center 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

$\displaystyle Y = \frac{1}{\nu(E)}\int\!\!\int_E y\,dx\,dy,$

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

$\displaystyle E = \{(x,\,y)\in\mathbb{R}^2\,\vdots\;\; -R\leqq x \leqq R,\; 0 \leqq y \leqq \sqrt{R^2-x^2}\}.$

Accordingly, we may write

$\displaystyle Y = \frac{2}{\pi R^2}\!\int_{-R}^R\!dx\int_0^{\sqrt{R^2-x^2}}\!y\...
...\!\!-R}^{\,\quad R}\left(\frac{R^2x}{2}-\frac{x^3}{6}\right) = \frac{4R}{3\pi}.$

Thus the centre of mass is the point $ (0,\,\frac{4R}{3\pi})$ .



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