Proof related to divergence theorem The exercise asks me to prove that if $u:D\cup\partial D \rightarrow \mathbb{R}^{2}$ is ${C}^{2}$ on $D$ and we define $B_{\rho}$ a circle of radius $\rho < R$ such that $B_{\rho} \subset D$. Then if we also define $I(\rho) = (1/\rho)\int_{\partial B_{\rho}} uds$ Show that:
$I'(\rho) = (1/\rho)\int_{B_{\rho}} {\nabla}^{2}{u}$ $dA$ (1)
There is a hint in the book that is, prove that following statement:
$ \int_{\partial B_{\rho}}  \frac{\partial u}{\partial n} ds =  \int_{B_{\rho}} {\nabla}^{2}{u}$ $dA$ (2). Where $n$ is a normal vector to the boundary of $B_{\rho}$ I've managed to prove that using the divergence theorem. but i can't relate the left side of the equality in (2) with $I'(\rho)$. Thanks in advance 
 A: Presumably $B_\rho$ is some ball of radius $\rho$ contained in $D$.   We will denote this ball more clearly as $B(x,\rho)$ where $x \in D$ is the center.
For the integral of $u$ over the circular boundary $\partial B(x,\rho)$ we can apply a change of variables where $y = x + \rho z$ and $ds_y = \rho ds_z$ to get 
$$I(\rho)=\frac{1}{\rho}\int_{\partial B(x, \rho)}u(y) \, ds_y = \int_{\partial B(0, 1)}u(x+\rho z) \, ds_z$$
As $u$ is sufficiently smooth on the boundary, we can interchange the $\rho$-derivative and integral on the RHS to obtain 
$$I'(\rho)=\frac{d}{d\rho}\int_{\partial B(0, 1)}u(x+\rho z) \, ds_z = \int_{\partial B(0, 1)}\nabla u(x + \rho z)\cdot z \, ds_z $$
On both circles $\partial B(0,1)$ and  $\partial B(x,\rho)$, the vector $z = \rho^{-1}(y-x)$ is a unit normal, and by transforming back to the $y$ variable we have
$$I'(\rho)= \int_{\partial B(0, 1)}\frac{\partial u}{\partial n}(x+\rho z) \, ds_z =  \frac{1}{\rho}\int_{\partial B(x, \rho)}\frac{\partial u}{\partial n}(y) \, ds_y $$
Finally, as you have already shown, the divergence theorem can be applied to obtain
$$I'(\rho) = \frac{1}{\rho}\int_{B(x, \rho)} \nabla \cdot \nabla u(y)\, dA = \frac{1}{\rho}\int_{B(x, \rho)} \nabla^2 u(y)\, dA$$
