$($IV$)$ An angle iron $($shown in the figure$)$ is made from two pieces of material which we model as thin rods with length $L_x=6m$ and $L_y=3m$ respectively, and with variable densities.

${}_x(x)=6(2-\frac{x}{3})k\frac{g}{m}$

${}_y(y)=6yk\frac{g}{m}$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}6(2-\frac{1}{3}x)=6(2x-\frac{1}{3}\frac{x^2}{2})$

$6(12-6)=36$

$($a$)$ First, consider the rod running along the $x-$axis from $(x,y)=(0,0)$ to $(6,0).$

$($i$)$ Determine the mass of this rod.

$(a)$ First, consider the rod running along the $x-$axis from $(x,y)=(0,0)to(0,0).$

$(i)$ Determine the mass $of$ this rod.

${\displaystyle \underset{0}{\overset{6}{}}}4(2-\frac{x}{3})dx=v_0*6{\displaystyle \underset{0}{\overset{6}{}}}(2-\frac{1}{3}x)dx={}_0*(2x-\frac{1}{3}\frac{x^2}{2}{)}_0^6=v_0^6$

$v_6{\displaystyle \underset{0}{\overset{6}{}}}6(2-\frac{x}{3})Lx=6{}_0*15\frac{g}{m}$

$($b$)$ Second, consider the rod running along the $y-$axis from $(x,y)=(0,0)$ to $(0,3).$

$($i$)$ Determine the mass of this rod.

$($:$v_0{\displaystyle \underset{0}{\overset{3}{}}}6ydy,v_0(6y^2){|}_0^3|={540}_6^{0}k\frac{s}{m}$