Calculus of a Single Variable, Chapter 7, 7.6, Section 7.6, Problem 14

For an irregularly shaped planar lamina of uniform density (rho) , bounded by graphs y=f(x).y=g(x) and a<=x<=b , the mass (m) of this region is given by,
m=rhoint_a^b[f(x)-g(x)]dx
m=rhoA , where A is the area of the region,
The moments about the x- and y-axes are given by:
M_x=rhoint_a^b 1/2([f(x)]^2-[g(x)]^2)dx
M_y=rhoint_a^bx(f(x)-g(x))dx
The center of mass (barx,bary) is given by:
barx=M_y/m
bary=M_x/m
We are given:y=6-x,y=0,x=0
Refer to the attached image for the bounded region.
Let's first evaluate the area of the region,
A=int_0^6(6-x)dx
A=[6x-x^2/2]_0^6
A=[6(6)-6^2/2]
A=[36-36/2]
A=18
Now let's find the moments about the x- and y-axes using the above stated formulas.
M_x=rhoint_0^6 1/2[(6-x)^2]dx
M_x=rhoint_0^6 1/2(6^2-2(6)x+x^2)dx
Take the constant out and simplify,
M_x=rho/2int_0^6(36-12x+x^2)dx
M_x=rho/2[36x-12x^2/2+x^3/3]_0^6
M_x=rho/2[36x-6x^2+x^3/3]_0^6
M_x=rho/2[36(6)-6(6)^2+6^3/3]
M_x=rho/2[216-216+216/3]
M_x=rho/2(72)
M_x=36rho
M_y=rhoint_0^6x(6-x)dx
M_y=rhoint_0^6(6x-x^2)dx
M_y=rho[6x^2/2-x^3/3]_0^6
M_y=rho[3x^2-x^3/3]_0^6
M_y=rho[3(6)^2-6^3/3]
M_y=rho[108-216/3]
M_y=rho(108-72)
M_y=36rho
Now let's find the coordinates of the center of mass,
barx=M_y/m=M_y/(rhoA)
barx=(36rho)/(rho18)
barx=2
bary=M_x/m=M_x/(rhoA)
bary=(36rho)/(rho18)
bary=2
The center of the mass is (2,2)