`x=-y , x=2y-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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gsarora17 eNotes educator| Certified Educator

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 `(barx,bary)` is given by:



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

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

Let's first evaluate the area of the bounded region,








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

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

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

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








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

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

`M_y=rhoint_0^3 1/2[4y^2-4y^3+y^4-y^2]dy`









Now let's find the center of the mass by plugging the moments and and the area evaluated above,









The coordinates of the center of mass are `((-3)/5,3/2)`

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