Question

How do you find the volume of `y=3/(x+1)`, `y=0`; `x=0`; `x= 8` rotated around the x-axis?

Answer 1

`8 pi`

Explanation:

The region to be rotated around x axis is shown, shaded blue, in the picture shown below

. In this region consider an element of width dx, at a distance x from y axis. Its length is y. If it is rotated about x axis, the volume of this elementary disc would be `pi y^2 dx`

The volume of the solid so generated by the rotation of the whole shaded region would be

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

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

= `9pi [1-1/9]` = `8 pi`