Triple integral and volume - please tell me if i'm wrong

[rado5]

Homework Statement

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 \frac{x^2}{4a^2} + \frac{4y^2}{a^2} + \frac{9z^2}{a^2} = 1

Homework Equations

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= \int \int \int r²sin\phi dr d\theta d\phi

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

0 \leq \theta \leq 2 \pi

0 \leq \phi \leq \pi

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

 

[clamtrox]

That can't be right since 1/sin is not bounded so you're integrating outside the sphere.

 

[rado5]

Please help me find the correct ranges. Are the ranges for \theta for \phi correct?

 

[clamtrox]

Well, I think your method should work... But you just calculated it wrong. Try again

\frac{x^2}{4a^2} + \frac{4y^2}{a^2} + \frac{9(a^2 - x^2 - y^2)}{a^2} = 1.

 

[rado5]

Dear clamtrox,
I calculated it again and sin\phi was in the same place. Because x = rsin\phi cos\theta & y = rsin\phi sin\theta

 

[rado5]

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.

0\leq \theta \leq 2\pi

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

-\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}

 

[clamtrox]

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.