Chapter 10 - Section 10.2 - Calculus with Parametric Curves - Exercises - Page 572: 30

$\ln 2$

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