Answer
$a^2 b+\dfrac{3}{2} ab^2$
Work Step by Step
The region $D$ using the point of intersection can be expressed as follows:
$
D=\left\{ (x, y) | 0 \leq x \leq 9, \ 0 \leq y \leq b \right\}
$
Now, $\iint_{D} f(x,y) dA=\iint_{D} (2x+3y) dA \\=\int_{0}^{a} \int_{0}^{b} f(x,y) (2x+3y) \ dy \ dx $
or, $=\int_0^a [2xy +\dfrac{3y^2}{2}]_0^b dx $
or, $= \int_0^a [2bx +\dfrac{3b^2}{2} ]dx $
or, $= [bx^2 +\dfrac{3b^2 x}{2}]_0^a$
or, $= a^2 b+\dfrac{3}{2} ab^2$
