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Homework Help: Setting up a triple integral to find volume of a region

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data

    I need to set up the triple integral to find the volume of the region bounded by the sphere: x^2+y^2+z^2=a^2 and the ellipsoid: (x^2/4a^2)+(4y^2/a^2)+(9z^2/a^2)=1.

    2. Relevant equations


    3. The attempt at a solution


    I'm not sure which interval I should be using here. I made a 3D graph of the region, but the a variable is really throwing me off. Can anyone point me in the right direction here? Thanks in advance!
  2. jcsd
  3. Nov 16, 2009 #2
    In what type of coordinates? Spherical coordinates looks good.
  4. Nov 16, 2009 #3
    I was considering spherical as well. That would work. I'm not picky on which coordinate system we use.
  5. Nov 16, 2009 #4
    To calculate the volume of the sphere which is x2 + y2 + z2 = a2 you should use the following for the cartesian coordinate.

    -a [tex]\leq[/tex] z [tex]\leq[/tex] a

    -[tex]\sqrt{a^2 - z^2}[/tex] [tex]\leq[/tex] y [tex]\leq[/tex] [tex]\sqrt{a^2 - z^2}[/tex]

    -[tex]\sqrt{a^2 - y^2 - z^2}[/tex] [tex]\leq[/tex] x [tex]\leq[/tex] [tex]\sqrt{a^2 - y^2 - z^2}[/tex]

    And you can do it the same for the ellipsoid!
  6. Nov 17, 2009 #5
    So, for the ellipsoid, should it be the following?:

    -sqrt[-x^2-4(9*z^2-a^2)]/4 <=y<= sqrt[-x^2-4(9*z^2-a^2)]/4 (shouldn't have x values...hmmm)

    I'm a bit lost here :confused: Would someone mind setting up the triple integral for me? I think having a look at the final product would open up some doors in my brain. No need to evaluate. Thanks again, people!
  7. Nov 17, 2009 #6
    Vsphere = [tex]\int[/tex][tex]^{a}_{-a}[/tex][tex]\int[/tex][tex]^{\sqrt{a^2 - z^2}}_{-\sqrt{a^2 - z^2}}[/tex][tex]\int[/tex][tex]^{\sqrt{a^2 - y^2 - z^2}}_{-\sqrt{a^2 - y^2 - z^2}}[/tex]dx dy dz

    and for the ellipsoid [tex]\frac{-a}{3}[/tex] [tex]\leq[/tex] z [tex]\leq[/tex] [tex]\frac{a}{3}[/tex] and ...
  8. Nov 17, 2009 #7
    Oh ok, I think I get that part now.

    -a/3 <= z <= a/3

    -sqrt[(a^2-9z^2)/4] <= y <= sqrt[(a^2-9z^2)/4]

    -sqrt[4a^2-16y^2-36z^2] <= x <= sqrt[4a^2-16y^2-36z^2]

    So, now that I have my intervals for both shapes, what comes next? I need the volume of the ellipsoid, but only the part that's within the sphere. Sorry to ask so many questions, but my book has no similar examples that I can work with. I really appreciate your help.
  9. Nov 18, 2009 #8
    I solved it in spherical coordination and I think it is correct, if it is not, somebody please tell me why?

    V= [tex]\int[/tex] [tex]\int[/tex] [tex]\int[/tex] r2sin[tex]\phi[/tex] dr d[tex]\theta[/tex] d[tex]\phi[/tex]

    0 [tex]\leq[/tex] r [tex]\leq[/tex] [tex]\frac{4 \sqrt{2}a}{sin\phi \sqrt{12 cos^2(\theta) +20}}[/tex]

    0 [tex]\leq[/tex] [tex]\theta[/tex] [tex]\leq[/tex] 2 [tex]\pi[/tex]

    0 [tex]\leq[/tex] [tex]\phi[/tex] [tex]\leq[/tex] [tex]\pi[/tex]
    Last edited: Nov 18, 2009
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