Unit 8 · Topic 8.11 · about 30 minutes · 7 parts · 11 scored checks, 3 of them multiple choice, plus 1 from earlier lessons

AP Calculus AB 8.11: Volume with Washer Method: Revolving Around the x- or y-Axis

Q: Why can't I subtract the radii first and then square?

A ring's area is a big disc minus a small one, R2 - r2. With R = 6 and r = 4 that is 20, but (6 - 4)2 is only 4, the area of a solid disc as wide as the ring.

Q: When the region turns about the y-axis, do I have to rewrite the curves?

Yes. The slices are horizontal, so each radius is an x-distance at a height y. Solve each curve for x, so that y = x3 becomes x = [3]y, and use y-limits.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 8: 10 to 15% of the multiple-choice section

What you'll be able to do

Calculate the volume of a solid formed by turning a region about the $x$ -axis or the $y$ -axis when its slices are rings, using a definite integral with the washer method.

Key vocabulary

Term	What it means	The exam trap
Solid of revolution	The solid swept out when a region in the plane is turned all the way around a line, called the axis of revolution.	One region gives a different solid about each axis, so read which axis the question names before you set anything up.
Washer	A ring-shaped cross section: a disc with a smaller disc removed from its center.	A washer is not a thin disc. Check whether a slice has a hole before you write its area.
Washer method	Finding the volume of a solid of revolution whose cross sections are rings by integrating the areas of those rings.	Look for a gap between the region and the axis before you choose between discs and washers.
Outer radius and inner radius	In a washer, the distances from the axis of revolution to the far edge and to the near edge of the region, written $R$ and $r$ .	Measure both from the axis, and check which boundary is farther from it at an input between the limits.

The region between the curve y = 4 minus x squared over 4 and the line y = 1, from x = 0 to about 3.46, shaded above the x-axis, with its mirror image shaded lightly below. A ring drawn at x = 1 is the washer there: its outer radius is 3.75, up to the curve, and its inner radius is 1, up to the line. The solid's volume is about 95.77 cubic units.

Part 1 Think first What is left after drilling?

Commit to a guess first. Even a wrong one helps the idea stick.

Predict first

A machinist starts with a solid steel rod 10 centimeters long with a radius of 3 centimeters. She drills a hole of radius 1 centimeter straight through it along its center line. How much steel is left?

In short. The washer method finds the volume of a solid of revolution with a hole along its axis by integrating the area of each ring-shaped slice, $π (R^{2} - r^{2})$ , across the region.

Each slice perpendicular to the axis is a washer: a disc of outer radius $R$ with a disc of inner radius $r$ taken out of its center.
About the $x$ -axis, $V = π \int_{a}^{b} (R (x)^{2} - r (x)^{2}) d x$ ; about the $y$ -axis, the radii and the limits are in terms of $y$ and the integral ends in $d y$ .
Both radii are measured from the axis of revolution, and the outer radius belongs to the boundary farther from it.

Part 2 Learn it From a ring to a volume

2a. A slice with a hole is a washer

Turn a region about the $x$ -axis and slice the solid straight across, perpendicular to the axis. If the region reaches down to the axis, every slice is a solid disc. If there is a gap between the region and the axis, the solid has a hole down its middle, and every slice is a ring, called a washer.

A washer is a big disc with a small disc taken out of its center. Its outer radius $R$ runs from the axis to the far edge of the region, and its inner radius $r$ runs from the axis to the near edge. So its area is

π R^{2} - π r^{2} = π (R^{2} - r^{2}) .

In the picture at the top, the region lies between $y = 4 - \frac{x^{2}}{4}$ and $y = 1$ , to the right of the $y$ -axis. The slice at $x = 1$ has $R = 3.75$ and $r = 1$ , so its area is $π ({3.75}^{2} - 1^{2}) = 13.0625 π$ square units.

Each slice is a thin washer of thickness $d x$ . Adding up those slices from $x = a$ to $x = b$ is a definite integral, the washer method:

V = π \int_{a}^{b} (R (x)^{2} - r (x)^{2}) d x .

Quick check · Calculate

No calculator. The region bounded by $y = 5 - x$ , the line $y = 1$ and the $y$ -axis is revolved about the $x$ -axis. The cross section of the solid at $x = 1$ is a washer with area $k π$ square units. What is $k$ ?

