- #1

Saladsamurai

- 3,019

- 6

## Homework Statement

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

[tex]\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)

[/tex]

where

*e*and [itex]\rho[/itex] are scalar quantities and we the define the operator

[tex]\frac{D}{Dt} \equiv \frac{\partial{}}{\partial{t}} + \mathbf{V}\cdot\nabla\qquad(2)[/tex]

## Homework Equations

Divergence theorem:

[tex]\int_S\mathbf{n}\cdot\mathbf{F}\,dA = \int_V \nabla\cdot\mathbf{F}\,dV \qquad(3)[/tex]

## The Attempt at a Solution

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

[tex]\int_S e\rho\mathbf{v}\cdot\mathbf{n}\,dA =

\int_S (e\rho\mathbf{v})\cdot\mathbf{n}\,dA \qquad(4)[/tex]

[tex]= \int_V\nabla\cdot(e\rho\mathbf{V})\,dV \qquad(5)[/tex]

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

[tex]

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

[/tex]

?