Answer
$$\iint_{D} f(x,y) dA=\int_{0}^{4} \int_{0}^{\sqrt y} f(x,y) \ dx \ dy $$
Work Step by Step
We can define the domain $D$ in the Type-1 using vertical cross-sections as follows:
$
D=\left\{ (x, y) | x^2 \leq y \leq 4, \ 0 \leq x \leq 2 \right\}
$
Therefore,
$$\iint_{D} f(x,y) dA=\int_{0}^{2} \int_{x^2}^{4} f(x,y) \ dy \ dx$$
and we can define the domain $D$ in the Type-II using horizontal cross-sections as follows:
$
D=\left\{ (x, y) | 0 \leq x \leq \sqrt y, \ 0 \leq y \leq 4 \right\}
$
Therefore, $$\iint_{D} f(x,y) dA=\int_{0}^{4} \int_{0}^{\sqrt y} f(x,y) \ dx \ dy $$
