I have the integral: ${\iiint} x^2 dx dy dz$ which is bounded from above by the elliptic paraboloid $z=2-x^2 - y^2$ and from below by the upper part of the cone $z^2 = x^2 + y^2$
I want to write this integral as a triple integral with cylindrical coordinates in the order $dz dr d\theta $
I know $r^2 = x^2 + y^2$, so by using this I can say $z^2 = x^2 + y^2$ becomes $z^2 = r^2$ (and so $z=r$) and $z=2-x^2 - y^2$ becomes $z=2-r^2$
I also set $r=2-r^2$ (since they both equal $z$) and found $r=1$ (since we want a positive $r$)
So I have what I believe is the correct set-up for the upper and lower bounds of my triple integration: ${\int}_0 ^{2\pi} {\int}_0 ^1 {\int}_r ^{2-r^2} *\text{something}* dz dr d\theta$
What I'm not sure of is what $*\text{something}*$ is, it was $x^2$ but I'm not sure how it changes now that my triple integral has cylindrical coordinates.
If someone could show me (or give me a hint to do it myself) what happens to $x^2$ when we convert to cylindrical coordinates it would really help.
I know what to do after I find out what $*\text{something}*$ is so there's no need to do the triple integral as well (unless you want to).
Thanks in advance
In cylindrical coordinates, $x^2$ becomes $r^2\cos^2\theta$. Besides, you have to multiply everything by the volume element, which is $r$. So, compute$$\int_0^{2\pi}\int_0^1\int_r^{2-r^2}r^3\cos^2(\theta)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.$$