Answer
$$V = 136$$
Work Step by Step
$$\eqalign{
& V = \int_{ - 1}^3 {\int_0^2 {\left( {24 - 3x - 4y} \right)} } dydx \cr
& {\text{Integrating with respect to }}y \cr
& V = \int_{ - 1}^3 {\left[ {24y - 3xy - 2{y^2}} \right]} _0^2dx \cr
& {\text{Evaluate the limits}} \cr
& V = \int_{ - 1}^3 {\left[ {24\left( 2 \right) - 3x\left( 2 \right) - 2{{\left( 2 \right)}^2}} \right]} dx \cr
& V = \int_{ - 1}^3 {\left( {40 - 6x} \right)} dx \cr
& {\text{Integrating}} \cr
& V = \left[ {40x - 3{x^2}} \right]_{ - 1}^3 \cr
& V = \left[ {40\left( 3 \right) - 3{{\left( 3 \right)}^2}} \right] - \left[ {40\left( { - 1} \right) - 3{{\left( { - 1} \right)}^2}} \right] \cr
& V = 93 + 43 \cr
& V = 136 \cr} $$
