Answer
$$4$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_0^{4 - {y^2}} {ydx} dy} \cr
& {\text{Integrate with respect to }}x \cr
& = \int_0^2 {\left[ {xy} \right]_0^{4 - {y^2}}dy} \cr
& {\text{Evaluating the limits}} \cr
& = \int_0^2 {\left[ {\left( {4 - {y^2}} \right)y - 0y} \right]dy} \cr
& = \int_0^2 {\left( {4y - {y^3}} \right)dy} \cr
& {\text{Integrate}} \cr
& = \left[ {2{y^2} - \frac{1}{4}{y^4}} \right]_0^2 \cr
& {\text{Evaluating}} \cr
& = \left[ {2{{\left( 2 \right)}^2} - \frac{1}{4}{{\left( 2 \right)}^4}} \right] - \left[ {2{{\left( 0 \right)}^2} - \frac{1}{4}{{\left( 0 \right)}^4}} \right] \cr
& = 4 - 0 \cr
& = 4 \cr} $$
