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Divergence Theorem

  • #1
3,003
2

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]

?
 

Answers and Replies

  • #2
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
That's fine. It's less confusing to use some parentheses

[tex]\nabla\cdot (e\rho\mathbf{V}) = e\rho\nabla\cdot\mathbf{V} + \mathbf{V}\cdot\nabla (e\rho), [/tex]

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

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

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