Answer
$$\left( {\overline x ,\overline y } \right) = \left( {\frac{8}{5},\frac{1}{2}} \right)$$
Work Step by Step
$$\eqalign{
& x = y + 2,{\text{ }}x = {y^2} \cr
& {\text{Using the graph shown below}} \cr
& {\text{Let }}h\left( y \right) = y + 2{\text{ and }}g\left( y \right) = {y^2},{\text{ on the interval }}\left[ { - 1,2} \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_{ - 1}^2 {\left[ {\left( {y + 2} \right) - \left( {{y^2}} \right)} \right]} dy \cr
& m = \rho \int_{ - 1}^2 {\left( {y + 2 - {y^2}} \right)} dy \cr
& m = \rho \left[ {\frac{1}{2}{y^2} + 2y - \frac{1}{3}{y^3}} \right]_{ - 1}^2 \cr
& m = \rho \left[ {\frac{1}{2}{{\left( 2 \right)}^2} + 2\left( 2 \right) - \frac{1}{3}{{\left( 2 \right)}^3}} \right] - \rho \left[ {\frac{1}{2}{{\left( { - 1} \right)}^2} + 2\left( { - 1} \right) - \frac{1}{3}{{\left( { - 1} \right)}^3}} \right] \cr
& m = \frac{{10}}{3}\rho + \frac{7}{6}\rho \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_{ - 1}^2 {\left[ {\frac{{\left( {y + 2} \right) + \left( {{y^2}} \right)}}{2}} \right]} \left[ {\left( {y + 2} \right) - \left( {{y^2}} \right)} \right]dy \cr
& {M_y} = \frac{1}{2}\rho \int_{ - 1}^2 {\left[ {{{\left( {y + 2} \right)}^2} - {y^4}} \right]} dy \cr
& {M_y} = \frac{1}{2}\rho \int_{ - 1}^2 {\left( { - {y^4} + {y^2} + 4y + 4} \right)} dy \cr
& {M_y} = \frac{1}{2}\rho \left[ { - \frac{1}{5}{y^5} + \frac{1}{3}{y^3} + 2{y^2} + 4y} \right]_{ - 1}^2 \cr
& {\text{Evaluating the limits}} \cr
& {M_y} = \frac{1}{2}\rho \left( {\frac{{184}}{{15}}} \right) - \frac{1}{2}\rho \left( { - \frac{{32}}{{15}}} \right) \cr
& {M_y} = \frac{{36}}{5}\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_{ - 1}^2 y \left[ {\left( {y + 2} \right) - \left( {{y^2}} \right)} \right]dy \cr
& {M_x} = \rho \int_{ - 1}^2 y \left( {y + 2 - {y^2}} \right)dy \cr
& {M_x} = \rho \int_{ - 1}^2 {\left( {{y^2} + 2y - {y^3}} \right)} dy \cr
& {M_x} = \rho \left[ {\frac{1}{3}{y^3} + {y^2} - \frac{1}{4}{y^4}} \right]_{ - 1}^2 \cr
& {\text{Evaluating the limits}} \cr
& {M_x} = \rho \left( {\frac{8}{3}} \right) - \rho \left( {\frac{5}{{12}}} \right) \cr
& {M_x} = \frac{9}{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{{36}}{5}\rho }}{{\frac{9}{2}\rho }} = \frac{8}{5} \cr
& \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{9}{4}\rho }}{{\frac{9}{2}\rho }} = \frac{1}{2} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{8}{5},\frac{1}{2}} \right) \cr} $$
