Answer
$$m = 4$$
Work Step by Step
$$\eqalign{
& 0 \leqslant x \leqslant 2,{\text{ }}0 \leqslant y \leqslant 2 \cr
& {\text{The mass }}m{\text{ of the lamina is given by}} \cr
& m = \iint\limits_R {\rho \left( {x,y} \right)dA} \cr
& {\text{The density }}\rho \left( {x,y} \right) = xy,{\text{ then}} \cr
& m = \int_0^2 {\int_0^2 {xy} } dydx \cr
& {\text{Integrating with respect to }}y \cr
& m = \int_0^2 {\left[ {\frac{{x{y^2}}}{2}} \right]} _0^2dx \cr
& m = \int_0^2 {\left[ {\frac{{x{{\left( 2 \right)}^2}}}{2} - \frac{{x{{\left( 0 \right)}^2}}}{2}} \right]} dx \cr
& m = \int_0^2 {2x} dx \cr
& {\text{Integrate and evaluate}} \cr
& m = \left[ {{x^2}} \right]_0^2 \cr
& m = {\left( 2 \right)^2} - {\left( 0 \right)^2} \cr
& m = 4 \cr} $$
