 Solution

Assume $r_0$ is an interior point and choose $\epsilon$ small enough that its closed ball lies inside $V$. Apply Green's second identity to the punctured region $V_{\epsilon }=V\setminus \overline{B_{\epsilon }(r_0)}$, where the potential $\psi =1/|r-r_0|$ is smooth and satisfies Laplace equation. Since ${\nabla }^2\varphi =-\rho$, this givesThe crucial orientation is the outward unit normal vector of the punctured region: on the inner sphere, it is $n_{\epsilon }=-e_r$, pointing toward $r_0$. Consequently $\psi =1/\epsilon$ and ${\partial }_{n_{\epsilon }}\psi =1/{\epsilon }^2$. The inner boundary term becomesWriting $r=r_0+\epsilon \omega$ and $dS={\epsilon }^{2}d\Omega$, its first part tends to ${\int }_{|\omega |=1}\varphi (r_0)d\Omega =4\pi \varphi (r_0)$ by continuity. The second part is $O(\epsilon )$ because the gradient of $\varphi$ is bounded near $r_0$, and hence tends to zero.

The omitted volume term also tends to zero: for bounded $\rho$ near $r_0$, its absolute value is bounded by a constant times $4\pi {\int }_0^{\epsilon }rdr=O({\epsilon }^2)$. Thus the limiting integral exists, and rearranging the outer boundary terms givesThis is Green's third identity for the Poisson equation with the stated sign convention. It retains both boundary terms; no homogeneous boundary condition has been assumed. The punctured-domain calculation accounts for the singular source without incorrectly applying the regular divergence theorem through $r_0$.