Divergence Theorem: Multiplied by Scalar Field

[YayMathYay]

Homework Statement

Homework Equations

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

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!

 

[LCKurtz]

Homework Statement

Homework Equations

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

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!

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.

 

[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.

 

[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.

 

[YayMathYay]

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.

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?

 

[YayMathYay]

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.

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 :(

 

[LCKurtz]

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?

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