x=-y , x=2y-y^2 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 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=-y,x=2y-y^2
Refer to the attached image. The plot of x=2y-y^2 is blue in color and plot of x=-y is red in color. The curves intersect at (0,0) and (-3,3) .
Let's first evaluate the area of the bounded region,
A=int_0^3((2y-y^2)-(-y))dy
A=int_0^3(2y-y^2+y)dy
A=int_0^3(3y-y^2)dy
A=[3y^2/2-y^3/3]_0^3
A=[3/2(3)^2-1/3(3)^3]
A=[27/2-9]
A=9/2
Now let' evaluate the moments about x- and y-axes using the above stated formulas:
M_x=rhoint_0^3 y((2y-y^2)-(-y))dy
M_x=rhoint_0^3 y(2y-y^2+y)dy
M_x=rhoint_0^3 y(3y-y^2)dy
M_x=rhoint_0^3(3y^2-y^3)dy
M_x=rho[3(y^3/3)-y^4/4]_0^3
M_x=rho[y^3-y^4/4]_0^3
M_x=rho[3^3-3^4/4]
M_x=rho[27-81/4]
M_x=rho[(108-81)/4]
M_x=27/4rho
M_y=rhoint_0^3 1/2[(2y-y^2)^2-(-y)^2]dy
M_y=rhoint_0^3 1/2[((2y)^2-2(2y)y^2+(y^2)^2)-(y^2)]dy
M_y=rhoint_0^3 1/2[4y^2-4y^3+y^4-y^2]dy
M_y=rho/2int_0^3(y^4-4y^3+3y^2)dy
M_y=rho/2[y^5/5-4(y^4/4)+3(y^3/3)]_0^3
M_y=rho/2[y^5/5-y^4+y^3]_0^3
M_y=rho/2[3^5/5-3^4+3^3]
M_y=rho/2[243/5-81+27]
M_y=rho/2[243/5-54]
M_y=rho/2[(243-270)/5]
M_y=-27/10rho
Now let's find the center of the mass by plugging the moments and and the area evaluated above,
barx=M_y/m=M_y/(rhoA)
barx=(-27/10rho)/(rho9/2)
barx=(-27/10)(2/9)
barx=(-3)/5
bary=M_x/m=M_x/(rhoA)
bary=(27/4rho)/(rho9/2)
bary=(27/4)(2/9)
bary=3/2
The coordinates of the center of mass are ((-3)/5,3/2)