Answer
$$V = \frac{{351}}{2}$$
Work Step by Step
$$\eqalign{
& z = 3x + 10y + 20;\,\,\,\,\,\,\,0 \leqslant x \leqslant 3,\,\,\,\,\, - 2 \leqslant y \leqslant 1 \cr
& {\text{by the equation just given}}{\text{, the volume is}} \cr
& V = \iint\limits_R {f\left( {x,y} \right)}dxdy \cr
& {\text{replacing the limits for the region }}R{\text{ and }}f\left( {x,y} \right) = z \cr
& V = \int_{ - 2}^1 {\int_0^3 {\left( {3x + 10y + 20} \right)dx} dy} \cr
& V = \int_{ - 2}^1 {\left[ {\int_0^3 {\left( {3x + 10y + 20} \right)dx} } \right]dy} \cr
& {\text{solve the inner integral, treating }}y{\text{ as a constant}} \cr
& = \int_0^3 {\left( {3x + 10y + 20} \right)dx} \cr
& {\text{integrate }} \cr
& = \left[ {\frac{{3{x^2}}}{2} + 10xy + 20x} \right]_0^3 \cr
& {\text{evaluating the limits in the variable }}x \cr
& = \left( {\frac{{3{{\left( 3 \right)}^2}}}{2} + 10\left( 3 \right)y + 20\left( 3 \right)} \right) - \left( {\frac{{3{{\left( 0 \right)}^2}}}{2} + 10\left( 0 \right)y + 20\left( 0 \right)} \right) \cr
& {\text{simplifying}} \cr
& = \left( {\frac{{27}}{2} + 30y + 60} \right) - \left( 0 \right) \cr
& = \frac{{147}}{2} + 30y \cr
& \cr
& V = \int_{ - 2}^1 {\left[ {\int_0^3 {\left( {3x + 10y + 20} \right)dx} } \right]dy} \cr
& V = \int_{ - 2}^1 {\left( {\frac{{147}}{2} + 30y} \right)} dy \cr
& {\text{integrating}} \cr
& V = \left( {\frac{{147}}{2}y + 15{y^2}} \right)_{ - 2}^1 \cr
& V = \left( {\frac{{147}}{2}\left( 1 \right) + 15{{\left( 1 \right)}^2}} \right) - \left( {\frac{{147}}{2}\left( { - 2} \right) + 15{{\left( { - 2} \right)}^2}} \right) \cr
& V = \frac{{177}}{2} + 87 \cr
& V = \frac{{351}}{2} \cr} $$
