Answer
$$\left( {\overline x ,\overline y } \right) = \left( {\frac{\pi }{8},\frac{L}{2}} \right)$$
Work Step by Step
$$\eqalign{
& y = \sin \frac{{\pi x}}{L},{\text{ }}y = 0,{\text{ }}x = 0,{\text{ }}x = L,{\text{ }}\rho \left( {x,y} \right) = k \cr
& {\text{The region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant y \leqslant \sin \frac{{\pi x}}{L},{\text{ }}0 \leqslant x \leqslant L} \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^L {\int_0^{\sin \frac{{\pi x}}{L}} k } dydx \cr
& m = k\int_0^L {\left[ {\int_0^{\sin \frac{{\pi x}}{L}} {dy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& m = k\int_0^L {\sin \left( {\frac{{\pi x}}{L}} \right)} dx \cr
& {\text{Integrate}} \cr
& m = - \frac{{kL}}{\pi }\left[ {\cos \left( {\frac{{\pi x}}{L}} \right)} \right]_0^L \cr
& m = - \frac{{kL}}{\pi }\left[ {\cos \left( {\frac{{\pi L}}{L}} \right) - \cos \left( {\frac{0}{L}} \right)} \right] \cr
& m = - \frac{{kL}}{\pi }\left( { - 1 - 1} \right) \cr
& m = \frac{{2kL}}{\pi } \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^L {\int_0^{\sin \frac{{\pi x}}{L}} {y\left( k \right)} } dydx \cr
& {M_x} = k\int_0^L {\left[ {\int_0^{\sin \frac{{\pi x}}{L}} {ydy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_x} = k\int_0^L {\left[ {\frac{1}{2}{y^2}} \right]} _0^{\sin \frac{{\pi x}}{L}}dx \cr
& {M_x} = \frac{k}{2}\int_0^L {{{\sin }^2}\left( {\frac{{\pi x}}{L}} \right)} dx \cr
& {M_x} = \frac{k}{2}\int_0^L {\left( {\frac{1}{2} - \frac{1}{2}\cos \left( {\frac{{2\pi x}}{L}} \right)} \right)} dx \cr
& {M_x} = \frac{k}{2}\left[ {\frac{1}{2}x - \frac{L}{{4\pi }}\sin \left( {\frac{{2\pi x}}{L}} \right)} \right]_0^L \cr
& {M_x} = \frac{k}{2}\left[ {\frac{1}{2}L - \frac{L}{{4\pi }}\sin \left( {\frac{{2\pi L}}{L}} \right)} \right] - \frac{k}{2}\left[ {\frac{1}{2}\left( 0 \right) - \frac{0}{{4\pi }}\sin \left( {\frac{0}{L}} \right)} \right] \cr
& {M_x} = \frac{{kL}}{4} \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^L {\int_0^{\sin \frac{{\pi x}}{L}} {x\left( k \right)} } dydx \cr
& {M_y} = k\int_0^L {x\left[ {\int_0^{\sin \frac{{\pi x}}{L}} {dy} } \right]} dx \cr
& {M_y} = k\int_0^L {x\left[ {\sin \frac{{\pi x}}{L}} \right]} dx \cr
& {M_y} = k\int_0^L {x\sin \left( {\frac{{\pi x}}{L}} \right)} dx \cr
& {\text{Integrating by parts}} \cr
& {M_y} = k\left[ {\frac{{{L^2}}}{{{\pi ^2}}}\sin \left( {\frac{{\pi x}}{L}} \right) - \frac{{Lx}}{\pi }\cos \left( {\frac{{\pi x}}{L}} \right)} \right]_0^L \cr
& {M_y} = k\left[ {0 - \frac{{{L^2}}}{\pi }\cos \left( \pi \right) - 0 + 0} \right] \cr
& {M_y} = k\left[ {0 - \frac{{{L^2}}}{\pi }\cos \left( \pi \right) - 0 + 0} \right] \cr
& {M_y} = \frac{{k{L^2}}}{\pi } \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{{kL}}{4}}}{{\frac{{2kL}}{\pi }}} = \left( {\frac{{kL}}{4}} \right)\left( {\frac{\pi }{{2kL}}} \right) = \frac{\pi }{8} \cr
& \frac{{{M_x}}}{m} = \frac{{\frac{{k{L^2}}}{\pi }}}{{\frac{{2kL}}{\pi }}} = \left( {\frac{{k{L^2}}}{\pi }} \right)\left( {\frac{{2kL}}{\pi }} \right) = \frac{L}{2} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{\pi }{8},\frac{L}{2}} \right) \cr} $$
