Calculus of a Single Variable, Chapter 7, 7.6, Section 7.6, Problem 24

For an irregularly shaped planar lamina of uniform density (rho) , bounded by graphs x=f(y),x=g(y) and c<=y<=d , the mass (m) of this region is given by:
m=rhoint_c^d[f(y)-g(y)]dy
m=rhoA , where A is the area of the region.
The moments about the x- and y-axes are given by:
M_x=rhoint_c^d y(f(y)-g(y))dy
M_y=rhoint_c^d 1/2([f(y)]^2-[g(y)]^2)dy
The center of mass is given by:
barx=M_y/m
bary=M_x/m
We are given:x=3y-y^2,x=0
Refer to the attached image, Plot of x=3y-y^2 is blue in color. The curves intersect at (0,0) and (0,3) .
First let's find the area of the bounded region,
A=int_0^3(3y-y^2)dy
A=[3y^2/2-y^3/3]_0^3
A=[3/2(3)^2-1/3(3)^3]
A=[27/2-9]
A=9/2
Now let's evaluate the moments about the x- and y-axes using the formulas stated above:
M_x=rhoint_0^3 y(3y-y^2)dy
M_x=rhoint_0^3(3y^2-y^3)dy
M_x=rho[3(y^3/3)-y^4/4]_0^3
M_x=rho[y^3-y^4/4]_0^3
M_x=rho[3^3-3^4/4]
M_x=rho[27-81/4]
M_x=rho[(108-81)/4]
M_x=27/4rho
M_y=rhoint_0^3 1/2(3y-y^2)^2dy
M_y=rho/2int_0^3((3y)^2-2(3y)y^2+(y^2)^2)dy
M_y=rho/2int_0^3(9y^2-6y^3+y^4)dy
M_y=rho/2[9(y^3/3)-6(y^4/4)+y^5/5]_0^3
M_y=rho/2[3y^3-3/2y^4+y^5/5]_0^3
M_y=rho/2[3(3)^3-3/2(3)^4+3^5/5]
M_y=rho/2[81-243/2+243/5]
M_y=rho/2[(810-1215+486)/10]
M_y=rho/2[81/10]
M_y=81/20rho
Now let's find the coordinates of the center of mass,
barx=M_y/m=M_y/(rhoA)
barx=(81/20rho)/(rho9/2)
barx=(81/20)(2/9)
barx=9/10
bary=M_x/m=M_x/(rhoA)
bary=(27/4rho)/(rho9/2)
bary=(27/4)(2/9)
bary=3/2
The center of mass is (9/10,3/2)