Calculus of a Single Variable, Chapter 7, 7.6, Section 7.6, Problem 18

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 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
We are given:y=sqrt(x),y=1/2x
Please refer to the attached image. Plot of y=sqrt(x) is red in color and plot of y=1/2x is blue in color. The curves intersect at (0,0) and (4,2) .
First let's find the area of the region,
A=int_0^4(sqrt(x)-1/2x)dx
A=int_0^4(x^(1/2)-1/2x)dx
A=[x^(1/2+1)/(1/2+1)-1/2(x^2)/2]_0^4
A=[2/3x^(3/2)-x^2/4]_0^4
A=[2/3(4)^(3/2)-4^2/4]
A=[2/3(2^2)^(3/2)-4]
A=[2/3(2)^3-4]
A=[16/3-4]
A=4/3
Now let's evaluate moments about the x- and y-axes,
M_x=rhoint_0^4 1/2((sqrt(x))^2-(1/2x)^2)dx
M_x=rhoint_0^4 1/2(x-x^2/4)dx
Take the constant out,
M_x=rho/2int_0^4(x-x^2/4)dx
M_x=rho/2[x^2/2-1/4(x^3/3)]_0^4
M_x=rho/2[x^2/2-x^3/12]_0^4
M_x=rho/2[4^2/2-4^3/12]
M_x=rho/2[16/2-64/12]
M_x=rho/2[8-16/3]
M_x=rho/2(8/3)
M_x=4/3rho
M_y=rhoint_0^4x(sqrt(x)-1/2x)dx
M_y=rhoint_0^4(xsqrt(x)-1/2x^2)dx
M_y=rhoint_0^4(x^(3/2)-1/2x^2)dx
M_y=rho[x^(3/2+1)/(3/2+1)-1/2(x^3/3)]_0^4
M_y=rho[2/5x^(5/2)-x^3/6]_0^4
M_y=rho[2/5(4)^(5/2)-4^3/6]
M_y=rho[2/5(2^2)^(5/2)-64/6]
M_y=rho[2/5(2)^5-32/3]
M_y=rho[64/5-32/3]
M_y=rho[(192-160)/15]
M_y=32/15rho
The center of mass can be evaluated by plugging in the moments and area as below:
barx=M_y/m=M_y/(rhoA)
barx=(32/15rho)/(rho4/3)
barx=(32/15)(3/4)
barx=8/5
bary=M_x/m=M_x/(rhoA)
bary=(4/3rho)/(rho4/3)
bary=1
The coordinates of the center of mass are (8/5,1)