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

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.
