# Help finding ellipsoid volume with antiderivative?

• Dec 22nd 2008, 07:19 PM
Jenberl
Help finding ellipsoid volume with antiderivative?
Hi, I'm stuck on a homework problem and would appreciate some help, because it's getting late. :-(

I have to integrate x^2/a^2=y^2/b^2=1 for the volume of this ellipsoid rotating about the x-axis. I have -a<x<a (supposed to be less than or equal to but I don't have that key?) and -b<y<b (also less than or equal to).

How do I do the Remiens sum to get to the integral problem? We're supposed to do this with the antiderivative method.
I've got a half ellipse drawn & trying to do the disks, but I'm really stuck.
I get y=sqare root of b^2(1-x^2/a^2).

Thanks,
Jen
• Dec 22nd 2008, 07:22 PM
mr fantastic
Quote:

Originally Posted by Jenberl
Hi, I'm stuck on a homework problem and would appreciate some help, because it's getting late. :-(

I have to integrate x^2/a^2=y^2/b^2=1 for the volume of this ellipsoid rotating about the x-axis. I have -a<x<a (supposed to be less than or equal to but I don't have that key?) and -b<y<b (also less than or equal to).

How do I do the Remiens sum to get to the integral problem? We're supposed to do this with the antiderivative method.
I've got a half ellipse drawn & trying to do the disks, but I'm really stuck.
I get y=sqare root of b^2(1-x^2/a^2).

Thanks,
Jen

Formula: $V = \pi \int_{-a}^{a} y^2 \, dx$

From symmetry:

$= 2 \pi \int_0^a y^2 \, dx$.

So you end up with quite a simple integral to do.
• Dec 22nd 2008, 07:33 PM
Jenberl
Thank you sir.
ok.
But I tried and I don't get it so simple.
Is 2*pi*b^2(x-x^3/3a^a) the right answer to plug into?
??? it looks funny, but can't see what I'm doing wrong.
I'm not seeing this right, I don't think.
• Dec 23rd 2008, 12:40 AM
mr fantastic
Quote:

Originally Posted by Jenberl
Thank you sir.
ok.
But I tried and I don't get it so simple.
Is 2*pi*b^2(x-x^3/3a^a) the right answer to plug into?
??? it looks funny, but can't see what I'm doing wrong.
I'm not seeing this right, I don't think.

$V = 2 \pi \int_0^a y^2 \, dx = 2 \pi b^2 \int_0^a 1 - \frac{x^2}{a^2} \, dx = 2 \pi b^2 \left[ x - \frac{x^3}{3 a^2}\right]_0^a = 2 \pi b^2 \left( a - \frac{a^3}{3a^2}\right)$

I hope I don't need to continue .....