Answer
$$\left( {\overline x ,\overline y } \right) = \left( {\frac{{36}}{7},0} \right)$$
Work Step by Step
$$\eqalign{
& x = 9 - {y^2},{\text{ }}x = 0,{\text{ }}\rho \left( {x,y} \right) = kx \cr
& {\text{The region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant x \leqslant 9 - {y^2},{\text{ }} - 3 \leqslant y \leqslant 3} \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_{ - 3}^3 {\int_0^{9 - {y^2}} {kx} } dxdy \cr
& m = k\int_{ - 3}^3 {\left[ {\int_0^{9 - {y^2}} x dx} \right]} dy \cr
& {\text{Integrating with respect to }}x \cr
& m = k\int_{ - 3}^3 {\left[ {\frac{1}{2}{x^2}} \right]_0^{9 - {y^2}}} dy \cr
& m = \frac{k}{2}\int_{ - 3}^3 {{{\left( {9 - {y^2}} \right)}^2}} dy \cr
& {\text{By symmetry}} \cr
& m = k\int_0^3 {{{\left( {9 - {y^2}} \right)}^2}} dy \cr
& m = k\int_0^3 {\left( {81 - 18{y^2} + {y^4}} \right)} dx \cr
& {\text{Integrate}} \cr
& m = k\left[ {81y - 6{y^3} + \frac{1}{5}{y^5}} \right]_0^3 \cr
& m = k\left[ {81\left( 3 \right) - 6{{\left( 3 \right)}^3} + \frac{1}{5}{{\left( 3 \right)}^5}} \right] - k\left[ 0 \right] \cr
& m = \frac{{648}}{5}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_{ - 3}^3 {\int_0^{9 - {y^2}} {y\left( {kx} \right)} } dxdy \cr
& {M_x} = k\int_{ - 2}^2 {y\left[ {\int_0^{9 - {y^2}} {xdx} } \right]} dy \cr
& {\text{Integrating with respect to }}x \cr
& {M_x} = k\int_{ - 3}^3 {y\left[ {\frac{1}{2}{x^2}} \right]} _0^{9 - {y^2}}dy \cr
& {M_x} = \frac{k}{2}\int_{ - 3}^3 {y{{\left( {9 - {y^2}} \right)}^2}} dx \cr
& {M_x} = - \frac{k}{4}\int_{ - 3}^3 {{{\left( {9 - {y^2}} \right)}^2}\left( { - 2y} \right)} dx \cr
& {M_x} = - \frac{k}{4}\left[ {\frac{{{{\left( {9 - {y^2}} \right)}^3}}}{3}} \right]_{ - 3}^3 \cr
& {M_x} = - \frac{k}{4}\left[ {\frac{{{{\left( {9 - {3^2}} \right)}^3}}}{3}} \right] + \frac{k}{4}\left[ {\frac{{{{\left( {9 - {{\left( { - 3} \right)}^2}} \right)}^3}}}{3}} \right] \cr
& {M_x} = - \frac{k}{4}\left[ {\frac{{{{\left( {9 - {3^2}} \right)}^3}}}{3}} \right] + \frac{k}{4}\left[ {\frac{{{{\left( {9 - {{\left( { - 3} \right)}^2}} \right)}^3}}}{3}} \right] \cr
& {M_x} = 0 \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_{ - 3}^3 {\int_0^{9 - {y^2}} {x\left( {kx} \right)} } dxdy \cr
& {M_y} = k\int_{ - 3}^3 {\left[ {\frac{1}{3}{x^3}} \right]_0^{9 - {y^2}}} dy \cr
& {M_y} = \frac{k}{3}\int_{ - 3}^3 {{{\left( {9 - {y^2}} \right)}^3}} dy \cr
& {\text{By symmetry}} \cr
& {M_y} = \frac{{2k}}{3}\int_0^3 {\left( {729 - 243{y^2} + 27{y^4} - {y^6}} \right)} dy \cr
& {M_y} = \frac{{2k}}{3}\left[ {729y - 81{y^3} + \frac{{27}}{5}{y^5} - \frac{{{y^7}}}{7}} \right]_0^3 \cr
& {M_y} = \frac{{2k}}{3}\left[ {729\left( 3 \right) - 81{{\left( 3 \right)}^3} + \frac{{27}}{5}{{\left( 3 \right)}^5} - \frac{{{{\left( 3 \right)}^7}}}{7}} \right] - 0 \cr
& {M_y} = \frac{{2k}}{3}\left( {\frac{{34992}}{{35}}} \right) \cr
& {M_y} = \frac{{23328}}{{35}}k \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{{\frac{{23328}}{{35}}k}}{{\frac{{648}}{5}k}} = 0 \cr
& \frac{{{M_x}}}{m} = \frac{0}{{\frac{{648}}{5}k}} = 0 \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{{36}}{7},0} \right) \cr} $$
