First lets find the bounds of integration. When looking at the graph the furthest that the lamina is bounded on the y-axis is where the curves interest. Lets find those points.
Therefore the y bounds are y=-1 and y=2. Then we will integrate between the furthest right curve (x=y) and the furthest left curve.
The center of Mass is:
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_-1 [int^(x=y+2)_(x=y^2) rho dx] dy`
`M=rho int^2_-1 [(y+2)-(y^2)] dy`
Now lets find the x moment of mass.
`M=int^2_-1 y*[int^(x=y+2)_(x=y^2) rho dx] dy`
`M=rho int^2_-1 y*[(y+2)-(y^2)] dy`
`M=rho int^2_-1 (y^2+2y-y^3) dy`
Now the y moment of mass.
`M=int^2_-1 [int^(x=y+2)_(x=y^2) rho x dx] dy`
`M=rho/2 int^2_-1 [x^2|^(y+2)_(y^2)] dy`
`M=rho/2 int^2_-1 (y^2+4y+4-y^4) dy`
Therefore the center of mass is:
`(x_(cm),y_(cm))=(M_y/M, M_x/M)=((36/5 rho)/(9/2 rho),(9/4 rho)/(9/2 rho))=(8/5,1/2)`
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: