Question

How do you find the volume of the solid obtained by rotating the region bounded by the curves `y=x^2` and `y=2-x^2` and `x=0` about the line `x=1`?

Answer 1

I would use shells (to avoid doing two integrals).

Explanation:

Here is a picture of the region. I have included the axis of rotation (`x=1`) as a dashed black line, a representative slice taken parallel to the axis of rotation (solid black segments inside the region) and the radius of the rotation (red line segment taken at about `y=1.7`).

When we rotate the slice, we'll get a cylindrical shell.

The representative cylindrical shell will have volume:

`2 pixx"radius"xx"height"xx"thickness"`.

Here we have:
`"radius"=r=(1-x)`.

The height will be the greater `y` value minus the lesser (the top y minus the bottom y).
`"height" = y_"top"-y_"bottom" = (2-x^2)-(x^2) = (2-2x^2)`

`"thickness" = dx`

We also not that in the region, `x` varies from `0` to `1`

Using shell, the volume of the solid is found by evaluating:

`V - int_a^b 2 pixx"radius"xx"height"xx"thickness"`

So we need

`V - int_0^1 2 pi (1-x)(2-2x^2) dx`

`= 2piint_0^1 (2-2x-2x^2+2x^3) dx`

`= 5/3 pi`