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 $\Omega$.
I am not even sure where to start with this problem:
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?
Any help with how to get started on this problem would be greatly appreciated.
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
$$\iint_{\partial\Omega}r\cdot n\;dS=\iiint_\Omega\nabla r\;dV=3\, V(\Omega)$$