PlanetPhysics/Centre of Mass of Half Disc

Let $$E$$ be the upper half-disc of the disc\, $$x^2+y^2 \leqq R$$\, in $$\mathbb{R}^2$$ with a constant surface-density 1. By the symmetry, its centre of mass locates 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 in this two-dimensional case instead a triple integral the double integral $$Y = \frac{1}{\nu(E)}\int\!\!\int_E y\,dx\,dy,$$ where\, $$\nu(E) = \frac{\pi R^2}{2}$$\, is the area (and the mass) of the half-disc. The region of integration is defined by $$E = \{(x,\,y)\in\mathbb{R}^2\,\vdots\;\; -R\leqq x \leqq R,\; 0 \leqq y \leqq \sqrt{R^2-x^2}\}.$$ Accordingly, we may write $$Y = \frac{2}{\pi R^2}\!\int_{-R}^R\!dx\int_0^{\sqrt{R^2-x^2}}\!y\,dy = \frac{2}{\pi R^2}\!\int_{-R}^R\frac{R^2\!-\!x^2}{2}\,dx = \frac{2}{\pi R^2}\sijoitus{-R}{\quad R}\left(\frac{R^2x}{2}-\frac{x^3}{6}\right) = \frac{4R}{3\pi}.$$ Thus the centre of mass is the point\, $$(0,\,\frac{4R}{3\pi})$$.