Find the centroid of a curve.

Question

Problem: Find the centroid of the part of the large loop of the limacon $r=1+2\cos(\theta)$ that does not include the small loop.

I know that in order to compute the centroid one needs to use following equations $$x=\frac{\iint_D xdydx}{m}$$ and $$y=\frac{\iint_D ydydx}{m},$$ where $$m=\iint_D dydx.$$ Note that we are assuming $\rho=1,$ which should work for this problem since I got the correct answer for a similar problem. The real issue that I am facing is that I am unable to set up the bounds on the integral because I do not what values of $r,\theta$ do I have to exclude in order to avoid the loop. Any help in this regard would be much appreciated.

Answer 1

[This question appears to be here two times: here's my previous answer.]

I believe that the answer to your problem is that the integration limits are $\theta\in [-2\pi/3,2\pi/3]$. I've gone ahead and solved the problem in the complex plane; you can compare results when you finish.

Given

$$ z=(2\cos\theta+1)e^{i\theta}\\ A=\frac{1}{2}\int_{-2\pi/3}^{2\pi/3}\Im\{z^*\dot z\}d\theta\\ Z_c=\frac{1}{3A}\int_{-2\pi/3}^{2\pi/3}z\ \Im\{z^*\dot z\}d\theta $$

Notice that an advantage of the complex plane is that you have single integrals rather than doubles. At any rate, I find (with the help pf WolframAlpha) that

$$ A=\frac{3\sqrt{3}}{2}+2\pi\\ Z_c=\frac{\frac{27\sqrt{3}}{4}+8\pi}{3A} $$

It's also true that the area and centroid are single integrals in polar coordinates, with

$$ x_c=\frac{\frac{1}{3}\int r^3\cos\theta d\theta}{\frac{1}{2}\int r^2 d\theta}\\ y_c=\frac{\frac{1}{3}\int r^3\sin\theta d\theta}{\frac{1}{2}\int r^2 d\theta} $$

Answer 2

$r = 1 - 2 \cos \theta$ is symmetric around the $\theta = 0$

$\theta = 0...y=0$ is a line of symmetry, and the $y$ coordinate of the centroid is $0.$

That means that we just need to find $\frac {\iint x\ dy\ dx}{\iint\ dy\ dx}$

converting to polar: $\iint\ dy\ dx = \int_0^{2\pi} \int_0^{1-2\cos\theta} r\ dr\ d\theta$

$\iint\ x\ dy\ dx = \int_0^{2\pi} \int_0^{1-2\cos\theta} (r\cos\theta)r\ dr\ d\theta$