## Centroid with Bezier

I did, but i wasn't too familiar with the notation. I'll take a look again.
 I'm having trouble understanding how xc = ∫Ax.dx.dy/∫Adx.dy is equal to x=1/A∫abxf(x)dx? Anyway here's my x derivation using x=1/A∫abxf(x)dx: x = 1/4.85∫01((0.9t3+5.1t2)(-1.8t3-3.9t2+13.2t)) = 2.656 I set it up the same way as the previous problem. I would think this is a fairly common problem, that is finding parametric equations to fit curves to things to calculate areas, centers, etc. It's strange I can't find more information about it out there. Obviously it's easily done with computers, but I just like to know how it works. Thanks

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 Quote by jjj888 I'm having trouble understanding how xc = ∫Ax.dx.dy/∫Adx.dy is equal to x=1/A∫abxf(x)dx?
Ax.dx.dy = ∫A.dy.x.dx = ∫xy.dy.x.dx = ∫xy.x.dx
 Anyway here's my x derivation using x=1/A∫abxf(x)dx: x=0.9t3+5.1t2 y=-1.8t3-3.9t2+13.2t x = 1/4.85∫01((0.9t3+5.1t2)(-1.8t3-3.9t2+13.2t)) = 2.656
What happened to dx? dx is (dx/dt)dt, not dt.
Let me illustrate just in calculating A:
A = ∫A.dy.dx = ∫xy.dy.dx = ∫xy.dx = ∫t y.(dx/tx).dt = ∫t (-1.8t3-3.9t2+13.2t).(2.7t2+10.2t).dt = 37.26
 Thanks, but before I try to decifer this I'm still having notation issues, what do these symbols mean: ∫A, ∫x, ∫y, ∫t. Are they definites of that variable, or is there something else. Is there a different way to write it? Thanks

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