Why is a washer's area R2 - r2 and not (R - r)2?

The face is a disc of radius R with a disc of radius r cut out of its middle, so its area is the difference of the two disc areas. (R - r)2 is the area of a disc only as wide as the ring's wall, which is much smaller.

Unit 8 · Topic 8.12 · about 30 minutes · 7 parts · 11 scored checks, 3 of them multiple choice, plus 1 from earlier lessons

AP Calculus AB 8.12: Volume with Washer Method: Revolving Around Other Axes

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 made by revolving a region about a horizontal or vertical line other than an axis, when its cross sections are ring shaped, using 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 revolved about a line.	Read which line the region spins about before you measure anything; a question can name a line that is not an axis.
Washer method	A way to find the volume of a solid of revolution whose cross sections are ring shaped, by integrating the area of each ring.	A ring's area is one disc's area minus another's, so keep the two radii separate until each is squared.
Outer radius and inner radius	For a ring-shaped cross section, the distances from the line of revolution to the boundary of the region farther from it and to the boundary nearer to it.	A height read off a graph is a distance to the $x$ -axis, which is a radius only when the $x$ -axis is the line.
Line of revolution	The horizontal or vertical line a region is revolved about.	Note whether it is above, below, left or right of the region before you decide which boundary is farther.

The region between y = 4x - x squared and y = x for x from 0 to 3, shaded, with its mirror image below the dashed line y = -1, the line it spins about. At x = 1 one slice is drawn as a ring: its outer radius runs from the line up to y = 3 and is 4, and its inner radius runs up to y = 1 and is 2.

Part 1 Think first Move the line, keep the wall

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A pipe fitting is modeled by spinning a flat rectangle about a line. The rectangle is 3 centimeters long, its near edge is 1 centimeter from the line and its far edge is 2 centimeters from it, and the fitting's volume is $9 π$ cubic centimeters. The designer moves the line 1 centimeter farther from the rectangle, so the edges are now 2 and 3 centimeters away. What is the new volume?

In short. When a region is revolved about a horizontal or vertical line and its cross sections are rings, the volume is $π$ times the integral of the outer radius squared minus the inner radius squared, with both radii measured from that line.

Each radius is a distance from the line to a boundary of the region: about $y = k$ , a height minus $k$ or $k$ minus a height.
The boundary farther from the line gives the outer radius, so the two curves can trade roles when the line is on the other side of the region.
A horizontal line calls for vertical slices and $d x$ ; a vertical line calls for horizontal slices, $d y$ , and boundaries written as $x$ in terms of $y$ .

Part 2 Learn it Three ideas, each checked as you go

12a. Radii measured from a line below the region

When a region spins about the $x$ -axis, a radius is just a height. When it spins about some other horizontal line, $y = k$ , a radius is a distance to that line instead.

Cut the region with a thin vertical slice at $x$ . Spun about $y = k$ , the slice sweeps out a washer, a disc with a hole in it. Say the region lies between an upper curve $y = f (x)$ and a lower curve $y = g (x)$ , and the line is below the region. Then

the outer radius reaches the upper curve: $R (x) = f (x) - k$ ;
the inner radius reaches the lower curve: $r (x) = g (x) - k$ .

The washer's face has area $π R (x)^{2} - π r (x)^{2}$ , and adding up thin washers from $x = a$ to $x = b$ gives

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

In the figure at the top, the region lies between $y = 4 x - x^{2}$ and $y = x$ for $0 \leq x \leq 3$ , and the line is $y = - 1$ . Subtracting $- 1$ means adding 1, so $R (x) = 4 x - x^{2} + 1$ and $r (x) = x + 1$ . At $x = 1$ those are 4 and 2, the two radii drawn. The volume is

V = π \int_{0}^{3} [(4 x - x^{2} + 1)^{2} - (x + 1)^{2}] d x .

Quick check · Multiple choice

No calculator. The region bounded by $y = 3 x$ and $y = x^{2}$ is revolved about the line $y = - 2$ . Which integral gives the volume of the solid?

12b. When the line is above the region

Now put the line above the region: $y = k$ , with $k$ at least as large as every height in the region. A distance is the larger coordinate minus the smaller, so each radius is $k$ minus a height.

What changes is which curve is farther away. The lower curve is now farther from the line, so it gives the outer radius:

R (x) = k - g (x), r (x) = k - f (x) .

