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,

`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...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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,

`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`

Now we are given `y=sqrt(x),y=0,x=4`

The attached image shows the region bounded by the functions,

Now let's find the area of the region,

`A=int_0^4sqrt(x)dx`

`A=[x^(1/2+1)/(1/2+1)]_0^4`

`A=[2/3x^(3/2)]_0^4`

`A=[2/3(4)^(3/2)]`

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

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

`A=16/3`

Now let's evaluate the moments about the x- and y-axes,

`M_x=rhoint_0^4 1/2(sqrt(x))^2dx`

`M_x=rho/2int_0^4xdx`

`M_x=rho/2[x^2/2]_0^4`

`M_x=rho/2[4^2/2]`

`M_x=rho/2(16/2)`

`M_x=4rho`

`M_y=rhoint_0^4xsqrt(x)dx`

`M_y=rhoint_0^4x^(3/2)dx`

`M_y=rho[x^(3/2+1)/(3/2+1)]_0^4`

`M_y=rho[2/5x^(5/2)]_0^4`

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

`M_y=rho[2/5(2^2)^(5/2)]`

`M_y=rho[2/5(2)^5]`

`M_y=rho[2/5(32)]`

`M_y=64/5rho`

The coordinates of the center of the mass are given by,

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

Plug in the values of `M_y,A`

`barx=(64/5rho)/(rho16/3)`

`barx=64/5(3/16)`

`barx=12/5`

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

`bary=(4rho)/(rho16/3)`

`bary=4(3/16)`

`bary=3/4`

The coordinates of the center of mass are `(12/5,3/4)`