Answer
$\dfrac{\pi}{48}$
Work Step by Step
The region of integration in cylindrical coordinates can be expressed as: $R=${$ (r,\theta, z) | 0 \lt r \leq 1, 0 \leq \theta \leq 2 \pi, 0 \leq z \leq 1$}
The function we want to integrate can be integrated in triple cylindrical coordinates as:
$ \iint_{R} x^2y^2z \ dV=\int^{0}_{1} \int_{0}^{2\pi} (r \cos \theta)^2 (r \sin \theta)^2 (z r dr d \theta \ dz) \\=\int_0^1 \int_0^{2 \pi} \int_0^1 r^5 \cos^2 \theta \sin^2 \theta (z dr d\theta dz) \\=\int_0^1 z dz \int_0^{2 \pi} \cos^2 \theta \sin^2 \theta d\theta \int_0^1 r^5 dr \\=[\dfrac{z^2}{2}]_0^1 \int_0^{2 \pi} \dfrac{\sin^2 2 \theta}{4} d\theta \times [\dfrac{r^6}{6}]_0^1 \\=\dfrac{1}{2} \times \dfrac{1}{6} [\dfrac{1}{8}-\dfrac{\cos 4 \theta}{8}]_0^{2 \pi} \\=\dfrac{\pi}{48}$
