$ is bounded , otherwise I don't think that line integral is well defined. – DonAntonio May 8 '16 at 5:47
Taking into account what I wrote in my comment below your question, and assuming it is true: we have that $\;\nabla r=1+1+1=3\;$ , so the divergence theorem gives at once
$$\iint_{\partial\Omega}r\cdot n\;dS=\iiint_\Omega\nabla r\;dV=3\, V(\Omega)$$