# 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 surface-density 1. By the symmetry, its centre of mass lies 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 the double integral

 $Y\;=\;\frac{1}{\nu(E)}\iint_{E}y\,dx\,dy,$

where  $\nu(E)=\frac{\pi R^{2}}{2}$  is the area 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

 $Y\;=\;\frac{2}{\pi R^{2}}\!\int_{-R}^{R}\!dx\int_{0}^{\sqrt{R^{2}-x^{2}}}\!y\,% dy\;=\;\frac{2}{\pi R^{2}}\!\int_{-R}^{R}\frac{R^{2}\!-\!x^{2}}{2}\,dx\;=\;% \frac{2}{\pi R^{2}}\operatornamewithlimits{\Big{/}}_{\!\!\!x=-R}^{\,\quad R}% \left(\frac{R^{2}x}{2}-\frac{x^{3}}{6}\right)\;=\;\frac{4R}{3\pi}.$

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

 Title centre of mass of half-disc Canonical name CentreOfMassOfHalfdisc Date of creation 2013-03-22 17:20:57 Last modified on 2013-03-22 17:20:57 Owner pahio (2872) Last modified by pahio (2872) Numerical id 10 Author pahio (2872) Entry type Example Classification msc 28A75 Classification msc 26B15 Synonym center of mass of half-disc Synonym centroid of half-disc Related topic SubstitutionNotation Related topic CentreOfMassOfPolygon Related topic CenterOfGravityOfCircularSector Related topic AreaOfSphericalZone