Answer
The mass of the lamina is
$$
\begin{aligned}
m &=\iint_{D} \rho(x, y) d A \\
&=\frac{7}{3} \pi k.
\end{aligned}
$$
The center of mass is
$$
(\bar{x}, \bar{y})=\left(0, \frac{45}{14 \pi}\right).
$$
Work Step by Step
The boundary of a lamina consists of the semicircles $y=\sqrt {1-x^{2}}$ and $y=\sqrt {4-x^{2}}$ together with the portions of the x-axis that joins them, if the density at any point is proportional to its distance from the origin.
Then the distance from a point $(x,y)$ to the center of the origin is $\sqrt {x^{2}+y^{2}}$. Therefore, the density function is
$$
\rho(x, y)=k\sqrt {x^{2}+y^{2}}
$$
where $k$ is some constant. Both the density function and the shape of the lamina suggest that we convert to polar coordinates. Then $\sqrt {x^{2}+y^{2}}=r $ and the region $D$ is given by:
$$ D=\left\{(r, \theta) | 1 \leq r \leq 2, \quad 0 \leq \theta \leq \pi \right\} .$$
The mass of the lamina is
$$
\begin{aligned}
m &=\iint_{D} \rho(x, y) d A \\
&=\int_{0}^{\pi} \int_{1}^{2} k r \cdot r d r d \theta \\
&=k \int_{0}^{\pi} d \theta \int_{1}^{2} r^{2} d r \\
&=k(\pi)\left[\frac{1}{3} r^{3}\right]_{1}^{2} \\
&=\frac{7}{3} \pi k.
\end{aligned}
$$
Now, we find:
$$
\begin{aligned}
M_{y} &=\iint_{D} x \rho(x, y) d A=\int_{0}^{\pi} \int_{1}^{2}(r \cos \theta)(k r) r d r d \theta \\
&=k \int_{0}^{\pi} \cos \theta d \theta \int_{1}^{2} r^{3} d r \\
&=k[\sin \theta]_{0}^{\pi}\left[\frac{1}{4} r^{4}\right]_{1}^{2} \\
&=k(0)\left(\frac{15}{4}\right)=0 ,
\end{aligned}
$$
and
$$
\begin{aligned}
M_{x} &=\iint_{D} y \rho(x, y) d A \\
&=\int_{0}^{\pi} \int_{1}^{2}(r \sin \theta)(k r) r d r d \theta \\
&=k \int_{0}^{\pi} \sin \theta d \theta \int_{1}^{2} r^{3} d r \\
&=k[-\cos \theta]_{0}^{\pi}\left[\frac{1}{4} r^{4}\right]_{1}^{2} \\
&=k(1+1)\left(\frac{15}{4}\right) \\
&=\frac{15}{2} k.
\end{aligned}
$$
Therefore, the center of mass is located at the point
$$
(\bar{x}, \bar{y})=\left(0, \frac{15 k / 2}{7 \pi k / 3}\right)=\left(0, \frac{45}{14 \pi}\right).
$$
