Question

How do you use the disk or shell method to find the volume of the solid generated by revolving the regions bounded by the graphs of `y = x^(1/2)`, `y = 2`, and `x = 0` about the line `x = -1`?

Answer 1

See the explanation section, below.

Explanation:

Here is a picture of the region with a thin slice taken vertically.

So thickness is `dx` and we will use cylindrical shells.

The representative shell will have volume
`2pi "radius" * "height" * "thickness" = 2pi(x+1)(2-sqrtx)dx`.

We see that the values of `x` vary from `0` to `4`

So the volume of the solid is found by evaluating:

`int_0^4 2pi(x+1)(2-sqrtx)dx = 2pi int_0^4 (x+1)(2-sqrtx)dx`

Expand the product and integrate term by term.

To use disks/washers , we need to take our slices perpendicular to the axis of rotation as shown below.

The curve is now expressed as `x=y^2`

The representative washer has volume: `pi(R^2-r^2)dy` Where `r` is the greater radius (`x=-1` to `x=y^2`) and `r` is the lesser radius (`x=-1` to `x=0`) and the thickness is `dy`.

`y` goes from `0` to `2`, so the Volume is

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

Again, expand and integrate term by term.