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Cone with spherical top triple integration

  1. May 1, 2014 #1
    1. The problem statement, all variables and given/known data

    QRWR5E8.png


    2. Relevant equations

    ∫∫∫dV

    3. The attempt at a solution

    Ok so I started by setting my bounds equal to √(200-x^2-y^2) ≥ z ≥ √(x^2+y^2), √(100-x^2) ≥ y ≥ -√(100-x^2), 10 ≥ x ≥ -10

    which I got from solving z^2 = (200-x^2-y^2) = x^2+y^2 => x^2+y^2 = 100 but it seems to be a very messy integral. Should I try integrating differently? Is there a way to do this using polar coordinates?
     
  2. jcsd
  3. May 1, 2014 #2

    LCKurtz

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    Set it up in spherical coordinates.
     
  4. May 4, 2014 #3
    sorry, I haven't learnt that yet and I think this problem was explicity set up in rectangular so I could practice learning these in rectangular I am simply unsure how to place pi in there also :/
     
  5. May 4, 2014 #4

    LCKurtz

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    If you have had polar coordinates but not spherical coordinates, take your xyz integral (which is set up correctly) and integrate the dz integral. That will leave you a double integral over the circle ##x^2+y^2 = 100## in the xy plane. Change that integral to polar coordinates using ##r^2=x^2+y^2,~dydx=rdrd\theta## and put ##r,\theta## limits for the circle on the integrals.
     
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