Spin the figure's region about $y = 5$ instead. The lower edge, $y = x$ , is now the far one, so $R (x) = 5 - x$ and $r (x) = 5 - (4 x - x^{2})$ . At $x = 2$ these are 3 and 1, and

V = π \int_{0}^{3} [(5 - x)^{2} - {(5 - 4 x + x^{2})}^{2}] d x .

Squaring hides one kind of slip and not another. Since $(f (x) - k)^{2} = (k - f (x))^{2}$ , writing a radius backward does no harm once it is squared. Giving the outer radius to the nearer curve does harm: the integrand comes out negative, and so does the volume.

So before you write anything, mark the line on a sketch, pick one $x$ inside the interval, and ask which edge of the region is farther from the line there.

Quick check · Match

No calculator. A region lies between $y = f (x)$ above and $y = g (x)$ below, and every point in it has $0 \leq y \leq 4$ . The region is revolved about one of two lines. Tap a radius, then the expression for it.

12c. A vertical line: slice across $y$

When the region spins about a vertical line $x = h$ , each washer must be perpendicular to that line, so the slices are horizontal and each has thickness $d y$ . Everything is then written in terms of $y$ :

write each boundary as $x$ in terms of $y$ , such as $x = \sqrt{y}$ for $y = x^{2}$ with $x \geq 0$ ;
use limits that are $y$ -values, the bottom and the top of the region;
measure each radius sideways, as a horizontal distance to $x = h$ .

If the line is to the left of the region, each radius is a boundary's $x$ minus $h$ , and the right boundary is the farther one. If the line is to the right, each radius is $h$ minus a boundary's $x$ , and the left boundary is the farther one. Then

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

For example, take the region bounded by $y = x^{3}$ , the line $y = 8$ and the $y$ -axis, spun about $x = - 2$ . Its right edge is $x = \sqrt[3]{y}$ and its left edge is $x = 0$ , both to the right of the line, so $R (y) = \sqrt[3]{y} + 2$ and $r (y) = 0 + 2 = 2$ . The region runs from $y = 0$ to $y = 8$ , and

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

Quick check · Sort it

No calculator. Each region is revolved about a line, and the volume is found with washers. Sort each one by the variable you integrate with respect to. Tap a card, then a bin.

Integrate with respect to $x$

Integrate with respect to $y$

Common mistakes

The mistake	Why it is wrong	Do this instead
Using the heights $f (x)$ and $g (x)$ as the radii when the region spins about $y = k$ rather than the $x$ -axis.	A height is a distance to the $x$ -axis. A radius is a distance to the line the region spins about, which here is a different line.	Subtract: $f (x) - k$ when the line is below the region, and $k - f (x)$ when it is above.
Giving the outer radius to the upper curve when the line is above the region.	Above the region, the lower curve is the farther one. With the radii swapped, the integrand is negative.	Pick one $x$ inside the interval, find both edges' distances to the line, and give the larger one to $R$ .
Writing $π \int_{a}^{b} {[R (x) - r (x)]}^{2} d x$ .	A washer's face is a disc of radius $R$ with a disc of radius $r$ removed, so its area is $π R^{2} - π r^{2}$ , not $π (R - r)^{2}$ .	Square each radius on its own, then subtract.

Part 3 Watch it done A region spun about a line above it

Watch each step, then do the last one yourself.

The region bounded by $y = 2 x$ and $y = x^{2}$ is revolved about the line $y = 4$ . Find the volume of the solid.

Step 1 · Find the interval and where the line sits
$2 x = x^{2}$ at $x = 0$ and $x = 2$ . On $[0, 2]$ , $2 x \geq x^{2}$ , and the highest point of the region is $(2, 4)$ , so the line $y = 4$ lies on or above the whole region.
Step 2 · Measure each radius from $y = 4$
The line is above the region, so each radius is 4 minus a height. $R (x) = 4 - x^{2}$ and $r (x) = 4 - 2 x$ .
Why this step?
Why does the outer radius come from $y = x^{2}$ , the lower curve?
The line is above the region, so the lower curve is the one farther from it. At $x = 1$ , for example, $y = x^{2}$ is 3 below the line and $y = 2 x$ is only 2 below it.
Step 3 · Write the integral
$V = π \int_{0}^{2} [(4 - x^{2})^{2} - (4 - 2 x)^{2}] d x$
Step 4 · Expand the integrand
$(4 - x^{2})^{2} - (4 - 2 x)^{2} = (16 - 8 x^{2} + x^{4}) - (16 - 16 x + 4 x^{2}) = x^{4} - 12 x^{2} + 16 x$
Step 5 · Find an antiderivative
$V = π \int_{0}^{2} (x^{4} - 12 x^{2} + 16 x) d x = π {[\frac{x^{5}}{5} - 4 x^{3} + 8 x^{2}]}_{0}^{2}$
Your step · Calculate

No calculator. Finish it. The volume is $π {[\frac{x^{5}}{5} - 4 x^{3} + 8 x^{2}]}_{0}^{2}$ , which equals $k π$ . What is $k$ ? Give an exact decimal.

