`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.

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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`

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