Centroid with Bezier

jjj888

I did, but i wasn't too familiar with the notation. I'll take a look again.

jjj888

I'm having trouble understanding how
xc = ∫_Ax.dx.dy/∫_Adx.dy is equal to x=1/A∫_a^bxf(x)dx?

Anyway here's my x derivation using x=1/A∫_a^bxf(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

haruspex

∫_Ax.dx.dy = ∫_A.dy.x.dx = ∫_x∫_y.dy.x.dx = ∫_xy.x.dx
What happened to dx? dx is (dx/dt)dt, not dt.
Let me illustrate just in calculating A:
A = ∫_A.dy.dx = ∫_x∫_y.dy.dx = ∫_xy.dx = ∫_t y.(dx/tx).dt = ∫_t (-1.8t³-3.9t²+13.2t).(2.7t²+10.2t).dt = 37.26

jjj888

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

haruspex

∫_A means the integral over the region A. This is a reasonably standard notation and has the merit of being meaningful independently of the choice of variable. In that respect, it's analogous to a path integral in complex analysis.
The other three I used merely to emphasise what the variable of integration was in each case. It's redundant here really because it should be clear from the .dx or whatever.

jjj888

So,

if dx = (dx/dt)dt then,

x=1/A∫_txf(x)(dx/dt)dt
y=1/A∫_t(1/2)[f(x)²](dx/dt)dt

xc = ∫_t[(0.9t³+5.1t²)(-1.8t³-3.9t²+13.2t)(2.7t²+10.2t)]dt/∫_t[(-1.8t³-3.9t²+13.2t)(2.7t²+10.2t)]dt = (125.59/37.26) = 3.37

yc = ∫_t(1/2)[(-1.8t³-3.9t²+13.2t)²(2.7t²+10.2t)]dt/∫_t[(-1.8t³-3.9t²+13.2t)(2.7t²+10.2t)]dt = (1/2)(244.94/37.26) = 3.28

BAM!

Thanks for the guidence.

I did find something in a text book regarding the substitution of t. But still a little unclear how the substitution works.