University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - Exercises - Page 572: 25



Work Step by Step

Since, $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Thus, $S=\int_{0}^{\pi} \sqrt{(-\sin t)^2+(1+\cos t)^2} dt=\int_{0}^{\pi} \sqrt{2+2 \cos t} dt$ or, $\int_{0}^{\pi} \sqrt{2+2 (2\cos^2 (t/2)-1)} dt=\int_{0}^{\pi} 2\cos (t/2) dt$ Thus, $S=[4 \sin (t/2)]_{0}^{\pi} =4$
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