Question

How do you find the volume of the solid generated by revolving the region bounded by the graphs `y=3/(x+1), y=0, x=0, x=8`, about the x axis?

Answer 1

The volume `V=8pi`

Explanation:

The unit volume `dV=piy^2dx`

`dV=pi(3^3/(x+1)^2)dx=(9pi)/(x+1)^2dx`

Therefore `V=9piint_0^8dx/(x+1)^2`

`=[-(9pi)/(x+1)]_0^8=9pi(-1/9+1)=8pi`