Answer
$$\left( {\overline x ,\overline y } \right) = \left( {\frac{8}{5},0} \right)$$
Work Step by Step
$$\eqalign{
& x = 4 - {y^2},{\text{ }}x = 0 \cr
& {\text{Using the graph shown below}} \cr
& {\text{Let }}h\left( y \right) = 4 - {y^2}{\text{ and }}g\left( y \right) = 0,{\text{ on the interval }}\left[ { - 2,2} \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_{ - 2}^2 {\left( {4 - {y^2}} \right)} dy \cr
& {\text{By symmetry}} \cr
& m = 2\rho \int_0^2 {\left( {4 - {y^2}} \right)} dy \cr
& m = 2\rho \left[ {4y - \frac{1}{3}{y^3}} \right]_0^2 \cr
& m = 2\rho \left[ {4\left( 2 \right) - \frac{1}{3}{{\left( 2 \right)}^3}} \right] - 2\rho \left[ 0 \right] \cr
& m = 2\rho \left( {\frac{{16}}{3}} \right) \cr
& m = \frac{{32}}{3}\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_{ - 2}^2 {\left[ {\frac{{4 - {y^2}}}{2}} \right]} \left[ {4 - {y^2}} \right]dy \cr
& {M_y} = 2\rho \int_0^2 {\left[ {\frac{{4 - {y^2}}}{2}} \right]} \left[ {4 - {y^2}} \right]dy \cr
& {M_y} = \rho \int_0^2 {{{\left( {4 - {y^2}} \right)}^2}} dy \cr
& {M_y} = \rho \int_0^2 {\left( {16 - 8{y^2} + {y^4}} \right)} dy \cr
& {M_y} = \rho \left[ {16y - \frac{8}{3}{y^3} + \frac{1}{5}{y^5}} \right]_0^2 \cr
& {M_y} = \rho \left[ {16\left( 2 \right) - \frac{8}{3}{{\left( 2 \right)}^3} + \frac{1}{5}{{\left( 2 \right)}^5}} \right] - \rho \left[ 0 \right] \cr
& {M_y} = \frac{{256}}{{15}}\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_{ - 2}^2 y \left[ {4 - {y^2}} \right]dy \cr
& {M_x} = \rho \int_{ - 2}^2 {\left( {4y - {y^3}} \right)} dy \cr
& {M_x} = \rho \left( 0 \right) \cr
& {M_x} = 0 \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{{256}}{{15}}\rho }}{{\frac{{32}}{3}\rho }} = \frac{8}{5} \cr
& \overline y = \frac{{{M_x}}}{m} = \frac{0}{{\frac{{32}}{3}\rho }} = 0 \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{8}{5},0} \right) \cr} $$
