# The area under the curve y=1/x from x=8 to x=10 is revolved about the x-axis. What is the volume of the solid formed?

Apr 1, 2017

$\frac{\pi}{40}$

#### Explanation:

The region bound by these curves can be thought of as the region bound by $y = \frac{1}{x}$ and the $x$ axis, between $x = 8$ and $x = 10$.

Hence, $V = \pi {\int}_{8}^{10} {\left(\frac{1}{x}\right)}^{2} \mathrm{dx}$

$= \pi {\int}_{8}^{10} \left(\frac{1}{{x}^{2}}\right) \mathrm{dx}$

$= \pi {\int}_{8}^{10} {x}^{-} 2 \mathrm{dx}$

$= \pi {\left[{x}^{-} \frac{1}{-} 1\right]}_{8}^{10}$

$= - \pi {\left[\frac{1}{x}\right]}_{8}^{10}$

$= - \pi \left(\frac{1}{10} - \frac{1}{8}\right)$

$= - \pi \left(- \frac{1}{40}\right)$
$= \frac{\pi}{40}$