Answer
$$\frac{{16}}{5}$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_0^y {\int_0^{xy} {ydz} } dx} dy \cr
& \int_0^2 {\int_0^y {\left[ {\int_0^{xy} {ydz} } \right]} dx} dy \cr
& {\text{Integrate with respect to }}z \cr
& \int_0^{xy} {ydz} = \left[ {yz} \right]_0^{xy} \cr
& \left[ {yz} \right]_0^{xy} = y\left( {xy} \right) - y\left( 0 \right) = x{y^2} \cr
& \int_0^2 {\int_0^y {\left[ {\int_0^{xy} {ydz} } \right]} dx} dy = \int_0^2 {\int_0^y {x{y^2}} dx} dy \cr
& = \int_0^2 {\left[ {\int_0^y {x{y^2}} dx} \right]} dy \cr
& {\text{Integrate with respect to }}x \cr
& \int_0^y {x{y^2}} dx = \left[ {\frac{1}{2}{x^2}{y^2}} \right]_0^y \cr
& = \frac{1}{2}{\left( y \right)^2}{y^2} - \frac{1}{2}{\left( 0 \right)^2}{y^2} \cr
& = \frac{1}{2}{y^4} \cr
& \int_0^2 {\left[ {\int_0^y {x{y^2}} dx} \right]} dy = \int_0^2 {\frac{1}{2}{y^4}} dy \cr
& {\text{Integrate}} \cr
& {\text{ = }}\left[ {\frac{1}{{10}}{y^5}} \right]_0^2 \cr
& = \frac{1}{{10}}{\left( 2 \right)^5} - \frac{1}{{10}}{\left( 0 \right)^5} \cr
& = \frac{{16}}{5} \cr} $$
