# Double Integrals - application to centre of mass.

• CAF123
The double integral is needed because the mass is not distributed evenly in the two dimensional case. The density is not uniform, so the C.O.M coordinates would not be accurate.

#### CAF123

Gold Member
I know that for a system of particles, the x, y and z coordinates of the centre of mass are given by $$(\frac{1}{M} \sum_{i = 1}^{n} m_i x_i, \frac{1}{M} \sum_{i = 1}^{n} m_i y_i,\frac{1}{M} \sum_{i = 1}^{n} m_i z_i).$$
For a solid body, we can treat this like a continuous distribution of matter of little mass elements $dm.$The above becomes $$(\frac{1}{M} \int x dm,\frac{1}{M} \int y dm,\frac{1}{M} \int z dm).$$
My question is: I also have a defintion of centre of mass using the double integral, that is, for the x coordinate say, $$x_{com} = \frac{1}{M} \int\int x \rho\, dA,$$ where $dm = \rho\, dA$ 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, $\small Area(R) = \int\int_R dA,$ but I also know that the volume of space below a curve $z = f(x,y)$ in a region $R$ is given by $\int\int_R f(x,y) dA.$ 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

Your question is a little confusing. At times you are talking about 3 dimensions, other times 2 dimensions.

In any case, using the sum over discrete masses being treated as points, space dimensions don't play a role. Your going from discrete to continuous misses the point in that dm = ρdA (2-d) or dm = ρdV (3-d) cannot be viewed as a 1-d integral.

Therefore when talking about real objects, spatially spread out, it is necessary to integrate over the area (2-d) or volume (3-d), so the integral is 2 or 3 dimensional.

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Ok thanks.
For a planar object with density $\rho(x,y)$, the C.O.M is given by, $$(\frac{1}{M} \int\int x\,\rho\, dA, \frac{1}{M} \int\int y\,\rho\, dA).$$
If this object is two dimensional, why the need for a double integral?. In my post previously, the C.O.M coordinates for a solid body (3D) were given as single integrals?As far as I understand from my text, the reason is to do with the density not being uniform in the former case?

The essentail point is that the integral with respect to dA means the integral over dxdy.
In three dimensions dV=dxdydz. When you use dm, it is an abstraction - you need to convert to ρdA or ρdV.