<s>[INST] 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.
 [/INST] 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 $\color{red}{\frac{\pi}{6}}$.
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