University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.3 - Trigonometric Substitutions - Exercises - Page 439: 53



Work Step by Step

Consider $x=3 \sin \theta \implies dx=3 \cos \theta d\theta$ Now, the given integral can be written as: $Area=\int_0^{\pi/2} \dfrac{\sqrt{9-9 \sin^2 \theta}}{3} d \theta=\int_0^{\pi/2} 3 \cos^2 \theta d\theta$ or, $=\dfrac{3}{2}[\theta+(1/2) \sin 2 \theta]_0^{\pi/2}+C$ Now, Area $=\dfrac{3}{2}(\dfrac{\pi}{2})=\dfrac{3\pi}{4}$
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