Answer
\[
\int_{0}^{\pi / 2} \int_{0}^{\sin x} f(x, y) d y d x
\]
Work Step by Step
\[
\iint_{D} f(x, y) d A=\int_{0}^{1} \int_{\sin ^{-1} y}^{\pi / 2} f(x, y) d x d y
\]
$D$ has been interpreted as a type II region (Using horizontal cross-sections).
We can also Interpret $D$ as type I region (using vertical cross-sections). Thus:
\[
\iint_{D} f(x, y) d A=\int_{0}^{\pi / 2} \int_{0}^{\sin x} f(x, y) d y d x
\]
