Answer
$$m = \frac{{243}}{4}$$
Work Step by Step
$$\eqalign{
& 0 \leqslant x \leqslant 3,{\text{ }}0 \leqslant y \leqslant 9 - {x^2} \cr
& {\text{The mass }}m{\text{ of the lamina is given by}} \cr
& m = \iint\limits_R {\rho \left( {x,y} \right)dA} \cr
& {\text{The density }}\rho \left( {x,y} \right) = xy,{\text{ then}} \cr
& m = \int_0^3 {\int_0^{9 - {x^2}} {xy} } dydx \cr
& {\text{Integrating with respect to }}y \cr
& m = \int_0^3 {\left[ {\frac{{x{y^2}}}{2}} \right]} _0^{9 - {x^2}}dx \cr
& m = \int_0^3 {\left[ {\frac{{x{{\left( {9 - {x^2}} \right)}^2}}}{2} - \frac{{x{{\left( 0 \right)}^2}}}{2}} \right]} dx \cr
& m = \frac{1}{2}\int_0^3 {x\left( {81 - 18{x^2} + {x^4}} \right)} dx \cr
& m = \frac{1}{2}\int_0^3 {\left( {81x - 18{x^3} + {x^5}} \right)} dx \cr
& {\text{Integrate and evaluate}} \cr
& m = \frac{1}{2}\left[ {\frac{{81{x^2}}}{2} - \frac{{18{x^4}}}{4} + \frac{{{x^6}}}{6}} \right]_0^3 \cr
& m = \frac{1}{2}\left[ {\frac{{81{{\left( 3 \right)}^2}}}{2} - \frac{{18{{\left( 3 \right)}^4}}}{4} + \frac{{{{\left( 3 \right)}^6}}}{6}} \right] - \frac{1}{2}\left[ 0 \right] \cr
& m = \frac{{243}}{4} \cr} $$
