Answer
$\dfrac{243}{8}$
Work Step by Step
The domain $D$ in type-1 can be expressed as follows:
$
D=\left\{ (x, y) | 0 \leq x\leq 3 , \ x^2 \leq y \leq 3x \right\}
$
Now, we can work out with the given integral as follows: $\iint_{D} xy dA=\int_{0}^{3} \int_{x^2}^{3x} (x)(y) \ dy \ dx \\= \int_0^3 [\dfrac{xy^2}{2}]_{x^2}^{3x} dx \\ = \int_0^3 \dfrac{x(3x)^2} {2}- \dfrac{x(x^2)^2}{2}dx \\ = \dfrac{9}{8} [x^4]_0^3 -\dfrac{1}{12}][x^6]_0^3 \\= \dfrac{243}{8}$
The domain $D$ in type-2 can be expressed as follows: $
D=\left\{ (x, y) | 0 \leq y \leq 9 , \ y/3 \leq x \leq \sqrt y \right\}
$
Now, we can work out with the given integral as follows: $\iint_{D} xy dA=\int_{0}^{9} \int_{y/3}^{\sqrt y} xy \ dx \ dy \\= \int_0^9 [\dfrac{xy^2}{2}]_{y/3}^{\sqrt y} dy \\ = \int_0^9 \dfrac{y^2} {2}- \dfrac{y^3}{18}dx \\ = \dfrac{1}{6}\times [y^3]_0^9 - \dfrac{1}{(4)(18)} \\= \dfrac{243}{8}$
