`y=-x^2+4x+2 , y=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.

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For an irregularly shaped planar lamina of uniform density `(rho)` bounded by graphs `y=f(x),y=g(x)` and `a<=x<=b` , the mass `(m)` of this region is given by:


`m=rhoA` , where A is the area of the region.

The moments about the x- and y-axes are given by:

`M_x=rhoint_a^b 1/2([f(x)]^2-[g(x)]^2)dx`


The center of mass `(barx,bary)` is given by:



We are given, `y=-x^2+4x+2,y=x+2`

Refer to the attached image. The plot of `y=-x^2+4x+2` is in red color and the plot of `y=x+2` is in blue color. The curves intersect at `(0,2)` and `(3,5)` .

Now let's evaluate the area (A) of the region,




Using basic integration properties:





Now let's evaluate the moments about the x- and y-axes using the formulas stated above,

`M_x=rhoint_0^3 1/2([-x^2+4x+2)]^2-[x+2]^2)dx`





Evaluate using the basic integration rules:




















Now evaluate the center of mass by plugging in the values of moments and area as below:









The center of mass `(barx,bary)` are `(3/2,22/5)`


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