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