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Area of portion of sphere

  1. Feb 16, 2008 #1
    For a sphere, the relation between steradians and the area they cover is O = A/(r^2), in which O is the measure of the solid angle, A is the area it covers, and r is the radius. If I were instead given the half-angle of the steradian...meaning that if there was a central axis running through the solid angle, connecting the surface of the sphere to its center, the half-angle would simply be the angle between the central axis and the edge of the solid angle....then how would I find the area covered by the solid angle in respect to the half angle.
  2. jcsd
  3. Feb 16, 2008 #2


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    You can integrate over the sphere using spherical coordinates by:

    [tex] \int_0^{2 \pi} d \phi \int_0^\pi d\theta R^2 \sin\theta [/tex]

    The region you're talking about then corresponds to the range [itex]0\leq \phi < 2 \pi [/itex] and [itex]0 \leq \theta < \theta_0[/itex], where [itex]\theta_0[/itex] is the half angle. So you can find the area by evaluating:

    [tex] A = \int_0^{2 \pi} d \phi \int_0^{\theta_0} d\theta R^2 \sin\theta [/tex]
  4. Feb 16, 2008 #3
    ...wait...wut? I sort of half get what you're saying. I was using earlier the integral with the sine in it...but where did the integral of dphi pop out of?
    Last edited: Feb 16, 2008
  5. Feb 16, 2008 #4


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    Look up spherical coordinates.
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