# `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. 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`

...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

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 `(barx,bary)` is given by:

`barx=M_y/m`

`bary=M_x/m`

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,

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

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

`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' 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_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[(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`

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

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

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

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

`M_y=rho/2[243/5-81+27]`

`M_y=rho/2[243/5-54]`

`M_y=rho/2[(243-270)/5]`

`M_y=-27/10rho`

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

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

`barx=(-27/10rho)/(rho9/2)`

`barx=(-27/10)(2/9)`

`barx=(-3)/5`

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

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

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

`bary=3/2`

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

Images:
This image has been Flagged as inappropriate Click to unflag
Approved by eNotes Editorial Team