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Triple Integrals: Spherical Coordinates - Finding the Bounds for ρ

  1. Jun 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid that lies above the cone z = root(x2 + y2) and below the sphere x2 + y2 + x2 = z.

    2. Relevant equations
    x2 + y2 + x2 = ρ2

    3. The attempt at a solution
    The main issue I have with this question is finding what the boundary of integration is for ρ. I tried to solve for it by:
    https://dl.dropbox.com/u/64325990/Photobook/Photo%202012-06-10%207%2041%2030%20PM.jpg [Broken]
    I end up getting 0 ≤ ρ ≤ root(2)sinΦ.

    However the answer says the 0 ≤ ρ ≤ cosΦ, what am I doing wrong?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 10, 2012 #2

    LCKurtz

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    The outer surface is the sphere ##x^2+y^2+z^2=z##. Writing that in spherical coordinates gives ##\rho^2=\rho \cos\phi##. Dividing by ##\rho## gives ##\rho = \cos\phi##. So ##\rho## goes from ##0## to ##\cos\phi##. You get the cone with an appropriate limits for ##\phi##.
     
    Last edited by a moderator: May 6, 2017
  4. Jun 10, 2012 #3
    Oh wow that makes so much sense now! Thanks!!! :DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD
     
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