Answer
$\dfrac{31}{8}$
Work Step by Step
The volume under the surface is given by :$z=f(x,y)$ and above the region $D$ in the xy-plane can be expressed as: $\ Volume ; V=\iint_{D} f(x,y) \space dA$
Our aim is to calculate the volume of the given surface.
The domain $D$ can be expressed as follows:
$D=\left\{ (x, y) | 1 \leq x \leq -3y+7 , \ 1 \leq y \leq 2 \right\}
$
Now, $ V =\iint_{D} f(x,y) \ dA \\=\int_{1}^{2} \int_{1}^{-3y+7} xy \ dx \ dy \\ =\int_{1}^{2} [x^2y/2]_{1}^{-(3y-7)} \ dy \\= \int_{1}^{2} [-\dfrac{(3y+7)^2y}{2}-[\dfrac{1}{2} \times y^2] \ dy \\ = \int_{1}^2 \dfrac{9y^3-42 \times y^2+48y dy}{2} \\= \dfrac{1}{2} [\dfrac{9}{4} y^4 -14\times y^3 +24 \times y^2 ]_1^2 \\=\dfrac{31}{8}$
