# Volume Enclosed by surfaces (Triple integration)

• Mar 18th 2012, 06:44 PM
ILoveTriangles
Volume Enclosed by surfaces (Triple integration)
Hello! I can't quite get this triple integral problem even after looking at the sticky for integration.

I have the surfaces z = x^2 + 3(y^2) and z = 8 - x*x - y*y and I need to find the volume of the region enclosed.
dV = dzdydx

All help is greatly appreciated!
• Mar 18th 2012, 08:38 PM
Ridley
Re: Volume Enclosed by surfaces (Triple integration)
The integral you want to solve is $\displaystyle \iiint_{x^2 + 3y^2}^{8 - x^2 - y^2}dzdxdy=\iint_D \left[8 - x^2 - y^2 - (x^2 + 3y^2)\right]dxdy$

The integration area D is given by the equation $\displaystyle x^2 + 3y^2 = 8 - x^2 - y^2 \Leftrightarrow x^2 + 2 y^2 = 4$

EDIT: Fixed notation issues.
• Mar 18th 2012, 08:45 PM
ILoveTriangles
Re: Volume Enclosed by surfaces (Triple integration)
I'm confused on why you only have one integral? How do I take this and turn it into three?
• Mar 18th 2012, 09:23 PM
Ridley
Re: Volume Enclosed by surfaces (Triple integration)
It's just a different notation. If it says dxdydz then you obviously have a triple integral.
• Mar 18th 2012, 09:44 PM
ILoveTriangles
Re: Volume Enclosed by surfaces (Triple integration)
Just to clarify, my limits are going to be for x: [-2, 2] and for y: [-sqrt(2), sqrt(2)]?
Is that right? or should the y be sqrt(2-x^2 / 2)?
• Mar 18th 2012, 10:15 PM
Ridley
Re: Volume Enclosed by surfaces (Triple integration)
It's easier if you switch to polar coordinates.

$\displaystyle x = r\cos\theta$
$\displaystyle \sqrt{2}y = r\sin\theta$

Don't forget to scale the new area element: $\displaystyle dxdy = \frac{r}{\sqrt{2}}drd\theta$

If you don't want to switch to polar coordinates then you can integrate from $\displaystyle x: \left[-\sqrt{4-2y^2}, \sqrt{4-2y^2}\right]$ and $\displaystyle y: \left[-\sqrt{2}, \sqrt{2}\right]$
• Mar 18th 2012, 10:22 PM
ILoveTriangles
Re: Volume Enclosed by surfaces (Triple integration)
Ah that makes so much more sense!
Thanks so much!