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Divergence theorem versions

  1. Sep 25, 2007 #1
    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. jcsd
  3. Sep 26, 2007 #2
    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.
  4. Sep 30, 2007 #3
    Thank you for your reply.
    I am working on a scattering problem, so part of my boundary lies infinitely far away. Moreover, another part of my boundary approaches pointlike sources, while yet another part of the boundary approaches virtual secondary sources on the boundary of the scattering object. But I think I got it right now.
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