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Solid angle

  1. Oct 8, 2005 #1
    So what i know that a solid angle is to sphere as the curve is to a circle...

    curve= rΘ, and (differential solid angle) dΩ=2ΠsinΘdΘ

    I need to prove it, and i'm a bit rusty and i donno where to start, i wonder if there's any usefull links or tips...

    Any more info about the use of a solid angle??

  2. jcsd
  3. Oct 8, 2005 #2


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  4. Oct 9, 2005 #3
    Astronuc what would my life be without u?? :D

  5. Oct 11, 2005 #4
    Well, i was thinking that doesn't the equation i wrote in the 1st post seem close to the parameter of a circle???And if that circle small, that its parameter would be almost equal to its area???
  6. Oct 11, 2005 #5


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    Area of a circle in polar coordinates is just the integral of "r d[itex]\theta[/itex] dr", with r limits of 0,r, and [itex]\theta[/itex] from 0, 2[itex]\pi[/itex] so one should end up with [itex]\pi[/itex]r2.

    Similarly in spherical coordinates the integrand is r2 sin [itex]\phi[/itex] d[itex]\phi[/itex] d[itex]\theta[/itex], and to find the area, one simply integrates over the two angle with r fixed, and the area should be 4[itex]\pi[/itex]r2

    [itex]\phi[/itex] limits -[itex]\pi[/itex], [itex]\pi[/itex] and [itex]\theta[/itex] limits 0, 2[itex]\pi[/itex]

    So in some sense, finding the circumference of a circle, is analogous to finding the area of a sphere.
    Last edited: Oct 11, 2005
  7. Oct 11, 2005 #6
    I do understand that, maybe i wasn't clear enough...I just thought it was quite similar to the relation i wanted to get to...
  8. Oct 11, 2005 #7


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    I just notice an error in one of the expression I posted.

    A = r2 d[itex]\Omega[/itex] should read

    dA = r2 d[itex]\Omega[/itex]


    Asphere = 4[itex]\pi[/itex]r2
  9. Oct 12, 2005 #8
    Alright that's what i proved, but after i proved it, i just thought something naive, so don't bother:wink:

    Thx Astronuc..

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