Your answer

Part 4 Try it Two new solids

Try it · Calculate

No calculator. The region in the first quadrant bounded by $y = x^{2}$ , the line $y = 4$ and the $y$ -axis is revolved about the line $x = 3$ . The volume of the solid is $k π$ . What is $k$ ?

Your answer

Try it · Decide and justify

No calculator. The region bounded by $y = 4 - x^{2}$ and the $x$ -axis for $0 \leq x \leq 2$ is revolved about the line $y = - 1$ . A classmate says this calls for the washer method with an inner radius of 1, even though the region touches the $x$ -axis. Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Set it up by hand

Free response is handwritten on the exam, so work this on paper. Nothing you type here leaves the page. Then open the model answer and score yourself point by point.

Free response is 50% of the exam score. This one is shaped like Question 4, from free-response Part B (calculator not permitted).

Free-response practice · Question 4 style

No calculator. A machined part is modeled by revolving a region about a line. Let $S$ be the region bounded by the graphs of $y = \sqrt{x}$ and $y = \frac{x}{2}$ , where $x$ and $y$ are measured in centimeters. The graphs meet at $(0, 0)$ and $(4, 2)$ .

(a) Write, but do not evaluate, an integral expression for the volume of the solid generated when $S$ is revolved about the vertical line $x = 5$ .

(b) Find the volume of the solid generated when $S$ is revolved about the horizontal line $y = - 2$ . Show the work that leads to your answer, and indicate units of measure.

(c) Explain why every slice of the solid in part (b) perpendicular to the $x$ -axis has a hole in it, so that the washer method applies.

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

Model answer

(a) The line is vertical, so slice horizontally from $y = 0$ to $y = 2$ . In terms of $y$ the boundaries are $x = y^{2}$ on the left and $x = 2 y$ on the right. The line $x = 5$ is to the right of $S$ , so the left boundary is farther: $R (y) = 5 - y^{2}$ and $r (y) = 5 - 2 y$ .

V = π \int_{0}^{2} [(5 - y^{2})^{2} - (5 - 2 y)^{2}] d y

(b) The line is below $S$ , so $R (x) = \sqrt{x} + 2$ and $r (x) = \frac{x}{2} + 2$ .

V = π \int_{0}^{4} [{(\sqrt{x} + 2)}^{2} - {(\frac{x}{2} + 2)}^{2}] d x = π \int_{0}^{4} (4 \sqrt{x} - x - \frac{x^{2}}{4}) d x

= π {[\frac{8}{3} x^{3 / 2} - \frac{x^{2}}{2} - \frac{x^{3}}{12}]}_{0}^{4} = π (\frac{64}{3} - 8 - \frac{16}{3}) = 8 π

The volume is $8 π$ cubic centimeters.

(c) The lower edge of $S$ is $y = \frac{x}{2}$ , which is never below $y = 0$ , so it is always at least 2 centimeters above the line $y = - 2$ . Every slice therefore has a hole of radius $r (x) = \frac{x}{2} + 2 \geq 2$ , and its cross section is a ring, not a full disc.

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

(a) Slices horizontally, with limits $y = 0$ and $y = 2$ , and writes the boundaries as $x = y^{2}$ and $x = 2 y$ .
(a) Writes $π \int_{0}^{2} [(5 - y^{2})^{2} - (5 - 2 y)^{2}] d y$ , with the outer radius from $x = y^{2}$ .
(b) Writes $π \int_{0}^{4} [{(\sqrt{x} + 2)}^{2} - {(\frac{x}{2} + 2)}^{2}] d x$ .
(b) Shows an antiderivative and gets $8 π$ cubic centimeters.
(c) Says the lower edge stays at least 2 units above $y = - 2$ , so the inner radius is never 0.

Part 6 Check yourself One bushing, spun two ways

Both case questions use the profile in the table. Answer them from it.

