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$$r=a\sin^3\left(\frac{\theta}{3}\right) $$ I tried solving it using the equation for arc length with $dr/d\theta$ and $r^2$. Comes out messy and complicated.

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  • $\begingroup$ Perhaps you could post your work, because it doesn't seem to me that it is very ugly at all. $\endgroup$ – rogerl May 18 '14 at 0:37
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    $\begingroup$ Be careful to apply the Chain Rule fully when calculating $ \ \frac{dr}{d\theta} \ $ . The Pythagorean Identity will help with simplifying the arclength element $ \ \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \ $ . You will get a "perfect square" under the radical. $\endgroup$ – colormegone May 18 '14 at 1:23
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The portion of the arclength calculation in which we determine the infinitesimal arclength element $ \ ds \ $ just requires applying the Chain Rule carefully and using the Pythagorean Identity:

$$ \frac{dr}{d\theta} \ = \ \frac{d}{d\theta} \left[ \ a \ \sin^3 \left(\frac{\theta}{3}\right) \right] \ = \ a \ \cdot \ 3 \ \sin^2\left(\frac{\theta}{3}\right) \cdot \cos\left(\frac{\theta}{3}\right) \cdot \frac{1}{3} \ = \ a \ \sin^2\left(\frac{\theta}{3}\right) \cos\left(\frac{\theta}{3}\right) $$

$$ \Rightarrow \ \ ds \ = \ \sqrt{ \ r^2 + \left(\frac{dr}{d\theta}\right)^2} \ \ d\theta \ = \ \sqrt{ \ \left[ \ a \ \sin^3 \left(\frac{\theta}{3}\right) \right]^2 + \left[ \ a \ \sin^2\left(\frac{\theta}{3}\right) \cos\left(\frac{\theta}{3}\right) \right]^2} \ \ d\theta $$

$$ = \ \sqrt{ \ a^2 \ \sin^6 \left(\frac{\theta}{3}\right) + \ \ a^2 \ \sin^4\left(\frac{\theta}{3}\right) \cos^2\left(\frac{\theta}{3}\right)} \ \ d\theta $$

$$ = \ \sqrt{ \ a^2 \ \sin^4 \left(\frac{\theta}{3}\right) \left[ \ \sin^2 \left(\frac{\theta}{3}\right) + \ \cos^2\left(\frac{\theta}{3}\right) \right]} \ \ d\theta $$

$$ = \ \sqrt{ \ a^2 \ \sin^4 \left(\frac{\theta}{3}\right) } \ \ d\theta \ = \ a \ \sin^2 \left(\frac{\theta}{3}\right) \ d\theta \ \ . $$

What turns out to be the tricky part is in finding the limits of integration in order to cover this curve fully. Here is a graph of the curve described by this polar equation (for $ \ a \ = \ 1 ) \ $ :

enter image description here

A graph of $ \ r \ $ as a function of $ \ \theta \ $ shows that, while the period of the function is $ \ 6 \pi \ $ , the radius is negative in the interval $ \ ( \ 3 \pi \ , \ 6 \ \pi \ ) \ $ .

enter image description here

The curve resembles a limaçon , but the interval of negative radii retraces the curve already "swept out" by positive radii in $ \ ( \ 0 \ , \ 3 \pi \ ) \ $ . Thus, we will find the arclength correctly by integrating arclength over this interval, rather than only up to $ \ 2 \pi \ $ or all the way to $ \ 6 \pi \ $ .

Our arclength integral is therefore

$$ \int_0^{3 \pi} \ ds \ \ = \ \ \int_0^{3 \pi} a \ \sin^2 \left(\frac{\theta}{3}\right) \ \ d\theta \ \ = \ \ a \ \int_0^{3 \pi} \ \frac{1}{2} \left[ \ 1 \ - \ \cos \left(\frac{2\theta}{3}\right) \right] \ \ d\theta $$

$$ = \ \ \frac{a}{2} \left[ \ \theta \ - \ \frac{3}{2} \sin \left(\frac{2\theta}{3}\right) \right] \vert_0^{3 \pi} \ = \ \frac{a}{2} \left[ \ \left(3 \pi \ - \ \frac{3}{2} \sin \left[\frac{6 \pi}{3}\right] \ \right) \ - \ \left(0 \ - \ \frac{3}{2} \sin \left[\frac{0}{3}\right] \right) \right] $$

$$ = \ \frac{a}{2} \ \left(3 \pi \ - \ \frac{3}{2} \sin \ 2 \pi \ \ - \ 0 \ + \ 0 \right) \ = \ \frac{3 \pi}{2} a \ \ . $$

As a check, we can see from the graph above that the arclength of the curve will be somewhat larger than the circumference of a circle of diameter $ \ \frac{5}{4} a \ $ , which is $ \ \frac{5 \pi}{4} a \ $ (the small "self-crossing" loop of this curve does not add much to the total arclength).

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This solution borders on trivial when solved in the complex plane. First, the range of $\theta$ must be determined. It was pointed out in an earlier that $\theta\in[0,3\pi]$. I found the same thing by plotting the equation. Now, the equation for the arc length in the complex plane is given by

$$s=\int|\dot z|du$$

Thus, for this problem we have (here we take $a=1$ without any loss in generality)

$$ z=\sin^3(\theta/3)e^{i\theta}\\ \dot z=\left[3\sin^2\left(\frac{\theta}{3}\right)\cos\left(\frac{\theta}{3}\right)\frac{1}{3}+i\sin^3\left(\frac{\theta}{3}\right)\right]e^{i\theta}\\ \begin{align} |\dot z| &=\sqrt{\sin^4\left(\frac{\theta}{3}\right)\cos^2\left(\frac{\theta}{3}\right)+\sin^6\left(\frac{\theta}{3}\right)}\\ &=\sin^2\left(\frac{\theta}{3}\right)\sqrt{\cos^2\left(\frac{\theta}{3}\right)+\sin^2\left(\frac{\theta}{3}\right)}\\ &=\sin^2\left(\frac{\theta}{3}\right)\\ \end{align}\\ $$

And finally, $$ s=\int_0^{3\pi}\sin^2\left(\frac{\theta}{3}\right)d\theta=\frac{3\pi}{2} $$

I have verified this result by numerical integration.

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