`x = y+2, x=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.

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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...

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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`

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