Answer
$\pi-3$
Work Step by Step
Given: $r=8 \sin^3 (\dfrac{\theta}{3})$
$r'=8 \sin^2 (\dfrac{\theta}{3}) \cos (\dfrac{\theta}{3})$
The length is given as follows:
$L= \int_{0}^{\pi/4} \sqrt{r^2+r'^2} d\theta=\int_{0}^{\pi/4} \sqrt {[64\sin^{4} (\dfrac{\theta}{3})} d\theta$
Then, $L=[4 \theta-6\sin(\dfrac{2\theta}{3})]_{0}^{\pi/4} $
After solving, we get $L=\pi-3$
