Find the centre of mass of a uniform:
$(a)$ half-disc,
$(b)$ quarter-disc.

(N/A) Let $M$ be the mass and $R$ be the radius of the half-disc.
Mass per unit area of the half-disc is $\sigma = \frac{M}{\frac{1}{2} \pi R^2} = \frac{2M}{\pi R^2}$.
$(a)$ Half-disc:
Consider a semicircular ring of radius $r$ and thickness $dr$. The area of this ring is $dA = \pi r dr$. The mass of this ring is $dm = \sigma dA = \frac{2M}{\pi R^2} \pi r dr = \frac{2M}{R^2} r dr$.
The centre of mass of this semicircular ring is at $(0, \frac{2r}{\pi})$.
$y_{CM} = \frac{1}{M} \int y dm = \frac{1}{M} \int_0^R \frac{2r}{\pi} \left( \frac{2M}{R^2} r dr \right) = \frac{4}{\pi R^2} \int_0^R r^2 dr = \frac{4}{\pi R^2} \left[ \frac{r^3}{3} \right]_0^R = \frac{4R}{3\pi}$.
So,the centre of mass is at $(0, \frac{4R}{3\pi})$.
$(b)$ Quarter-disc:
Consider a quarter-circular ring of radius $r$ and thickness $dr$. The area is $dA = \frac{1}{2} \pi r dr$. The mass is $dm = \sigma dA = \frac{M}{\frac{1}{4} \pi R^2} \frac{1}{2} \pi r dr = \frac{2M}{R^2} r dr$.
The centre of mass of a quarter-circular ring is at $(\frac{2r}{\pi}, \frac{2r}{\pi})$.
$x_{CM} = \frac{1}{M} \int x dm = \frac{1}{M} \int_0^R \frac{2r}{\pi} \left( \frac{2M}{R^2} r dr \right) = \frac{4R}{3\pi}$.
By symmetry,$y_{CM} = \frac{4R}{3\pi}$.
So,the centre of mass is at $(\frac{4R}{3\pi}, \frac{4R}{3\pi})$.