# Finding the centroid of a polar curve

I have absolutely no idea how to find the area centroid of this problem. I have been working at this one for ages but can't seem to get anywhere.

Any first steps? How would one go about solving this?

Find the centroid of the area by direct integration. • Do you know the general formula for the center of mass? It's sum of the positions vectors times their mass element there. Does that help? Oct 29 '13 at 5:15
• I understand how to do these problems in general, but this one has me stumped. I am not sure where to begin. Oct 29 '13 at 5:16

Here we go. Now we want to calculate the centroid $(\overline{r}, \overline{\theta})$ of the area that was defined by a polar function $r = r(\theta), (\alpha \leqslant \theta \leqslant \beta)$.

We know the general formula for centroid: \left\{ \begin{aligned} \overline{x} = \dfrac{1}{A}\int_A x \,\mathrm{d}A\\ \overline{y} = \dfrac{1}{A}\int_A y \,\mathrm{d}A \end{aligned} \right. For each polar point $(r(\theta), \theta)$ on the curve, we can take a fan-shaped surface element just like the following figure. The centroid of this surface element is $\left(\frac{2}{3}r(\theta), \theta\right)$(Why there is a $\frac{2}{3}$ just like you ask? We can consider the surface element as an triangle, and the centroid of this triangle is obviously at here.) The cartesian coordinate of it's centroid is $\left(\frac{2}{3}r(\theta)\cos\theta, \frac{2}{3}r(\theta)\sin\theta\right)$. And the area of this surface element $\mathrm{d}A = \frac{1}{2}r^2(\theta)\mathrm{d}\theta$. So we have \left\{ \begin{aligned} \overline{x} = \dfrac{1}{3A}\int_{\alpha}^{\beta} r^3(\theta)\cos\theta \,\mathrm{d}\theta\\ \overline{y} = \dfrac{1}{3A}\int_{\alpha}^{\beta} r^3(\theta)\sin\theta \,\mathrm{d}\theta \end{aligned} \right.\\ \mbox{with}~A = \dfrac{1}{2}\int_{\alpha}^{\beta}r^2(\theta)\,\mathrm{d}\theta Now we get the cartesian coordinate $(\overline{x}, \overline{y})$ of the centroid of the hole form. But we want the polar coordinate $(\overline{r}, \overline{\theta})$. Notice that \left\{ \begin{aligned} & \overline{r} = \left|\overline{x}+\mathrm{i}\,\overline{y}\right|\\ & \overline{\theta} = \arg(\overline{x}+\mathrm{i}\,\overline{y}) \end{aligned} \right. with \begin{aligned} \overline{x}+\mathrm{i}\,\overline{y} & = \dfrac{1}{3A}\int_{\alpha}^{\beta} r^3(\theta)\cos\theta \,\mathrm{d}\theta + \mathrm{i}\,\left(\dfrac{1}{3A}\int_{\alpha}^{\beta} r^3(\theta)\sin\theta \,\mathrm{d}\theta\right)\\ & = \dfrac{1}{3A}\int_{\alpha}^{\beta} r^3(\theta)(\cos\theta + \mathrm{i}\,\sin\theta) \,\mathrm{d}\theta\\ & = \dfrac{1}{3A}\int_{\alpha}^{\beta} r^3(\theta)\,\mathrm{e}^{\mathrm{i}\theta} \,\mathrm{d}\theta \end{aligned} Finally we have \left\{ \begin{aligned} & \overline{r} = \dfrac{1}{3A}\left|\underline{Z}\right|\\ & \overline{\theta} = \arg\left(\underline{Z}\right) \end{aligned} \right.\\ ~\\ \mbox{with}~A = \dfrac{1}{2}\int_{\alpha}^{\beta}r^2(\theta)\,\mathrm{d}\theta\\ \mbox{and}~\underline{Z} = \int_{\alpha}^{\beta} r^3(\theta)\,\mathrm{e}^{\mathrm{i}\theta} \,\mathrm{d}\theta

