# Divergence Theorem: Multiplied by Scalar Field

1. Oct 28, 2012

### YayMathYay

1. The problem statement, all variables and given/known data

2. Relevant equations

Definitely related to the divergence theorem (we're working on it):

3. The attempt at a solution

I'm a bit confused about multiplying a scalar field f into those integrals on the RHS, and I'm not sure if they can be taken out or not. If they can be, I evaluated the RHS out to be 0 (zero), which doesn't make sense with my evaluation of the LHS, which is just grad f dotted into F.

On the other hand, if it CAN'T be taken out of the integral, I'm at a loss as of how this relates to the divergence theorem..
I'm not sure what I'm missing here :( Help would be very much appreciated!

2. Oct 28, 2012

### LCKurtz

Start by using the divergence theorem on the first term on the right$$\iint_{\partial R}(f\vec F)\cdot \hat n\, dA = \iiint_R\nabla \cdot (f\vec F)\, dV$$Work out that $\nabla \cdot (f\vec F)$ in the integrand and go from there.

3. Oct 28, 2012

### Dick

First show,
$$\nabla \cdot (fF)=\nabla f \cdot F+f \nabla \cdot F$$
If you write it out in components it's just the product rule.

4. Oct 28, 2012

### lurflurf

This is the product rule
$$\nabla \cdot (\psi\mathbf{A}) = \mathbf{A} \cdot\nabla\psi + \psi\nabla \cdot \mathbf{A}$$
wrapped up with the divergence theorem.

5. Oct 28, 2012

### YayMathYay

For the integrand, I'm getting:
$\partial (f A) / \partial x + \partial (f B) / \partial y + \partial (f C) / \partial z$, where $\vec F = (A, B, C)$

Am I on the right track?

6. Oct 28, 2012

### YayMathYay

Double Post, but:

Ahh so if I use the Divergence Thm as LCKurtz suggested on the first term on the RHS, I get the product rule in the form:

$$\nabla f \cdot F = \nabla \cdot (fF) - f \nabla \cdot F$$

Except with the terms as integrands. I'm not sure if this is sufficient to prove the validity of the equation though? I'm sorry guys, I feel like you guys are putting the answer right in my face but I'm just not getting it :(

7. Oct 28, 2012

### LCKurtz

Assuming $\vec F = \langle A,B,C\rangle$, Yes. Keep going...