Hint for HW#2, Question 1-i

It looks nasty, but it has the general form "a closed surface integral= a volume integral involving derivatives". Whenever I see that, I immediately think "Divergence theorem"! (You should start making that connection too!)

So go look again at the Divergence Theorem, this is Griffiths Eq 1.56. Now, that theorem has an arbitrary field "v" in it (on both sides).
On the "surface integral" side, it's simply v dot da. So, when I look again at the homework problem, I look at the term with the surface integral and ALSO see "something dot da". So I naturally ask myself, what if that WHOLE EXPRESSION dotting da is just called "v". That's it, that's the trick!

(Can you procede from here? If not, keep reading...)

Try it! Let v = T grad(U). (Note that T is a scalar, and grad(U) is a vector, so indeed v is a vector here)

Now all we have to do to "plug this in" to the divergence theorem is take div(V) to insert on the volume integral side.... That may seem nasty, but take a look now at the front flyleaf of Griffiths text. It's really handy, all the "vector calculus" trickery you need this term and next is likely to be there. For this problem, we need to take
div( T grad U)
But, generically, that's of the form div(some scalar * some vector), and that's one of those formulas given in Griffiths! Just find the right one, plug it in, the answer should fall out. (You do need to remember one more thing, namely what the definition of "Del^2" is. That's in Chapter 1, it's Del^2(something) = Div(grad (something))

In the end, after all that, when you write it on paper, this is really a "quicky" (It is, if you knew the right places to go!)