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.
 A: As mentioned in comments, this is easier in  cylindrical coordinates.
$x = r \cos\theta, y = r \sin\theta, z = z$
Paraboloid surface is $z = x^2 + y^2 = r^2$ and surface of the cone is $z = \sqrt{x^2+y^2} = r$
i) going in the order $dr$ first,
$z \leq r \leq \sqrt z, 0 \leq z \leq 1, 0 \leq \theta \leq 2\pi$
$\displaystyle \int_0^{2\pi} \left[\int_0^1 \left[\int_z^{\sqrt z} r \ dr \right] \ dz \right]\ d\theta $
ii) going in the order $dz$ first,
you can set up the integral using bounds,
$r^2 \leq z \leq r, 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi \ $.
Both integrals are straightforward to evaluate.
A: 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}]
A: The surfaces intersect at the origin and at the point such that $z=1$, so the wanted volume is the difference between the volume of a paraboloid and the volume of a cone, which equals $\frac{\pi}{3}$. The volume of the paraboloid equals (by integrating the areas of the $z$-sections)
$$ \int_{0}^{1}\pi z\,dz = \frac{\pi}{2} $$
so the outcome is $\color{red}{\frac{\pi}{6}}$.
