Calculus of a Single Variable, Chapter 7, 7.6, Section 7.6, Problem 17

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,
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
Now we are given y=x^2,y=x^3
Refer the attached image. Plot in red color is of y=x^2 and blue color is of y=x^3
Curves intersect at (1,1)
Now let's evaluate the area of the region,
A=int_0^1(x^2-x^3)dx
A=[x^3/3-x^4/4]_0^1
A=[1^3/3-1^4/4]
A=(1/3-1/4)=(4-3)/12
A=1/12
Now let's evaluate the moments about the x- and y-axes,
M_x=rhoint_0^1 1/2[(x^2)^2-(x^3)^2]dx
M_x=rho/2int_0^1(x^4-x^6)dx
M_x=rho/2[x^5/5-x^7/7]_0^1
M_x=rho/2[1^5/5-1^7/7]
M_x=rho/2(1/5-1/7)
M_x=rho/2(7-5)/(35)
M_x=rho/35
M_y=rhoint_0^1x(x^2-x^3)dx
M_y=rhoint_0^1(x^3-x^4)dx
M_y=rho[x^4/4-x^5/5]_0^1
M_y=rho[1^4/4-1^5/5]
M_y=rho[5-4)/(20)
M_y=rho/20
barx=M_y/m=M_y/(rhoA)
Plug in the values of M_y and A ,
barx=(rho/20)/(rho1/12)
barx=12/20
barx=3/5
bary=M_x/m=M_x/(rhoA)
bary=(rho/35)/(rho1/12)
bary=12/35
The coordinates of the center of mass are (3/5,12/35)