# 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

$$Y=\frac{1}{\nu (E)}{\iint}_{E}y\mathit{d}x\mathit{d}y,$$ |

where $\nu (E)=\frac{\pi {R}^{2}}{2}$ is the area of the half-disc. The region of integration is defined by

$$E=\{(x,y)\in {\mathbb{R}}^{2}\mathrm{\vdots}-R\leqq x\leqq R,\mathrm{\hspace{0.33em}0}\leqq y\leqq \sqrt{{R}^{2}-{x}^{2}}\}.$$ |

Accordingly we may write

$$Y=\frac{2}{\pi {R}^{2}}{\int}_{-R}^{R}\mathit{d}x{\int}_{0}^{\sqrt{{R}^{2}-{x}^{2}}}y\mathit{d}y=\frac{2}{\pi {R}^{2}}{\int}_{-R}^{R}\frac{{R}^{2}-{x}^{2}}{2}\mathit{d}x=\frac{2}{\pi {R}^{2}}\underset{x=-R}{\overset{R}{/}}\left(\frac{{R}^{2}x}{2}-\frac{{x}^{3}}{6}\right)=\frac{4R}{3\pi}.$$ |

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 |