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Hi, I'm trying to follow the proof for the statement

[tex]

\nabla . u = 0

[/tex]

I'm basing it off this paper:

http://delivery.acm.org/10.1145/119...GUIDE&dl=GUIDE&CFID=25582973&CFTOKEN=82107744

(page 7, 8)

In case thats not accessable (I'm in university just now, and I'm not sure if thats a subscriber only paper) I'll write what I've got.

so they start off with defining a fluid volume [tex]\Omega[/tex], and it's boundary surface as [tex]\partial \Omega[/tex], then defining the rate of change around this volume as

[tex] \frac{d}{dt} Volume(\Omega ) = \int \int_{\partial \Omega} u.n [/tex]

The volume should stay constant, thus

[tex] \int \int_{\partial \Omega} u.n = 0 [/tex]

from this step they mention the divergence theorem, then jump to

[tex] \int \int \int_{\Omega} \nabla .u = 0 [/tex]

It's this last jump I don't follow. From http://mathworld.wolfram.com/DivergenceTheorem.html" [Broken], I figured the divergence theorem changed to fit this problem would be..

[tex] \int_{\Omega } (\nabla .u) d\Omega = \int_{\partial \Omega} u.da [/tex]

they've dropped [tex] d\Omega [/tex] and gained two integrals, and I don't follow how they did this.

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# Homework Help: Divergence Theorem and Incompressible Fluids

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