Finding the Centroid Using Green's Theorem

[bodensee9]

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.

 

[Dick]

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.

 

[calorimetry]

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?

 

[Dick]

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.