1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Triple integral and volume - please tell me if i'm wrong!
  1. Triple integral volume (Replies: 5)

Loading...