## Thomas' Calculus 13th Edition

$2$
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 $A=(2)[- \cos (2\theta) ]_{0}^{(\pi/2)} =2$