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Question on solid angle of sphere.

  1. Jul 17, 2011 #1
    I understand the surface of the sphere is [itex]4\pi [/itex] sr. where the area of one sr is [itex]r^2[/itex].

    My question is why [itex] d\Omega = sin \theta \;d \theta \;d\phi[/itex]? Can anyone show me how to derive this.

    Is it because surface area [itex]dS = (Rd\theta)(R\; sin\;\theta\;d\phi)\;\hbox { so if }\; \Omega = \frac S {R^2} \;\Rightarrow \; d\Omega = \frac {d\;S}{R^2}= sin \theta d\theta d \phi[/itex]
     
    Last edited: Jul 17, 2011
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  3. Jul 17, 2011 #2

    BruceW

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    [tex] \vec{r} = r( cos(\phi)sin(\theta) \hat{i} + sin(\phi)sin(\theta) \hat{j} + cos(\theta) \hat{k} ) [/tex]
    and the surface element perpendicular to [itex] \hat{r} [/itex] is:
    [tex] d\Omega = \frac{\partial \vec{r}}{\partial \theta} \ d\theta \wedge \frac{\partial \vec{r}}{\partial \phi} \ d\phi [/tex]
    which gives:
    [tex] d\Omega = sin(\theta) d\theta d\phi \hat{r} [/tex]

    EDIT: I can't remember which order the cross product should go, but it should be such that [itex] d\Omega [/itex] is positive.
     
    Last edited: Jul 17, 2011
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