Chapter 10 - Section 10.2 - Calculus with Parametric Curves - Exercises - Page 572: 34

$\pi$

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