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Converting triple integral to spherical

  1. Nov 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate this integral using spherical coordinates: http://img262.imageshack.us/img262/9361/xyzm.th.jpg [Broken]

    2. Relevant equations

    http://img40.imageshack.us/img40/9508/conss.th.jpg [Broken]

    3. The attempt at a solution

    http://img264.imageshack.us/img264/9457/attempt.th.jpg [Broken]

    Can anyone check and see if this is right? Specifically the limits of integration, thats what I'm having trouble with.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 9, 2009 #2
    Well the best way of checking the limits in a case like this is to draw the regions you are integrating over (ignoring the integrand).

    What is the integration region of the first integral?

    I'll help you with this one: x goes from 0 to 1, z goes from -sqrt(1-x^2) to sqrt(1+x^2) so these two integrals trace out a semicircle in the x-z plane (in the x>0 region). Since y goes from -sqrt(1-x^2-z^2) to sqrt(1-x^2-z^2) this rotates the semicircle along the y-axis to give a...


    What is the second (transformed) integration region?

    Are these the same? If so good, if not change the limits so they are.
     
  4. Nov 9, 2009 #3
    I don't really get what you are saying.

    I know that sqrt(1-x^2-z^2) and -sqrt(1-x^2-z^2) makes a sphere with radius 1, which can be turned into rho=1.

    I just cant figure out what phi and theta run to. Since I think it's a sphere (right?) can I assume phi and theta run complete cycles? Is that right?

    Can someone please tell me these two limits (and if rho does indeed =1) so that I can see if I'm doing this right?
     
  5. Nov 9, 2009 #4
    Wait, if it's a sphere of rho=1, it then projects a shadow on the x-z plane of a circle with radius one, but the only area of interest is from x=0 to x=1.

    So a semi-sphere that exists only in the positive-x?

    Is that the shape and the correspondingly correct limits?
     
  6. Nov 9, 2009 #5

    lanedance

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    hemi-sphere in positive x sounds good to me, now what do your spherical limits become...
     
  7. Nov 9, 2009 #6
    rho from 0 to 1

    phi from 0 to pi

    and theta from 0 to pi

    Right?
     
  8. Nov 9, 2009 #7
    Right!?
     
  9. Nov 10, 2009 #8

    lanedance

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    depends on your convention if phi is teh angle around teh horizontal & the x axis aligns with phi = pi/2, which i think it does, then yes
     
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