Finding center of mass of a solid with non-uniform density 
Find the center of mass of the solid $$ R = {(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 \leq 1^2, 0 \leq z \leq \frac \pi2 }$$ if the density is given by $$\rho(x,y,z) = 5\bigg(\sqrt{x^2 + y^2} + sin(z)\bigg). $$

First i tried solving for the mass with the triple integral $$\int_0^{2\pi} \int_0^{\frac\pi2} \int_0^15r\big(r+sin(z)\big) = \frac{10\pi^2}{6}  $$ Which gave me the $z$ coordinate $\frac{3+\pi}{\pi}$ Not sure if this is correct though. Also, i don't know how to find the other coordinates now seeing as the density is not uniform.
 A: Note that the method is correct since $\rho(x,y,z)$ is symmetric with respect to z axis the center of mass has coordinates $(0,0,z_G)$ with (I didn't check the detail of calculation):
$$z_G=\frac{\int_V z\cdot \rho(x,y,z)dV}{M}$$
A: In general, the coordinates of the center of mass of the region $\Omega$ are given by
$$
\overline{{\bf x}} = \frac{\int_\Omega{\rm d}{\bf x}~{\bf x}\rho({\bf x})}{\int_\Omega{\rm d}{\bf x}~\rho({\bf x})} \tag{1}
$$
In your case $\Omega = \{(x,y,z) | x^2+y^2 < 1, 0 < z < \pi/2 \}$
We can then calculate 

$\int_\Omega{\rm d}{\bf x}~\rho({\bf x})$

\begin{eqnarray}
\int_\Omega{\rm d}{\bf x}~\rho({\bf x}) &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R\rho(R,\phi,z) \\
&=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R [5(R + \sin z)] \\
&=& 10\pi \int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R (R + \sin z) \\
&=& (\dots)
\end{eqnarray}

$\int_\Omega{\rm d}{\bf x}~x\rho({\bf x})$

\begin{eqnarray}
\int_\Omega{\rm d}{\bf x}~x\rho({\bf x}) &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~Rx\rho(R,\phi,z) \\
&=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R [R\cos\phi][5(R + \sin z)] 
&=& (\dots)
\end{eqnarray}

$\int_\Omega{\rm d}{\bf x}~y\rho({\bf x})$

\begin{eqnarray}
\int_\Omega{\rm d}{\bf x}~y\rho({\bf x}) &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~Ry\rho(R,\phi,z) \\
&=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R [R\sin\phi][5(R + \sin z)] 
&=& (\dots)
\end{eqnarray}

$\int_\Omega{\rm d}{\bf x}~z\rho({\bf x})$

\begin{eqnarray}
\int_\Omega{\rm d}{\bf x}~z\rho({\bf x}) &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~Rz\rho(R,\phi,z) \\
&=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R [z][5(R + \sin z)] 
&=& (\dots)
\end{eqnarray}

Everything together

When you calculate these integrals, just use Eq. (1)
