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Verify pulling out the partial derivative.

  1. Dec 9, 2013 #1
    For spherical coordinates, [itex]u(r,\theta,\phi)[/itex] is function of [itex]r,\theta,\phi[/itex]. [itex]a[/itex] is constant and is the radius of the spherical region. Is:

    [tex]\int_{0}^{2\pi}\int_{0}^{\pi}\frac{\partial\;u(r,\theta,\phi)}{\partial {r}}a^2\sin\theta d\theta d\phi=\frac{\partial}{\partial {r}}\left[\int_{0}^{2\pi}\int_{0}^{\pi}u(r,\theta,\phi)a^2\sin\theta d\theta d\phi\right][/tex]

    Last edited: Dec 9, 2013
  2. jcsd
  3. Dec 10, 2013 #2


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    This is true as long as ##\partial u/\partial r## exists and is continuous on the appropriate domain. This is known as the Leibniz Integral Rule and a proof is given at the link.
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