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Evaluate the integral inside domain V, where V is bounded by the planes

  1. Apr 11, 2012 #1
    1. Evaluate the integral

    VxdV

    inside domain V, where V is bounded by the planes x=0, y=x, z=0, and the surface x2+y2+z2=1



    Answer given: 1/8 - √2/16 (which is NOT what I got.. :yuck:)
    2. The attempt at a solution

    Ok, it's a triple integral, I know this.

    ∫dx runs from 0 to 1

    ∫dy runs from -√1-x2 to x

    ∫dz runs from 0 to √1-x2-y2

    So the order of integrals go ∫∫∫dz dy dx

    Yeah... it's not working. I keep getting stuck with things like x(1-y2-x2)1/2 which I can't integrate onwards... :cry:
     
  2. jcsd
  3. Apr 12, 2012 #2

    tiny-tim

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    Hi Cloudless! :smile:
    this is a vertically sliced wedge from θ = 45° to 90°, so it would be much easier to use spherical coordinates :wink:

    of course, you can use x y z, but …

    no

    no
     
  4. Apr 12, 2012 #3

    HallsofIvy

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    In the xy-plane, z= 0 so [itex]x^2+ y^2+ z^2= 1[/itex] projects to the quarter circle [itex]x^2+ y^2= 1[/itex]. For each x, y ranges for 0 up to [itex]y= \sqrt{1- x^2}[/itex], NOT [itex]y= -\sqrt{1- x^2}[/itex] because the bounding surface is above the xy-plane.
     
  5. Apr 12, 2012 #4
    I redid it in spherical coordinates. Much easier, but still missing a pi/2. I think the conclusion is the answer key was wrong.


    The question was ambiguous in its bounded area. I thought it was from negative pi/2 to pi/4.
     
  6. Apr 12, 2012 #5

    tiny-tim

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    If you want us to check your answer, you'll have to show us your calculations. :wink:
     
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