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 `y= sqrt(5x)`, `x=5` rotated about the y axis?

Answer 1

Please see the explanation section, below.

Explanation:

`y=sqrt(5x)` and `x=5` (and, presumably, the `x` axis)

The region bounded by the functions is shown below in blue.
We are using the shell method, so we take a representative slice parallel to the axis of revolution. In this case, that is the `y`-axis. So the thickness of the slice is `dx`.
The dashed line shows the radius of revolution for the representative.

We will be integrating with respect to `x`, so we need the minimum and maximum `x` values.

From the given information, we conclude that `x` varies from `0` to `5`.

The volume of a cylindrical shell is the surface area of the cylinder times the thickness:

`2pirh xx "thickness"`

In the picture, the radius is shown by the dashed line, it is `x`.

The height of the slice is the upper `y` value minus the lower: `sqrt(5x)`

The volume of the representative shell is: `2pixsqrt(5x)dx`

The volume of the resulting solid is:

`V=int_0^5 2pixsqrt(5x)dxdx`

`= 2pisqrt5 int_0^5 x^(3/2)dx`

`= 2pisqrt5[(2x^(5/2))/5]_0^5= 100pi`