The disk and washer methods are useful for finding volumes of solids of revolution. In this article, we’ll review the methods and work out a number of example problems. By the end, you’ll be prepared for any disk and washer methods problems you encounter on the AP Calculus AB/BC exam!

## Solids of Revolution

The disk and washer methods are specialized tools for finding volumes of certain kinds of solids — **solids of revolution**. So what is a solid of revolution?

Starting with a flat region of the plane, generate the solid that would be “swept out” as that region revolves around a fixed axis.

For example, if you start with a right triangle, and then revolve it around a vertical axis through its upright leg, then you get a cone.

Here’s another cool example of a solid of revolution that you might have seen hanging up as a decoration!

## The Disk and Washer Methods: Formulas

So now that you know a bit more about solids of revolution, let’s talk about their volumes.

Suppose *S* is a solid of revolution generated by a region *R* in the plane. There are two related formulas, depending on how complicated the region *R* is.

### Disk Method

The simplest case is when *R* is the area under a curve *y* = *f*(*x*) between *x* = *a* and *x* = *b*, revolved around the *x*-axis.

Now imagine cutting the solid into thin slices perpendicular to the *x*-axis. Each slice looks like a disk or cylinder, except that the outer surface of the disk may have a curve or slant. Let’s approximate each slice by a cylinder of height *dx*, where *dx* is very small.

In fact, I like to think of each disk as being generated by revolving a thin rectangle around the *x*-axis. Then you can see that the height of the rectangle, *y*, is the same as the radius of the disk.

Now let’s compute the volume of a typical disk located at position *x*. The radius is *y*, which itself is just the function value at *x*. That is, *r* = *y* = *f*(*x*). The height of the disk is equal to *dx* (think of the disk as a cylinder standing on edge).

Therefore, the volume of a single cylindrical disk is: *V* = π *r*^{2}*h* = π *f*(*x*)^{2}*dx*.

This calculation gives the approximate volume of a thin slice of *S*. Next, to approximate the volume of the entirety of *S*, we have to add up all of the disk volumes throughout the solid. For simplicity, assume that the thickness of each slice is constant (*dx*). Also, for technical reasons, we have to keep track of the various *x*-values along the interval from *a* to *b* using the notation *x _{k}* for a “generic” sample point.

Finally, by letting the number of slices go to infinity (by taking a limit as *n* → ∞), we develop a useful formula for volume as an integral.

### Example 1: Disk Method

Let *R* be the region under the curve *y* = 2*x*^{3/2} between *x* = 0 and *x* = 4. Find the volume of the solid of revolution generated by revolving *R* around the *x*-axis.

#### Solution

Let’s set up the disk method for this problem.

The volume of the solid is 256π (roughly 804.25) cubic units.

### Washer Method

Now suppose the generating region *R* is bounded by two functions, *y* = *f*(*x*) on the top and *y* = *g*(*x*) on the bottom.

This time, when you revolve *R* around an axis, the slices perpendicular to that axis will look like washers.

A **washer** is like a disk but with a center hole cut out. The formula for the volume of a washer requires both an **inner radius** *r*_{1} and **outer radius** *r*_{2}.

We’ll need to know the volume formula for a single washer.

*V* = π (*r*_{2}^{2} – *r*_{1}^{2}) *h* = π (*f*(*x*)^{2} – *g*(*x*)^{2}) *dx*.

As before, the exact volume formula arises from taking the limit as the number of slices becomes infinite.

### Example 2: Washer Method

Determine the volume of the solid. Here, the bounding curves for the generating region are outlined in red. The top curve is *y* = *x* and the bottom one is *y* = *x*^{2}

#### Solution

This is definitely a solid of revolution. We’ll set up the formula with *f*(*x*) = *x* (top) and *g*(*x*) = *x*^{2} (bottom). But what should we use as *a* and *b*?

Well, just as in some area problems, you may have to solve for the bounds. Clearly the region is bounded by the two curves between their common intersection points. Set *f*(*x*) equal to *g*(*x*) and solve to locate these points of intersection.

*x* = *x*^{2} → *x* – *x*^{2} = 0 → *x*(1 – *x*) = 0.

We find two such points: *x* = 0 and 1. So set *a* = 0 and *b* = 1 in the formula.

### Example 3: Different Axes

Set up an integral that computes the volume of the solid generated by revolving he region bounded by the curves *y* = *x*^{2} and *x* = *y*^{3} around the line *x* = -1.

#### Solution

Be careful not to blindly apply the formula without analyzing the situation first!

This time, the axis of rotation is a vertical line *x* = -1 (rather than the horizontal *x*-axis). The radii will be horizontal segments, so think of *x*_{1} and *x*_{2} (rather than *y*-values).

Furthermore, because everything is turned on its side compared to previous problems, we have to make sure both boundary functions are solved for *x*. The thickness of the washer is now *dy* (instead of *dx*).

Finally, because the axis of revolution is one unit to the left of the *y*-axis, that adds another unit to each radius. (The further away the axis, the longer the radius must be to reach the figure, right?) Take a look at the graph below to help visualize what’s going on.

- Inner Radius:
*x*=*y*^{3}+ 1 - Outer Radius:
*x*=*y*^{1/2}+ 1

As before, set the functions equal and solve for points of intersection. Those are again at *x* = 0 and 1.

Using the Washer Method formula for volume, we obtain:

The problem only asks for setup, so we are done at this point.