Chapter 7 - Applications of Integration - 7.6 Exercises - Page 494: 30

$$\left( {\overline x ,\overline y } \right) \approx \left( {0{\text{ }},{\text{ }}0.818} \right)$$

$$\eqalign{ & y = \frac{8}{{{x^2} + 4}},{\text{ }}y = 0,{\text{ }}x = - 2,{\text{ }}x = 2 \cr & {\text{Using the graph shown below}} \cr & {\text{Let }}f\left( x \right) = \frac{8}{{{x^2} + 4}}{\text{ and }}g\left( x \right) = 0,{\text{ on }}\left[ { - 2,2} \right] \cr & \cr & {\text{*The mass of the lamina is }} \cr & m = \rho \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr & m = \rho \int_{ - 2}^2 {\frac{8}{{{x^2} + 4}}} dx \cr & {\text{Using a graphing utility to approximate the integral}} \cr & m = \rho \left( {2\pi } \right) \cr & m = 2\pi \rho \cr & \cr & *{\text{The moment about the }}x{\text{ - axis is}} \cr & {M_x} = \rho \int_a^b {\left[ {\frac{{f\left( x \right) + g\left( x \right)}}{2}} \right]} \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr & {M_x} = \rho \int_{ - 2}^2 {\left[ {\frac{4}{{{x^2} + 4}}} \right]} \left[ {\frac{8}{{{x^2} + 4}}} \right]dx \cr & {M_x} = 32\rho \int_{ - 2}^2 {\frac{1}{{{{\left( {{x^2} + 4} \right)}^2}}}} dx \cr & {\text{Using a graphing utility to approximate the integral}} \cr & {M_x} \approx 32\rho \left( {0.1606} \right) \cr & {M_x} \approx 5.1416\rho \cr & \cr & *{\text{The moment about the }}y{\text{ - axis is}} \cr & {M_y} = \rho \int_a^b x \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr & {M_y} = \rho \int_0^4 x \left[ {\frac{8}{{{x^2} + 4}}} \right]dx \cr & {M_y} = \rho \int_0^4 {\frac{{8x}}{{{x^2} + 4}}} dx \cr & {M_y} = \rho \left( 0 \right) \cr & {M_y} = 0 \cr & \cr & *{\text{The center of mass }}\left( {\overline x ,\overline y } \right){\text{ is given by:}} \cr & \overline x = \frac{{{M_y}}}{m} = \frac{0}{{2\pi \rho }} = 0 \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{5.1416\rho }}{{2\pi \rho }} = 0.818 \cr & \left( {\overline x ,\overline y } \right) \approx \left( {0{\text{ }},{\text{ }}0.818} \right) \cr} $$