Answer
$\frac{\pi}{2}-\frac{3\sqrt 3}{8}$
Work Step by Step
$r=sin^3(\theta/3)$
$r'=sin^2(\theta/3)cos(\theta/3)$
Length of the curve is given as:
$L=\int_{0}^{\pi}sin^2(\theta/3) d \theta$
$=\int_{0}^{\pi}\frac{1}{2}-(\frac{cos(2\theta/3)}{2}) d \theta$
$=[\frac{\theta}{2}-\frac{3}{2}\frac{sin(2\theta/3)}{2}]_{0}^{\pi}$
$=\frac{\pi}{2}-\frac{3\sqrt 3}{8}$
