Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - 10.4 Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 713: 25

Answer

$$A=(4\sqrt 3-\frac{4}{3}\pi)$$

Work Step by Step

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