Answer
$$\left( {\overline x ,\overline y } \right) = \left( {\frac{6}{7},2} \right)$$
Work Step by Step
$$\eqalign{
& y = {x^2},{\text{ }}y = 0,{\text{ }}x = 2,{\text{ }}\rho \left( {x,y} \right) = kxy \cr
& {\text{The region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant y \leqslant {x^2},{\text{ }}0 \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_0^2 {\int_0^{{x^2}} {kxy} } dydx \cr
& m = k\int_0^2 {x\left[ {\int_0^{{x^2}} y } \right]} dydx \cr
& {\text{Integrating with respect to }}y \cr
& m = k\int_0^2 {x\left[ {\frac{1}{2}{y^2}} \right]_0^{{x^2}}} dx \cr
& m = \frac{k}{2}\int_0^2 {x\left[ {{{\left( {{x^2}} \right)}^2}} \right]} dx \cr
& m = \frac{k}{2}\int_0^2 {{x^5}} dx \cr
& {\text{Integrate}} \cr
& m = \frac{k}{{12}}\left[ {{x^6}} \right]_0^2 \cr
& m = \frac{k}{{12}}\left( {64} \right) \cr
& m = \frac{{16}}{3}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_0^2 {\int_0^{{x^2}} {y\left( {kxy} \right)} } dydx \cr
& {M_x} = k\int_0^2 {\left[ {\int_0^{{x^2}} {x{y^2}dy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_x} = k\int_0^2 {x\left[ {\frac{1}{3}{y^3}} \right]_0^{{x^2}}} dx \cr
& {M_x} = \frac{k}{3}\int_0^2 {x\left[ {{{\left( {{x^2}} \right)}^3} - 0} \right]} dx \cr
& {M_x} = \frac{k}{3}\int_0^2 {{x^7}} dx \cr
& {\text{Integrate}} \cr
& {M_x} = \frac{k}{{24}}\left[ {{x^8}} \right]_0^2 \cr
& {M_x} = \frac{{32}}{3}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_0^2 {\int_0^{{x^2}} {x\left( {kxy} \right)} } dydx \cr
& {M_y} = k\int_0^2 {{x^2}\left[ {\int_0^{{x^2}} {ydy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_y} = k\int_0^2 {{x^2}\left[ {\frac{1}{2}{y^2}} \right]_0^{{x^2}}} dx \cr
& {M_y} = \frac{k}{2}\int_0^2 {{x^2}\left[ {\frac{1}{2}{{\left( {{x^2}} \right)}^2} - 0} \right]} dx \cr
& {M_y} = \frac{k}{4}\int_0^2 {{x^6}} dx \cr
& {\text{Integrate}} \cr
& {M_y} = \frac{k}{{28}}\left[ {{x^7}} \right]_0^2 \cr
& {M_y} = \frac{{32}}{7}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{{\left( {32/7} \right)k}}{{16k/3}} = \frac{6}{7} \cr
& \frac{{{M_x}}}{m} = \frac{{\left( {32/3} \right)k}}{{16k/3}} = 2 \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{6}{7},2} \right) \cr} $$
