# How do you find the volume of the region bounded by y=6x y=x and y=18 is revolved about the y axis?

Aug 21, 2015

Use washers. to get $V = 1890 \pi$

#### Explanation:

The region is the bounded region in:

graph{(y-6x)(y-x)(y-0.0001x-18) sqrt(81-(x-9)^2)sqrt(85-(y-9)^2)/sqrt(81-(x-9)^2)sqrt(85-(y-9)^2) = 0 [-28.96, 44.06, -7.7, 28.83]}

Taking vertical slices and integrating over $x$ would require two integrals, so take horizontal slices.

Rewrite the region: $x = \frac{1}{6} y$, $x = y$ and $y = 18$

As $y$ goes from $0$ to $18$, x goes from $x = \frac{1}{6} y$ on the left, to $x = y$ on the right.
The greater radius is $R = y$ and the lesser is $r = \frac{1}{6} y$

Evaluate
$\pi {\int}_{0}^{18} \left({R}^{2} - {r}^{2}\right) \mathrm{dy} = \pi {\int}_{0}^{18} \left({y}^{2} - {\left(\frac{y}{6}\right)}^{2}\right) \mathrm{dy}$

$= \frac{35 \pi}{36} {\int}_{0}^{18} {y}^{2} \mathrm{dy}$

$= 1890 \pi$

(Steps omitted because once it is set up, I think this is a straightforward integration.)