For an irregularly shaped planar lamina of uniform density `(rho)` , bounded by graphs `y=f(x).y=g(x)` and `a<=x<=b` , the mass `(m)` of this region is given by,

`m=rhoint_a^b[f(x)-g(x)]dx`

`m=rhoA` , where A is the area of the region,

The moments about the x- and y-axes are given by:

`M_x=rhoint_a^b 1/2([f(x)]^2-[g(x)]^2)dx`

`M_y=rhoint_a^bx(f(x)-g(x))dx`

The center of mass `(barx,bary)` is given by:

`barx=M_y/m`

`bary=M_x/m`

We are given:`y=6-x,y=0,x=0`

Refer to the attached image for the bounded region.

Let's first evaluate the area of the region,

`A=int_0^6(6-x)dx`

`A=[6x-x^2/2]_0^6`

`A=[6(6)-6^2/2]`

`A=[36-36/2]`

`A=18`

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

`M_x=rhoint_0^6 1/2[(6-x)^2]dx`

`M_x=rhoint_0^6 1/2(6^2-2(6)x+x^2)dx`

Take the constant out and simplify,

`M_x=rho/2int_0^6(36-12x+x^2)dx`

`M_x=rho/2[36x-12x^2/2+x^3/3]_0^6`

`M_x=rho/2[36x-6x^2+x^3/3]_0^6`

`M_x=rho/2[36(6)-6(6)^2+6^3/3]`

`M_x=rho/2[216-216+216/3]`

`M_x=rho/2(72)`

`M_x=36rho`

`M_y=rhoint_0^6x(6-x)dx`

`M_y=rhoint_0^6(6x-x^2)dx`

`M_y=rho[6x^2/2-x^3/3]_0^6`

`M_y=rho[3x^2-x^3/3]_0^6`

`M_y=rho[3(6)^2-6^3/3]`

`M_y=rho[108-216/3]`

`M_y=rho(108-72)`

`M_y=36rho`

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

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

`barx=(36rho)/(rho18)`

`barx=2`

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

`bary=(36rho)/(rho18)`

`bary=2`

The center of the mass is `(2,2)`

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