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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.

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  • $\begingroup$ I think the problem's giving $\;\Omega\;$ is bounded , otherwise I don't think that line integral is well defined. $\endgroup$ – DonAntonio May 8 '16 at 5:47
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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

$$\iint_{\partial\Omega}r\cdot n\;dS=\iiint_\Omega\nabla r\;dV=3\, V(\Omega)$$

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