Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 688: 19

$\dfrac{9 \pi}{2}$

Here, we have $L=\int_{0}^{3\pi/2}\sqrt{(\dfrac{dx}{d\theta})^2+(\dfrac{dy}{d\theta})^2}d\theta=\int_{0}^{(3\pi/2)} \sqrt{3(\sin^2 \theta+\cos^2 \theta)} d\theta$ This implies that $L= ( \sqrt{3}) \int_{0}^{(3\pi/2)} d\theta$ Therefore, $L =(3) (\dfrac{3\pi}{2}-0)=\dfrac{9 \pi}{2}$