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: 7

Answer

$2$

Work Step by Step

Consdier $A=(\dfrac{1}{2}) \int_m^n (r^2) d \theta=\int_{0}^{\pi/2} (\dfrac{1}{2}) 4 \sin 2 \theta d \theta$ $\implies A= \int_{0}^{\pi/2} (\dfrac{1}{2}) 4 \sin 2 \theta d \theta$ This implies that $A=(2)[- \cos (2\theta) ]_{0}^{(\pi/2)} =2$
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