# Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. `y = 3/(x+1)` , `y=0` , `x = 0` , `x = 8`

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First, sketch the region bounded by the graphs of the given equations.

`y=3/(x+1)` (Green curve)

`y=0` (Yellow line)

`x=0` (Red line)

`x=8` (Green line)

Since there is no gap between the bounded region and the axis of rotation, to solve for the volume of the solid formed by revolving it around the x-axis, apply disk method. Its formula when the axis of revolution is horizontal is:

`V=int_a^b pi r^2 dx`

The radius is the distance of the upper edge of the bounded region from the axis of revolution. In the graph above, the upper edge is the blue curve.

`r=y-0=3/(x+1)-0=3/(x+1)`

And the limits of the integral are the end x values of the bounded region.

`a=0`

`b=8`

Plugging them to the formula above, we would have:

`V=int_0^8 pi (3/(x+1))^2dx`

Simplifying the integrand, it becomes:

`V=pi int_0^8 9/(x+1)^2 dx`

To take the integral of it, apply u-substitution method.

Let,

`u= x + 1`

And, differentiate it.

`du = dx`

So the integral becomes:

`V= pi int 9/u^2 du`

`V=pi int 9u^(-2)du`

`V= pi ((9u^-1)/(-1))+C`

`V=pi(-9/u)+ C`

Then, substitute back u =x + 1. So the volume of the solid is:

`V=pi(-9/(x+1))|_0^8`

`V=pi[(-9/(8+1))-(-9/(0+1))]`

`V=pi(-9/9+9/1)`

`V=pi(-1+9)`

`V=8pi`

**Therefore, the volume of the solid formed by revolving the bounded region around the x-axis is `8pi` cubic units.**