Given Polar Coordinates - Find the rectangular coordinates of a centroid.

Question

I'm having trouble with this question on my Calculus III homework. I'm not really sure how to go about this one honestly.

A lamina is bounded by the curve $r = 2\cos(3θ)\space$ when $-π/6 <= θ <= π/6$. Assuming a constant density, $ρ = k$, give the rectangular coordinates of the centroid.

For clarity of the question, and clarity of what type of answer I am looking for:

My professor taught me the following: $$M_x = \iint y\rho(x,y)dA$$ $$M_y = \iint x\rho(x,y)dA$$ $$m = \iint \rho(x,y)dA$$ The answer I need is, $$(\bar x, \bar y) = \left(\frac{M_y}{m},\frac{M_x}{m}\right)$$ In addition, I was taught that $$ x = r\cos(\theta) $$ $$ y = r\sin(\theta) $$ when converting between polar and rectangular coordinates.

Any assistance would be greatly appreciated. Thanks.

Answer 1

Let's first calculate the total mass of the centroid. The total mass $M$ is given by

$$\begin{align} M&=\int_{-\pi/6}^{\pi/6} \int_0^{2\cos(3\theta)}\rho \,r\,dr\,d\theta\\\\ &=4\rho\int_0^{\pi/6}\cos^2(3\theta)\,d\theta\\\\ &=\frac\pi3\rho \end{align}$$

Because of the symmetry of the centroid, we expect the center of mass to lie on the $x$-axis. In fact, the center of mass, $\vec r_{CM}=\hat x \bar x+\hat y \bar y$, is given by

$$\begin{align} \vec r_{CM}&=\frac1M\int_{-\pi/6}^{\pi/6}\int_0^{2\cos(3\theta)} \vec r\,\rho\,r\,dr\,d\theta\\\\ &=\frac3\pi\left(\hat x \int_{-\pi/6}^{\pi/6}\int_0^{2\cos(3\theta)} r^2\cos(\theta)\,dr\,d\theta+\hat y \underbrace{\int_{-\pi/6}^{\pi/6}\int_0^{2\cos(3\theta)} r^2\sin(\theta)\,dr\,d\theta}_{=0\,\text{due to odd symmetry}}\right)\\\\ &=\hat x \frac{16}{\pi}\int_0^{\pi/6}\cos^3(3\theta)\cos(\theta)\,d\theta \tag 1 \end{align}$$

The evaluation of the integral on the right-hand side of $(1)$ is left as an exercise.

Answer 2

EDIT1

$$ {\bar x} = \frac{\int r ^3\cos \theta d\theta}{\int r^2 d\theta},\, {\bar y} = \frac{\int r ^3\sin \theta d\theta}{\int r^2 d\theta} $$

Plugin $r$ and integrate between given $\theta$ limits. The latter vanishes as for an odd function.

Answer 3

Let me demonstrate how much simpler this problem is in the complex plane. First of all, we note that curve can be expressed as

$$z=2\cos(3\theta)e^{i\theta}$$

The arc length, area, and centroid are given by

$$ s=\int |\dot z| d\theta\\ A=\frac{1}{2}\int \mathfrak{Im}\{z^*\dot z\}d\theta\\ z_c=\frac{1}{3A}\int z\ \mathfrak{Im}\{z^*\dot z\}d\theta $$

Thus, we need $\dot z$, $z^*$, $\mathfrak{Im}\{z^*\dot z\}$ as follows

$$\dot z=[3\cdot 2\sin(3\theta)+i\cdot2\cos(3\theta)]e^{i\theta}\\ z^*=2\cos(3\theta)e^{-i\theta}\\ \mathfrak{Im}\{z^*\dot z\}=4\cos^2(3\theta) $$

Then we can readily show that

$$ A=\frac{1}{2}\int_{-\pi/6}^{\pi/6}4\cos^2(3\theta)d\theta=\frac{2}{3}\int_{-\pi/2}^{\pi/2}\cos^2\varphi d\varphi=\frac{\pi}{3}\\ z_c=\frac{8}{3A}\int_{-\pi/6}^{\pi/6}\cos^3(3\theta)e^{i\theta}d\theta=\frac{8}{3A}\int_{-\pi/6}^{\pi/6}\cos^3(3\theta)\cos\theta d\theta=\frac{8}{\pi}\left(\frac{81\sqrt{3}}{320}+0i\right) $$

Notice the cosine arrangement in the last integral.