Chapter 11: Parametric Equations and Polar Coordinates - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 658: 34

$\pi$

Since, $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt= \sqrt{(\sec t- \cos t)^2+(\sin t)^2}$ or, $dS= \sqrt{\sec^2 t+-2+\cos^2 t+\sin^2 t} dt= \sqrt {\sec^2t-1} dt$ or, $dS= \tan t dt$ This implies that $S=\int_{0}^{\pi/3} (2 \pi) y ds= (2 \pi) \int_{0}^{\pi/3}(\cos t)( \tan t) dt$ or, $S= (2 \pi) \int_{0}^{\pi/3} \sin t dt$ Thus, $S=[-2 \pi \cos t]_{0}^{(\pi/3)} =-\pi+2 \pi=\pi$