y=sqrt(x) +1 , y=1/3x + 1 Find the x and y moments of inertia and center of mass for the laminas of uniform density p bounded by the graphs of the equations.

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)+1,y=1/3x+1
Refer to the attached image, plot of y=sqrt(x)+1 is red in color and plot of y=1/3x+1 is blue in color.The curves intersect at (0,1) and (9,4) .
Now let's evaluate the area of the region bounded by the graphs of the given equations,
A=int_0^9((sqrt(x)+1)-(1/3x+1))dx
A=int_0^9(sqrt(x)+1-1/3x-1)dx
A=int_0^9(sqrt(x)-1/3x)dx
Evaluate using power rule,
A=[x^(1/2+1)/(1/2+1)-1/3(x^2/2)]_0^9
A=[2/3x^(3/2)-1/6x^2]_0^9
A=[2/3(9)^(3/2)-1/6(9)^2]
A=[2/3(3^2)^(3/2)-1/6(81)]
A=[2/3(3)^3-27/2]
A=[18-27/2]
A=9/2
Now let's find the moments about the x- and y-axes using the formulas stated above,
M_x=rhoint_0^9 1/2([sqrt(x)+1]^2-[1/3x+1]^2)dx
M_x=rhoint_0^9 1/2([x+2sqrt(x)+1]-[(1/3x)^2+2(1/3x)(1)+1^2])dx
M_x=rhoint_0^9 1/2(x+2sqrt(x)+1-x^2/9-2/3x-1)dx
Take the constant out,
M_x=rho/2int_0^9(x-2/3x+2sqrt(x)-x^2/9)dx
M_x=rho/2int_0^9(x/3+2(x)^(1/2)-x^2/9)dx
Apply the basic integration rules i.e sum and power rules,
M_x=rho/2[1/3(x^2/2)+2x^(1/2+1)/(1/2+1)-1/9(x^3/3)]_0^9
M_x=rho/2[x^2/6+2(2/3)x^(3/2)-x^3/27]_0^9
M_x=rho/2[x^2/6+4/3x^(3/2)-x^3/27]_0^9
M_x=rho/2[9^2/6+4/3(9)^(3/2)-9^3/27]
M_x=rho/2[81/6+4/3(3^2)^(3/2)-9^3/(9*3)]
M_x=rho/2[27/2+4/3(3^3)-27]
M_x=rho/2[27/2+36-27]
M_x=rho/2[27/2+9]
M_x=rho/2((27+18)/2)
M_x=45/4rho
M_y=rhoint_0^9x((sqrt(x)+1)-(1/3x+1))dx
M_y=rhoint_0^9x(sqrt(x)+1-1/3x-1)dx
M_y=rhoint_0^9x(x^(1/2)-1/3x)dx
M_y=rhoint_0^9(x^(3/2)-1/3x^2)dx
M_y=rho[x^(3/2+1)/(3/2+1)-1/3(x^3/3)]_0^9
M_y=rho[2/5x^(5/2)-1/9x^3]_0^9
M_y=rho[2/5(9)^(5/2)-1/9(9)^3]
M_y=rho[2/5(3^2)^(5/2)-81]
M_y=rho[2/5(3)^5-81]
M_y=rho[2/5(243)-81]
M_y=rho[486/5-81]
M_y=rho[(486-405)/5]
M_y=81/5rho
Now let's evaluate the coordinates of the center of mass by plugging in the values of the moments and area,
barx=M_y/m=M_y/(rhoA)  
barx=(81/5rho)/(rho9/2)
barx=(81/5)(2/9)
barx=18/5
bary=M_x/m=M_x/(rhoA)
bary=(45/4rho)/(rho9/2)
bary=(45/4)(2/9)
bary=5/2
The center of mass is (18/5,5/2)