Calculate the volume between $z=\sqrt{x^2+y^2}$ and $z=x^2+y^2$. Calculate the volume between $z=\sqrt{x^2+y^2}$ and $z=x^2+y^2$.
Attempt
We project on the $xy$ plane the intersection between $z=\sqrt{x^2+y^2}$ and $z=x^2+y^2$, which is the circle $x^2+y^2=1, z=1$.
We can conclude that the region between $z=\sqrt{x^2+y^2}$ and $z=x^2+y^2$ can be described by
$$-1\leq x\leq 1, -\sqrt{1-x^2}\leq y \leq \sqrt{1-x^2}, x^2+y^2\leq z \leq \sqrt{x^2+y^2}$$
The volume is given by
$$V=\int \int \int_W dxdydz=\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{x^2+y^2}^{\sqrt{x^2+y^2}} dxdydz$$
When I try to solve this, I get a difficult expression and cannot calculate it. So I think, everything I have done is wrong.

I obtained $\frac{\pi}{6}$ by using Integrate[ Integrate[ Integrate[1, {z, x^2 + y^2, Sqrt[x^2 + y^2]}], {y, -Sqrt[1 - x^2],  Sqrt[1 - x^2]}], {x, -1, 1}]
