Answer
$$44$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_0^{3y} {\left( {{x^2} + y} \right)} dxdy} \cr
& = \int_0^2 {\left[ {\int_0^{3y} {\left( {{x^2} + y} \right)} dx} \right]dy} \cr
& {\text{solve the inner integral by treating }}y{\text{ as a constant}} \cr
& = \int_0^{3y} {\left( {{x^2} + y} \right)} dx \cr
& = \left( {\frac{{{x^3}}}{3} + xy} \right)_0^{3y} \cr
& {\text{evaluating the limits in the variable }}x \cr
& = \left( {\frac{{{{\left( {3y} \right)}^3}}}{3} + \left( {3y} \right)y} \right) - \left( {\frac{{{{\left( 0 \right)}^3}}}{3} + \left( 0 \right)y} \right) \cr
& = \frac{{27{y^3}}}{3} + 3{y^2} \cr
& = 9{y^3} + 3{y^2} \cr
& \cr
& \int_0^2 {\left[ {\int_0^{3y} {\left( {{x^2} + y} \right)} dx} \right]dy} = \int_0^2 {\left( {9{y^3} + 3{y^2}} \right)dy} \cr
& {\text{integrating by the power rule}} \cr
& = \left( {\frac{{9{y^4}}}{4} + {y^3}} \right)_0^2 \cr
& {\text{evaluating}} \cr
& = \left( {\frac{{9{{\left( 2 \right)}^4}}}{4} + {{\left( 2 \right)}^3}} \right) - \left( {\frac{{9{{\left( 0 \right)}^4}}}{4} + {{\left( 0 \right)}^3}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{{9\left( {16} \right)}}{4} + 8 \cr
& = 44 \cr} $$
