Answer
$$\frac{1}{8}$$
Work Step by Step
Using the rules of integrals, we obtain:
$\int_{0}^{\pi / 4} \int_{\sin y}^{1 / \sqrt{2}} x d x d y=\iint_{R} x d A$
$=\int_{0}^{\pi / 4}\left[\frac{1}{2} x^{2}\right]_{\sin y}^{1 / \sqrt{2}} d y$
$=\int_{0}^{\pi / 4} \frac{1}{4}-\frac{1}{2} \sin ^{2} y d y$
$=\frac{1}{4} \int_{0}^{\pi / 4} 1-2 \sin ^{2} y d y$
$=\frac{1}{4} \int_{0}^{\pi / 4} \cos 2 y d y$
$\frac{1}{8}=\frac{1}{4}\left[\frac{1}{2} \sin 2 y\right]_{0}^{\pi / 4}$
