centre of mass of half-disc

Let $E$ be the upper half-disc of the disc  $x^2+y^2\leqq R$  in ${ℝ}^2$ with a surface-density 1. By the symmetry, its centre of mass lies on its medium radius, and therefore we only have to calculate the ordinate $Y$ of the centre of mass. For doing that, one can use the double integral

$$Y=\frac{1}{\nu (E)}{\iint }_{E}y𝑑x𝑑y,$$

where  $\nu (E)=\frac{\pi R^2}{2}$  is the area of the half-disc. The region of integration is defined by

$$E=\{(x,y)\in {ℝ}^2\mathrm{⋮}-R\leqq x\leqq R,\mathrm{\hspace{0.33em}0}\leqq y\leqq \sqrt{R^2-x^2}\}.$$

Accordingly we may write

$$Y=\frac{2}{\pi R^2}{\int }_{-R}^R𝑑x{\int }_0^{\sqrt{R^2-x^2}}y𝑑y=\frac{2}{\pi R^2}{\int }_{-R}^R\frac{R^2-x^2}{2}𝑑x=\frac{2}{\pi R^2}\underset{x=-R}{\overset{R}{/}}\left(\frac{R^{2}x}{2}-\frac{x^3}{6}\right)=\frac{4R}{3\pi }.$$

Thus the centre of mass is the point  $(0,\frac{4R}{3\pi })$.