# Divergence theorem versions

1. Sep 25, 2007

### geonat

Are there versions of the divergence theorem that don't require a compact domain?

In my reference literature the divergence theorem is proved under the assumption that the domain is compact.

2. Sep 26, 2007

### easttiger2007

I think divergence theorem require a domain that has "boundary", which is necessary for a "compact domain". Basically, divergence thm, along with the 1D Newton-Leibniz version, and the 2D Stoke's version, can be regarded as the basic thm of calculus in the first 3 dimensions. They apparently give a method of degrading an integral of a certain dimension into one of a lower dimension, with the latter be defined on the boundary of the domain of the former. Therefore, a compact domain should be sufficient to enable this process to happen.

3. Sep 30, 2007