Answer
$\left( {\frac{\pi }{8},1} \right)$
Work Step by Step
$$\eqalign{
& {\text{Let the functions }}y = \arcsin x,{\text{ }}x = 0,{\text{ }}y = \frac{\pi }{2} \cr
& {\text{Graph shown below}} \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_0^1 {\left( {\frac{\pi }{2} - \arcsin x} \right)} dx \cr
& \cr
& *{\text{Integrating }}\int {\arcsin x} dx{\text{ by parts}} \cr
& {\text{Let }}u = \arcsin x,{\text{ }}dv = dx \cr
& du = \frac{1}{{\sqrt {1 - {x^2}} }}dx{\text{ and }}v = x,{\text{ then}} \cr
& \int {\arcsin x} dx = x\arcsin x - \int {\frac{x}{{\sqrt {1 - {x^2}} }}} dx \cr
& = x\arcsin x + \sqrt {1 - {x^2}} + C \cr
& \cr
& m = \rho \int_0^1 {\left( {\frac{\pi }{2} - \arcsin x} \right)} dx \cr
& m = \rho \left[ {\frac{\pi }{2}x - x\arcsin x - \sqrt {1 - {x^2}} } \right]_0^1 \cr
& {\text{Evaluating}} \cr
& m = \rho \left[ {\frac{\pi }{2}\left( 1 \right) - \left( 1 \right)\arcsin \left( 1 \right) - \sqrt {1 - {{\left( 1 \right)}^2}} } \right] - \rho \left[ {0 - \sqrt {1 - 0} } \right] \cr
& m = \rho \left( {\frac{\pi }{2} - \frac{\pi }{2} - 0} \right) - \rho \left( {0 - 1} \right) \cr
& m = \rho \cr
& \cr
& *{\text{The moment about the }}x{\text{ - axis is}} \cr
& {M_x} = \rho \int_a^b {\left[ {\frac{{\frac{\pi }{2} + \arcsin x}}{2}} \right]} \left[ {\frac{\pi }{2} - \arcsin x} \right]dx \cr
& {M_x} = \frac{\rho }{2}\int_0^1 {\left( {\frac{{{\pi ^2}}}{4} - {{\left( {\arcsin x} \right)}^2}} \right)} dx \cr
& {M_x} = \frac{{\rho {\pi ^2}}}{8}\int_0^1 {dx} - \frac{\rho }{2}\int_0^1 {{{\left( {\arcsin x} \right)}^2}} dx{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& *{\text{Integrating }}\int {{{\left( {\arcsin x} \right)}^2}} dx{\text{ by parts}} \cr
& {\text{Let }}u = {\left( {\arcsin x} \right)^2} \to du = 2\arcsin x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)dx \cr
& dv = dx \to v = x \cr
& \int {{{\left( {\arcsin x} \right)}^2}} dx = x{\left( {\arcsin x} \right)^2} - \int {2x\arcsin x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)dx} \cr
& \int {{{\left( {\arcsin x} \right)}^2}} dx = x{\left( {\arcsin x} \right)^2} - 2\underbrace {\int {\arcsin x\left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right)dx} }_{{\text{Integrate by parts}}} \cr
& u = \arcsin x \to du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr
& dv = \frac{x}{{\sqrt {1 - {x^2}} }} \to v = - \sqrt {1 - {x^2}} \cr
& \int {\arcsin x\left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right)dx} = - \sqrt {1 - {x^2}} \arcsin x + \int {dx} \cr
& \int {\arcsin x\left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right)dx} = - \sqrt {1 - {x^2}} \arcsin x + x + C \cr
& \int {{{\left( {\arcsin x} \right)}^2}} dx = x{\left( {\arcsin x} \right)^2} - 2\left( { - \sqrt {1 - {x^2}} \arcsin x + x} \right) + C \cr
& \int {{{\left( {\arcsin x} \right)}^2}} dx = x{\left( {\arcsin x} \right)^2} + 2\sqrt {1 - {x^2}} \arcsin x - 2x + C \cr
& {\text{Substituting the previous result into }}\left( {\bf{1}} \right) \cr
& {M_x} = \frac{{\rho {\pi ^2}}}{8}\left[ x \right]_0^1 - \frac{\rho }{2}\left[ {x{{\left( {\arcsin x} \right)}^2} + 2\sqrt {1 - {x^2}} \arcsin x - 2x} \right]_0^1 \cr
& {\text{Evaluating we obtain}} \cr
& {M_x} = \frac{{\rho {\pi ^2}}}{8} - \frac{\rho }{2}\left[ {\left( 1 \right){{\left( {\frac{\pi }{2}} \right)}^2} + 2\sqrt 0 \arcsin \left( 1 \right) - 2} \right] - \frac{\rho }{2}\left[ 0 \right] \cr
& {M_x} = \frac{{\rho {\pi ^2}}}{8} - \frac{\rho }{2}\left( {\frac{{{\pi ^2}}}{4} - 2} \right) \cr
& {M_x} = \frac{{\rho {\pi ^2}}}{8} - \frac{{\rho {\pi ^2}}}{8} + \rho \cr
& {M_x} = \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_0^1 x \left[ {\frac{\pi }{2} - \arcsin x} \right]dx \cr
& {M_y} = \frac{\pi }{2}\rho \int_0^1 x dx - \rho \int_0^1 {x\arcsin x} dx{\text{ }}\left( {\bf{2}} \right) \cr
& \cr
& *{\text{Integrating }}\int {x\arcsin x} dx{\text{ by parts}} \cr
& {\text{Let }}u = \arcsin x \to du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr
& dv = xdx \to v = \frac{{{x^2}}}{2} \cr
& \int {x\arcsin x} dx = \frac{{{x^2}}}{2}\arcsin x - \int {\frac{{{x^2}}}{{2\sqrt {1 - {x^2}} }}} dx \cr
& {\text{Integrating by a graphing calculator or computer we obtain}} \cr
& \int {x\arcsin x} dx = \frac{{{x^2}}}{2}\arcsin x - \frac{1}{4}\arcsin x + \frac{1}{4}x\sqrt {1 - {x^2}} + C \cr
& {\text{Substituting the previous result into }}\left( {\bf{2}} \right) \cr
& {M_y} = \frac{\pi }{2}\rho \left[ {\frac{{{x^2}}}{2}} \right]_0^1 - \rho \left[ {\frac{{{x^2}}}{2}\arcsin x - \frac{1}{4}\arcsin x + \frac{1}{4}x\sqrt {1 - {x^2}} } \right]_0^1 \cr
& {\text{Evaluating we obtain}} \cr
& {M_y} = \frac{\pi }{2}\rho \left( {\frac{1}{2}} \right) - \rho \left( {\frac{\pi }{4} - \frac{1}{4}\left( {\frac{\pi }{2}} \right)} \right) \cr
& {M_y} = \frac{\pi }{4}\rho - \frac{\pi }{4}\rho + \frac{\pi }{8}\rho \cr
& {M_y} = \frac{\pi }{8}\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{\pi }{8}\rho }}{\rho } = \frac{\pi }{8} \cr
& \overline y = \frac{{{M_x}}}{m} = \frac{\rho }{\rho } = 1 \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{\pi }{8},1} \right) \cr} $$
