$Z = \{(x,y,z)\in \mathbb{R}^3 | -1 < z < 1, \,\ x^2+y^2 \leq 1\}$
Calculate $\int_Z z^2x^2$.
My approach:
Polar coordinates: $x = r \sin(\theta),\,\,y = r \cos(\theta)$. Then:
$\int_Z z^2x^2 = $
$\int_{-1}^1 \int_0^{2\pi} \int_0^1 z^2 r^2 \sin(\theta)^2 \,\,dr \,d\theta \, dz = $
$4 \int_{0}^1 \int_0^{\pi} \int_0^1 z^2 r^2 \sin(\theta)^2 \,\,dr \,d\theta \, dz = $
$4 \int_{0}^1 z^2 \int_0^{\pi} \int_0^1 r^2 \sin(\theta)^2 \,\,dr \,d\theta \, dz = $
$4 \int_{0}^1 z^2 (\int_0^{\pi} \sin(\theta)^2 \,d\theta)(\int_0^1 r^2 \,\,dr) \, dz = $
$\frac{4}{3} \int_{0}^1 z^2 (\int_0^{\pi} \sin(\theta)^2 \,d\theta) \, dz = $
$\frac{4}{3} \int_{0}^1 z^2 \frac{\pi}{2} \, dz = $
$\frac{2}{3} \pi \int_{0}^1 z^2 \, dz = $
$\frac{2}{9} \pi$
This is the wrong result though and I cannot find my mistake...
You forgot the factor $r$. After the first equality, you should have$$\int_{-1}^1\int_0^{2\pi}\int_0^1r^{\color{red}3}\sin^2(\theta)z^2\,\mathrm dr\,\mathrm d\theta\,\mathrm dz=\frac\pi6.$$
As a side note, when we use cylindrical coordinates it is much more common to do $x=r\cos\theta$ and $y=r\sin\theta$.