Answer
$$\left( {\overline x ,\overline y } \right) = \left( {0,\frac{{16}}{7}} \right)$$
Work Step by Step
$$\eqalign{
& y = 4 - {x^2},{\text{ }}y = 0,{\text{ }}\rho \left( {x,y} \right) = ky \cr
& {\text{The region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant y \leqslant 4 - {x^2},{\text{ }} - 2 \leqslant x \leqslant 2} \right\} \cr
& {\text{The mass }}m{\text{ of the lamina is given by}} \cr
& m = \iint\limits_R {\rho \left( {x,y} \right)dA} \cr
& m = \int_{ - 2}^2 {\int_0^{4 - {x^2}} {ky} } dydx \cr
& m = k\int_{ - 2}^2 {\left[ {\int_0^{4 - {x^2}} {ydy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& m = k\int_{ - 2}^2 {\left[ {\frac{1}{2}{y^2}} \right]_0^{4 - {x^2}}} dx \cr
& m = \frac{k}{2}\int_{ - 2}^2 {{{\left( {4 - {x^2}} \right)}^2}} dx \cr
& {\text{By symmetry}} \cr
& m = k\int_0^2 {{{\left( {4 - {x^2}} \right)}^2}} dx \cr
& m = k\int_0^2 {\left( {16 - 8{x^2} + {x^4}} \right)} dx \cr
& {\text{Integrate}} \cr
& m = k\left[ {16x - \frac{8}{3}{x^3} + \frac{1}{5}{x^5}} \right]_0^2 \cr
& m = k\left[ {16\left( 2 \right) - \frac{8}{3}{{\left( 2 \right)}^3} + \frac{1}{5}{{\left( 2 \right)}^5}} \right] - k\left[ 0 \right] \cr
& m = \frac{{256}}{{15}}k \cr
& \cr
& {\text{*The moment of mass with respect to the }}x{\text{ axis is:}} \cr
& {M_x} = \iint\limits_R {y\rho \left( {x,y} \right)dA},{\text{ then}} \cr
& {M_x} = \int_{ - 2}^2 {\int_0^{4 - {x^2}} {y\left( {ky} \right)} } dydx \cr
& {M_x} = k\int_{ - 2}^2 {\left[ {\int_0^{4 - {x^2}} {{y^2}dy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_x} = k\int_{ - 2}^2 {\left[ {\frac{1}{3}{y^3}} \right]} _0^{4 - {x^2}}dx \cr
& {M_x} = \frac{k}{3}\int_{ - 2}^2 {{{\left( {4 - {x^2}} \right)}^3}} dx \cr
& {M_x} = \frac{k}{3}\int_{ - 2}^2 {\left( {64 - 48{x^2} + 12{x^4} - {x^6}} \right)} dx \cr
& {\text{By symmetry}} \cr
& {M_x} = \frac{{2k}}{3}\int_0^2 {\left( {64 - 48{x^2} + 12{x^4} - {x^6}} \right)} dx \cr
& {M_x} = \frac{{2k}}{3}\left[ {64x - 16{x^3} + \frac{{12}}{5}{x^5} - \frac{1}{7}{x^7}} \right]_0^2 \cr
& {M_x} = \frac{{2k}}{3}\left[ {\frac{{2048}}{{35}}} \right] - \frac{{2k}}{3}\left[ 0 \right] \cr
& {M_x} = \frac{{4096}}{{105}}k \cr
& \cr
& {\text{*The moment of mass with respect to the }}y{\text{ axis is:}} \cr
& {M_y} = \iint\limits_R {x\rho \left( {x,y} \right)dA},{\text{ then}} \cr
& {M_y} = \int_{ - 2}^2 {\int_0^{4 - {x^2}} {x\left( {ky} \right)} } dydx \cr
& {\text{By symmetry}} \cr
& {M_y} = 0 \cr
& \cr
& {\text{The center of mass is}} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{{{M_y}}}{m},\frac{{{M_x}}}{m}} \right) \cr
& \frac{{{M_y}}}{m} = \frac{0}{{\frac{{256}}{{15}}k}} = 0 \cr
& \frac{{{M_x}}}{m} = \frac{{\frac{{4096}}{{105}}k}}{{\frac{{256}}{{15}}k}} = \frac{{16}}{7} \cr
& \left( {\overline x ,\overline y } \right) = \left( {0,\frac{{16}}{7}} \right) \cr} $$
