Chapter 7 - Applications of Integration - 7.6 Exercises - Page 494: 27

$$\left( {\overline x ,\overline y } \right) \approx \left( {3.0{\text{ }},{\text{ }}126.0} \right)$$

$$\eqalign{ & y = 10x\sqrt {125 - {x^3}} ,{\text{ }}y = 0 \cr & {\text{Using the graph shown below}} \cr & {\text{Let }}f\left( x \right) = 10x\sqrt {125 - {x^3}} {\text{ and }}g\left( x \right) = 0,{\text{ on }}\left[ {0,5} \right] \cr & \cr & {\text{*The mass of the lamina is }} \cr & m = \rho \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr & m = \rho \int_0^5 {\left( {10x\sqrt {125 - {x^3}} } \right)} dx \cr & {\text{Using a graphing utility to approximate the integral}} \cr & m \approx \rho \left( {1033.03} \right) \cr & m \approx 1033.03\rho \cr & \cr & *{\text{The moment about the }}x{\text{ - axis is}} \cr & {M_x} = \rho \int_a^b {\left[ {\frac{{f\left( x \right) + g\left( x \right)}}{2}} \right]} \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr & {M_x} = \rho \int_0^5 {\left[ {\frac{{10x\sqrt {125 - {x^3}} }}{2}} \right]} \left[ {10x\sqrt {125 - {x^3}} } \right]dx \cr & {M_x} = 50\rho \int_0^5 {{x^2}{{\left( {\sqrt {125 - {x^3}} } \right)}^2}} dx \cr & {M_x} = 50\rho \int_0^5 {{x^2}\left( {125 - {x^3}} \right)} dx \cr & {\text{Using a graphing utility to approximate the integral}} \cr & {M_x} = 50\rho \left( {\frac{{15625}}{6}} \right) \cr & {M_x} \approx 130208\rho \cr & \cr & *{\text{The moment about the }}y{\text{ - axis is}} \cr & {M_y} = \rho \int_a^b x \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr & {M_y} = \rho \int_0^5 x \left[ {10x\sqrt {125 - {x^3}} } \right]dx \cr & {M_y} = 10\rho \int_0^5 {{x^2}\sqrt {125 - {x^3}} } dx \cr & {M_y} \approx 10\rho \left( {310.565} \right) \cr & {M_y} = 3105.65\rho \cr & \cr & *{\text{The center of mass }}\left( {\overline x ,\overline y } \right){\text{ is given by:}} \cr & \overline x = \frac{{{M_y}}}{m} = \frac{{3105.65\rho }}{{1033.03\rho }} \approx 3.0 \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{130208\rho }}{{1033.03\rho }} \approx 126.0 \cr & \left( {\overline x ,\overline y } \right) \approx \left( {3.0{\text{ }},{\text{ }}126.0} \right) \cr} $$