Answer
$=4\sqrt 3-\frac{4}{3}\pi$
Work Step by Step
$A=4\times \int_{0}^{\pi/6}\frac{1}{2}(8cos(2\theta)-4)d \theta$
$=2[4sin(2\theta)-4\theta]_{0}^{\pi/6}$
$=4\sqrt 3-\frac{4}{3}\pi$
Calculus: Early Transcendentals 9th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1337613924
ISBN 13:
978-1-33761-392-7
Chapter 10 - Section 10.4 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 700: 25
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