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Volume Triple Integral

  1. May 7, 2009 #1

    danago

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    Find the volume of the region above the sphere [tex]x^2+y^2+z^2=4[/tex] and below the
    paraboloid [tex]z = 6-x^2- y^2[/tex].


    Ok so the first thing i did was to find out if the two surfaces ever intersect by substituting [tex]x^2+y^2=6-z[/tex] into the first equation and solving for z. I got only complex solutions, hence they never intersect.

    If i am picturing the region correctly, i am dealing with the volume of a region who's base is bounded by the equator of the sphere, all the way up to the paraboloid.

    My initial thought was to use a cylindrical coordinate system. If i have set up my bounds correctly, the integral i would need to evaluate would be:

    [tex]V=\int^{2\pi}_{0}\int^{2}_{0}\int^{6-r^2}_{\sqrt{4-r^2}}dz\:r\:dr\: d\theta=\frac{32\pi}{3}[/tex]

    However, according to the solutions, the answer is [tex]\frac{2\pi}{3} (28-3\pi)[/tex]. Im sure that i evaluated my integral correctly because i checked it with Mathematica, but i just cant seem to get the same answer as the solution guide.

    Any help would be greatly appreciated,
    Thanks,
    Dan.
     
  2. jcsd
  3. May 7, 2009 #2
    I agree with you. Furthermore, I can't even think of reasonable related question that would end up with a pi squared in the answer, as they have.
     
  4. May 7, 2009 #3

    danago

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    Ok thats good then :smile: I think i found how the books solution came about; If the integration is carried out with the same bounds, but with [tex]dV =dz dr d\theta[/tex] i.e. without the 'r', then the solution is as they stated.
     
  5. May 7, 2009 #4

    Mark44

    Staff: Mentor

    Are you sure you've given us the right problem?
    It's disconcerting to me that the two surfaces don't intersect, since that makes the volume region ambiguous. You have arbitrarily chosen to take "above the sphere" to mean "from its equator up." I think one could reasonably interpret "above the sphere" to mean from the top of its north pole up to the paraboloid. If the two surfaces intersected, then the region bounded by them would be nonambiguous.

    If you have given us the problem exactly as worded in your book, then maybe the problem itself is poorly constructed.
     
  6. May 7, 2009 #5

    danago

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    Yea thats exactly as it was stated.

    I do see what you mean and i understand how it can be interpreted in more than 1 way, but the way i did it just seemed most natural to me. I didnt really even give it a second thought when interpreting it, though maybe i should have.
     
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