Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.2 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 673: 25

Answer

$=4\sqrt 3-\frac{4}{3}\pi$

Work Step by Step

$A=4\times \int_{0}^{\pi/6}\frac{1}{2}(8cos(2\theta)-4)d \theta$ $=2[4sin(2\theta)-4\theta]_{0}^{\pi/6}$ $=4\sqrt 3-\frac{4}{3}\pi$
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