A metal bushing is made by spinning a flat profile about a line. The profile is the region between two curves, with $x$ and $y$ in centimeters.

The edges of the bushing's profile
Edge	Equation
Upper edge	$y = x + 2$
Lower edge	$y = x^{2}$
Where the edges meet	$(- 1, 1)$ and $(2, 4)$

Case question 1 · Calculate

No calculator. The profile is spun about the line $y = - 1$ . The volume of the bushing is $k π$ cubic centimeters. What is $k$ ? Give an exact decimal.

Your answer

Case question 2 · Multiple choice

No calculator. Instead, the profile is spun about the line $y = 5$ . Which integral gives the volume of the bushing, in cubic centimeters?

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

Check yourself · Spot the error

No calculator. A student finds the volume of the solid generated when the region bounded by $x = y^{2}$ and $x = 4$ is revolved about the line $x = 6$ . Tap the first line with a mistake.

Check yourself · Fill the statement

No calculator. The region bounded by $x = y^{2}$ and $x = y + 2$ is revolved about the line $x = - 1$ . Tap a blank, then tap an answer.

The line is vertical, so the slices are and each one has thickness . The outer radius of the slice at height $y$ is .

Part 7 Keep it Bring back earlier ideas

This lesson stands on the washer method about an axis and on evaluating a definite integral, here with a substitution, which is how every washer problem ends. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. The region bounded by $y = x$ and $y = x^{3}$ for $0 \leq x \leq 1$ is revolved about the $x$ -axis. Which integral gives the volume of the solid?

From 6.14 Selecting Techniques for Antidifferentiation · Calculate

No calculator. Evaluate $\int_{0}^{1} 6 x (x^{2} + 1)^{2} d x$ .

Your answer

In three lines

About a line other than an axis, each radius is the distance from that line to a boundary of the region.
The boundary farther from the line gives the outer radius, so the curves trade roles when the line is on the other side of the region.
A horizontal line takes $d x$ and $V = π \int_{a}^{b} [R (x)^{2} - r (x)^{2}] d x$ ; a vertical line takes $d y$ , with each boundary written as $x$ in terms of $y$ .

Unit 8 check: 12 questions from every lesson in Unit 8

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.

Before you write the integral, write the two radii on a line of their own, for example $R (x) = 5 - x^{2}$ and $r (x) = 3 - x$ , each measured from the given line. That keeps your setup visible in your work, and you can check at one value of $x$ that $R \geq r$ . Put $π$ , both limits and $d x$ or $d y$ in the integral, and give units on a volume in context.

Questions students ask

Do I always subtract

k

from each height?

Only when the line is below the region. A radius is a distance, the larger coordinate minus the smaller. Below the region that is a height minus $k$ ; above it, it is $k$ minus a height. For a line such as $y = - 2$ , subtracting $- 2$ means adding 2.

How do I know which curve gives the outer radius?

Pick one value inside the interval, find each edge's distance to the line there, and give the larger one to $R$ . As long as the line stays on one side of the region, the same edge is farther all the way across.

When is it a disc instead of a washer?

When the region's edge lies along the line for the whole interval. Then the inner radius is 0 and the formula becomes the disc method. If there is a gap anywhere between the region and the line, the slices there have holes.

Why is a washer's area

π R^{2} - π r^{2}

and not

π (R - r)^{2}

The face is a disc of radius $R$ with a disc of radius $r$ cut out of its middle, so its area is the difference of the two disc areas. $π (R - r)^{2}$ is the area of a disc only as wide as the ring's wall, which is much smaller.

Can I leave a volume in terms of

π

Yes, unless a question asks for a decimal. An exact answer such as $8 π$ is complete. When a decimal is asked for, give three places after the decimal point.

Course alignment, for teachers

AP Calculus AB topic 8.12, Unit 8: Applications of Integration.

Skill 2.D: Identify how mathematical characteristics or properties of functions are related in different representations.

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.

What you'll be able to do

Key vocabulary

Part 1 Think first Move the line, keep the wall

Part 2 Learn it Three ideas, each checked as you go

12a. Radii measured from a line below the region

12b. When the line is above the region

12c. A vertical line: slice across y

Part 3 Watch it done A region spun about a line above it

Part 4 Try it Two new solids

Part 5 Explain it Set it up by hand

Part 6 Check yourself One bushing, spun two ways

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic

12c. A vertical line: slice across $y$