This is taken from the bottom of page 37 of Griffith's Introduction to Electrodynamics
Using integration by parts,
$\nabla \cdot (f {\bf A})$ = $ f (\nabla \cdot {\bf A}) + {\bf A} \cdot (\nabla f)$,
and therefore,
$\int \nabla \cdot (f {\bf A})$ = $ \int f (\nabla \cdot {\bf A}) + \int {\bf A} \cdot (\nabla f)$
The divergence theorem states that
$\int_V (\nabla \cdot {\bf A}) d \tau $ = $ \oint_S {\bf A} \cdot dA$.
In the textbook, by invoking the divergence theorem, the obtained result was that
$\int \nabla \cdot (f {\bf A}) d \tau $ = $\int f(\nabla \cdot {\bf A}) d \tau $ + $\int {\bf A}\cdot (\nabla f) d \tau $ = $\oint f {\bf A} \cdot d{\bf a}$.
My confusion begins when substituting the divergence theorem into the equation in the 2nd line.
I think that
$\int \nabla \cdot (f {\bf A})$ = $ \oint_S {\bf A} \cdot dA$ + $\int {\bf A} \cdot (\nabla f)$,
so I am not sure what happens to the $\int {\bf A} \cdot (\nabla f)$ in the textbook's result.
Also, any clarification on when exactly to use the $v, s, c$ and all the other integral subscripts are to be used would be appreciated. It's easy to know when the questions asks for a specific volume integral or surface integral etc, but is it possible to tell just from the equations given?
With the 2 questions that follow after the example in the book, I am having trouble understanding their solutions.
The question is given with its solution here where I am to use product rules and integral theorems to prove equality of the integrals. However, like my problem in the example, I cannot seem to be able to follow when trying to substitute.
Thanks.
I don't think it's a good idea to drop the $dV$ and $d\mathbf S$ that should be present in integrals (which also tell you what you're integrating over). You seem to have done this in several places, which is probably why the formulas all look the same.
Anyway, you know that, for some vector field $\mathbf B$,
$$\int \nabla \cdot \mathbf B \, dV = \oint \mathbf B \cdot d\mathbf S$$
That's the divergence theorem. Just let $\mathbf B = f \mathbf A$.
$$\int \nabla \cdot (f\mathbf A) \, dV = \oint f \mathbf A \cdot d\mathbf S$$
Using the product rule on the left side gives
$$\int f (\nabla \cdot \mathbf A) \, dV + \int \mathbf A \cdot \nabla f \, dV = \oint f \mathbf A \cdot d\mathbf S$$