Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 514: 92

$\overline x = \frac{2}{{\arcsin \left( {\frac{4}{5}} \right)}}$

$$\eqalign{ & {\text{Let the functions}} \cr & y = \frac{5}{{\sqrt {25 - {x^2}} }},{\text{ }}y = 0,{\text{ }}x = 0{\text{ and }}x = 4 \cr & {\text{Graph shown below}} \cr & \cr & {\text{*The mass of the lamina is }} \cr & m = \rho \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr & m = \rho \int_0^4 {\left( {\frac{5}{{\sqrt {25 - {x^2}} }} - 0} \right)} dx \cr & m = \rho \int_0^4 {\frac{5}{{\sqrt {25 - {x^2}} }}} dx \cr & {\text{Integrate}}{\text{, use }}\int {\frac{1}{{\sqrt {{a^2} - {x^2}} }}dx = \arcsin \left( {\frac{x}{a}} \right) + C,{\text{ then}}} \cr & m = 5\rho \left[ {\arcsin \left( {\frac{x}{5}} \right)} \right]_0^4 \cr & m = 5\rho \left[ {\arcsin \left( {\frac{4}{5}} \right) - \arcsin \left( {\frac{0}{5}} \right)} \right] \cr & m = 5\rho \arcsin \left( {\frac{4}{5}} \right) \cr & \cr & *{\text{The moment about the }}y{\text{ - axis is}} \cr & {M_y} = \rho \int_a^b x \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr & {M_y} = \rho \int_0^4 x \left[ {\frac{5}{{\sqrt {25 - {x^2}} }} - 0} \right]dx \cr & {M_y} = \rho \int_0^4 {\frac{{5x}}{{\sqrt {25 - {x^2}} }}} dx \cr & {\text{Rewrite the integrand}} \cr & {M_y} = \frac{{5\rho }}{2}\int_4^0 {\frac{{ - 2x}}{{\sqrt {25 - {x^2}} }}} dx \cr & {\text{Integrate}} \cr & {M_y} = \frac{{5\rho }}{2}\left[ {2\sqrt {25 - {x^2}} } \right]_4^0 \cr & {M_y} = 5\rho \left[ {\sqrt {25 - {0^2}} - \sqrt {25 - {4^2}} } \right] \cr & {M_y} = 5\rho \left( 2 \right) \cr & {M_y} = 10\rho \cr & \cr & *{\text{The coordinate }}\overline x {\text{ of the center of mass is given by:}} \cr & \overline x = \frac{{{M_y}}}{m} = \frac{{10\rho }}{{5\rho \arcsin \left( {\frac{4}{5}} \right)}} = \frac{2}{{\arcsin \left( {\frac{4}{5}} \right)}} \cr & \overline x = \frac{2}{{\arcsin \left( {\frac{4}{5}} \right)}} \cr} $$