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Spherical integration

  1. Jul 22, 2008 #1
    if i want to find the volume of a sphere why can't i switch the limits of integration on [itex]\theta[/itex] and [itex]\phi[/itex]

    i.e. why does this work

    [tex] \int^a_0 \int^{2\pi}_0 \int^{\pi}_0 r^2 \sin(\theta) d\theta d\phi dr [/tex]

    but this doesn't

    [tex] \int^a_0 \int^{\pi}_0 \int^{2\pi}_0 r^2 \sin(\theta) d\theta d\phi dr [/tex]
     
  2. jcsd
  3. Jul 22, 2008 #2

    mathman

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    The second integral=0. You are integrating sin over one complete cycle.
     
  4. Jul 22, 2008 #3
    Although both parameterizations cover the whole sphere, the second one turns inside out when it crosses the south pole and goes back up the opposite side.
     
  5. Jul 22, 2008 #4

    Defennder

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    You might have confused [tex]\theta[/tex] which is the angle on the x-y plane, with [tex]\phi[/tex] which is the azimuthal angle. Some textbooks interpret the notation the other way round. But given the differential volume integral in sphercal coordinates (I'm using the convention stated in the first sentece) which is [tex]r^2 \sin \phi dr d\phi d\theta[/tex], the limits for [tex]\theta[/tex] go from 0 to 2pi while [tex]\phi[/tex] goes from 0 to pi.
     
  6. Jul 22, 2008 #5
    yes i know
    can you be a little more clear because this pertains precisely to what i'm wondering about
    no i redefined the angles.
     
  7. Jul 22, 2008 #6

    Defennder

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    Well what is your redefinition? It appears in
    , you have taken the azimuthal angle to be [tex]\theta[/tex] and that should go from 0 to pi. Whereas the other planar angle phi ranges from 0 to 2pi across the x-y plane. So your limits are not correct.
     
  8. Jul 22, 2008 #7

    Defennder

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    He means to say that the azimuthal line which in your incorrect formulation ranges from 0 to 2pi sweeps from the north to south, and then back to the north again. But this covers the same region twice whereas the other planar line associated with phi in your formulation goes from 0 to pi, which doesn't cover the plane. You might want to convince yourself that the volume differential is [tex]r^2 \sin \theta drd\theta d\phi[/tex] by using the Jacobian to evaluate the integral differential variables for dxdydz.
     
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