Answer
$\left( {\overline x ,\overline y } \right) = \left( {\frac{{18}}{5},\frac{5}{2}} \right)$
Work Step by Step
$$\eqalign{
& y = \sqrt x + 1,{\text{ }}y = \frac{1}{3}x + 1 \cr
& \sqrt x + 1 = \frac{1}{3}x + 1 \cr
& \sqrt x - \frac{1}{3}x = 0 \cr
& \sqrt x \left( {1 - \frac{1}{3}\sqrt x } \right) = 0 \cr
& x = 0,{\text{ }}x = 9 \cr
& \sqrt x + 1 \geqslant \frac{1}{3}x + 1{\text{ on the interval }}\left[ {0,9} \right],{\text{ then}} \cr
& {\text{Let }}f\left( x \right) = \sqrt x + 1{\text{ and }}g\left( x \right) = \frac{1}{3}x + 1,{\text{ on }}\left[ {0,9} \right] \cr
& \cr
& {\text{*The mass of the lamina is }} \cr
& m = \rho \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& m = \rho \int_0^9 {\left[ {\left( {\sqrt x + 1} \right) - \left( {\frac{1}{3}x + 1} \right)} \right]} dx \cr
& m = \rho \int_0^9 {\left( {\sqrt x - \frac{1}{3}x} \right)} dx \cr
& m = \rho \left[ {\frac{2}{3}{x^{3/2}} - \frac{1}{6}{x^2}} \right]_0^9 = \rho \left[ {\frac{2}{3}{{\left( 9 \right)}^{3/2}} - \frac{1}{6}{{\left( 9 \right)}^2}} \right] \cr
& m = \frac{9}{2}\rho \cr
& \cr
& *{\text{The moment about the }}x{\text{ - axis is}} \cr
& {M_x} = \rho \int_a^b {\left[ {\frac{{f\left( x \right) + g\left( x \right)}}{2}} \right]} \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr
& {M_x} = \rho \int_0^9 {\left( {\left[ {\frac{{\left( {\sqrt x + 1} \right) + \left( {\frac{1}{3}x + 1} \right)}}{2}} \right]} \right)} \left[ {\left( {\sqrt x + 1} \right) - \left( {\frac{1}{3}x + 1} \right)} \right]dx \cr
& {M_x} = \rho \int_0^9 {\left( {\frac{{\sqrt x + \frac{1}{3}x + 2}}{2}} \right)} \left( {\sqrt x - \frac{1}{3}x} \right)dx \cr
& {M_x} = \frac{1}{6}\rho \int_0^9 {\left( {\sqrt x + \frac{1}{3}x + 2} \right)} \left( {3\sqrt x - x} \right)dx \cr
& {M_x} = \frac{1}{{18}}\rho \int_0^9 {\left( {3\sqrt x + x + 6} \right)\left( {3\sqrt x - x} \right)} dx \cr
& {M_x} = \frac{1}{{18}}\rho \int_0^9 {\left( { - {x^2} + 3x + 18\sqrt x } \right)} dx \cr
& {M_x} = \frac{1}{{18}}\rho \left[ { - \frac{{{x^3}}}{3} + \frac{{3{x^2}}}{2} + 12{x^{3/2}}} \right]_0^9 \cr
& {M_x} = \frac{1}{{18}}\rho \left[ { - \frac{{{{\left( 9 \right)}^3}}}{3} + \frac{{3{{\left( 9 \right)}^2}}}{2} + 12{{\left( 9 \right)}^{3/2}}} \right] - \frac{1}{{18}}\rho \left[ 0 \right] \cr
& {M_x} = \frac{{45}}{4}\rho \cr
& \cr
& *{\text{The moment about the }}y{\text{ - axis is}} \cr
& {M_y} = \rho \int_a^b x \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr
& {M_y} = \rho \int_0^9 x \left[ {\left( {\sqrt x + 1} \right) - \left( {\frac{1}{3}x + 1} \right)} \right]dx \cr
& {M_y} = \rho \int_0^9 x \left( {\sqrt x + 1 - \frac{1}{3}x - 1} \right)dx \cr
& {M_y} = \rho \int_0^9 x \left( {\sqrt x - \frac{1}{3}x} \right)dx \cr
& {M_y} = \rho \int_0^9 {\left( {{x^{3/2}} - \frac{1}{3}{x^2}} \right)} dx \cr
& {M_y} = \rho \int_0^9 {\left( {\frac{2}{5}{x^{5/2}} - \frac{1}{9}{x^3}} \right)} dx \cr
& {M_y} = \rho \left[ {\frac{2}{5}{{\left( 9 \right)}^{5/2}} - \frac{1}{9}{{\left( 9 \right)}^3}} \right] - \rho \left[ {\frac{2}{5}{{\left( 0 \right)}^{5/2}} - \frac{1}{9}{{\left( 0 \right)}^3}} \right] \cr
& {M_y} = \frac{{81}}{5}\rho \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{{81}}{5}\rho}}{{\frac{9}{2}\rho }} = \frac{{18}}{5} \cr
& \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{{45}}{4}\rho }}{{\frac{9}{2}\rho }} = \frac{5}{2} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{{18}}{5},\frac{5}{2}} \right) \cr
& \cr
& {\text{Summary}} \cr
& m = \frac{9}{2}\rho \cr
& {M_x} = \frac{{18}}{5}\rho \cr
& {M_y} = \frac{5}{2}\rho \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{{18}}{5},\frac{5}{2}} \right) \cr} $$
