Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises - Page 523: 86

$\left( {3.09,9.332} \right)$

$$\eqalign{ & {\text{Let the functions }} \cr & f\left( x \right) = {x^2},{\text{ and }}g\left( x \right) = {2^x}{\text{ on the interval }}2 \leqslant x \leqslant 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_2^4 {\left( {{x^2} - {2^x}} \right)} dx \cr & {\text{Integrating }} \cr & m = \rho \left[ {\frac{1}{3}{x^3} - \frac{{{2^x}}}{{\ln 2}}} \right]_2^4 \cr & m = \rho \left[ {\frac{1}{3}{{\left( 4 \right)}^3} - \frac{{{2^4}}}{{\ln 2}}} \right] - \rho \left[ {\frac{1}{3}{{\left( 2 \right)}^3} - \frac{{{2^2}}}{{\ln 2}}} \right] \cr & m = \rho \left( {\frac{{56}}{3} - \frac{{12}}{{\ln 2}}} \right) \cr & {\text{Simplifying}} \cr & m \approx 1.3543\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} = \frac{\rho }{2}\int_2^4 {\left[ {{x^2} + {2^x}} \right]} \left[ {{x^2} - {2^x}} \right]dx \cr & {M_x} = \frac{\rho }{2}\int_2^4 {\left( {{x^4} - {2^{2x}}} \right)} dx \cr & {M_x} = \frac{\rho }{2}\int_2^4 {\left( {{x^4} - {4^x}} \right)} dx \cr & {\text{Integrating}} \cr & {M_x} = \frac{\rho }{2}\left[ {\frac{1}{5}{x^5} - \frac{{{4^x}}}{{\ln 4}}} \right]_2^4 \cr & {M_x} = \frac{\rho }{2}\left[ {\frac{1}{5}{{\left( 4 \right)}^5} - \frac{{{4^4}}}{{\ln 4}}} \right] - \frac{\rho }{2}\left[ {\frac{1}{5}{{\left( 2 \right)}^5} - \frac{{{4^2}}}{{\ln 4}}} \right] \cr & {\text{Simplifying}} \cr & {M_x} = 12.6383\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_2^4 x \left[ {{x^2} - {2^x}} \right]dx \cr & {M_y} = \rho \int_2^4 {\left( {{x^3} - x \cdot {2^x}} \right)} dx \cr & {\text{Integrating}} \cr & {M_y} = \rho \left[ {\frac{1}{4}{x^4} - \frac{{x \cdot {2^x}}}{{\ln 2}} + \frac{{{2^x}}}{{{{\left( {\ln 2} \right)}^2}}}} \right]_2^4 \cr & {M_y} = \rho \left[ {\frac{1}{4}{{\left( 4 \right)}^4} - \frac{{\left( 4 \right) \cdot {2^4}}}{{\ln 2}} + \frac{{{2^4}}}{{{{\left( {\ln 2} \right)}^2}}}} \right] \cr & - \rho \left[ {\frac{1}{4}{{\left( 2 \right)}^4} - \frac{{\left( 2 \right) \cdot {2^2}}}{{\ln 2}} + \frac{{{2^2}}}{{{{\left( {\ln 2} \right)}^2}}}} \right] \cr & {M_y} \approx 4.1855\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{{4.1855\rho }}{{1.3543\rho }} \approx 3.09 \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{12.6383\rho }}{{1.3543\rho }} = 9.332 \cr & \left( {\overline x ,\overline y } \right) = \left( {3.09,9.332} \right) \cr} $$