PlanetPhysics/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 constant surface-density 1. By the symmetry, its [[../CenterOfGravity/|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 [[../CoriolisEffect/|two-dimensional]] case instead a triple integral the double integral $Y=\frac{1}{\nu (E)}\int \underset{E}{\int }ydxdy,$ where\, $\nu (E)=\frac{\pi R^2}{2}$\, is the area (and the [[../Mass/|mass]]) of the half-disc. The region of integration is defined by $E=\{(x,y)\in {ℝ}^2⋮-R\leqq x\leqq R,0\leqq y\leqq \sqrt{R^2-x^2}\}.$ Accordingly, we may write Failed to parse (unknown function "\sijoitus"): {\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\, $(0,\frac{4R}{3\pi })$.