Answer
$\dfrac{\pi}{48}$
Work Step by Step
Region of integration in cylindrical coordinates:
$R=${$ (r,\theta, z) | 0 \lt r \leq 1, 0 \leq \theta \leq 2 \pi, 0 \leq z \leq 1$}
Consider $I= \iint_{R} x^2y^2z \ dV$
or, $=\int^{0}_{1} \int_{0}^{2\pi} (r \cos \theta)^2 (r \sin \theta)^2 (z) (r dr) \ d \theta \ dz)$
or, $=\int_0^1 \int_0^{2 \pi} \int_0^1 r^5 \cos^2 \theta \sin^2 \theta z dr d\theta dz $
or, $=\int_0^1 (z) dz \times \int_0^{2 \pi} \cos^2 \theta \sin^2 \theta d\theta \times \int_0^1 r^5 \ dr $
or, $=[\dfrac{z^2}{2}]_0^1 \int_0^{2 \pi} \dfrac{1}{4} \times [\sin^2 2 \theta ]d\theta [\dfrac{r^6}{6}]_0^1$
or, $=\dfrac{1}{2} \times \dfrac{1}{6} [\dfrac{1}{8}-\dfrac{\cos 4 \theta}{8}]_0^{2 \pi} $
or, $=\dfrac{\pi}{48}$
