Proving the equivalency of the potential energy of a system of charges and the work required to assemble a system of charges This is a very cool, and highly beneficial problem in my opinion. I feel as though truly understanding this proof would broaden anyone's conceptual understanding of electric potential.
My textbook asks me to utilize the identity: 
$\bigtriangledown(\phi\bigtriangledown\phi) = (\phi\bigtriangledown)^2 + \phi\bigtriangledown^2\phi$ 
and the divergence theorem to prove that the potential energy of a system of charges $U_E = \cfrac \pi8\int_{entire-field}E^2dv$ and the work that it takes to assemble a charge distribution $U_w = \cfrac 12\int \rho\phi dv$ are "not different" for all charge distributions of finite values.
So far, I have decided to substitute $-\bigtriangledown^2\cfrac{1}{8\pi}$ in for $\rho$ so that I can use the given identity [ I also had to rearrange the ordering to $\phi\bigtriangledown^2\phi = \bigtriangledown(\phi\bigtriangledown\phi) -(\phi\bigtriangledown)^2 $] and ascertain that $U = \cfrac \pi8\int \bigtriangledown(\phi\bigtriangledown\phi)dv - \cfrac \pi8\int \phi\bigtriangledown^2\phi dv,  $ but now I am stuck. I know that the next step must have to do with using the divergence theorem to simplify this thing though. Any help would be greatly appreciated.
 A: Hint:
The divergence theorem tells us that the divergence of a vector field integrated over a region $R$ with boundary $\partial R$ equals the integral of that vector field dotted with the outward-pointing normal along the boundary;
\begin{align}
  \int_R dV\, \nabla\cdot \mathbf v=\int_{\partial R} dS\, \mathbf v\cdot\mathbf n.
\end{align}
If we are integrating over all space, then this is like integrating over the inside of a sphere but in the limit that this sphere is infinitely big.  So we have
\begin{align}
  \int_{\mathbb R^3} dV\, \nabla\cdot\mathbf v = \lim_{r\to\infty}\int_{B_r}dV\,\nabla\cdot \mathbf v = \lim_{r\to\infty} \int_{S_r}dS \,\mathbf v\cdot \mathbf n
\end{align}
where $B_r$ denotes the closed ball of radius $r$ (namely the inside of a sphere of radius $r$ along with its boundary) and $S_r$ denotes the sphere of radius $r$.  Now, on the surface of a sphere of radius $r$, the area element and outward-pointing normal are
\begin{align}
  dS = r^2\sin\theta d\theta d\phi, \qquad \mathbf n = \hat{\mathbf r},
\end{align}
so we get
\begin{align}
  \int_{\mathbb R^3} dV\, \nabla\cdot\mathbf v = \lim_{r\to\infty} \int_{S_r} d\theta\,d\phi\,\sin\theta\,r^2 v_r
\end{align}
If $v_r$ decreases sufficiently rapidly with $r$, then the expression on the right will be zero.  In other words, we have found that

Provided the radial component of a vector field falls off sufficiently rapidly with $r$, the integral of the divergence of the vector field over all of space vanishes.

Try to use this fact along with things you know about how the electric field of a finite distribution of charge behaves very far from the distribution.
