I know that for a system of particles, the x, y and z coordinates of the centre of mass are given by [tex] (\frac{1}{M} \sum_{i = 1}^{n} m_i x_i, \frac{1}{M} \sum_{i = 1}^{n} m_i y_i,\frac{1}{M} \sum_{i = 1}^{n} m_i z_i). [/tex](adsbygoogle = window.adsbygoogle || []).push({});

For a solid body, we can treat this like a continuous distribution of matter of little mass elements [itex] dm. [/itex]The above becomes [tex] (\frac{1}{M} \int x dm,\frac{1}{M} \int y dm,\frac{1}{M} \int z dm). [/tex]

My question is: I also have a defintion of centre of mass using the double integral, that is, for the x coordinate say, [tex] x_{com} = \frac{1}{M} \int\int x \rho\, dA, [/tex] where [itex] dm = \rho\, dA [/itex] for 2 dimensional objects.

Why do we need the double integral?

As an aside, I also have in my notes that the area of a region R, [itex]\small Area(R) = \int\int_R dA, [/itex] but I also know that the volume of space below a curve [itex] z = f(x,y) [/itex] in a region [itex] R [/itex] is given by [itex] \int\int_R f(x,y) dA. [/itex] Just to check: The reason why the former describes an area, while the latter a volume is because we have some projection up the z axis, ie f(x,y).

Many thanks

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Double Integrals - application to centre of mass.

Loading...

Similar Threads - Double Integrals application | Date |
---|---|

A Maximization Problem | Jan 31, 2018 |

I Q about finding area with double/volume with triple integral | Sep 13, 2017 |

I Find total charge (using double integration) | Apr 15, 2017 |

I Integration Limits Changing in Double Integral Order Change | Sep 7, 2016 |

I Change of variables in double integral | Jul 9, 2016 |

**Physics Forums - The Fusion of Science and Community**