Answer
$\left( {\overline x ,\overline y } \right) = \left( {\frac{9}{{10}},\frac{3}{2}} \right)$
Work Step by Step
$$\eqalign{
& x = 3y - {y^2},{\text{ }}x = 0 \cr
& {\text{Let }}h\left( y \right) = 3y - {y^2}{\text{ and }}g\left( y \right) = 0,{\text{ on the interval }}\left[ {0,3} \right] \cr
& \cr
& {\text{*The mass of the lamina is }} \cr
& m = \rho \int_c^d {\left[ {h\left( y \right) - g\left( y \right)} \right]} dy \cr
& m = \rho \int_0^3 {\left( {3y - {y^2}} \right)} dy \cr
& m = \rho \left[ {\frac{3}{2}{y^2} - \frac{1}{3}{y^3}} \right]_0^3 \cr
& m = \rho \left[ {\frac{3}{2}{{\left( 3 \right)}^2} - \frac{1}{3}{{\left( 3 \right)}^3}} \right] - \rho \left[ 0 \right] \cr
& m = \frac{9}{2}\rho \cr
& \cr
& *{\text{The moment about the }}y{\text{ - axis is}} \cr
& {M_y} = \rho \int_c^d {\left[ {\frac{{h\left( y \right) + g\left( y \right)}}{2}} \right]} \left[ {h\left( y \right) - g\left( y \right)} \right]dy \cr
& {M_y} = \rho \int_0^3 {\left[ {\frac{{3y - {y^2}}}{2}} \right]} \left[ {3y - {y^2}} \right]dy \cr
& {M_y} = \frac{1}{2}\rho \int_0^3 {{{\left( {3y - {y^2}} \right)}^2}} dy \cr
& {M_y} = \frac{1}{2}\rho \int_0^3 {\left( {9{y^2} - 6{y^3} + {y^4}} \right)} dy \cr
& {M_y} = \frac{1}{2}\rho \left[ {3{y^3} - \frac{3}{2}{y^4} + \frac{1}{5}{y^5}} \right]_0^3 \cr
& {M_y} = \frac{1}{2}\rho \left[ {3{{\left( 3 \right)}^3} - \frac{3}{2}{{\left( 3 \right)}^4} + \frac{1}{5}{{\left( 3 \right)}^5}} \right] - \frac{1}{2}\rho \left[ 0 \right] \cr
& {M_y} = \frac{{81}}{{20}}\rho \cr
& \cr
& *{\text{The moment about the }}x{\text{ - axis is}} \cr
& {M_x} = \rho \int_c^d y \left[ {h\left( y \right) - g\left( y \right)} \right]dy \cr
& {M_x} = \rho \int_0^3 y \left[ {3y - {y^2}} \right]dy \cr
& {M_x} = \rho \int_0^3 {\left( {3{y^2} - {y^3}} \right)} dy \cr
& {M_x} = \rho \left[ {{y^3} - \frac{1}{4}{y^4}} \right]_0^3 \cr
& {M_x} = \rho \left[ {{{\left( 3 \right)}^3} - \frac{1}{4}{{\left( 3 \right)}^4}} \right] - \rho \left[ 0 \right] \cr
& {M_x} = \frac{{27}}{4}\rho \cr
& \cr
& *{\text{The center of mass }}\left( {\overline x ,\overline y } \right){\text{ is given by:}} \cr
& \overline x = \frac{{{M_y}}}{m} = \frac{{\frac{{81}}{{20}}\rho }}{{\frac{9}{2}\rho }} = \frac{9}{{10}} \cr
& \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{{27}}{4}\rho }}{{\frac{9}{2}\rho }} = \frac{3}{2} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{9}{{10}},\frac{3}{2}} \right) \cr} $$
