Let $G$ be a vector field defined in $A = \mathbb{R}^n -{0}$ with $\operatorname{div} G = 0$ in $A$. let $M_1$ and $M_2$ be compact n-manifolds in $\mathbb{R}^n$, such that the origin is contained in both $M_1 - \partial M_1$ and $M_2 - \partial M_2$. Let $N_i$ be the unit outward normal vector field to $\partial M_i$, for $i = 1,2$. Show that
$\int_{M_1} \langle G,N_1 \rangle dV = \int_{M_2} \langle G,N_2 \rangle \,dV$
The answer key that I saw used a $\epsilon$-ball, $M_3$ and had the following expression:
$\int_{M_1 - Int M_3} \operatorname{div} G \, dV = \int_{\partial M_1}\langle G,N_1 \rangle - \int_{\partial M_3} \langle G,N_3 \rangle\, dV$
I understand that $M_1- Int(M_3)$ is compact, but I don't quite know how this statement is justified. Any help will be welcomed. Thanks in advance.
Let me attempt to answer my own question, but I may be wrong.
Let $K$ be the unit normal vector field to $\partial M_1 \cup \partial M_3$ pointing ourwards from M $\int_{M_1 - Int M_3} div G \ dV = \int_{\partial M_1 - IntM3}divG \ dV = \int_{\partial M_1 - IntM3} \langle G, K \rangle \ dV $
$= \int_{\partial M_1} \langle G, K \rangle \ dV - \int_{\partial M3} \langle G, K \rangle \ dV $
$= \int_{\partial M_1} \langle G, N_1 \rangle \ dV - \int_{\partial M3} \langle G, N_3 \rangle \ dV $
As the $N_1$ part of $K$ is not on $\partial M_1$, and $N_3$ part of $K$ is not on $\partial M3$. The last equality will be a result of linearity of inner product and integral.