Answer
$\frac{1}{3}$
Work Step by Step
$\int_{R} \int 2 y d A$
$R=\left\{(r, \theta): \frac{\pi}{4} \leqslant \theta \leqslant \frac{\pi}{2}, 0 \leqslant r \leqslant 2 \cos \theta\right\}$
$\int_{R} \int 2 y d A=\int_{\pi / 4}^{\pi / 2} \int_{0}^{2 \cos \theta} 2(r \sin \theta) r d r d \theta$
simplify
$=\int_{\pi / 4}^{\pi / 2} \int_{0}^{2 \cos \theta} 2 r^{2} \sin \theta d r d \theta$
integrate
$=\int_{\pi / 4}^{\pi / 2}\left[\frac{2 r^{3}}{3} \sin \theta\right]_{0}^{2 \cos \theta} d \theta$
$=\int_{\pi / 4}^{\pi / 2} \frac{16}{3} \cos ^{3} \theta \sin \theta d \theta$
$=-\frac{16}{3}\left[\frac{\cos ^{4} \theta}{4}\right]_{\pi / 4}^{\pi / 2}$
$=-\left[-\cos ^{4} \frac{\pi}{4}+\cos ^{4} \frac{\pi}{2}\right]\frac{4}{3}$
$\frac{1}{3}=-\frac{4}{3}\left(-\frac{1}{4}\right)$
