Chapter 8 - Section 8.3 - Applications to Physics and Engineering. - 8.3 Exercises - Page 585: 25

$\left( {\overline x ,\overline y } \right) = \left( {\frac{2}{3},\frac{4}{3}} \right)$

$$\eqalign{ & {\text{From the given graph}}{\text{, we can estimate the centroid at}} \cr & \left( {0.5,1.5} \right) \cr & \cr & 2x + y = 4 \to y = 4 - 2x \cr & {\text{Let }}y = 4 - 2x,{\text{ }}y = 0,{\text{ }}x = 2 \cr & {\text{Let }}f\left( x \right) = 4 - 2x{\text{ and }}g\left( x \right) = 0 \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_0^2 {\left( {4 - 2x - 0} \right)} dx \cr & m = \rho \int_0^2 {\left( {4 - 2x} \right)} dx \cr & m = \rho \left[ {4x - {x^2}} \right]_0^2 \cr & m = 4\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_0^2 {\left[ {\frac{{4 - 2x}}{2}} \right]} \left[ {4 - 2x} \right]dx \cr & {M_x} = \frac{1}{2}\rho \int_0^2 {{{\left( {4 - 2x} \right)}^2}} dx \cr & {M_x} = - \frac{1}{4}\rho \int_0^2 {{{\left( {4 - 2x} \right)}^2}} \left( { - 2} \right)dx \cr & {M_x} = - \frac{1}{4}\rho \left[ {\frac{{{{\left( {4 - 2x} \right)}^3}}}{3}} \right]_0^2 \cr & {M_x} = - \frac{1}{{12}}\rho \left[ {{{\left( {4 - 4} \right)}^3} - {{\left( {4 - 0} \right)}^3}} \right] \cr & {M_x} = \frac{{16}}{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_0^2 x \left[ {4 - 2x} \right]dx \cr & {M_y} = \rho \int_0^2 {\left( {4x - 2{x^2}} \right)} dx \cr & {M_y} = \rho \left[ {2{x^2} - \frac{2}{3}{x^3}} \right]_0^2 \cr & {M_y} = \rho \left[ {2{{\left( 2 \right)}^2} - \frac{2}{3}{{\left( 2 \right)}^3}} \right] - \rho \left[ 0 \right] \cr & {M_y} = \frac{8}{3}\rho \cr & \cr & *{\text{The coordinates }}\left( {\overline x ,\overline y } \right){\text{ of the centroid are:}} \cr & \overline x = \frac{{{M_y}}}{m} = \frac{{\frac{8}{3}\rho }}{{4\rho }} = \frac{2}{3} \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{{16}}{3}\rho }}{{4\rho }} = \frac{4}{3} \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{2}{3},\frac{4}{3}} \right) \cr} $$