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Homework Help: Triple integral and volume - please tell me if i'm wrong!

  1. Nov 18, 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 x2 + y2 + z2 = a2 and the ellipsoid [tex]\frac{x^2}{4a^2}[/tex] + [tex]\frac{4y^2}{a^2}[/tex] + [tex]\frac{9z^2}{a^2}[/tex] = 1

    2. Relevant equations

    3. The attempt at a solution

    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]

    How I found r? Well we have two z2s here, one for the ellipsoid and the other for the sphere. I actually found r by equating the tow z2s! And so I got 0 [tex]\leq[/tex] r [tex]\leq[/tex] [tex]\frac{4 \sqrt{2}a}{sin\phi \sqrt{12 cos^2(\theta) +20}}[/tex]

    Attached Files:

  2. jcsd
  3. Nov 19, 2009 #2
    That can't be right since 1/sin is not bounded so you're integrating outside the sphere.
  4. Nov 19, 2009 #3
    Please help me find the correct ranges. Are the ranges for [tex]\theta[/tex] for [tex]\phi[/tex] correct?
  5. Nov 19, 2009 #4
    Well, I think your method should work... But you just calculated it wrong. Try again

    [tex] \frac{x^2}{4a^2} + \frac{4y^2}{a^2} + \frac{9(a^2 - x^2 - y^2)}{a^2} = 1. [/tex]
  6. Nov 20, 2009 #5
    Dear clamtrox,
    I calculated it again and [tex]sin\phi[/tex] was in the same place. Because [tex]x = rsin\phi cos\theta[/tex] & [tex]y = rsin\phi sin\theta[/tex]
  7. Nov 20, 2009 #6
    I tried to solve it in cylindrical coordinates and I found the bounderies. But I couldn't calculate the triple integral and it is too difficult. But would you please tell me that my boundries are correct in cylindrical coordinates.

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

    [tex]0\leq r \leq \frac{4\sqrt{2}a}{\sqrt{15 cos^2(\theta)+1}}[/tex]

    [tex]-\frac{\sqrt{4a^2-r^2-15r^2sin^2\theta}}{6} \leq z \leq \frac{\sqrt{4a^2-r^2-15r^2sin^2\theta}}{6}[/tex]
  8. Nov 20, 2009 #7
    Yeah sorry about that, that was completely wrong by me :) That obviously does not work... I can't figure out an easy way to do it, but what would work is that you just solve the value of r from the ellipsoid equation and then figure out for which angles is r larger than a. That looks really messy though.
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