Question

Write an integral which represents the volume of the solid obtained by revolving this region around the line x=0. Help?

Consider the region bounded by the lines x=3, x=6, the horizontal axis,

and the function f(x)= cos`sqrt(10+x)`

Answer 1

Please see below

Explanation:

Here is a graph of the region. The region is shaded in blue and the rotation around the `y` axis is shown by the small black circular arrow.

We'll use cylindrical shells.

So the volume of a represenative shell is `2pirh * "thickness"`
`"thickness"` will be a small change in `x`, denoted `dx`
Radius will be `x`

The only truly important fact to note in the graph is that for `x` in `(3,6)`, we have `cossqrt(x+10) <0`.
The significance of that is that the height of a representative slice is `h = 0-sqrt(cos(10+x)) = -sqrtcos(10-x)`
(Rather than the other way around.)

The values of `x` vary from `3` to `6` so the volume of the solid is

`V = int_3^6 pi ( x(-sqrt(cos(x+10))) dx = -pi int_3^6 ( xsqrt(cos(x+10))`