`x=3y-y^2 , x=0` 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 `x=f(y),x=g(y)` and `c<=y<=d` , 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_c^d y(f(y)-g(y))dy`

`M_y=rhoint_c^d 1/2([f(y)]^2-[g(y)]^2)dy`

The center of mass is given by:



We are given:`x=3y-y^2,x=0`

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

First let's find the area of the bounded region,






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

`M_x=rhoint_0^3 y(3y-y^2)dy`








`M_y=rhoint_0^3 1/2(3y-y^2)^2dy`










Now let's find the coordinates of the center of mass,









The center of mass is `(9/10,3/2)`


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