# How do you find the volume of the solid generated by revolving the region bounded by the graphs xy=6, y=2, y=6, x=6, about the line x=6?

Apr 11, 2017

#### Explanation:

Find the volume of the solid generated by revolving the region bounded by $x y = 6$, $y = 2$, $y = 6$, and $x = 6$, about the line $x = 6$.

Here is a graph. The region is in blue. A representative slice (black line segments) has been taken at a $y$ value. The thickness is $\mathrm{dy}$.

The axis of revolution is the line $x = 6$ and is indicated with a red circular arrow.

Because the thickness is the differential of $y$ (rather than $x$), we'll want the curve in terms of $y$.
$x y = 6$ if and only if $x = \frac{6}{y}$

When rotated, the radius is the $x$ on the right minus the $x$ on the left. (greater minus lesser values of $x$)

$r = \left(6 - \frac{6}{y}\right)$

The volume of the representative disk is

$\pi {r}^{2} \mathrm{dy} = {\left(6 - \frac{6}{y}\right)}^{2} \mathrm{dy}$

$= \pi \left(36 - \frac{72}{y} + \frac{36}{y} ^ 2\right) \mathrm{dy}$

$y$ varies from $2$ to $6$, so the volume of the solid is

$V = \pi {\int}_{2}^{6} \left(36 - \frac{72}{y} + \frac{36}{y} ^ 2\right) \mathrm{dy}$

$= \pi \left(156 - 72 \ln \left(3\right)\right)$