`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.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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)`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Image (1 of 1)
Approved by eNotes Editorial