Question

How do you find the volume of the solid obtained by rotating the region bounded by the curves `y=sqrtx` and `y=x/3` rotated around the `x=-1`?

Answer 1

See the explanation section, below.

Explanation:

Here is a picture of the region (in blue) and the line `x=-1`.

It shows a representative slice of thickness `dx`. The slice is taken parallel to the axis of rotation, so when we rotate we'll get cylindrical shell.
The volume of the shell is `2pirhdx`. The radius, `r`, is shown by the dotted red horizontal line. Its length is `x+1`.
The height of the shell is the greater `y` value minus the lesser `y` value. That is: `h = sqrtx-x/3`

We also need the point of intersection where `x/3=sqrt3`. Which happens at `0` and at `9`

The integral we need to evaluate is:

`int_0^9 2 pi (x+1)(sqrtx-x/3) dx`

Expand the integrand to get 4 terms in powers of `x` which is straightforward to evaluate.

`2 pi int_0^9 (x+1)(sqrtx-x/3) dx = 207/5pi`