I found it easiest to use cylindrical coordinates to set up the integrals needed for the center of mass. Before we do so, however, I set my coordinate system up as follows. I have positive $x$ coming out of the screen, positive $y$ going to the right, and positive $z$ up. In cylindrical coordinates $(r,\phi,z)$:
$$x = r \cos{\phi}$$
$$y=r \sin{\phi}$$
where we have the limits defining the region $\Omega$:
$$\phi \in \left [ -\frac{\pi}{2},\frac{\pi}{2} \right]$$
$$z \in [-R_2,R_2]$$
$$r \in \left [ R_1 - \sqrt{R_2^2-z^2}, R_1 + \sqrt{R_2^2-z^2}\right]$$
Also, for an object of constant mass density, the expression for the $x$ component of the center of mass is
$$\bar{x} = \frac{\displaystyle \int_{\Omega} d^3 \vec{x} \, x}{\displaystyle \int_{\Omega} d^3 \vec{x}}$$
(Note that, by symmetry, we have $\bar{y}=0$ and $\bar{z}=0$.) Let's first evaluate the denominator:
$$\begin{align} \int_{\Omega} d^3 \vec{x} &= \int_{-\pi/2}^{\pi/2} d\phi \, \int_{-R_2}^{R_2} dz \, \int_{R_1 - \sqrt{R_2^2-z^2}}^{R_1 + \sqrt{R_2^2-z^2}} dr \, r \\ &= \frac{\pi}{2}\int_{-R_2}^{R_2} dz \left [\left (R_1 + \sqrt{R_2^2-z^2} \right )^2 - \left (R_1 - \sqrt{R_2^2-z^2} \right )^2 \right ]\\ &= 4 \pi R_1 \int_0^{R_2} dz \, \sqrt{R_2^2-z^2}\\ &= \pi^2 R_1 R_2^2 \end{align}$$
Now we evaluate the center of mass:
$$\begin{align}\bar{x} &= \frac{1}{\pi^2 R_1 R_2^2} \int_{-\pi/2}^{\pi/2} d\phi \, \int_{-R_2}^{R_2} dz \, \int_{R_1 - \sqrt{R_2^2-z^2}}^{R_1 + \sqrt{R_2^2-z^2}} dr \, r^2 \cos{\phi} \\ &= \frac{4}{3 \pi^2 R_1 R_2^2} \int_0^{R_2} dz \, \left [\left (R_1 + \sqrt{R_2^2-z^2} \right )^3 - \left (R_1 - \sqrt{R_2^2-z^2} \right )^3 \right ]\\ &= \frac{8}{3 \pi^2 R_1 R_2^2} \int_0^{R_2} dz \,\left [3 R_1^2 \sqrt{R_2^2-z^2} + \left (R_2^2-z^2 \right )^{3/2} \right ] \\ &= \frac{8}{3 \pi^2 R_1 R_2} \left ( \frac{3 \pi}{4} R_1^2 R_2 + \frac{3 \pi}{16} R_2^3 \right )\end{align}$$
Simplifying, I get
$$\bar{x} = \frac{4 R_1^2+R_2^2}{2 \pi R_1}$$
ADDENDUM
As a quick note, in the limits as $R_2 \to 0$, we find that the center of mass becomes
$$\bar{x}=\frac{2}{\pi} R_1$$
which agrees with the center of mass of a uniform wire bent into a semicircle.
Best Answer
Two of the "coordinates" of the centre of mass are obvious by symmetry. Put the thing flat (open) side down. So the cross-section is a semi-circle, say of radius $r$, though I would prefer $1$.
There are two things to worry about: (i) the two ends and (ii) the rest. We find the height of the centroid of each part, say $h_1$ and $h_2$. Then the height of the centroid of the combination is the weighted average of the centroids of the two parts. The combined weight of the two ends is $w_1=\pi r^2$, and the weight of the rest is $w_2=\pi r l$, where $l$ is the length. The centroid of the combination is at height $$\frac{w_1h_1+w_2h_2}{w_1+w_2}.$$
Now we need $h_1$ and $h_2$. I will leave $h_1$ to you, standard formula.
Calculating $h_2$ is a bit more interesting. As usual we divide the moment about the $x$-axis by the circumference of the rounded part of the semicircle, assuming unit linear density.
To find the moment, look at the piece that is above the little interval from $x$ to $x+dx$. This is at height $\sqrt{r^2-x^2}$. Its length is roughly $\sqrt{1+\left(\frac{dy}{dx} \right)^2\,dx}$. So the moment is equal to $$\int_{-r}^r \sqrt{r^2-x^2}\sqrt{1+\left(\frac{dy}{dx} \right)^2}\,dx.$$ When you calculate $\sqrt{1+\left(\frac{dy}{dx} \right)^2}$, you will have an extremely pleasant surprise!