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!