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Double Integrals - application to centre of mass.

  1. Sep 3, 2012 #1

    CAF123

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    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]
    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
     
  2. jcsd
  3. Sep 3, 2012 #2

    mathman

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    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.
     
    Last edited: Sep 3, 2012
  4. Sep 4, 2012 #3

    CAF123

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    Ok thanks.
    For a planar object with density [itex] \rho(x,y) [/itex], the C.O.M is given by, [tex] (\frac{1}{M} \int\int x\,\rho\, dA, \frac{1}{M} \int\int y\,\rho\, dA). [/tex]
    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?
     
  5. Sep 4, 2012 #4

    mathman

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    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.
     
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