Calculus of a Single Variable, Chapter 7, 7.6, Section 7.6, Problem 22

Consider an irregularly shaped planar lamina of uniform density rho , bounded by graphs y=f(x),y=g(x) , and am=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
Given:y=x^(2/3),y=4
Refer to attached image. The plot of y=x^(2/3) is red in color.
The graphs intersect at the coordinates (-8,4),(8,4)
Let's evaluate the area of the bounded region,
A=int_(-8)^8(4-x^(2/3))dx
A=2int_0^8(4-x^(2/3))dx
A=2[4x-x^(2/3+1)/(2/3+1)]_0^8
A=2[4x-3/5x^(5/3)]_0^8
A=2[4*8-3/5(8)^(5/3)]
A=2[32-3/5(2^3)^(5/3)]
A=2[32-3/5(2)^5]
A=2[32-96/5]
A=2[(160-96)/5]
A=(2(64))/5
A=128/5
Now let's evaluate the moments about the x- and y-axes,
Since the graph is symmetrical about the y-axis,
So, M_y=0 and barx=0 ,
M_x=rhoint_(-8)^8 1/2([4^2]-[x^(2/3)]^2)dx
M_x=rho/2int_(-8)^8(16-x^(4/3))dx
M_x=rho/2(2)int_0^8(16-x^(4/3))dx
M_x=rho[16x-(x^(4/3+1)/(4/3+1))]_0^8
M_x=rho[16x-3/7x^(7/3)]_0^8
M_x=rho[16*8-3/7(8)^(7/3)]
M_x=rho[128-3/7(2^7)]
M_x=rho[128-3/7(128)]
M_x=512/7rho
bary=M_x/m=M_x/(rhoA)
bary=(512/7rho)/(rho128/5)
bary=(512/7)(5/128)
bary=20/7
The coordinates of the center of mass are (0,20/7)