Hi. Was hoping someone could help em out.

I have to calculate the volume between the two functions.

$\displaystyle x^2+y^2 \leq z \leq sqrt(3-x^2-y^2)$

So i figured I try to find the area D:

$\displaystyle z^2-3=-x^2-y^2 $

$\displaystyle z=x^2+x^2 $

Wich gives: $\displaystyle 3-z^2=z$ => $\displaystyle {(sqrt(13)-1} )/{ 2}$

I get a integral:

$\displaystyle \int_{0}^{2\pi}\int_{0}^{((sqrt(13)-1)/2}sqrt(3-r^2)-r^2) rdrd\theta $

However solving the integral I get a rather nasty algebra. Is there a more clever way to solve the equation?

Very thankful for a little help.