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Homework Help: Centroid with Green's Theorem

  1. Feb 5, 2007 #1
    Can someone help me with the following? I'm supposed to find the centroid of a region D using Green's Theorem. Assume that this density function is constant.


    ∫Pdx + ∫Qdy = ∫∫(dQ/dx)-(dP/dy)
    A = ∫xdy = -∫ydx = ½*∫xdy - ydx


    I know that the mass of a region D with constant density function is ∫kdA (which is the area times some constant K). Let's make it easy and assume that k = 1 with the area A. So, the centroid of the region D would be located at (1/A*∫∫xdA) and (1/A*∫∫ydA). So, if I set Pdx as -ydx, and Qdy as xdy, I would get from Green's Theorem that ∫Pdx = -∫∫ydxdy and ∫Qdy = ∫∫xdxdy. But if you divide by A this is the expression for the coordinates of the centroid, since ∫∫xdxdy = ∫∫xdA and -∫∫ydxdy = ∫∫ydA. So you have the coordinates as 1/2A∫x^2/dy and 1/2A∫-y^2dx as the coordinates? Thanks.
     
  2. jcsd
  3. Feb 5, 2007 #2

    Dick

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    Yes. Setting P=0, Q=x^2/2 gives you x integrated over the surface and P=(-y)^2/2, Q=0 gives you y. So your final conclusion is correct.
     
  4. Nov 17, 2009 #3
    I know this is an old thread, but I need to understand the derived centroid coordinates from Green's theorem.

    I basically got lost when he said "So, if I set Pdx as -ydx, and Qdy as xdy, I would get from Green's Theorem that∫Pdx = -∫∫ydxdy and ∫Qdy = ∫∫xdxdy."

    By my understanding, ∫Pdx = ∫-ydx, using Green's theorem,
    ∫∫(dQ/dx - dP/dy)dA = ∫∫(--1)dxdy = ∫∫dA = A?
     
  5. Nov 17, 2009 #4

    Dick

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    I know you know it's an old thread, but parasitizing old threads is an almost sure way to get completely ignored. If Q=x^2/2 then dQ/dx is x. That's all I'm going to say until you do the right thing and post your own problems.
     
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