# Centroid ( center of mass ) of cardioid

1. May 3, 2006

### jbusc

centroid ("center of mass") of cardioid

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

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

$$\int_{0}^{2\pi}\int_{0}^{1+\cos\theta}r \cos\theta rdrd\theta$$ =
$$\int_{0}^{2\pi} \frac{1}{3} (1+\cos\theta)^3 \cos\theta d\theta$$

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 $$\int \int x dA$$ 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... :)

Last edited: May 3, 2006
2. May 4, 2006

### Tom Mattson

Staff Emeritus
Not at all. I would expand the $(1+\cos(\theta))^3$ out and then distribute the remaining factor $\cos(\theta)$. You wouldn't want to do that for arbitrarily large exponents, but what the heck this one is a mere '3'.

You then will have 4 terms that should be easy to integrate.

3. May 4, 2006

### jbusc

True, but this is for a final exam and it's one of many problems expected to solve in a short time span...so most problems don't involve much calculation (rather sort of "concept" oriented), in fact, many of the problems he asked on the final are ones that the direct solution is quite involved to calculate but you can use clever tricks to solve quickly.

I see how to do it now, it's just to expand it out and use some identities to solve it. I thought it was difficult because Mathematica did not do a good job integrating it... But the prof didn't ask any centroid problems so...I did well, hopefully :)

4. May 4, 2006

### Tom Mattson

Staff Emeritus
OK, here's a concept to shorten the calculation: The odd terms go to zero when integrated from 0 to 2pi. That cuts your integration in half. :D