Calculus of a Single Variable, Chapter 7, 7.6, Section 7.6, Problem 21

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=rhoA
where A is the area of the region.
The moments about the x and y-axes are given by the formula:
M_x=rhoint_a^b1/2([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 and bary=M_x/m ,
barx=1/Aint_a^bx(f(x)-g(x))dx
bary=1/Aint_a^b1/2[f(x)]^2-[g(x)]^2)dx
Now we given y=x^(2/3),y=0,x=8
The plot of the functions is attached as image and the bounds of the limits can be found from the same.
Area of the region A =int_0^8x^(2/3)dx
Use the power rule,
A=[x^(2/3+1)/(2/3+1)]_0^8
A=[3/5x^(5/3)]_0^8
A=[3/5(8)^(5/3)]
A=[3/5(2^3)^(5/3)]
A=[3/5(2)^5]
A=(3/5(32))
A=96/5
Now let's evaluate the moments about the x and y-axes,
M_x=rhoint_a^b1/2([f(x)]^2-[g(x)]^2)dx
=rhoint_0^8[1/2(x^(2/3))^2]dx
=rhoint_0^8 1/2x^(4/3)dx
Take the constant out and apply the power rule,
=rho/2int_0^8x^(4/3)dx
=rho/2[x^(4/3+1)/(4/3+1)]_0^8
=rho/2[3/7x^(7/3)]_0^8
=rho/2[3/7(8)^(7/3)]
=rho/2[3/7(2^3)^(7/3)]
=rho/2[3/7(2)^7]
=rho(3/7)(2)^6
=rho(3/7)(64)
=192/7rho
M_y=rhoint_a^bx(f(x)-g(x))dx
=rhoint_0^8x(x)^(2/3)dx
=rhoint_0^8x^(5/3)dx
=rho[x^(5/3+1)/(5/3+1)]_0^8
=rho[3/8x^(8/3)]_0^8
=rho[3/8(8)^(8/3)]
=rho[3/8(2^3)^(8/3)]
=rho[3/8(2^8)]
=rho(3/8)(256)
=96rho
Now let's find the center of mass,
barx=M_y/m=M_y/(rhoA)
Plug in the value of M_y and A ,
barx=(96rho)/(rho(96/5))
barx=5
bary=M_x/m=M_x/(rhoA)
Plug in the values of M_x and A ,
bary=(192/7rho)/(rho(96/5))
bary=(192/7)(5/96)
bary=10/7
The coordinates of the center of mass are,(5,10/7)