Prove that $(div \ w)(p) = \lim \limits_{r\rightarrow 0} \frac{1}{V_n r^n}\int_{S_r(p)} \langle w,\nu \rangle dS$ using Gauss's theorem In a book I study I found this exercise:
Let $U \subset \mathbb{R}^n$ open and $w: U \rightarrow \mathbb{R}^n$ vector field. Let $p \in U$.
$V_n$ is the volume of the n-dimensional unit sphere and $S_r(p) \subset \mathbb{R}^n$ the $n-1$-dimensional sphere around $p$ with radius $r > 0$ and $\nu$ the outer vector field of normals. 
Now I want to show using Gauss's theorem that:
$(div \ w)(p) = \lim \limits_{r\rightarrow 0} \frac{1}{V_n r^n}\int_{S_r(p)} \langle w,\nu \rangle dS$
I proofed a similar problem in three dimensions. But now I don't even know how to start showing this relation. 
Can I use the closed form for an n-sphere?:
$V_n=\begin{cases}
\frac{\pi^k}{k!} & \text{for} \ \ \ n=2k\\
\frac{2k!(4\pi)^k}{(2k+1)!} & \text{for} \ \ \ n=2k+1
\end{cases}$
I hope you can help!
 A: Instead of the sphere $S_\epsilon(p) = \{p:  ||p' - p|| < \epsilon \}$ we are going to use cube: $ p + \tfrac{\epsilon}{2} [-1,1]^n$ with volume $\epsilon^n$?  This will simplify calculations.  I am claiming:
$$ 
(\nabla \cdot w)(p) = \lim \limits_{\epsilon \rightarrow 0} \frac{1}{V_n \epsilon^n}\int_{S_\epsilon(p)} \langle w,\nu \rangle dS
= \lim_{\epsilon \to 0} \frac{1}{\epsilon^n} \int_{p + \tfrac{\epsilon}{2} [-1,1]^n} (w \cdot \nu) \,dS
$$


*

*The normal is always pointing outwards from the cube, so we get contributions from opposite faces where $w \cdot v$ is both positive and negative.

*The volume of each face has volume $|dS| = \epsilon^{n-1}$.  And we need to sum over each face. Integrating over the surfaces (and remembering the normal vector points outwards from the cube):
$$ \int_{p + \tfrac{\epsilon}{2} [-1,1]^n} (w \cdot \nu) \,dS
\approx
 \sum_{k=1}^n \bigg[w_k(p + \epsilon_k) -  w_k(p - \epsilon_k)\bigg]\epsilon^{n-1}
$$
At this point we recover the divergence formula using infinitesimal analysis:
\begin{eqnarray} \frac{1}{\epsilon^n} \sum_{k=1}^n \bigg[w_k(p + \epsilon_k) -  w_k(p - \epsilon_k)\bigg]\epsilon^{n-1} &\approx& 
\sum_{k=1}^n \frac{1}{\epsilon}\bigg[w_k(p + \epsilon_k) -  w_k(p - \epsilon_k)\bigg]  \\\\
&=&\nabla \cdot w\; + \;O(\epsilon)\end{eqnarray}

See the nice physics question: What is divergence?
