Answer
$$m = \frac{1}{8}$$
Work Step by Step
$$\eqalign{
& 0 \leqslant x \leqslant 1,{\text{ }}0 \leqslant y \leqslant \sqrt {1 - {x^2}} \cr
& {\text{The mass }}m{\text{ of the lamina is given by}} \cr
& m = \iint\limits_R {\rho \left( {x,y} \right)dA} \cr
& {\text{The density }}\rho \left( {x,y} \right) = xy,{\text{ then}} \cr
& m = \int_0^1 {\int_0^{\sqrt {1 - {x^2}} } {xy} } dydx \cr
& {\text{Integrating with respect to }}y \cr
& m = \int_0^1 {\left[ {\frac{{x{y^2}}}{2}} \right]} _0^{\sqrt {1 - {x^2}} }dx \cr
& m = \int_0^1 {\left[ {\frac{{x{{\left( {\sqrt {1 - {x^2}} } \right)}^2}}}{2} - \frac{{x{{\left( 0 \right)}^2}}}{2}} \right]} dx \cr
& m = \frac{1}{2}\int_0^1 {x\left( {1 - {x^2}} \right)} dx \cr
& m = \frac{1}{2}\int_0^1 {\left( {x - {x^3}} \right)} dx \cr
& {\text{Integrate and evaluate}} \cr
& m = \frac{1}{2}\left[ {\frac{1}{2}{x^2} - \frac{1}{4}{x^4}} \right]_0^1 \cr
& m = \frac{1}{2}\left[ {\frac{1}{2}{{\left( 1 \right)}^2} - \frac{1}{4}{{\left( 1 \right)}^4}} \right] - \frac{1}{2}\left[ 0 \right] \cr
& m = \frac{1}{8} \cr} $$
