Question

How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by `y=5x^2`, y=5x revolved about the x-axis?

Answer 1

Please see below.

Explanation:

In order to use shells, we need to take our representative slice parallel to the axis of rotation. So we will take a horizontal slice at a value of `y` and of thickness `dy`

Below is a partial picture. The curves intersect at `(0,0)` and `1,5)`.

The horizontal slice is shown in black and the radius of revolution is the dotted black line. Its length is `y`.

The volume of a representative shell is `2pirh*"thickness"`

In this problem, `"thickness" = dy` and `r = y`

`h` is the greater `x` value - the lesser `x` value.
The greater `x` value is the one on the right.

We need to rewrite the equation `y = 5x^2` so that we have `x` as a function of `y`. (`y` is the variable we'll be using for the integration.)

`x = sqrt(y/5)` (We ignore the negative square root because `x` it positive in this set-up.)

The lesser `x` is on the left where `x = y/5`

So the reprentative shell has volume

`2piy(sqrt(y/5)-y/5) dy`

`y` varies from `0` to `5`, so the solid has volume

`V = int_0^5 pi y(sqrt(y/5)-y/5) dy`

Details omitted

`= (5pi)/3`