y= 1/2x^2 , y=0, x=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.

The center of Mass is:
(x_(cm),y_(cm))=(M_y/M, M_x/M)
Where the moments of mass are defined as:
M_x=int int_A rho(x,y)*y dy dx
M_y=int int_A rho(x,y)*x dy dx
The total mass is defined as:
M=int int_A rho(x,y)dy dx
First, lets find the total mass.
M=int^2_0 [int^(1/2x^2)_0 rho dy] dx
M=rho int^2_0 [y|^(1/2x^2)_0] dx
M=rho int^2_0 1/2x^2 dx
M=rho/2 (1/3)x^3|^2_0
M=rho/2 (1/3)2^3
M=4/3 rho
Now lets find the x moment of mass.
M_x=int^2_0 [int^(1/2x^2)_0 rho*y dy] dx
M_x=rho int^2_0 [(1/2)y^2|^(1/2x^2)_0] dx
M_x=rho int^2_0 [(1/2)(1/2x^2)^2] dx
M_x=rho int^2_0 (1/8)x^4 dx
M_x=(rho/8)(1/5)x^5|^2_0
M_x=rho/40*2^5=32/40 rho=4/5 rho
Now the y moment of mass.
M_y=int^2_0 [int^(1/2x^2)_0 rho*x dy] dx
M_y=rho int^2_0 x[int^(1/2x^2)_0 dy] dx
M_y=rho int^2_0 x[y|^(1/2x^2)_0] dx
M_y=rho int^2_0 x[1/2x^2] dx
M_y=rho/2 int^2_0 x^3 dx
M_y=rho/2 (1/4)x^4|^2_0
M_y=rho/2 (1/4)2^4=2rho
Therefore the center of mass is:
(x_(cm),y_(cm))=(M_y/M, M_x/M)=((2rho)/(4/3 rho),(4/5 rho)/(4/3 rho))=(3/2,3/5)
The moments of inerita or the second moments of the lamina are:
I_x=int int_A rho(x,y)*y^2 dy dx
I_y=int int_A rho(x,y)*x^2 dy dx
I won't solve these integrals step by step since they are very similar to the others, but you will find that:
I_x=16/21 rho
I_y=16/5 rho