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 - Practice Exercises - Page 688: 50



Work Step by Step

Here, $A=2 \int_{-(\pi/2)}^{(\pi/2)} (\dfrac{1}{2}) [2(1+\sin \theta)]^2 d\theta-\pi$ This gives: $A=[6 (\theta)-8 \cos (\theta)-\sin (2\theta)]_{-(\pi/2)}^{(\pi/2)}-\pi$ Hence, $A=3\pi -(-3 \pi)-\pi=6 \pi-\pi=5\pi$
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