Answer
$$V = \frac{{105}}{2}$$
Work Step by Step
$$\eqalign{
& z = x + 8y + 4;\,\,\,\,\,\,\,0 \leqslant x \leqslant 3,\,\,\,\,\,1 \leqslant y \leqslant 2 \cr
& {\text{by the equation just given}}{\text{, the volume is}} \cr
& V = \iint\limits_R {f\left( {x,y} \right)}dxdy \cr
& {\text{replacing the limits for the region }}R{\text{ and }}f\left( {x,y} \right) = z \cr
& V = \int_1^2 {\int_0^3 {\left( {x + 8y + 4} \right)dx} dy} \cr
& V = \int_1^2 {\left[ {\int_0^3 {\left( {x + 8y + 4} \right)dx} } \right]dy} \cr
& {\text{solve the inner integral by treating }}y{\text{ as a constant}} \cr
& = \int_0^3 {\left( {x + 8y + 4} \right)dx} \cr
& {\text{integrate }} \cr
& = \left[ {\frac{{{x^2}}}{2} + 8xy + 4x} \right]_0^3 \cr
& {\text{evaluating the limits in the variable }}x \cr
& = \left( {\frac{{{{\left( 3 \right)}^2}}}{2} + 8\left( 3 \right)y + 4\left( 3 \right)} \right) - \left( {\frac{{{{\left( 0 \right)}^2}}}{2} + 8\left( 0 \right)y + 4\left( 0 \right)} \right) \cr
& {\text{simplifying}} \cr
& = \left( {\frac{9}{2} + 24y + 12} \right) - \left( 0 \right) \cr
& = \frac{{33}}{2} + 24y \cr
& \cr
& V = \int_1^2 {\left[ {\int_0^3 {\left( {x + 8y + 4} \right)dx} } \right]dy} \cr
& V = \int_1^2 {\left( {\frac{{33}}{2} + 24y} \right)dy} \cr
& {\text{integrating}} \cr
& V = \left( {\frac{{33}}{2}y + 12{y^2}} \right)_1^2 \cr
& V = \left( {\frac{{33}}{2}\left( 2 \right) + 12{{\left( 2 \right)}^2}} \right) - \left( {\frac{{33}}{2}\left( 1 \right) + 12{{\left( 1 \right)}^2}} \right) \cr
& V = 81 - \frac{{57}}{2} \cr
& V = \frac{{105}}{2} \cr} $$
