Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - Review Exercises - Page 580: 69

$$\left( {\overline x ,\overline y } \right) = \left( {0,\frac{4}{{3\pi }}} \right)$$

$$\eqalign{ & y = \sqrt {1 - {x^2}} ,{\text{ }}y = 0 \cr & {\text{Using the graph shown below}} \cr & \cr & {\text{*The mass of the lamina is }} \cr & m = \rho \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr & m = \rho \int_{ - 1}^1 {\sqrt {1 - {x^2}} } dx \cr & {\text{The area is a semicircle of radius 1, so}} \cr & m = \rho \left( {\frac{1}{2}\pi {{\left( 1 \right)}^2}} \right) \cr & m = \frac{1}{2}\pi \rho \cr & \cr & *{\text{The moment about the }}x{\text{ - axis is}} \cr & {M_x} = \rho \int_a^b {\left[ {\frac{{f\left( x \right) + g\left( x \right)}}{2}} \right]} \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr & {M_x} = \rho \int_{ - 1}^1 {\left[ {\sqrt {1 - {x^2}} } \right]} \left[ {\sqrt {1 - {x^2}} } \right]dx \cr & {M_x} = \rho \int_{ - 1}^1 {\left( {1 - {x^2}} \right)} dx \cr & {M_x} = 2\rho \int_0^1 {\left( {1 - {x^2}} \right)} dx \cr & {M_x} = 2\rho \left[ {x - \frac{1}{3}{x^3}} \right]_0^1 \cr & {M_x} = 2\rho \left( {\frac{1}{3}} \right) = \frac{2}{3}\rho \cr & \cr & *{\text{The moment about the }}y{\text{ - axis is}} \cr & {M_y} = \rho \int_a^b x \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr & {M_y} = \rho \int_0^1 x \left[ {\sqrt {1 - {x^2}} } \right]dx \cr & {M_y} = \rho \int_{ - 1}^1 {x\sqrt {1 - {x^2}} } dx \cr & {\text{By symmetry}} \cr & {M_y} = 0 \cr & \cr & *{\text{The center of mass }}\left( {\overline x ,\overline y } \right){\text{ is given by:}} \cr & \overline x = \frac{{{M_y}}}{m} = \frac{0}{{\frac{1}{2}\pi \rho }} = 0 \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{2}{3}\rho }}{{\frac{1}{2}\pi \rho }} = \frac{4}{{3\pi }} \cr & \left( {\overline x ,\overline y } \right) = \left( {0,\frac{4}{{3\pi }}} \right) \cr} $$