Question

How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region `x= y^2` `x= y+2` rotated about the y-axis?

Answer 1

See explanation below.

Explanation:

To use shells when revoling about a vertical line, we need to take our representative slices vertically, so the thickness is `dx`. And we need the boundaries written as functions of `x`.

`y = sqrtx`, `y=-sqrtx`, `y=x-2`

Here is a picture of the region to be revolved about the `y` axis. I have included two representative slices of the region.

The curves intersect at `(1,-1)` and at `(4,2)`

The radii of our cylindrical shells will be `x` from `x=0` to `x=4`.
The thickness is `dx` and
the height of each shell is the greater `y` at `x` - lesser `y` at `x`.
(The `y` above - `y` below#)

The greater value of `y` can be found using `y=sqrtx` for all `x` from `0` to `4`.
The lesser value of `y` changes at `x=3`. So we'll need to evaluate two integrals.

From `x=0` to `x=1`, the lesser `y` value is `-sqrtx`, so the height is `sqrtx-(-sqrtx) = 2sqrtx`

To get that part of the volume, we need to evaluate:

`int_0^1 2pirhdx = 2pi int_0^1 x(2sqrtx)dx = 4pi int_0^1 x^(3/2) dx = 8/5pi`

From `x=1` to `x=4`, the lesser `y` value is `x-2`, so the height is `sqrtx-(x-2) = sqrtx-x+2`

To get that part of the volume, we need to evaluate:

`int_1^4 2pirhdx = 2pi int_1^4 x(sqrtx-x+2)dx = 2pi int_1^4 (x^(3/2) - x^2+2x) dx = 64/5pi`

The total volume is found by adding these two results, to get:

`72/5pi`