Calculus of a Single Variable, Chapter 7, 7.6, Section 7.6, Problem 26

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.
y+2=y^2
0=(y+1)(y-2)
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:
(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_-1 [int^(x=y+2)_(x=y^2) rho dx] dy
M=rho int^2_-1 [(y+2)-(y^2)] dy
M=rho [(1/2)y^2+2y-(1/3)y^3]|^2_-1
M=9/2 rho
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
M=rho ((1/3)y^3+y^2-(1/4)y^4)|^2_-1
M_x=9/4 rho
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
M=rho/2((1/3)y^3+2y^2+4y-(1/5)y^5)|^2_-1
M_y=36/5 rho
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:
I_x=63/20 rho
I_y=423/28 rho