Question

The first quadrant region enclosed by y=2x, the x-axis and the line x=1 is resolved about the line y=0. How do you find the resulting volume?

Answer 1

Evaluate `piint_0^1 (4x^2) dx` to get `V = (4pi)/3`

Explanation:

The region is shown in the graph below.

The graph shows a representative slice of the region. The slice is taken at some value of `x` and had width (or thickness) `dx`.

When rotated about the `x` axis, the slice generates a disk of radius `r = 2x`.

The volume of the representative disk is

`pir^2*"thickness"`.

In this case we have:

the volume of the slice is `pi(2x)^2 dx`.

`x` varies from `0` to `1`, so the solid has volume

`V = int_0^1 (pi(2x)^2) dx = pi int_0^1 4x^2 dx = (4pi)/3`

Note

Because the question was posted under the topic "Calculating Volumes using Integrals", I have used the integral.

With some imagination we might notice that the solid will be a right circular cone.
The cone will have radius `r = 2` and height `h = 1`. (It's on its side with the vertex or apex at the origin).

The volume of a right circular cone is

`V = 1/3 pir^2`.

Using this formula we also get

`V = (4pi)/3`.