Your answer

2b. Which radius is which

Both radii are distances from the axis of revolution. For a region above the $x$ -axis, the upper curve is farther from the axis, so it gives $R (x)$ , and the lower curve gives $r (x)$ . The limits $a$ and $b$ are where the region starts and ends, often where the two curves meet.

Take the region between $y = x + 3$ and $y = x^{2} + 1$ . The curves meet where $x^{2} + 1 = x + 3$ , at $x = - 1$ and $x = 2$ . At $x = 0$ the line is at height 3 and the parabola at height 1, so the line is the outer edge:

V = π \int_{- 1}^{2} ((x + 3)^{2} - {(x^{2} + 1)}^{2}) d x .

Two slips are common. One is squaring the difference, $(R - r)^{2}$ . With $R = 5$ and $r = 3$ the ring's area is $π (25 - 9) = 16 π$ , but $π (5 - 3)^{2}$ is only $4 π$ , the area of a solid disc as wide as the ring. The other is swapping the radii. Then $R^{2} - r^{2}$ is negative, and a negative volume is a sure sign the curves are the wrong way round.

Quick check · Spot the error

No calculator. A student finds the volume of the solid formed when the region between $y = \sqrt{x}$ and $y = \frac{x}{2}$ is revolved about the $x$ -axis. Tap the first line with a mistake.

2c. Turning about the y-axis

When the region turns about the $y$ -axis, slice it perpendicular to that axis, which means horizontally. Each slice is a washer of thickness $d y$ , so the whole setup moves into $y$ :

each radius is a horizontal distance from the $y$ -axis, so write each boundary as $x$ in terms of $y$ ;
the limits are the lowest and highest $y$ -values of the region.

V = π \int_{c}^{d} (R (y)^{2} - r (y)^{2}) d y

Take the region in the first quadrant between $y = 4 x$ and $y = x^{3}$ . They meet at $(0, 0)$ and $(2, 8)$ , so the region runs from $y = 0$ to $y = 8$ , not from 0 to 2. Solved for $x$ , the line is $x = \frac{y}{4}$ and the cubic is $x = \sqrt[3]{y}$ . At $y = 1$ these are $\frac{1}{4}$ and 1, so the cubic is farther from the axis:

V = π \int_{0}^{8} ({(\sqrt[3]{y})}^{2} - {(\frac{y}{4})}^{2}) d y .

Let the axis tell you the orientation. An axis that runs up and down means horizontal slices, $d y$ and $y$ -limits.

Quick check · Multiple choice

No calculator. The region in the first quadrant between $y = x$ and $y = x^{2}$ is revolved about the $y$ -axis. Which integral gives the volume of the solid?

Common mistakes

The mistake	Why it is wrong	Do this instead
Writing $π \int_{a}^{b} {(R (x) - r (x))}^{2} d x$ , which squares the difference of the radii.	$π (R - r)^{2}$ is the area of a solid disc as wide as the ring, which is much smaller than the ring itself.	Square each radius on its own, then subtract: the integrand is $π (R (x)^{2} - r (x)^{2})$ .
Using $x$ -limits and $d x$ when the region turns about the $y$ -axis.	Slices perpendicular to the $y$ -axis are horizontal, so their thickness is $d y$ and they run between $y$ -values.	Rewrite each boundary as $x$ in terms of $y$ , and read the limits off the $y$ -axis.
Putting the boundary nearer the axis in as the outer radius.	Then $R^{2} - r^{2}$ is negative, and the integral gives a negative volume.	Before you integrate, test one input between the limits to see which boundary is farther from the axis.

Part 3 Watch it done A washer volume from start to finish

Read each step as it opens, then do the last one yourself.

The region bounded by $y = 2 \sqrt{x}$ , the line $y = 2$ and the line $x = 4$ is revolved about the $x$ -axis. Find the volume of the solid.

