Set $D:=\{(x,y,z)\in \Bbb R^2:x^2+y^2+z^2 \le 1\}$. Denote $\Delta:= \frac{\partial f^2}{\partial x^2}+\frac{\partial f^2}{\partial y^2}+\frac{\partial f^2}{\partial z^2}$. Suppose that a $C^2$-function $\phi(x,y,z)$ vanishes in the neighborhood of $\partial D$. I want to prove that $$\iiint_D (x^2+y^2+z^2)^{-1/2}(\Delta \phi)(x,y,z)dxdydz=-4\pi\phi(0,0,0).$$ Using the spherical coordinates I can write $$\iiint_D (x^2+y^2+z^2)^{-1/2}(\Delta \phi)(x,y,z)dxdydz=\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^1 (\Delta \phi)(\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi)\rho \sin \varphi d\rho d\theta d\varphi.$$ However I don't know where to use the assumption $\phi$ vanishes in the neighborhood of $\partial D$.
Any help would be much appreciated.
Use Green's second identity, $$\iiint_D (\psi\Delta\phi-\phi\Delta\psi)dV=\iint_{\partial D^+}\left(\psi\frac{\partial \phi}{\partial {\bf n}}-\phi\frac{\partial \psi}{\partial {\bf n}}\right)dS$$ with $\psi(x,y,z)=(x^2+y^2+z^2)^{-1/2}$.
Note that $\Delta\psi(x,y,z)=-4\pi\delta(x,y,z)$ and the integral on the right-hand side is zero because $\phi$ and $\frac{\partial \phi}{\partial {\bf n}}$ vanish in the neighborhood of $\partial D$.