I have been trying this, but I do not understand how to get the gradient of the triple integration.
If a region $V$ bounded by a surface $S$ has a continuous charge (or mass) distribution of density $\rho$, the potential $\phi(P)$ at a point $P$ is defined by $\phi = \iiint_V \frac{\rho dV}{r}$. Deduce the following under suitable assumptions:
$\iint_S \mathbf E \cdot d\mathbf S = 4 \pi \iiint_V \rho d V$, where $\mathbf E = - \nabla \phi$.
$\nabla^2\phi = - 4 \pi \rho$ (Poisson's equation) at all points $P$ where charges exist, and $\nabla^2 \phi = 0$ (Laplace's equation) where no charges exist.
Now, I tried to use Divergence theorem. But I do not understand how do I find gradient of triple integration. Thanking you all.
Here is how I tried.Assuming $\rho$ to be a function of r.
$\phi = \iiint_V \frac{\rho dV}{r}$
$\nabla\phi = \iiint_V \nabla(\frac{\rho }{r})dV$
$\nabla\phi = \iiint_V (\frac{\rho'\mathbf r}{r}+\rho\frac{-\mathbf r}{r^3})dV$
$\nabla^2\phi = \iiint_V \nabla(\frac{\rho'\mathbf r}{r}+\rho\frac{-\mathbf r}{r^3})dV$
$\nabla^2\phi = \iiint_V (\rho''\frac{\mathbf r . \mathbf r}{r^2}+(\rho'\frac{\mathbf r}{r}+\rho(\frac{3}{r^3}+\mathbf r\frac{-3\mathbf r}{r^5})))dV$
$\nabla^2\phi = \iiint_V (\rho''+(\rho'\frac{\mathbf r}{r}))dV$
$\nabla^2\phi = \iiint_V (\rho''+\rho'\hat{r}))dV$
Here, if I assume $\rho$ to be constant then $\nabla^2\phi$ is equal to zero. But my result are not upto the expectation of the question. I followed as I was advised but results are unsatisfactory. If there is any mistake in my calculation ,please correct me.
Thanking you
Hint: For part 1, the idea is to use the divergence theorem to write $$ \iint_S \mathbf E \cdot d \mathbf S = \iiint_V \nabla \cdot \mathbf E\,dV. $$ You should be able to simplify the expression $\nabla \cdot \mathbf E = \nabla \cdot (-\nabla \phi) = -\nabla^2 \phi$ to end up with the integrand $4 \pi \rho$.
For part 2, you will have already found that $-\nabla^2 \phi = 4 \pi \rho$ in part 1. Multiplying both sides gives you the desired result $\nabla^2 \phi = -4 \pi \rho$, which simply becomes $\nabla^2\phi = 0$ at points $P$ for which $\rho(P) = 0$.