Answer
The answer is below.
Work Step by Step
(a) We know that $x$ goes from $0 \rightarrow 2$ and $y$ goes from $0 \rightarrow x^{2}$ so
\[
\int_{0}^{2} \int_{0}^{x^{2}} f(x, y) d y d x=\iint_{r} f(x, y) d A
\]
(b) The upper bound of $y$ is given by $4=2^{2} .$ For the bound of $x,$ we see that $\sqrt{y} =x$ for the upper function, so
\[
\int_{0}^{4} \int_{\sqrt{y}}^{2} f(x, y) d x d y=\iint_{r} f(x, y) d A
\]
