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I'm having trouble calculating the centroid of the cardioid (and various other polar-coordinate-defined lamina), i.e., [tex]r = 1 +\cos \theta[/tex]

I can see how it's symmetric over the x-axis, so y-bar is zero. So to calculate x-bar, then I do:

[tex]\int_{0}^{2\pi}\int_{0}^{1+\cos\theta}r \cos\theta rdrd\theta[/tex] =

[tex]\int_{0}^{2\pi} \frac{1}{3} (1+\cos\theta)^3 \cos\theta d\theta[/tex]

which seems to be particularly difficult to integrate, even Mathematica returns a nasty antiderivative involving things like sec ^6 theta which are obviously not necessary for solving the problem.

So, obviously I'm missing the main point here. I thought it would just be [tex]\int \int x dA[/tex] but in polar coordinates... I followed the integral above and calculated it by hand but got a weird value, 3.75*pi, which cannot obviously be x-bar as is isn't even near the lamina. Plus, Mathematica got the much more complicated integral than I did, suggesting I made a mistake in the integration.

If I just made a minor mistake in integration and the above single integral for x-bar is correct...then I will try again.

Thanks for any help...

Edit: I reworked it and got (5/4)*Pi...which makes more sense...ahh I see now!! you have to divide by the area, which gives you 5/6 which is the provided answer. Oh well :) Hope he doesn't ask such a question on the final... :)

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# Homework Help: Centroid ( center of mass ) of cardioid

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