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Setting up a triple integral with spherical coordinates

  1. Jan 6, 2012 #1
    1. The problem statement, all variables and given/known data
    http://img28.imageshack.us/img28/7118/capturenbc.jpg [Broken]


    2. Relevant equations

    x2 + y2 + z2 = p2

    http://img684.imageshack.us/img684/3370/eq0006m.gif [Broken]

    3. The attempt at a solution

    Using the relevant equations I converted the given equation to:

    ∫∫∫e(p3/2) * p2 * sin(∅) dp dθ d∅

    Then the bounds of the first integral, dp, would be from 1 to 2, because of the radius of the spheres.

    I cannot figure out what the bounds for the second and third integrals would be though.

    Thanks for your help!
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 6, 2012 #2
    What shape is defined by z2 = 2(x2 + y2) ?
     
  4. Jan 6, 2012 #3
    Last edited by a moderator: May 5, 2017
  5. Jan 6, 2012 #4

    SammyS

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    You have [itex]x^2+y^2+z^2\le1\,.[/itex]

    I think you mean [itex]x^2+y^2+z^2\ge1\,.[/itex]



    To find the limits of integration for θ and φ , you need to know how x, y, and z are related to ρ, θ, and φ.
     
    Last edited by a moderator: May 5, 2017
  6. Jan 6, 2012 #5
    OK , it's a cone, so which integration does that imply a limit for, in spherical coordinates?
     
  7. Jan 7, 2012 #6
    So, normally for a cone would have a θ from 0 <= θ <= 2∏, but the question says first octant, so would that mean 0 <= θ <= ∏/4?

    And ∅ would be: 0 <= ∅ <= ∏/4?

    EDIT: wait, wouldn't θ be limited by 3x <= y <= 4x? So, then how could i find those limits?
     
    Last edited: Jan 7, 2012
  8. Jan 7, 2012 #7

    SammyS

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    "... but the question says first octant, so would that mean 0 <= θ <= ∏/4?"

    No. This is three dimensions. The coordinate planes cut R3 into 8 Octants. In the first Octant x, y, and z are all positive.
    Therefore, in the first Octant, 0 <= θ <= π/2 .​

    The cone determines the limits for φ . z2 tan2(φ) = x2 + y2.

    The more difficult task will be to express y=3x and y = 4x in spherical coordinates. But it's not all that difficult .
     
  9. Jan 7, 2012 #8
    [itex]\phi[/itex] is the one I was looking for limited by the cone, yes. The limit angle is not [itex]\pi/4[/itex], though.

    And the other two planes do indeed limit [itex]\theta[/itex]. Just need a little trig in each case to work out the angles (or express them as inverse trig functions, I guess).
     
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