Double Integrals - application to centre of mass.

[CAF123]

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

 

[mathman]

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.

 

[CAF123]

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?

 

[mathman]

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.