Chapter 10 - Parametric Equations and Polar Coordinates - 10.4 Exercises - Page 693: 8

$\dfrac {\pi }{8}\approx 0.39$

$A=\int ^{\pi /2}_{0}\dfrac {1}{2}\left[ r2\right] d\theta =\int ^{\pi /2}_{0}\dfrac {1}{2}\sin ^{2}2\theta d\theta =\int ^{\pi /2}_{0}\dfrac {1}{2}\times \dfrac {1-\cos 4\theta }{2}d\theta =\left[ \dfrac {1}{4}\theta -\dfrac {1}{16}\sin 4\theta \right] ^{\pi /2}_{0}=\dfrac {\pi }{8}\approx 0.39$