# 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?

Nov 9, 2016

The volume $V = 8 \pi$

#### Explanation:

The unit volume $\mathrm{dV} = \pi {y}^{2} \mathrm{dx}$

$\mathrm{dV} = \pi \left({3}^{3} / {\left(x + 1\right)}^{2}\right) \mathrm{dx} = \frac{9 \pi}{x + 1} ^ 2 \mathrm{dx}$

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

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