Divergence Theorem Problem Using Multiple Arbitrary Fields

[JackofTrades]

Homework Statement

Letting a(r) and b(r) be arbitrary scalar fields, use the Divergence theorem(for an arbitrary closed surface S, enclosing a volume V) to show: $$ \int dS · (a∇b−b∇a) = \int V \,dV (a(∇^2)b − b(∇^2)a).$$

Show that, where u and v are arbitrary vector fields: $$ \int S dS·[u×(∇×v)−v×(∇×u)] =\int VdV [v·(∇×[∇×u])−u·(∇×[∇×v])], $$ where S is an arbitrary closed surface, enclosing a volume V.

Relevant Equations

$\int dS · A = \int dV ∇·A$

My main issue with this question is the manipulation of the two arbitrary fields into a single one which can then be substituted into the divergence theorem and worked through to the given algebraic forms.

My attempt:

$$ ∇(ab) = a∇b + b∇a $$

Subsituting into the Eq. gives $$ \int dS · (a∇b+b∇a) = \int V \, dV ∇·(a∇b+b∇a) $$ $$ \int dS · (a∇b+b∇a) = \int V \, dV ( ∇a∇b + a ∇^2 b + ∇b∇a + b ∇^2 a) $$ $$ \int dS · (a∇b+b∇a) = \int V \, dV ( 2∇a∇b + a ∇^2 b + b ∇^2 a) $$
Which would give me the required expression if my initial combination of the two arbitrary fields had a negative in, but I can't figure out how to get to that point.

The same problem exists for the vector portion of the question

$$ ∇(u·v) = u x (∇ x v) + v x (∇ x u) + (u · ∇)v + (v · ∇)u $$

Where I believe that it would cancel to the form required if the intial combination of the two fields produced a negative rather than a positive.

 

[TSny]

show: $$ \int dS · (a∇b−b∇a) = \int V \,dV (a(∇^2)b − b(∇^2)a).$$
My attempt:
Subsituting into the Eq. gives $\int dS · (a∇b+b∇a) = \int V \, dV ∇·(a∇b+b∇a)$

Don't you want to consider $\int dS · (a∇b-b∇a) = \int_{\small{V}} \, dV ∇·(a∇b-b∇a)$?