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Homework Help: Triple Integral

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  1. Dec 7, 2017 #1
    1. The problem statement, all variables and given/known data
    Evaluate the triple integral y^2z^2dv. Where E is bounded by the paraboloid x=1-y^2-z^2 and the place x=0.
    2. Relevant equations
    x=r^2cos(theta) y=r^2sin(theta)
    3. The attempt at a solution
    I understand how to find these three limits, -1 to 1 , -sqrt(1-y^2) to sqrt(1-y^2) , 0 to 1-y^2-z^2. I then solved the first integral and got y^2z(1-y^2-z^2) I know I need to convert to polar coordinates. I am able to do that 0 to 2pi, 0 to 1.
    I am having trouble converting the equation y^2z(1-y^2-z^2) to polar coordinates.
     
    Last edited by a moderator: Dec 8, 2017
  2. jcsd
  3. Dec 8, 2017 #2

    Mark44

    Staff: Mentor

    What are these equations? They aren't the equations to convert from cartesian to polar.
     
  4. Dec 8, 2017 #3
    yes they are I am having trouble converting it do I leave the z that left over in?
     
  5. Dec 8, 2017 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    It looks like the x-axis in your problem plays the role usually reserved for the z-axis in similar problems. So, you will find it easier to use un-standard polar coordinates ##y = r \cos \theta## and ##z = r \sin \theta##, with volume element ##dV = r\, dr \,d\theta \, dx##.
     
  6. Dec 8, 2017 #5
    Thank you :).
     
  7. Dec 8, 2017 #6

    Mark44

    Staff: Mentor

    No, they aren't. In the usual conversions, ##x = r\cos(\theta)## and ##y = r\sin(\theta)##, not with ##r^2## as you show. Use Ray's advice to treat x as you would normally do for z
     
  8. Dec 8, 2017 #7
    that was a mistype .
     
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