`y=sqrt(x) , y =1/2x` Find the x and y moments of inertia and center of mass for the laminas of uniform density `p` bounded by the graphs of...

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=sqrt(x),y=1/2x`

Please refer to the attached image. Plot of `y=sqrt(x)` is red in...

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=sqrt(x),y=1/2x`

Please refer to the attached image. Plot of `y=sqrt(x)` is red in color and plot of `y=1/2x` is blue in color. The curves intersect at `(0,0)` and `(4,2)` .

First let's find the area of the region,

`A=int_0^4(sqrt(x)-1/2x)dx`

`A=int_0^4(x^(1/2)-1/2x)dx`

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

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

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

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

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

`A=[16/3-4]`

`A=4/3`

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

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

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

Take the constant out,

`M_x=rho/2int_0^4(x-x^2/4)dx`

`M_x=rho/2[x^2/2-1/4(x^3/3)]_0^4`

`M_x=rho/2[x^2/2-x^3/12]_0^4`

`M_x=rho/2[4^2/2-4^3/12]`

`M_x=rho/2[16/2-64/12]`

`M_x=rho/2[8-16/3]`

`M_x=rho/2(8/3)`

`M_x=4/3rho`

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

`M_y=rhoint_0^4(xsqrt(x)-1/2x^2)dx`

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

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

`M_y=rho[2/5x^(5/2)-x^3/6]_0^4`

`M_y=rho[2/5(4)^(5/2)-4^3/6]`

`M_y=rho[2/5(2^2)^(5/2)-64/6]`

`M_y=rho[2/5(2)^5-32/3]`

`M_y=rho[64/5-32/3]`

`M_y=rho[(192-160)/15]`

`M_y=32/15rho`

The center of mass can be evaluated by plugging in the moments and area as below:

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

`barx=(32/15rho)/(rho4/3)`

`barx=(32/15)(3/4)`

`barx=8/5`

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

`bary=(4/3rho)/(rho4/3)`

`bary=1`

The coordinates of the center of mass are `(8/5,1)`