Originally Posted by

**arbolis** Today a friend of mine told me the right answer is $\displaystyle \frac{7\pi}{48}$. I had shown him your (NCA) post, he copied it in his draft. He went to university and checked the result with a teacher who said it is wrong. My friend didn't understand all what the teacher said, he just came up with $\displaystyle V=\int_0^{2\pi} \int_0^{\frac{1}{2}} \int _{2r^2}^{\frac{1}{2}+\sqrt{\frac{1}{4}-r^2}} r dzdrd\theta = \frac{7\pi}{48}$.

As I trust you and that your post makes sense to me, I wonder where's the error of the teacher. I know he used cylindrical coordinates, but I don't really know what he did. It seems that for him $\displaystyle z$ goes from $\displaystyle 2r^2$ (so the paraboloid) to $\displaystyle \frac{1}{2}+\sqrt{\frac{1}{4}-r^2}$, which I don't know what is it (it has no other choice than being the sphere, but as you pointed out, the paraboloid is inside it. I'm quite confused!)

Does someone has anything to say?