What will be the position of centre of mass of a half disc of uniform mass density as shown?

Fact that COM of semicircular disc is $\frac{4a}{3\mathrm{\pi }}$ from centre

alternate method

(distance moved by COM) X area of the semicircle= volume generated

$$\begin{gathered} 2\pi x_{cm}\times \frac{\pi a^2}{2}=\frac{4}{3}\pi a^3 \\ {x}_{cm}=\frac{4a}{3\pi } \end{gathered}$$