Answer
The coordinates of the center of mass: $\left( {0,0,\frac{2}{3}} \right)$.
Work Step by Step
We have a mass density of $\delta \left( {x,y,z} \right) = z$.
Let ${\cal W}$ denote the region defined by the cylinder ${x^2} + {y^2} = 1$ for $0 \le z \le 1$. We choose to describe ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 1,0 \le \theta \le 2\pi ,0 \le z \le 1} \right\}$
Evaluate the mass of ${\cal W}$:
$M = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \delta \left( {x,y,z} \right){\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^1 zr{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^1 r{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = 0}^1 z{\rm{d}}z} \right)$
$ = 2\pi \left( {\frac{1}{2}{r^2}|_0^1} \right)\left( {\frac{1}{2}{z^2}|_0^1} \right) = \frac{\pi }{2}$
Evaluate ${x_{CM}}$:
${x_{CM}} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} x\delta \left( {x,y,z} \right){\rm{d}}V$
$ = \frac{2}{\pi }\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^1 z{r^2}\cos \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{2}{\pi }\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^1 {r^2}{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = 0}^1 z{\rm{d}}z} \right)$
Since $\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta = \sin \theta |_0^{2\pi } = 0$, so ${x_{CM}} = 0$.
Evaluate ${y_{CM}}$:
${y_{CM}} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y\delta \left( {x,y,z} \right){\rm{d}}V$
$ = \frac{2}{\pi }\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^1 z{r^2}\sin \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{2}{\pi }\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^1 {r^2}{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = 0}^1 z{\rm{d}}z} \right)$
Since $\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta = - \cos \theta |_0^{2\pi } = 0$, so ${y_{CM}} = 0$.
Evaluate ${z_{CM}} = 0$:
${z_{CM}} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z\delta \left( {x,y,z} \right){\rm{d}}V$
$ = \frac{2}{\pi }\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^1 {z^2}r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{2}{\pi }\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^1 r{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = 0}^1 {z^2}{\rm{d}}z} \right)$
$ = 4\left( {\frac{1}{2}{r^2}|_0^1} \right)\left( {\frac{1}{3}{z^3}|_0^1} \right) = \frac{2}{3}$
So, the coordinates of the center of mass: $\left( {0,0,\frac{2}{3}} \right)$.
