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`
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.
And the limits of the integral are the end x values of the bounded region.
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.
`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`
Then, substitute back u =x + 1. So the volume of the solid is:
Therefore, the volume of the solid formed by revolving the bounded region around the x-axis is `8pi` cubic units.