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Solving for dA in Gauss's Law

  1. Sep 20, 2007 #1
    Hey, I'm having some trouble solving for the dA portion of Guss's Law for a sphere as the Gaussian surface and a point charge on the inside.

    According to my book, Integral(1dA) = 4(pi)r^2

    So when I try to integrate it myself I get 2(pi^2)r^2
    The integral I solve is [tex]\int[/tex][tex]\int r\phi r\theta[/tex] where theta = [0,2pi] and phi = [0,pi]

    Can anyone set me straight please?
    Last edited: Sep 20, 2007
  2. jcsd
  3. Sep 20, 2007 #2


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    Staff: Mentor

    On the surface of a sphere, the element of area bounded by [itex]d\theta[/itex] and [itex]d\phi[/itex] has area [itex](r d\theta)(r \sin \theta d\phi) = r^2 \sin \theta d\theta d\phi[/itex].
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