Calculus of a Single Variable, Chapter 7, 7.6, Section 7.6, Problem 19

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=-x^2+4x+2,y=x+2
Refer to the attached image. The plot of y=-x^2+4x+2 is in red color and the plot of y=x+2 is in blue color. The curves intersect at (0,2) and (3,5) .
Now let's evaluate the area (A) of the region,
A=int_0^3((-x^2+4x+2)-(x+2))dx
A=int_0^3(-x^2+4x+2-x-2)dx
A=int_0^3(-x^2+3x)dx
Using basic integration properties:
A=[-x^3/3+3x^2/2]_0^3
A=[-(3)^3/3+3/2(3)^2]
A=[-9+27/2]
A=9/2
Now let's evaluate the moments about the x- and y-axes using the formulas stated above,
M_x=rhoint_0^3 1/2([-x^2+4x+2)]^2-[x+2]^2)dx
M_x=1/2rhoint_0^3{[(-x^2+4x+2)+(x+2)][(-x^2+4x+2)-(x+2)]}dx
M_x=1/2rhoint_0^3{[-x^2+5x+4][-x^2+3x]}dx
M_x=1/2rhoint_0^3(x^4-3x^3-5x^3+15x^2-4x^2+12x)dx
M_x=1/2rhoint_0^3(x^4-8x^3+11x^2+12x)dx
Evaluate using the basic integration rules:
M_x=1/2rho[x^5/5-8(x^4/4)+11(x^3/3)+12(x^2/2)]_0^3
M_x=1/2rho[x^5/5-2x^4+11/3x^3+6x^2]_0^3
M_x=1/2rho[(3)^5/5-2(3)^4+11/3(3)^3+6(3)^2]
M_x=1/2rho[243/5-162+99+54]
M_x=1/2rho[243/5-9]
M_x=1/2rho[(243-45)/5]
M_x=1/2rho(198/5)
M_x=99/5rho
M_y=rhoint_0^3x((-x^2+4x+2)-(x+2))dx
M_y=rhoint_0^3x(-x^2+4x+2-x-2)dx
M_y=rhoint_0^3x(-x^2+3x)dx
M_y=rhoint_0^3(-x^3+3x^2)dx
M_y=rho[-x^4/4+3(x^3/3)]_0^3
M_y=rho[-x^4/4+x^3]_0^3
M_y=rho[-1/4(3)^4+3^3]
M_y=rho[-1/4(81)+27]
M_y=rho[(-81+108)/4]
M_y=rho(27/4)
M_y=27/4rho
Now evaluate the center of mass by plugging in the values of moments and area as below:
barx=M_y/m=M_y/(rhoA)
barx=(27/4rho)/(rho9/2)
barx=(27/4)(2/9)
barx=3/2
bary=M_x/m=M_x/(rhoA)
bary=(99/5rho)/(rho9/2)
bary=(99/5)(2/9)
bary=22/5
The center of mass (barx,bary) are (3/2,22/5)