[SOLVED] Double Integrals - Volume Between Paraboloids

I'm studying for my calculus exam and having difficulty with this question.

Find the volume lying between the paraboloids $\displaystyle z = x^2+y^2$ and $\displaystyle 3z=4-x^2-y^2$.

I know to switch to polar coordinates, so $\displaystyle z=r^2$ and $\displaystyle 3z=4-r^2$. The solution to the question says that they intersect where $\displaystyle 3(x^2+y^2)=4-(x^2+y^2) $, which I understand is just because you're setting z=z and solving, and that r can also be subbed in to be $\displaystyle 3r^2=4-r^2$. But then the solution says they intersect at that equation where z=z, "i.e. on the cylinder $\displaystyle x^2+y^2=1$. Is this just solving for $\displaystyle r$ from the $\displaystyle 3r^2=4-r^2$?

I suppose my question is, what is the process I follow to solve this question? The solution integrates over the disc of radius 1, sets the intersection equation to zero with everything on one side, and integrates. But I don't know why.

The final answer is $\displaystyle \frac{2pi}{3}$ cubic units.

Thanks so much for your help.