centre of mass of half-discCentreOfMassOfHalfDisc2007-07-02 12:39:302009-04-18 13:57:22Examplecentre of mass% this is the default PlanetMath preamble. as your knowledge
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{\operatornamewithlimits{\Big/}_{\!\!\!#1}^{\,#2}}Let $E$ be the upper half-disc of the disc\, $x^2+y^2 \leqq R$\, in $\mathbb{R}^2$ with a \PMlinkescapetext{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\, $(0,\,\frac{4R}{3\pi})$.