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

Mar 30, 2016

$8 \pi$

#### 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 $\pi {y}^{2} \mathrm{dx}$

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

${\int}_{0}^{8} \pi {y}^{2} \mathrm{dx}$ = $9 \pi {\int}_{0}^{8} \frac{1}{x + 1} ^ 2 \mathrm{dx}$

=$9 \pi {\left[- \frac{1}{x + 1}\right]}_{0}^{8}$

= $9 \pi \left[1 - \frac{1}{9}\right]$ = $8 \pi$