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HallsofIvy
Homework Helper
It's a little hard to distinguish between $\nabla$ and V!!
What you are asking is how they went from
$$\int\int\int (\nabla\cdot E)V d\tau$$
to
$$\int\int\int E\cdot\nabla V d\tau+ \int\int V (E\cdot dA)$$
(some people prefer $E\cdot n dA$ rather than $E\cdot dA$ where "n" is the unit normal to the surface.) V here is the scalar potential and E is a vector function.

There are actually two steps in there. First they are using the "product" rule:
[tex]\nabla\cdot(VE)= V\nabla\cdot E+ (\nabla V)\cdot E[/itex]
where $\nabla\cdot E$ and $\nabla\cdot (VE)$ are the divergence (div) of the vectors and $\nabla V$ is grad V.

so
[tex]\int\int\int\nabla\cdot(VE)d\tau= \int\int\int V\nabla\cdot Ed\tau +\int\int\int(\nabla V)\cdot E d\tau[/itex]

Now use the divergence theorem to convert that first integral on the right to
[tex]\int\int VE\cdot dA[/itex]

But I'm not at all clear why the $\epsilon_0/2$ only multiplies the first integral!