`y=sqrt(x) +1 , y=1/3x + 1` Find the x and y moments of inertia and center of mass for the laminas of uniform density `p` bounded by the...

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)+1,y=1/3x+1`

Refer to the attached image, plot of `y=sqrt(x)+1` is red in color and...

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)+1,y=1/3x+1`

Refer to the attached image, plot of `y=sqrt(x)+1` is red in color and plot of `y=1/3x+1` is blue in color.The curves intersect at `(0,1)` and `(9,4)` .

Now let's evaluate the area of the region bounded by the graphs of the given equations,

`A=int_0^9((sqrt(x)+1)-(1/3x+1))dx`

`A=int_0^9(sqrt(x)+1-1/3x-1)dx`

`A=int_0^9(sqrt(x)-1/3x)dx`

Evaluate using power rule,

`A=[x^(1/2+1)/(1/2+1)-1/3(x^2/2)]_0^9`

`A=[2/3x^(3/2)-1/6x^2]_0^9`

`A=[2/3(9)^(3/2)-1/6(9)^2]`

`A=[2/3(3^2)^(3/2)-1/6(81)]`

`A=[2/3(3)^3-27/2]`

`A=[18-27/2]`

`A=9/2`

Now let's find the moments about the x- and y-axes using the formulas stated above,

`M_x=rhoint_0^9 1/2([sqrt(x)+1]^2-[1/3x+1]^2)dx`

`M_x=rhoint_0^9 1/2([x+2sqrt(x)+1]-[(1/3x)^2+2(1/3x)(1)+1^2])dx`

`M_x=rhoint_0^9 1/2(x+2sqrt(x)+1-x^2/9-2/3x-1)dx`

Take the constant out,

`M_x=rho/2int_0^9(x-2/3x+2sqrt(x)-x^2/9)dx`

`M_x=rho/2int_0^9(x/3+2(x)^(1/2)-x^2/9)dx`

Apply the basic integration rules i.e sum and power rules,

`M_x=rho/2[1/3(x^2/2)+2x^(1/2+1)/(1/2+1)-1/9(x^3/3)]_0^9`

`M_x=rho/2[x^2/6+2(2/3)x^(3/2)-x^3/27]_0^9`

`M_x=rho/2[x^2/6+4/3x^(3/2)-x^3/27]_0^9`

`M_x=rho/2[9^2/6+4/3(9)^(3/2)-9^3/27]`

`M_x=rho/2[81/6+4/3(3^2)^(3/2)-9^3/(9*3)]`

`M_x=rho/2[27/2+4/3(3^3)-27]`

`M_x=rho/2[27/2+36-27]`

`M_x=rho/2[27/2+9]`

`M_x=rho/2((27+18)/2)`

`M_x=45/4rho`

`M_y=rhoint_0^9x((sqrt(x)+1)-(1/3x+1))dx`

`M_y=rhoint_0^9x(sqrt(x)+1-1/3x-1)dx`

`M_y=rhoint_0^9x(x^(1/2)-1/3x)dx`

`M_y=rhoint_0^9(x^(3/2)-1/3x^2)dx`

`M_y=rho[x^(3/2+1)/(3/2+1)-1/3(x^3/3)]_0^9`

`M_y=rho[2/5x^(5/2)-1/9x^3]_0^9`

`M_y=rho[2/5(9)^(5/2)-1/9(9)^3]`

`M_y=rho[2/5(3^2)^(5/2)-81]`

`M_y=rho[2/5(3)^5-81]`

`M_y=rho[2/5(243)-81]`

`M_y=rho[486/5-81]`

`M_y=rho[(486-405)/5]`

`M_y=81/5rho`

Now let's evaluate the coordinates of the center of mass by plugging in the values of the moments and area,

`barx=M_y/m=M_y/(rhoA)`  

`barx=(81/5rho)/(rho9/2)`

`barx=(81/5)(2/9)`

`barx=18/5`

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

`bary=(45/4rho)/(rho9/2)`

`bary=(45/4)(2/9)`

`bary=5/2`

The center of mass is `(18/5,5/2)`