Answer
$$\left( {\overline x ,\overline y } \right) = \left( {\frac{2}{3},\frac{8}{{15}}} \right)$$
Work Step by Step
$$\eqalign{
& y = \sqrt x ,{\text{ }}y = 0,{\text{ }}x = 1,{\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 \sqrt x ,{\text{ }}0 \leqslant x \leqslant 1} \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_0^1 {\int_0^{\sqrt x } {ky} } dydx \cr
& {\text{Integrating with respect to }}y \cr
& m = \int_0^1 {\left[ {\frac{{k{y^2}}}{2}} \right]} _0^{\sqrt x }dx \cr
& m = \frac{k}{2}\int_0^1 {\left[ {{{\left( {\sqrt x } \right)}^2} - {{\left( 0 \right)}^2}} \right]} dx \cr
& m = \frac{k}{2}\int_0^1 {\left( x \right)} dx \cr
& {\text{Integrate}} \cr
& m = \frac{k}{2}\left[ {\frac{{{x^2}}}{2}} \right]_0^1 \cr
& m = \frac{k}{2}\left[ {\frac{1}{2} - 0} \right] \cr
& m = \frac{k}{4} \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_0^1 {\int_0^{\sqrt x } {y\left( {ky} \right)} } dydx \cr
& {M_x} = k\int_0^1 {\left[ {\int_0^{\sqrt x } {{y^2}dy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_x} = k\int_0^1 {\left[ {\frac{1}{3}{y^3}} \right]_0^{\sqrt x }} dx \cr
& {M_x} = \frac{k}{3}\int_0^1 {\left[ {{{\left( {\sqrt x } \right)}^3} - 0} \right]} dx \cr
& {M_x} = \frac{k}{3}\int_0^1 {{x^{3/2}}} dx \cr
& {\text{Integrate}} \cr
& {M_x} = \frac{k}{3}\left[ {\frac{2}{5}{x^{5/2}}} \right]_0^1 \cr
& {M_x} = \frac{{2k}}{{15}} \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_0^1 {\int_0^{\sqrt x } {x\left( {ky} \right)} } dydx \cr
& {M_y} = k\int_0^1 {\left[ {\int_0^{\sqrt x } {xydy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_y} = k\int_0^1 {\left[ {\frac{x}{2}{y^2}} \right]_0^{\sqrt x }} dx \cr
& {M_y} = \frac{k}{2}\int_0^1 {\left[ {x\left( {\sqrt x } \right) - 0} \right]} dx \cr
& {M_y} = \frac{k}{2}\int_0^1 {{x^2}} dx \cr
& {\text{Integrate}} \cr
& {M_y} = \frac{k}{2}\left[ {\frac{1}{3}{x^3}} \right]_0^1 \cr
& {M_y} = \frac{k}{6} \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{{k/6}}{{k/4}} = \frac{2}{3} \cr
& \frac{{{M_x}}}{m} = \frac{{2k/15}}{{k/4}} = \frac{8}{{15}} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{2}{3},\frac{8}{{15}}} \right) \cr} $$
