PlanetPhysics/Centre of Mass of Half Disc

Let be the upper half-disc of the disc\, \, in 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 of the centre of mass. For doing that, one can use in this two-dimensional case instead a triple integral the double integral where\, \, is the area (and the mass) of the half-disc. The region of integration is defined by 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\, .