Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 10 - Review - Exercises - Page 720: 32

Answer

$6a^2 \pi$

Work Step by Step

$\int_{0}^{2 \pi}dy=a\int_{0}^{2 \pi}(2cost-cos2t)(cost-cos2t)dt$ $=2a^2\int_{0}^{2 \pi}[cos2t+1+(\dfrac{cos4t+1}{4})-3(cost)(cos2t)]dt$ $=6a^2 \pi$
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