Answer
$\frac{\pi}{16}$
Work Step by Step
$\sin 2 \theta=r, 1=r$
$\frac{\pi}{4} \leqslant \theta \leqslant \frac{\pi}{2}$
so
$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{\sin 2 \theta}^{1} r d r d \theta=A$
integrate with respect to $r$
$A=\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\left[\frac{r^{2}}{2}\right]_{\sin \theta}^{1} d \theta$
$A=\frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\left(-\sin ^{2} 2 \theta+1\right) d \theta$
and then
$A=\frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\left(1-\frac{1}{2}+\frac{\cos 4 \theta}{2}\right) d \theta$
$A=\frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\left(\frac{1}{2}+\frac{\cos 4 \theta}{2}\right) d \theta$
integrate
$A=\frac{1}{2}\left[\frac{\theta}{2}-\frac{\sin 4 \theta}{8}\right]_{\frac{\pi}{2}}^{\frac{\pi}{2}}$
evaluate
$-\frac{1}{2}\left[\frac{\pi}{4}-\frac{\sin 2 \pi}{8}\right]+\frac{1}{2}\left[\frac{\pi}{8}-\frac{\sin \pi}{8}\right]=A$
$A=-\frac{\pi}{8}+\frac{\pi}{16}=\frac{\pi}{16}$