Answer
$$V = 8$$
Work Step by Step
$$\eqalign{
& {\text{The volume of the solid is given by}} \cr
& V = \int_0^1 {\int_0^2 {\left( {6 - x - 2y} \right)} } dxdy \cr
& {\text{Integrating}} \cr
& V = \int_0^1 {\left[ {6x - \frac{1}{2}{x^2} - 2xy} \right]_0^2} dy \cr
& {\text{Evaluate the limits}} \cr
& V = \int_0^1 {\left[ {6\left( 2 \right) - \frac{1}{2}{{\left( 2 \right)}^2} - 2\left( 2 \right)y} \right]} dy \cr
& V = \int_0^1 {\left( {10 - 4y} \right)} dy \cr
& {\text{Integrating}} \cr
& V = \left[ {10y - 2{y^2}} \right]_0^1 \cr
& V = \left[ {10\left( 1 \right) - 2{{\left( 1 \right)}^2}} \right] - \left[ {10\left( 0 \right) - 2{{\left( 0 \right)}^2}} \right] \cr
& V = 8 \cr} $$
