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

1 Answer
Sep 11, 2015

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

Explanation:

Graph of the two curvesGraph of the two curves

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.