University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Section 10.5 - Areas and Lengths in Polar Coordinates - Exercises - Page 585: 10


$\dfrac{2\pi}{3}-\dfrac{\sqrt 3}{2}$

Work Step by Step

The area of a shaded region can be found as: $A=\dfrac{1}{2}\int_p^q r^2 d \theta$ Here, we have $A_1=\int_{\pi/6}^{3\pi/6} \dfrac{1}{2}1 d\theta =\dfrac{\pi}{3}$ $A_2=(1/2) \int_{0}^{\pi/6} (2\sin \theta)^2 d \theta= \int_{0}^{\pi/6}(1- \cos 2\theta) =\dfrac{\pi}{6}-\dfrac{\sqrt 3}{4}$ Now, $A_3=(1/2) \int_{3 \pi/6}^{\pi} (2\sin \theta)^2 d \theta=\dfrac{\pi}{6}-\dfrac{\sqrt 3}{4}$ $A=\dfrac{\pi}{3}+2-(\dfrac{\pi}{6}-\dfrac{\sqrt 3}{4})=\dfrac{2\pi}{3}-\dfrac{\sqrt 3}{2}$
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