Answer
$\int_{4}^{9} \int_{\sqrt y}^{3} f(x,y) \ dx \ dy = \int_{2}^{3} \int_{4}^{x^2} f(x,y) \ dy \ dx$
Work Step by Step
We are given the domain $0 \leq x \leq 4$ and $x \leq y \leq 4$.
The iterated integral can be written for the domain $4 \leq y \leq x^2$ and $2 \leq x \leq 3$ as:
$\iint_{D} f(x,y) d A= \int_{4}^{9} \int_{\sqrt y}^{3} f(x,y) \ dx \ dy = \int_{2}^{3} \int_{4}^{x^2} f(x,y) \ dy \ dx$
