So basically I need to find the volume of a solid bounded below by the upper nappe of the cone $z^2=x^2+y^2$ and bounded above by the sphere $x^2+y^2+z^2=9$
I know it looks like this:
But I'm not sure how to setup the double integral to find the volume. I do understand what my limits of integration will be and that I will need to do some conversions to polar coordinates, but I don't understand how to find the functions to integrate.
Any hints or tips are greatly appreciated.
EDIT
After thinking about the problem a little more, is it a matter of just adding the two functions together?
For example:
$$\iint \sqrt{x^2+y^2}+\sqrt{9-x^2-y^2}\;dy\;dx$$
Both surfaces intersect at $z=3/\sqrt{2},$ and the projection of this intersection in the $xy$ plane in a circle $C$ centered at $(0,0)$ with radius $r=3/\sqrt{2}$. Let $D$ be the disc with border $C$.
It follows that
$$V=\iint_D \sqrt{9-x^2-y^2}-\sqrt{x^2+y^2} dA,$$
which should be easy to integrate with polar coordinates given the nature of the function to integrate, and the nature of $D$.
Alternatively, spherical coordinates should work too here. Your solid can be written as
$$ E=\{(\rho,\theta,\phi)\;|\; 0\le \rho\le 3, 0\le \theta\le 2\pi, 0\le \phi \le \pi/4 \} $$ You can most surely take it from there.