Step 1 · Find the limits
The curve meets the line $y = 2$ where $2 \sqrt{x} = 2$ , so at $x = 1$ . The region runs from $x = 1$ to $x = 4$ .
Step 2 · Name the radii
For $1 < x < 4$ , $2 \sqrt{x} > 2$ , so the curve is farther from the $x$ -axis: $R (x) = 2 \sqrt{x}$ and $r (x) = 2$ .
Step 3 · Write the integral
$V = π \int_{1}^{4} ({(2 \sqrt{x})}^{2} - 2^{2}) d x = π \int_{1}^{4} (4 x - 4) d x$
Why this step?
Why is the integrand $4 x - 4$ and not ${(2 \sqrt{x} - 2)}^{2}$ ?
Each slice is a disc of radius $2 \sqrt{x}$ with a disc of radius 2 removed from its center, so its area is $π {(2 \sqrt{x})}^{2} - π (2)^{2}$ . Squaring the difference $2 \sqrt{x} - 2$ would give the area of a solid disc as wide as the ring, which is much smaller.
Step 4 · Find an antiderivative
$V = π {[2 x^{2} - 4 x]}_{1}^{4}$
Your step · Calculate

No calculator. Finish it. The volume is $π {[2 x^{2} - 4 x]}_{1}^{4}$ , which equals $k π$ . What is $k$ ?

Your answer

Part 4 Try it A different axis, then a calculator

Try it · Calculate

No calculator. The region to the right of the line $x = 1$ , bounded by $y = x^{2}$ and the line $y = 4$ , is revolved about the $y$ -axis. The volume of the solid is $k π$ . What is $k$ ? Give an exact decimal.

Your answer

Try it · Calculate

Graphing calculator required. The region enclosed by the graphs of $y = \cos x$ and $y = x^{2}$ is revolved about the $x$ -axis. What is the volume of the solid? Use radian mode, and give your answer to three decimal places.

Your answer

Part 5 Explain it Write it the way the exam asks

Free response is written by hand on the exam, so do this one on paper. Nothing you type on this page is sent anywhere. Then open the model answer and check your work against each point.

Free response is 50% of the exam score. This one is shaped like Question 6, from free-response Part B (calculator not permitted).

Free-response practice · Question 6 style

No calculator. A glass ornament is made by revolving a region about an axis. In a coordinate plane measured in centimeters, the region $S$ lies between the graphs of $y = 5 - x$ and $y = \frac{4}{x}$ , which meet at $(1, 4)$ and $(4, 1)$ .

(a) Write, but do not evaluate, an integral expression for the volume of the solid formed when $S$ is revolved about the $x$ -axis.

(b) Find the volume of the solid in part (a). Show the antiderivative you use, and include units.

(c) A second ornament is made by revolving $S$ about the $y$ -axis. Write, but do not evaluate, an integral expression in terms of $y$ for its volume. Explain why it has the same volume as the ornament in part (a).

Your answer stays on this page. It is never sent or saved.

Model answer

(a) For $1 < x < 4$ , $5 - x > \frac{4}{x} > 0$ , so the outer radius is $5 - x$ and the inner radius is $\frac{4}{x}$ . The volume is $π \int_{1}^{4} ((5 - x)^{2} - {(\frac{4}{x})}^{2}) d x$ .

(b) The integrand is $25 - 10 x + x^{2} - 16 x^{- 2}$ , with antiderivative $25 x - 5 x^{2} + \frac{x^{3}}{3} + \frac{16}{x}$ . So $V = π {[25 x - 5 x^{2} + \frac{x^{3}}{3} + \frac{16}{x}]}_{1}^{4} = π (\frac{136}{3} - \frac{109}{3}) = 9 π$ cubic centimeters.

(c) Solved for $x$ , the boundaries are $x = 5 - y$ and $x = \frac{4}{y}$ for $1 \leq y \leq 4$ , and $5 - y$ is farther from the $y$ -axis. The volume is $π \int_{1}^{4} ((5 - y)^{2} - {(\frac{4}{y})}^{2}) d y$ . This is the integral from part (a) with $y$ in place of $x$ , and renaming the variable does not change the value of a definite integral, so this ornament also has volume $9 π$ cubic centimeters.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Writes $π \int_{1}^{4} ((5 - x)^{2} - {(\frac{4}{x})}^{2}) d x$ , with limits 1 and 4 and $5 - x$ as the outer radius.
Shows an antiderivative of the integrand, such as $25 x - 5 x^{2} + \frac{x^{3}}{3} + \frac{16}{x}$ .
Gets $9 π$ cubic centimeters, with units.
Writes $π \int_{1}^{4} ((5 - y)^{2} - {(\frac{4}{y})}^{2}) d y$ , with $y$ -limits 1 and 4.
Explains that the integral in part (c) is the one in part (a) with the variable renamed, so the two volumes are equal.

