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=rhoint_c^d[f(y)-g(y)]dy`

`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:

`barx=M_y/m`

`bary=M_x/m`

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,

`A=int_0^3(3y-y^2)dy`

`A=[3y^2/2-y^3/3]_0^3`

`A=[3/2(3)^2-1/3(3)^3]`

`A=[27/2-9]`

`A=9/2`

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

`M_x=rho[3(y^3/3)-y^4/4]_0^3`

`M_x=rho[y^3-y^4/4]_0^3`

`M_x=rho[3^3-3^4/4]`

`M_x=rho[27-81/4]`

`M_x=rho[(108-81)/4]`

`M_x=27/4rho`

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

`M_y=rho/2int_0^3((3y)^2-2(3y)y^2+(y^2)^2)dy`

`M_y=rho/2int_0^3(9y^2-6y^3+y^4)dy`

`M_y=rho/2[9(y^3/3)-6(y^4/4)+y^5/5]_0^3`

`M_y=rho/2[3y^3-3/2y^4+y^5/5]_0^3`

`M_y=rho/2[3(3)^3-3/2(3)^4+3^5/5]`

`M_y=rho/2[81-243/2+243/5]`

`M_y=rho/2[(810-1215+486)/10]`

`M_y=rho/2[81/10]`

`M_y=81/20rho`

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

`barx=M_y/m=M_y/(rhoA)`

`barx=(81/20rho)/(rho9/2)`

`barx=(81/20)(2/9)`

`barx=9/10`

`bary=M_x/m=M_x/(rhoA)`

`bary=(27/4rho)/(rho9/2)`

`bary=(27/4)(2/9)`

`bary=3/2`

The center of mass is `(9/10,3/2)`

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