Proof of Green's third identity I don't completely understand the proof of the following result, which is also called Green's third identity:

Let $D\subset{\mathbb R^m}$ be a bounded domain of class $C^1$ and let $u\in C^2(\bar{D})$ be harmonic in $D$. Then
  $$u(x)=\int_{\partial D}\bigg\{\frac{\partial u}{\partial\nu}(y)\Phi(x,y)-u(y)\frac{\partial \Phi(x,y)}{\partial\nu}\bigg\}ds(y),\quad x\in D$$
  where
  $$\Phi(x,y):=
\begin{cases}
\dfrac{1}{2\pi}\ln\dfrac{1}{|x-y|},\quad m=2\\
\dfrac{1}{4\pi}\dfrac{1}{|x-y|},\quad m=3.
\end{cases}$$

Proof. For $x\in D$ choose a sphere $\Omega(x;r):=\{y\in{\mathbb R}^m:|y-x|=r\}$ of radius $r$ such that $\Omega(x;r)\subset D$ and direct the unit normal $\nu$ to $\Omega(x;r)$ into the interior of $\Omega(x;r)$. Apply Green's second identity to the harmonic functions $u$ and $\Phi(x,\cdot)$ in the domain $\{y\in D:|y-x|>r\}$ to obtain
$$
\int_{\partial D\cup\Omega(x;r)}\bigg\{u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}-\frac{\partial u}{\partial\nu}(y)\Phi(x,y)\bigg\}ds(y)=0.
$$
Now we have
$$
\int_{\Omega(x;r)}\bigg\{u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}-\frac{\partial u}{\partial\nu}(y)\Phi(x,y)\bigg\}ds(y)=\int_{\partial D}\bigg\{\frac{\partial u}{\partial\nu}(y)\Phi(x,y)-u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}\bigg\}ds(y)
$$
Here is my question:
How can I show that 
$$
\lim_{r\to 0}\int_{\Omega(x;r)}\bigg\{u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}-\frac{\partial u}{\partial\nu}(y)\Phi(x,y)\bigg\}ds(y)=u(x)?
$$
 A: For the second term, since $u\in C^2(\bar{D})$, $\partial u/\partial \nu$ is bounded, say by $M$. Then
$$
\begin{eqnarray}
\left|\int_{\Omega(x;r)}\frac{\partial u}{\partial\nu}(y)\Phi(x,y)\mathrm ds(y)\right|
&\le&
M\int_{\Omega(x;r)}|\Phi(x,y)|\mathrm ds(y)
\\
&=&
Mcr^{m-1}\Phi(r)\;,
\end{eqnarray}$$
where $c$ is the surface area of the unit sphere ($2\pi$ and $4\pi$, respectively) and $\Phi(r)$ is the respective function such that $\Phi(x,y)=\Phi(|x-y|)$. This product goes to zero as $r\to0$, since $r\ln r\to0$.
For the first term, we have $\partial\Phi(x,y)/\partial\nu(y)=-\Phi'(r)=1/(cr^{m-1})$, where the minus sign arises because the unit normal is directed into the interior. Thus
$$
\begin{eqnarray}
\lim_{r\to 0}\int_{\Omega(x;r)}u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}ds(y)
&=&
\lim_{r\to 0}\int_{\Omega(x;r)}\frac{u(y)}{cr^{m-1}}ds(y)
\\
&=&
\lim_{r\to 0}\frac1c\int_{\Omega(x;r)}u(y)\mathrm d\Omega
\\
&=&
u(x)
\end{eqnarray}
$$
(where $\mathrm d\Omega$ denotes integration over the solid angle), since $u$ is continuous.
