Question

How do you find the volume of the region enclosed by the curves `y=2x`, `y=x^2` rotated about the x-axis?

Answer 1

The volume of the solid generated by `y = 2x`, `y = x^2` revolved about the x-axis is `(64pi)/15`.

Explanation:

Revolving the area between these two curves about the x-axis, we end up with something that looks sort of like a cone... a hollow cone, with a curved inside. Now, imagine for a second taking a cross section parallel to the y-z plane, cutting the cone down the middle. The cross-section is going to look like a washer, with an inner radius equal to the height of `y = x^2` wherever we are in the solid, and the outer radius equal to `y = 2x` at that same `x`.

For reference, the area of a washer is equal to `pi*(r_"outer")^2 - pi*(r_"inner")^2`, if `r_"outer"` is the outer radius and `r_"inner"` is the inner radius. This fact should be immediately obvious - we're just subtracting the area of a circle from the area of another circle.

Now, to find the volume of the solid, we need to sum (integrate) the area of each of cross-sectional washer in the solid. This procedure is commonly called the method of washers .

So, we'll use the method of washers to find the volume of this solid.

The general formula is

`V = int_a^b pi (f(x))^2 dx - int_a^b pi (g(x))^2 dx`

where `f(x)` is a function giving the outer radius of the washer at any x, and `g(x)` is a function giving the inner radius of the washer. Note that the integrands here represent the areas of circles - and together, they give the area of each infinitesimal washer.

In our case, `f(x)` is equal to `2x` and `g(x)` is equal to `x^2`.

`V = int_a^b pi (2x)^2 dx - int_a^b pi (x^2)^2 dx`

Next, we need to worry about the limits of the integrals. Since we're only concerned with rotating the region between `x^2` and `2x`, the limits should involve the locations where `y = x^2` and `y = 2x` intersect. Intersection occurs at `x = 0` and `x = 2`.

`V = int_0^2 pi (2x)^2 dx - int_0^2 pi (x^2)^2 dx`

The difficult part - setting up the appropriate integral with correct limits - is done, so I won't walk through evaluating the integral. It evaluates to `(64pi)/15`, but if you don't trust me you can evaluate it yourself as an exercise.