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

1. Nov 18, 2009

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 $$\frac{x^2}{4a^2}$$ + $$\frac{4y^2}{a^2}$$ + $$\frac{9z^2}{a^2}$$ = 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= $$\int$$ $$\int$$ $$\int$$ 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 $$\leq$$ $$\frac{4 \sqrt{2}a}{sin\phi \sqrt{12 cos^2(\theta) +20}}$$

#### Attached Files:

• ###### Sphere - Ellipsoid.jpg
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2. Nov 19, 2009

### clamtrox

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

3. Nov 19, 2009

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

4. Nov 19, 2009

### clamtrox

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

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

5. Nov 20, 2009

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

6. Nov 20, 2009

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.

$$0\leq \theta \leq 2\pi$$

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

$$-\frac{\sqrt{4a^2-r^2-15r^2sin^2\theta}}{6} \leq z \leq \frac{\sqrt{4a^2-r^2-15r^2sin^2\theta}}{6}$$

7. Nov 20, 2009

### 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.