Divergence Theorem: Show e\rho Integral Equality

[Saladsamurai]

Homework Statement

This is from a fluid mechanics text. There are no assumptions being made (i.e., no constants):

Show that

$$\frac{\partial{}}{\partial{t}}\int_V e\rho \,dV + \int_S e\rho\mathbf{v}\cdot\mathbf{n}\,dA = \rho\frac{De}{Dt}\,dV\qquad(1)$$

where e and $\rho$ are scalar quantities and we the define the operator

$$\frac{D}{Dt} \equiv \frac{\partial{}}{\partial{t}} + \mathbf{V}\cdot\nabla\qquad(2)$$

Homework Equations

Divergence theorem:

$$\int_S\mathbf{n}\cdot\mathbf{F}\,dA = \int_V \nabla\cdot\mathbf{F}\,dV \qquad(3)$$

The Attempt at a Solution

I tried to use (3) on the surface integral in (1):

$$\int_S e\rho\mathbf{v}\cdot\mathbf{n}\,dA = \int_S (e\rho\mathbf{v})\cdot\mathbf{n}\,dA \qquad(4)$$

$$= \int_V\nabla\cdot(e\rho\mathbf{V})\,dV \qquad(5)$$

Now in (5) I used the vector identity: $\nabla\cdot (\phi\mathbf{F}) = \mathbf{F}\cdot\nabla\phi + \phi\nabla\cdot\mathbf{F} \qquad(6)$ however, I am not sure if the way I did it was legal. I let $\phi = e\rho$. Is that a legal move? That is, is this true:

$$\nabla\cdot (e\rho\mathbf{V}) = e\rho\nabla\cdot\mathbf{V} + \mathbf{V}\cdot\nabla e\rho$$

?

 

[fzero]

That's fine. It's less confusing to use some parentheses

$$\nabla\cdot (e\rho\mathbf{V}) = e\rho\nabla\cdot\mathbf{V} + \mathbf{V}\cdot\nabla (e\rho),$$

so that it's clear on what the gradient acts.

I think you have an integral sign missing on the RHS of equ (1).