Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Section 11.5 - Areas and Lengths in Polar Coordinates - Exercises 11.5 - Page 670: 10

Answer

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

Work Step by Step

Consider $A_1=(\dfrac{1}{2}) \int_{(\pi/6)}^{(3\pi/6)} (1) d\theta =\dfrac{\pi}{3}$ and $A_2=(\dfrac{1}{2}) \int_{0}^{\pi/6} (2\sin \theta)^2 d \theta= \int_{0}^{\pi/6}(1- \cos 2\theta) =\dfrac{\pi}{6}-\dfrac{\sqrt 3}{4}$ Now, $A_3=(\dfrac{1}{2}) \int_{(3 \pi/6)}^{\pi} (2)^2 (\sin \theta)^2 d \theta=\dfrac{\pi}{6}-\dfrac{\sqrt 3}{4}$ Total area: $A=A_1+A_2+A_3=\dfrac{\pi}{3}+2-(\dfrac{\pi}{6}-\dfrac{\sqrt 3}{4})$ or, $A=\dfrac{2\pi}{3}-\dfrac{\sqrt 3}{2}$
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