Now apply this conclusion in this problem. Here, $r(\theta) = a\mathrm{e}^{\theta},(0 \leqslant \theta \leqslant \pi)$. So \begin{aligned} A & = \dfrac{1}{2}\int_{0}^{\pi}a^2\mathrm{e}^{2\theta}\,\mathrm{d}\theta\\ & = \dfrac{1}{4}a^2(\mathrm{e}^{2\pi}-1) \end{aligned}

\begin{aligned} \underline{Z} & = \int_{0}^{\pi} a^3\mathrm{e}^{(3+\mathrm{i})\theta}\,\mathrm{d}\theta\\ & = -\dfrac{a^3(\mathrm{e}^{3\pi}+1)}{3+\mathrm{i}} \end{aligned} So \begin{aligned} \overline{r} & = \dfrac{1}{3A}\left|\underline{Z}\right|\\ & = \dfrac{2\sqrt{10}}{15}\dfrac{\mathrm{e}^{3\pi}+1}{\mathrm{e}^{2\pi}-1}\,a \approx 9.78\,a \end{aligned}

\begin{aligned} \overline{\theta} & = \arg(\underline{Z})\\ & = \pi - \arctan \dfrac{1}{3} \approx 161.57^\circ \end{aligned} Plot the centroid $C = (\overline{r}, \overline{\theta})$ on the figure: So the center of mass is given by $$\int \vec{r} dm$$ Your mass density is uniform. Your limits of integration will be $\theta: (0,\pi)$ and $r: (0,a e^{\theta})$. The Jacobian will be simply $r$. Remember that $\vec{r}=r \cos(\theta) \hat{i} + r\sin(\theta) \hat{j}$. You'll be left with a tricky enough integration, I would advise integrating with respect to $dr$ first.

Can you handle it from here?

• I am not quite sure on the application of the Jacobian here. The solution to this problem is actually in the text. It shows the conversion into polar coordinates first, but X(bar) = 2/3 r *cos(theta) and Y(bar) = 2/3 r* sin(theta). Any idea on where the 2/3 scalers came from? Oct 29 '13 at 5:36
• I see no reason why those $2/3$s should be there... After all the definition is pretty clear. The jacobian arises because you are taking an integral over an area. Oct 29 '13 at 5:41
• Isn't $dm = dx.dy/A = r.dr.d\theta/A$, where $A$ is the total area? Then $A$ has to be calculated as well. Oct 29 '13 at 16:05
• @HandeBruijn You are right. Oct 29 '13 at 18:20

You have already seen the solution to the problem and I see that they have made recourse to complex variables. I would like to demonstrate that the solution might be somewhat simpler if one were to fully embrace the complex plane.

To begin with, for $r=ae^{\theta}$ the curve can be expressed in the complex plane simply as

$$z=ae^{(1+i)\theta}$$

This is seen to be a logarithmic spiral with a flair coefficient of unity, which means that the radius increases by factor $e$ per radian. You'll notice that at $\theta=\pi$ the radius is $r(\pi)=e^{\pi}$. Now, we know that the area and centroid are given by (see reference)

$$A=\frac{1}{2}\int\Im\{z^*\dot z\}d\theta\\ Z_c=\frac{1}{3A}\int z\ \Im\{z^*\dot z\}d\theta$$

where $()^*$ is the conjugate and $(^\dot \,)$ is the derivative with respect to $\theta$. Therefore we can calculate the centroid as follows:

$$\dot z=a(1+i)e^{(1+i)\theta}\\ z^*=ae^{(1-i)\theta}\\ z^*\dot z=a^2(1+i)e^{2\theta}\\ \Im\{z^*\dot z\}=a^2e^{2\theta}\\$$

Then

A=\frac{a^2}{2}\int_0^{\pi}e^{2\theta}\,d\theta=\frac{a^2}{4}(e^{2\pi}-1)\\ \begin{align} Z_c &=\frac{a^2}{3A}\int_0^{\pi}e^{(1+i)\theta}e^{2\theta}\,d\theta=\frac{a^2}{3A}\frac{(e^{(3+i)\pi}-1)}{(3+i)}\\ &=\frac{4a}{3}\frac{1}{(3+i)}\frac{(e^{(3+i)\pi}-1)}{(e^{2\pi}-1)}\approx-9.274a+3.091ai \end{align}

An finally, we can the centroid in polar coordinates from

$$r_c=|Z_c|\approx9.776a\\ \theta_c=\arg Z_c\approx161.87^{\circ}$$

It's been my experience that it's much simpler to solve problems like this in the complex plane.

The reference for this work is: Zwikker, C. (1968). The Advanced Geometry of Plane Curves and Their Applications, Dover Press.