${\mathrm{y}}_{\mathrm{CM}}=\frac{1}{\mathrm{M}}\int \mathrm{ydm}=\frac{1}{\mathrm{M}}{\int }_0^{\mathrm{R}}\left(\frac{2\mathrm{r}}{\mathrm{\pi }}\right)\left(\frac{2\mathrm{Mr}}{{\mathrm{R}}^2}\mathrm{dr}\right)\left[\text{ as }\mathrm{y}=\frac{2\mathrm{r}}{\mathrm{\pi }}\right]$

 $=\frac{4}{{\mathrm{\pi R}}^2}{\int }_0^{\mathrm{R}}{\mathrm{r}}^2\mathrm{dr}=\frac{4}{{\mathrm{\pi R}}^2}\left(\frac{{\mathrm{R}}^3}{3}\right)=\frac{4\mathrm{R}}{3\mathrm{\pi }}$ 

∴  Centre of mass of the semicircular disc $=\left(0,\frac{4\mathrm{R}}{3\mathrm{\pi }}\right)$ 

Centre of mass of a uniform quarter disc :

Mass per unit area of the quartr disc:

$\mathrm{\sigma }=\frac{\mathrm{M}}{{\mathrm{\pi R}}^2/4}=\frac{4\mathrm{M}}{{\mathrm{\pi R}}^2}$ 

Area of the element $=\frac{1}{2}\mathrm{\pi rdr}$ 

$\mathrm{dm}=\left(\frac{1}{2}\mathrm{\pi rdr}\right)\mathrm{\sigma }=\frac{2\mathrm{Mr}}{{\mathrm{R}}^2}\mathrm{dr}$ 

${\mathrm{x}}_{\mathrm{CM}}=\frac{1}{\mathrm{M}}\int \mathrm{xdm}=\frac{1}{\mathrm{M}}{\int }_0^{\mathrm{R}}\left(\frac{2\mathrm{r}}{\mathrm{\pi }}\right)\left(\frac{2\mathrm{Mr}}{{\mathrm{R}}^2}\mathrm{dr}\right)=\frac{4\mathrm{R}}{3\mathrm{\pi }} \left[\text{ as }\mathrm{x}=\mathrm{y}=\frac{2\mathrm{r}}{\mathrm{y}}\right]$ 

Similary, ${\mathrm{y}}_{\mathrm{CM}}=\frac{1}{\mathrm{M}}\int \mathrm{ydm}=\frac{4\mathrm{R}}{3\mathrm{\pi }}$ 

Hence for the quarter disc, centre of mass $=\left(\frac{4\mathrm{R}}{3\mathrm{\pi }},\frac{4\mathrm{R}}{3\mathrm{\pi }}\right)$