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: 31


$8 \pi^2$

Work Step by Step

Since, $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Thus, $dS=\int_{0}^{2\pi} \sqrt{(-\sin t)^2+( \cos t)^2}=1$ or, $S=\int_{0}^{2\pi} (2 \pi)(2+\sin t)(1) dt$ Thus, $S=[(2 \pi)(2t-\cos t)]_{0}^{2\pi} =8 \pi^2$
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