Chapter 7 - Applications of Integration - 7.6 Exercises - Page 494: 14

$$\eqalign{ & m = 18\rho \cr & {M_x} = 36\rho \cr & {M_y} = 36\rho \cr & \left( {\overline x ,\overline y } \right) = \left( {2,2} \right) \cr} $$

$$\eqalign{ & y = 6 - x,{\text{ }}y = 0,{\text{ }}x = 0 \cr & {\text{For }}x = 0 \to y = 6 \cr & {\text{Let }}f\left( x \right) = \frac{1}{2}x{\text{ and }}g\left( x \right) = 0,{\text{ on the interval }}\left[ {0,6} \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_0^6 {\left( {6 - x - 0} \right)} dx \cr & m = \rho \int_0^6 {\left( {6 - x} \right)} dx \cr & m = \rho \left[ {6x - \frac{1}{2}{x^2}} \right]_0^6 = \rho \left[ {6\left( 6 \right) - \frac{1}{2}{{\left( 6 \right)}^2}} \right] - 0 \cr & m = 18\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_0^6 {\left[ {\frac{{6 - x}}{2}} \right]} \left[ {6 - x} \right]dx \cr & {M_x} = \frac{1}{2}\rho \int_0^6 {{{\left( {6 - x} \right)}^2}} dx \cr & {M_x} = - \frac{1}{2}\rho \left[ {\frac{{{{\left( {6 - x} \right)}^3}}}{3}} \right]_0^6 \cr & {M_x} = - \frac{1}{2}\rho \left[ {\frac{{{{\left( {6 - 6} \right)}^3}}}{3}} \right] + \frac{1}{2}\rho \left[ {\frac{{{{\left( {6 - 0} \right)}^3}}}{3}} \right]_0^6 \cr & {M_x} = 36\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_0^6 x \left[ {6 - x} \right]dx \cr & {M_y} = \rho \int_0^6 {\left( {6x - {x^2}} \right)} dx \cr & {M_y} = \rho \left[ {3{x^2} - \frac{1}{3}{x^3}} \right]_0^6 \cr & {M_y} = \rho \left[ {3{{\left( 6 \right)}^2} - \frac{1}{3}{{\left( 6 \right)}^3}} \right] - \rho \left[ {3{{\left( 0 \right)}^2} - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr & {M_y} = 36\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{{36\rho }}{{18\rho }} = 2 \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{36\rho }}{{18\rho }} = 2 \cr & \left( {\overline x ,\overline y } \right) = \left( {2,2} \right) \cr & \cr & {\text{Summary}} \cr & m = 18\rho \cr & {M_x} = 36\rho \cr & {M_y} = 36\rho \cr & \left( {\overline x ,\overline y } \right) = \left( {2,2} \right) \cr} $$