Consider 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=4-y^2` ,`x=0`
Refer to the attached image. The...
Consider 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=4-y^2` ,`x=0`
Refer to the attached image. The plot of `x=4-y^2` is red in color.
Let's evaluate the area of the region,
`A=int_(-2)^2(4-y^2)dy`
`A=2int_0^2(4-y^2)dy`
`A=2[4y-y^3/3]_0^2`
`A=2[4*2-2^3/3]`
`A=2[8-8/3]`
`A=32/3`
`M_y=2rhoint_0^2 1/2(4-y^2)^2dy`
`M_y=(2rho)/2int_0^2(4^2-2(4)y^2+(y^2)^2)dy`
`M_y=rhoint_0^2(16-8y^2+y^4)dy`
`M_y=rho[16y-8(y^3/3)+y^5/5]_0^2`
`M_y=rho[16*2-8/3(2^3)+1/5(2^5)]`
`M_y=rho[32-64/3+32/5]`
`M_y=rho[(480-320+96)/15]`
`M_y=256/15rho`
By symmetry, `M_x=0,bary=0`
`barx=M_y/m=M_y/(rhoA)`
`barx=(256/15rho)/(rho32/3)`
`barx=(256/15)(3/32)`
`barx=8/5`
The coordinates of the center of mass are `(barx,bary)` are `(8/5,0)`
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