## Calculus: Early Transcendentals 8th Edition

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