y = 1/x , y=0 , x=1 , x=3 Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis.

For the region bounded by y=1/x ,y=0 , x=1 and x=3 and revolved about the x-axis, we may apply Disk method. For the Disk method, we consider a perpendicular rectangular strip with the axis of revolution.
As shown on the attached image, the thickness of the rectangular strip is "dx" with a vertical orientation perpendicular to the x-axis (axis  of revolution).
We follow the formula for the Disk method:V = int_a^b A(x) dx  where disk base area is A= pi r^2 with.
 Note: r = length of the rectangular strip. We may apply r = y_(above)-y_(below).
Then r = f(x)= 1/x-0
        r =1/x
The boundary values of x is a=1 to b=3 .
Plug-in the f(x) and the boundary values to integral formula, we get: 
V = int_1^3 pi (1/x)^2 dx
V = int_1^3 pi 1/x^2 dx
Apply basic integration property: intc*f(x) dx = c int f(x) dx .
V = pi int_1^3 1/x^2 dx
Apply Law of Exponent: 1/x^n =x^(-n) and Power rule for integration: int x^n dy= x^(n+1)/(n+1) .
V = pi int_1^3 x^(-2) dx
V = pi*x^((-2+1))/((-2+1)) |_1^3
V = pi*x^(-1)/(-1) |_1^3
V = pi*-1/x |_1^3 or -pi/x|_1^3
Apply definite integration formula: int_a^b f(y) dy= F(b)-F(a) .
V = (-pi/3) - (-pi/1)
V = -pi/3+pi
V = 2pi/3