Deriving the Center of Mass of a semi-circular disk with cylindrical coordinates

The center of mass of a uniform half-disk obviously lies on the perpendicular bisector of the base diameter, at distance $d$ from the centre of the disk. By the Pappus centroid theorem,

$$ 2\pi d \cdot \frac{\pi}{2}R^2 = \frac{4\pi}{3}R^3, $$ hence $d=\color{red}{\large\frac{4R}{3\pi}}$.

Notice, your formula $Y_{CM} = \displaystyle{\int y\ dm}$ is not correct.

the center of mass is given as $$Y_{CM}=\frac{\displaystyle{\int y\ dm}}{\displaystyle{\int dm}}$$ Now, substituting the values $y=r\sin \theta$ & $dm=\sigma rdrd\theta$, we get

$$Y_{CM}=\frac{\displaystyle{\int_{0}^{R}\int_{0}^{\pi} \sigma r^2\sin\theta d\theta\ dr}}{\displaystyle{\int_{0}^{R}\int_{0}^{\pi}\sigma r d\theta\ dr}}$$

$$=\frac{\displaystyle{\int_{0}^{R}\left(\int_{0}^{\pi}\sin\theta d\theta\right)r^2\ dr}}{\displaystyle{\int_{0}^{R}\left(\int_{0}^{\pi}d\theta\right)r\ dr}}$$

$$ = \frac{\displaystyle{\int_{0}^{R}\left(2\right)r^2\ dr}}{\displaystyle{\int_{0}^{R}\left(\pi\right)r\ dr}} = \frac{\displaystyle{2\int_{0}^{R}r^2\ dr}}{\displaystyle{\pi\int_{0}^{R}r\ dr}}$$

$$ = \frac{2\left[\frac{r^3}{3}\right]_{0}^{R}}{\pi\left[\frac{r^2}{2}\right]_{0}^{R}}$$

$$ = \frac{4R^3}{3\pi R^2} = \frac{4R}{3\pi}$$

$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{Y_{CM}=\frac{4R}{3\pi}}}$$