## Divergence Theorem and Incompressible Fluids

1. The problem statement, all variables and given/known data

Hi, I'm trying to follow the proof for the statement
$$\nabla . u = 0$$

I'm basing it off this paper:
http://delivery.acm.org/10.1145/1190...TOKEN=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 $$\Omega$$, and it's boundary surface as $$\partial \Omega$$, then defining the rate of change around this volume as

$$\frac{d}{dt} Volume(\Omega ) = \int \int_{\partial \Omega} u.n$$

The volume should stay constant, thus

$$\int \int_{\partial \Omega} u.n = 0$$

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

$$\int \int \int_{\Omega} \nabla .u = 0$$

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

$$\int_{\Omega } (\nabla .u) d\Omega = \int_{\partial \Omega} u.da$$

they've dropped $$d\Omega$$ and gained two integrals, and I don't follow how they did this.
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 Quote by Bucky $$\int_{\Omega } (\nabla .u) d\Omega = \int_{\partial \Omega} u.da$$
 Quote by Bucky $$\int \int \int_{\Omega} \nabla .u = 0$$
 Quote by Bucky $$\int \int_{\partial \Omega} u.n = 0$$
As I said previously, they have simply chosen not to write $d\Omega$ and $dA$, which is acceptable but can be confusing.