Calculus of a Single Variable, Chapter 7, 7.6, Section 7.6, Problem 23

Consider 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 (barx,bary) is given by:
barx=M_y/m
bary=M_x/m
We are given x=4-y^2 ,x=0
Refer to the attached image. The plot of x=4-y^2 is red in color.
Let's evaluate the area of the region,
A=int_(-2)^2(4-y^2)dy
A=2int_0^2(4-y^2)dy
A=2[4y-y^3/3]_0^2
A=2[4*2-2^3/3]
A=2[8-8/3]
A=32/3
M_y=2rhoint_0^2 1/2(4-y^2)^2dy
M_y=(2rho)/2int_0^2(4^2-2(4)y^2+(y^2)^2)dy
M_y=rhoint_0^2(16-8y^2+y^4)dy
M_y=rho[16y-8(y^3/3)+y^5/5]_0^2
M_y=rho[16*2-8/3(2^3)+1/5(2^5)]
M_y=rho[32-64/3+32/5]
M_y=rho[(480-320+96)/15]
M_y=256/15rho
By symmetry, M_x=0,bary=0
barx=M_y/m=M_y/(rhoA)
barx=(256/15rho)/(rho32/3)
barx=(256/15)(3/32)
barx=8/5
The coordinates of the center of mass are (barx,bary) are (8/5,0)