Chapter 11: Parametric Equations and Polar Coordinates - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 657: 30

$\ln 2$

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