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Proof of Divergence Formula in Spherical Coordinates

  1. Nov 18, 2007 #1
    Hello - I'm supposed to derive the divergence formula for spherical coordinates by carrying out the surface integrals of the surface of the volume in the figure (the figure is a piece of a sphere similar to a box but with curves). The radial coord is r. The polar angle is [tex]\varphi[/tex] and the azimuthal angle is [tex]\theta[/tex].

    The divergence formula is easy enought to look up: DIV(F) = [tex]\nabla\bullet[/tex]F =

    [tex]\frac{1}{r^{2}}\frac{\partial}{\partial r}r^{2}F_{r}[/tex]+[tex]\frac{1}{rsin\varphi}\frac{\partial}{\partial \varphi}\left( sin\varphi F_{\varphi}\right)[/tex] + [tex]\frac{1}{rsin\varphi}[/tex][tex]\frac{\partial F_{\theta}}{\partial\theta}[/tex]

    And the volume of the little piece of a sphere is easy enough:
    [tex]r^{2}sin\varphi \Delta r \Delta\varphi\Delta\theta[/tex]

    But when I try to set up the limits for each side as the volume goes to zero I never end up with the first and second [tex]sin\varphi[/tex] in the equation. Supposedly I'm supposed to multiply by a [tex]sin\theta[/tex] but I don't see why.

    What I end up with is:
    [tex]\frac{\partial}{\partial r}F_{r}[/tex]+[tex]\frac{1}{r}\frac{\partial}{\partial \varphi}\left( F_{\varphi}\right)[/tex] + [tex]\frac{1}{rsin\varphi}[/tex][tex]\frac{\partial F_{\theta}}{\partial\theta}[/tex]
    Last edited: Nov 19, 2007
  2. jcsd
  3. Nov 18, 2007 #2


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    How is polar angle different from azimuth angle?
  4. Nov 19, 2007 #3
    Phi is the angle in the page, and the theta is the angle out of the page.
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