Triple integration $\iiint_{x^2+y^2+z^2 \leq 2z} x^2y^2 dxdydz$ $$\iiint_{x^2+y^2+z^2 \leq 2z} x^2y^2 dxdydz$$
I want to solve it with cylindrical coordinates:
So we get:
$$\iiint_{r^2+z^2 \leq 2z} r^5\sin^2(\phi)\cos^2(\phi) dxdydz$$
I see that the bounds are actually a circle with radius of $\sqrt{2z}$.
How can I use this fact to solve the above integral?
 A: It's better to use spherical coordinates
The region is $ x^2 + y^2 + z^2 \le 2 z $, i.e., $x^2 + y^2 + (z - 1)^2 \le 1 $
So we can define the spherical coordinates $r, \theta, \phi $ such that
$ x = r \sin \theta \cos \phi $
$ y = r \sin \theta \sin \phi $
$ z = 1 + r \cos \theta $
And the integral becomes
$ I = \displaystyle \int_{r = 0 }^ 1 \int_{\theta = 0 }^ \pi \int_{\phi = 0}^{2 \pi}
r^6 \sin^5 \theta \cos^2 \phi \sin^2 \phi \text{d}\phi \text{d}\theta \text{d} r $
Integrating with respect $r $ and with respect to $\phi$ is straight forward, and the integral reduces to
$ I = \dfrac{\pi}{28} \displaystyle \int_{\theta = 0}^\pi sin^5 \theta \text{d}\theta $
The integral with respect to $\theta$ is solved using the substitution $ u = \cos \theta$, then
$ \displaystyle \int \sin^5 \theta \text{d} \theta = -\int (1 - u^2)^2 du =-( u -\dfrac{2}{3} u^3 +\dfrac{u^5}{5}) $
Hence,
$I = \dfrac{4 \pi}{105}$
A: Let's at least define $Z:=z-1$ first, so we want $\\\int_{x^2+y^2+Z^2\le1}x^2y^2dxdydZ$. Then you can define cylindrical coordinates from $x,\,y,\,Z$ instead of $x,\,y,\,z$: in particular, the position along the axis is given by $Z$. (Spherical coordinates would be a little easier but, as you said, you want cylindrical.) Since $dxdydZ=rdrd\phi dZ$, our integral is$$\underbrace{\int_0^{2\pi}\cos^2\phi\sin^2\phi d\phi}_{\pi/4}\int_{-1}^1dZ\underbrace{\int_0^{\sqrt{1-Z^2}}r^5dr}_{\frac16(1-Z^2)^3}=\frac{\pi}{12}\int_0^1(1-Z^2)^3dZ.$$You can do the rest yourself.
