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Divergence theorem on non compact sets of R3

  1. Jun 29, 2015 #1
    So my question here is: the divergence theorem literally states that
    Let [itex]\Omega[/itex] be a compact subset of [itex] \mathbb{R}^3 [/itex] with a piecewise smooth boundary surface [itex]S[/itex]. Let [itex]\vec{F}: D \mapsto \mathbb{R}^3[/itex] a continously differentiable vector field defined on a neighborhood D of [itex]\Omega[/itex].
    Then:
    [itex]\int_{\Omega} \nabla \cdot \vec{F} dxdydz = \oint_S \vec{F} \cdot \vec{n} dS [/itex]

    My problem here is: why people (and with which argument) use this divergence or Gauss theorem to compute the electric field of some NOT bound set (for example, the typical infinte cylinder) of surface charge.
     
  2. jcsd
  3. Jun 29, 2015 #2

    Orodruin

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    Doing so is fine if you can show that the integral over any surface closing the area goes to zero when that closing surface is placed further and further away. This may put additional requirements on the integrand, such as vanishing sufficiently fast.

    There are also cases where the volume is not infinite although you are studying an infinite setup, but, for example, the end caps of thr cylinder do not contribute. In other cases, symmetry may reduce the dimensionality of the problem and it is sufficient to consider it in one dimension less where the volume is bounded.
     
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