Calculus of a Single Variable, Chapter 7, 7.6, Section 7.6, Problem 15

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=sqrt(x),y=0,x=4
The attached image shows the region bounded by the functions,
Now let's find the area of the region,
A=int_0^4sqrt(x)dx
A=[x^(1/2+1)/(1/2+1)]_0^4
A=[2/3x^(3/2)]_0^4
A=[2/3(4)^(3/2)]
A=[2/3(2^2)^(3/2)]
A=[2/3(2)^3]
A=16/3
Now let's evaluate the moments about the x- and y-axes,
M_x=rhoint_0^4 1/2(sqrt(x))^2dx
M_x=rho/2int_0^4xdx
M_x=rho/2[x^2/2]_0^4
M_x=rho/2[4^2/2]
M_x=rho/2(16/2)
M_x=4rho
M_y=rhoint_0^4xsqrt(x)dx
M_y=rhoint_0^4x^(3/2)dx
M_y=rho[x^(3/2+1)/(3/2+1)]_0^4
M_y=rho[2/5x^(5/2)]_0^4
M_y=rho[2/5(4)^(5/2)]
M_y=rho[2/5(2^2)^(5/2)]
M_y=rho[2/5(2)^5]
M_y=rho[2/5(32)]
M_y=64/5rho
The coordinates of the center of the mass are given by,
barx=M_y/m=M_y/(rhoA)
Plug in the values of M_y,A
barx=(64/5rho)/(rho16/3)
barx=64/5(3/16)
barx=12/5
bary=M_x/m=M_x/(rhoA)
bary=(4rho)/(rho16/3)
bary=4(3/16)
bary=3/4
The coordinates of the center of mass are (12/5,3/4)