Answer
$ \dfrac{243}{8}$
Work Step by Step
We can define the domain $D$ in the Type-1 as follows: $
D=\left\{ (x, y) | 0 \leq x\leq 3, \ x^2 \leq y \leq 3x \right\}
$
Therefore, $\iint_{D} xy dA=\int_{0}^{3} \int_{x^2}^{3x} xy \ dy \ dx \\= \int_0^3 [\dfrac{xy^2}{2}]_{x^2}{3x} dx \\ = \int_0^3 \dfrac{x(3x)^2} {2}- \dfrac{x(x^2)^2}{2}dx \\ = [\dfrac{9x^4}{8}-\dfrac{x^{6}}{12}]_0^3 \\= \dfrac{243}{8}$
Now, we need to define the domain $D$ in the Type-2 as follows: $
D=\left\{ (x, y) | 0 \leq y \leq 9, \ y/3 \leq x \leq \sqrt y \right\}
$
Therefore, $\iint_{D} xy dA=\int_{0}^{9} \int_{y/3}^{\sqrt y} xy \ dx \ dy \\= \int_0^9 [\dfrac{xy^2}{2}]_{y/3}^{\sqrt y} dy \\ = \int_0^9 \dfrac{y^2} {2}- \dfrac{y^3}{18}dx \\ = [\dfrac{y^3}{6}-\dfrac{y^4}{(4)(18)}]_0^9 \\= \dfrac{243}{8}$
