Consider 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`
Given:`y=x^(2/3),y=4`
Refer to attached image. The plot of `y=x^(2/3)` is red in color.
The graphs...
Consider 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`
Given:`y=x^(2/3),y=4`
Refer to attached image. The plot of `y=x^(2/3)` is red in color.
The graphs intersect at the coordinates `(-8,4),(8,4)`
Let's evaluate the area of the bounded region,
`A=int_(-8)^8(4-x^(2/3))dx`
`A=2int_0^8(4-x^(2/3))dx`
`A=2[4x-x^(2/3+1)/(2/3+1)]_0^8`
`A=2[4x-3/5x^(5/3)]_0^8`
`A=2[4*8-3/5(8)^(5/3)]`
`A=2[32-3/5(2^3)^(5/3)]`
`A=2[32-3/5(2)^5]`
`A=2[32-96/5]`
`A=2[(160-96)/5]`
`A=(2(64))/5`
`A=128/5`
Now let's evaluate the moments about the x- and y-axes,
Since the graph is symmetrical about the y-axis,
So, `M_y=0` and `barx=0` ,
`M_x=rhoint_(-8)^8 1/2([4^2]-[x^(2/3)]^2)dx`
`M_x=rho/2int_(-8)^8(16-x^(4/3))dx`
`M_x=rho/2(2)int_0^8(16-x^(4/3))dx`
`M_x=rho[16x-(x^(4/3+1)/(4/3+1))]_0^8`
`M_x=rho[16x-3/7x^(7/3)]_0^8`
`M_x=rho[16*8-3/7(8)^(7/3)]`
`M_x=rho[128-3/7(2^7)]`
`M_x=rho[128-3/7(128)]`
`M_x=512/7rho`
`bary=M_x/m=M_x/(rhoA)`
`bary=(512/7rho)/(rho128/5)`
`bary=(512/7)(5/128)`
`bary=20/7`
The coordinates of the center of mass are `(0,20/7)`
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