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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikiversity.org/v1/":): {\displaystyle 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\, $(0,\,{\frac {4R}{3\pi }})$.