Answer
After reversing the order of integration we get
\[
\int_{0}^{1} \int_{x^{2}}^{\sqrt{x}} f(x, y) d y d x
\]
Work Step by Step
\[
\iint_{D} f(x, y) d A=\int_{0}^{1} \int_{y^{2}}^{\sqrt{y}} f(x, y) d x d y
\]
$D$ 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). So:
\[
\int_{0}^{1} \int_{x^{2}}^{\sqrt{x}} f(x, y) d y d x=\iint_{D} f(x, y) d A
\]