`y=x^(2/3), y=0, x=8` 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 `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=rhoA`

where A is the area of the region.

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

`M_x=rhoint_a^b1/2([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`  and `bary=M_x/m` ,

`barx=1/Aint_a^bx(f(x)-g(x))dx`

`bary=1/Aint_a^b1/2[f(x)]^2-[g(x)]^2)dx`

Now we given `y=x^(2/3),y=0,x=8`

The plot of the functions is attached as image and the bounds of the limits can be found from the same.

Area of the region A =`int_0^8x^(2/3)dx`

Use the power rule,

`A=[x^(2/3+1)/(2/3+1)]_0^8` 

`A=[3/5x^(5/3)]_0^8`

`A=[3/5(8)^(5/3)]`

`A=[3/5(2^3)^(5/3)]`

`A=[3/5(2)^5]`

`A=(3/5(32))`

`A=96/5`

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

`M_x=rhoint_a^b1/2([f(x)]^2-[g(x)]^2)dx`

`=rhoint_0^8[1/2(x^(2/3))^2]dx`

`=rhoint_0^8 1/2x^(4/3)dx`

Take the constant out and apply the power rule,

`=rho/2int_0^8x^(4/3)dx`

`=rho/2[x^(4/3+1)/(4/3+1)]_0^8`

`=rho/2[3/7x^(7/3)]_0^8`

`=rho/2[3/7(8)^(7/3)]`

`=rho/2[3/7(2^3)^(7/3)]`

`=rho/2[3/7(2)^7]`

` ` `=rho(3/7)(2)^6`

`=rho(3/7)(64)`

`=192/7rho`

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

`=rhoint_0^8x(x)^(2/3)dx`

`=rhoint_0^8x^(5/3)dx`

`=rho[x^(5/3+1)/(5/3+1)]_0^8`

`=rho[3/8x^(8/3)]_0^8`

`=rho[3/8(8)^(8/3)]`

`=rho[3/8(2^3)^(8/3)]`

`=rho[3/8(2^8)]`

`=rho(3/8)(256)`

`=96rho`

Now let's find the center of mass,

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

Plug in the value of `M_y` and  `A` ,

`barx=(96rho)/(rho(96/5))`

`barx=5`

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

Plug in the values of `M_x` and `A` ,

`bary=(192/7rho)/(rho(96/5))`

`bary=(192/7)(5/96)`

`bary=10/7`

The coordinates of the center of mass are,`(5,10/7)`

 

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