Question

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

Answer 1

Please see below.

Explanation:

Here is a graph of the region.

I've taken a slice perpendicular to the axis of rotation. The slice is taken at a variable value of `y`.

The thickness of the slice is `dy`, so we need the equations in the form `x =` a function of `y`.

The curve on the left (`y = sqrtx`) is `x = y^2`

on the right is the line `x = y`

Rotating the slice will generate a washer of thickness `dy` and
volume `pi(R^2-r^2) dy`
where `r` is the outer radius and `r` the inner.

The outer radius of the washer is the distance between the curve on the left and the line `x=2`.
So `R = 2-y^2`

The inner radius is the distance from the line on the right and the line `x=2`.
So `R = 2-y`

`y` varies from `0` to `1`.

The volume of the representative washer is

`int_0^1 pi((2-y^2)^2-(2-y)^2) dy = pi int_0^1 (y^4-5y^2+4y) dy`

evaluate to get

`= pi (8/15) = (8pi)/15`