Answer
$$\frac{{32}}{3}$$
Work Step by Step
$$\eqalign{
& \int_1^3 {\int_0^2 {{x^2}y} } dxdy \cr
& = \int_1^3 {\left[ {\int_0^2 {{x^2}y} dx} \right]} dy \cr
& = \int_1^3 {y\left[ {\int_0^2 {{x^2}} dx} \right]} dy \cr
& {\text{solve the inner integral}} \cr
& \int_0^2 {{x^2}} dx = \left[ {\frac{{{x^3}}}{3}} \right]_0^2 \cr
& = \frac{1}{3}\left[ {{x^3}} \right]_0^2 \cr
& {\text{evaluating the limits}} \cr
& = \frac{1}{3}\left( {{2^3} - {0^3}} \right) \cr
& = \frac{8}{3} \cr
& \cr
& = \int_1^3 {y\left[ {\int_0^2 {{x^2}} dx} \right]} dy = \int_1^3 {y\left( {\frac{8}{3}} \right)} dy \cr
& = \frac{8}{3}\int_1^3 y dx \cr
& {\text{integrating}} \cr
& = \frac{8}{3}\left( {\frac{{{y^2}}}{2}} \right)_1^3 \cr
& = \frac{4}{3}\left( {{y^2}} \right)_1^3 \cr
& {\text{evaluate}} \cr
& = \frac{4}{3}\left( {{3^2} - {1^2}} \right) \cr
& = \frac{4}{3}\left( 8 \right) \cr
& = \frac{{32}}{3} \cr} $$
