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Washers and Discs

This original textbook reading covers part of the AP topic. Follow the AP course outline for complete topic instruction and practice.

A different concept

This time, rather than summing up cylindrical shells, we are summing up (integrating) discs, or washers if there is a bound function on the bottom (you can think of washers as really thin donut slices).

Whereas for cylindrical shells we were using the area of the side of a cylinder, now we will be using the area of circles, or more specifically discs.

This method was covered in section 9.3 using a cylinder. For more complex shapes such as cones, the concept stays the same, except the radius of each disc varies with dx or dy.

This time, the direction of integration will be perpendicular to the radius of each disc. Conceptually, whether we choose dy or dx depends on which direction leads us to the next infinitely thin shell or disc or whichever shape we are integrating (yes, it’s possible to integrate using other shapes as well, ex: spherical shells to find the volume of a sphere).

In general disc or washer integration is easier to do in the x-direction, but it is possible to do so in the y-direction as well. We will practice with both methods to increase our proficiency.

Interactive 10.2.1 A familiar shape

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Same curve

The same cone was used in an earlier example problem using the method of cylindrical shells, when we had to integrate with respect to x. Here, we would have to integrate with respect to y, but a problem like this would be too easy.

Infinitely thin discs

In order to jump from one disc to the next, we have to integrate with respect to the height of the cone, dy. Would we use dx or dy if the cone was centered around the x-axis using the method of discs?

The core concept of summing up an infinite number of very thin shapes stays the same. We are simply changing the shape, and thus the equations we integrate.

Example Problem 10.2.1

On the interval of 0 to 1, the function $y = \sqrt{x}$ is bound by $y = x^{2}$ on the bottom. The area between them is rotated around the x-axis. Find the volume of the resulting solid using the method of discs (also known as washers).

I wonder where I have seen this problem before... This is an interesting problem though. This time, rather than using cylinders changing with respect to y, we will use discs that change in area with respect to x. Similarly, instead of integrating from 0 to 1 in the y-direction, we will integrate from 0 to 1 in the x-direction. Refer to the interactive image on the next page.

Interactive 10.2.2 Example 10.2.1

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Area of Orange washer/disc

The area of the orange disc is equal to the area of the outer shape minus the area of the inner shape because only the area between the two curves counts.

Summation along x-axis

You can imagine if there were infinitely many, infinitely thin discs like the blue disc and the orange disc, that the summations of their areas along the x-axis (their integral with respect to dx) would equal to the total volume of the solid the question asks for. Let’s go ahead and solve this integral.

Let’s call the outer green curve $y_{2} = \sqrt{x}$ and the inner, dotted red curve, $y_{1} = x^{2}$ . Then, the area of each disc is represented by $A = π \cdot y_{2}^{2} - π \cdot y_{1}^{2}$ . Integrating these discs, or infinitely summing them in the x-direction, gives:

$\begin{aligned} V & = \int_{0}^{1} π (y_{2}^{2} - y_{1}^{2}) d x \\ = \int_{0}^{1} π ({(\sqrt{x})}^{2} - {(x^{2})}^{2}) d x \\ = \int_{0}^{1} π (x - x^{4}) d x \\ = π ({[\frac{x^{2}}{2}]}_{0}^{1} - {[\frac{x^{5}}{5}]}_{0}^{1}) \\ = π (\frac{1}{2} - \frac{1}{5}) \end{aligned}$

$= \frac{3 π}{10}$ We got the same answer! Amazing!

Interactive 10.2.3 A Hint for Example 10.2.2

Magnification 1×

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Interval of Integration

Since we are having r be as general as possible, what is the interval of integration? It is not a number, but a variable.

Radius of each disc

I won’t show discs this time, but we can by now probably imagine that the radius of each disc corresponds to the y-value of the graph of the semi-circle.

We can also imagine that the area of each disc will be equal to A=π•y²

Equation of a circle

y=√(r² - x²)

We want to develop a general formula so we refrain from plugging in r=3, and we will have r in our integral all the way until the final formula.

Example Problem 10.2.2

Using the disc method, prove that the volume of a sphere is $V = \frac{4}{3} π r^{3}$

We have no numbers in this problem, because everything is relative to the radius. Our interval of integration is from -r to r because the sphere and cross-sectional circle are both centered at the origin. In this case, it doesn’t matter whether we integrate with respect to y with horizontal discs or if we integrate with respect to x with vertical discs. Our final result should still be the same. Keep in mind that the equation of a circle is $x^{2} + y^{2} = r^{2}$ .

$\begin{aligned} V & = \int_{- r}^{r} π y^{2} d x \\ = \int_{- r}^{r} π (r^{2} - x^{2}) d x \end{aligned}$

We can go ahead and evaluate the integral now:

$= {[π r^{2} x]}_{- r}^{r} - {[π \frac{x^{3}}{3}]}_{- r}^{r}$
$\begin{aligned} = & [(π r^{2} \cdot r) - (π r^{2} \cdot (- r))] \\ - [(π \frac{r^{3}}{3}) - (π \frac{(- r)^{3}}{3})] \end{aligned}$
$= 2 π r^{3} - 2 π \frac{r^{3}}{3}$
$= \frac{6 π r^{3}}{3} - \frac{2 π r^{3}}{3}$

$= \frac{4}{3} π r^{3}$

Glossary and deeper reading · 3 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Cylindrical shell method
The cylindrical shell method computes volume by accumulating thin cylindrical layers parallel to a rotation axis. The volume contribution is circumference times shell height times thickness. For example, rotating a vertical strip at x>0 around the vertical axis gives a contribution 2πx·height·dx. Choose the radius as a distance from the axis and check the shell height against the bounding region.
Read more: Original source · Cylindrical Shells (beyond AB) →
Disc method
The disc method finds a solid’s volume using circular slices perpendicular to the rotation axis. A slice with radius R has area πR². For example, rotating the region under y=x on [0,1] around the horizontal axis gives volume equal to the integral of πx². If the region leaves a hole around the axis, a washer rather than a full disc is needed.
Read more: 8.9 · Disc method →
Washer method
The washer method uses circular slices with a central hole to compute volume. A slice with outer radius R and inner radius r has area π(R²−r²). For example, rotating the region 1≤y≤2 on a horizontal interval around the horizontal axis gives washer area 3π. Both radii are distances from the rotation axis and must be expressed using the same slice variable.
Read more: 8.11 · Washer method →

Interactive 10.2.1 A familiar shape

Same curve

Infinitely thin discs

Interactive 10.2.2 Example 10.2.1

Area of Orange washer/disc

Summation along x-axis

Interactive 10.2.3 A Hint for Example 10.2.2

Interval of Integration

Radius of each disc

Equation of a circle

Pop Quiz 10.2.1 Method of discs

Cylindrical shell method

Disc method

Washer method

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