Question

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

Answer 1

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=1/6y`, `x=y` and `y=18`

As `y` goes from `0` to `18`, x goes from `x = 1/6y` on the left, to `x=y` on the right.
The greater radius is `R = y` and the lesser is `r = 1/6y`

Evaluate
`pi int_0^18 (R^2-r^2) dy = pi int_0^18 (y^2 - (y/6)^2) dy`

`= (35pi)/36 int_0^18 y^2 dy`

`= 1890pi`

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