Part 6 Check yourself A turned candle holder

The first two questions use the table. The others stand on their own.

A woodworker turns a candle holder 8 centimeters tall on a lathe, so every cross section perpendicular to its center line is a circle. Then she bores a hole down the center line, wider near the top where the candle sits. At a height of $x$ centimeters above the base, the outer radius of the holder is $R (x)$ centimeters and the radius of the hole is $r (x)$ centimeters. Some values are in the table.

Radii of the candle holder
$x$ (centimeters)	$R (x)$ (centimeters)	$r (x)$ (centimeters)
0	4	1
2	3.5	1
4	3	1.5
6	3	2
8	3.5	2

Case question 1 · Multiple choice

No calculator. Which expression gives the volume of the candle holder, in cubic centimeters?

Case question 2 · Calculate

No calculator. Use a left Riemann sum with the four subintervals in the table to approximate the volume of the candle holder. The approximation is $k π$ cubic centimeters. What is $k$ ?

Your answer

The rest stand on their own and do not use the case.

Check yourself · Sort it

No calculator. Each region is turned about the axis named. Sort it by the shape of its slices perpendicular to that axis.

Washers: the slices are rings

Discs: the slices are solid

Check yourself · Decide and justify

No calculator. On $[a, b]$ , $0 \leq g (x) \leq f (x)$ . A classmate says that turning the region between the graphs of $f$ and $g$ about the $x$ -axis gives the volume of the solid formed by turning the region under $f$ , minus the volume of the solid formed by turning the region under $g$ . Is the classmate right?

Which reason supports your answer?

Part 7 Keep it Bring back earlier ideas

Washers are built from the discs of 8.9, and the volumes you work by hand need antiderivatives, sometimes found by substitution, from Unit 6. Both are worth a quick look back.

Bring it back · Multiple choice

No calculator. The region bounded by $y = \frac{1}{x}$ , the $x$ -axis and the lines $x = 1$ and $x = 3$ is revolved about the $x$ -axis. Which integral gives the volume of the solid?

From 6.9 Integrating Using Substitution · Calculate

No calculator. For some positive integer $n$ , $\int x^{3} (x^{4} + 1)^{4} d x = \frac{1}{n} (x^{4} + 1)^{5} + C$ . What is $n$ ?

Your answer

In three lines

A solid with a hole along its axis has ring-shaped slices, called washers.
A washer's area is $π (R^{2} - r^{2})$ , with both radii measured from the axis of revolution.
Integrate that area across the region: in $x$ for the $x$ -axis, in $y$ for the $y$ -axis.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

Write the whole integral, with $π$ , each radius squared on its own, both limits and the differential, before you look for an antiderivative, so a reader can follow your setup even if the arithmetic slips. Give a volume in cubic units, and when the axis is the $y$ -axis, check that the limits are $y$ -values.

Questions students ask

How do I know whether to use discs or washers?

Look at a slice perpendicular to the axis. If the region reaches the axis there, the slice is a solid disc. If there is a gap between the region and the axis, the slice is a ring, and you use washers.

Why can't I subtract the radii first and then square?

A ring's area is a big disc minus a small one, $π R^{2} - π r^{2}$ . With $R = 6$ and $r = 4$ that is $20 π$ , but $π (6 - 4)^{2}$ is only $4 π$ , the area of a solid disc as wide as the ring.

When the region turns about the

y

-axis, do I have to rewrite the curves?

Yes. The slices are horizontal, so each radius is an $x$ -distance at a height $y$ . Solve each curve for $x$ , so that $y = x^{3}$ becomes $x = \sqrt[3]{y}$ , and use $y$ -limits.

What does a negative volume mean?

The inner and outer radii are swapped. Test an input between the limits to see which boundary is farther from the axis, and put that one in as $R$ .

Can the inner radius be 0 at a single point?

Yes. If the region touches the axis at only one point, such as where two curves meet on the axis, the slices are rings everywhere else, and the washer integral still gives the volume.

Course alignment, for teachers

AP Calculus AB topic 8.11, Unit 8: Applications of Integration.

Skill 4.E: Apply appropriate rounding procedures.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.