Answer
The center of mass:
$\left( {{x_{CM}},{y_{CM}},{z_{CM}}} \right) \simeq \left( {0,0,0.925} \right)$
Work Step by Step
We have the region ${\cal W}$, defined by a cylinder of radius $2$ and height $4$.
In cylindrical coordinates, the mass density $\delta \left( {x,y,z} \right) = {{\rm{e}}^{ - z}}$ is
$\delta \left( {r\cos \theta ,r\sin \theta ,z} \right) = {{\rm{e}}^{ - z}}$
The description of ${\cal W}$:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 2,0 \le \theta \le 2\pi ,0 \le z \le 4} \right\}$
Step 1. 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}^2 \mathop \smallint \limits_{z = 0}^4 {{\rm{e}}^{ - z}}r{\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}^2 r{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = 0}^4 {{\rm{e}}^{ - z}}{\rm{d}}z} \right)$
$ = 2\pi \left( {\frac{1}{2}{r^2}|_0^2} \right)\left( { - {{\rm{e}}^{ - z}}|_0^4} \right)$
$M = 4\pi \left( {1 - {{\rm{e}}^{ - 4}}} \right)$
Step 2. Evaluate the $x$-coordinate of the center of mass:
${x_{CM}} = \frac{{{M_{yz}}}}{M} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} x\delta \left( {x,y,z} \right){\rm{d}}V$
$ = \frac{1}{{4\pi \left( {1 - {{\rm{e}}^{ - 4}}} \right)}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 \mathop \smallint \limits_{z = 0}^4 {{\rm{e}}^{ - z}}{r^2}\cos \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{4\pi \left( {1 - {{\rm{e}}^{ - 4}}} \right)}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^2 {r^2}{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = 0}^4 {{\rm{e}}^{ - z}}{\rm{d}}z} \right)$
Since $\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta = \left( {\sin \theta |_0^{2\pi }} \right) = 0$, so ${x_{CM}} = 0$.
Step 3. Evaluate the $y$-coordinate of the center of mass:
${y_{CM}} = \frac{{{M_{xz}}}}{M} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y\delta \left( {x,y,z} \right){\rm{d}}V$
$ = \frac{1}{{4\pi \left( {1 - {{\rm{e}}^{ - 4}}} \right)}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 \mathop \smallint \limits_{z = 0}^4 {{\rm{e}}^{ - z}}{r^2}\sin \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{4\pi \left( {1 - {{\rm{e}}^{ - 4}}} \right)}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^2 {r^2}{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = 0}^4 {{\rm{e}}^{ - z}}{\rm{d}}z} \right)$
Since $\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta = - \left( {\cos \theta |_0^{2\pi }} \right) = 0$, so ${y_{CM}} = 0$.
Step 4. Evaluate the $z$-coordinate of the center of mass:
${z_{CM}} = \frac{{{M_{xy}}}}{M} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z\delta \left( {x,y,z} \right){\rm{d}}V$
$ = \frac{1}{{4\pi \left( {1 - {{\rm{e}}^{ - 4}}} \right)}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 \mathop \smallint \limits_{z = 0}^4 {{\rm{e}}^{ - z}}rz{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{4\pi \left( {1 - {{\rm{e}}^{ - 4}}} \right)}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^2 r{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = 0}^4 z{{\rm{e}}^{ - z}}{\rm{d}}z} \right)$
Write $u=z$ and $dv = {{\rm{e}}^{ - z}}dz$. So, $du = dz$ and $v = - {{\rm{e}}^{ - z}}$. Using Integration by Parts Formula (Section 8.1),
$\smallint u{\rm{d}}v = uv - \smallint v{\rm{d}}u$
we get
$ = \frac{1}{{4\pi \left( {1 - {{\rm{e}}^{ - 4}}} \right)}}\left( {2\pi } \right)\left( {\frac{1}{2}{r^2}|_0^2} \right)\left( { - \left( {z{{\rm{e}}^{ - z}}|_0^4} \right) + \mathop \smallint \limits_{z = 0}^4 {{\rm{e}}^{ - z}}{\rm{d}}z} \right)$
$ = \frac{1}{{1 - {{\rm{e}}^{ - 4}}}}\left( { - 4{{\rm{e}}^{ - 4}} - \left( {{{\rm{e}}^{ - z}}|_0^4} \right)} \right)$
$ = \frac{1}{{1 - {{\rm{e}}^{ - 4}}}}\left( { - 4{{\rm{e}}^{ - 4}} - {{\rm{e}}^{ - 4}} + 1} \right) = \frac{{1 - 5{{\rm{e}}^{ - 4}}}}{{1 - {{\rm{e}}^{ - 4}}}}$
Thus, the center of mass: $\left( {{x_{CM}},{y_{CM}},{z_{CM}}} \right) = \left( {0,0,\frac{{1 - 5{{\rm{e}}^{ - 4}}}}{{1 - {{\rm{e}}^{ - 4}}}}} \right) \simeq \left( {0,0,0.925} \right)$.
