Unit 8 · Topic 8.9 · about 30 minutes · 7 parts · 11 scored checks, 3 of them multiple choice, plus 1 from earlier lessons

AP Calculus AB 8.9: Volume with Disc Method: Revolving Around the x- or y-Axis

Q: Do I have to expand the square before integrating?

Only if you need to. A polynomial like (2 + x2)2 is easiest to expand and integrate term by term. Something like (4 - y)2 simplifies to 4 - y at once. On a calculator question you can integrate the square as it is.

Q: Should my answer have in it?

Without a calculator, yes: an exact answer such as 163 is the volume. With a calculator, give the decimal to the places the question asks for, after writing the integral that produced it.

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

Find the volume of a solid formed by revolving a region around the x-axis or the y-axis, using a definite integral with the disc method.

Key vocabulary

Term	What it means	The exam trap
Solid of revolution	The solid swept out when a flat region is revolved around a line.	Read which line the region turns around before you set anything up; the same region can make more than one solid.
Axis of revolution	The line a region is revolved around; in this topic, the x-axis or the y-axis.	Note whether the axis is horizontal or vertical. That orientation decides which variable you integrate with respect to.
Disc method	Finding the volume of a solid of revolution with a definite integral that adds the volumes of thin circular slices cut perpendicular to the axis.	Check what you wrote inside the integral: it has to be the area of one disc, not some other measurement of the circle.
Radius of a disc	The distance from the axis of revolution to the edge of the region, measured along one slice.	Measure it from the axis the region actually turns around, and write it in the variable you integrate with respect to.

The region under y = 1 + x squared over 4 from x = 0 to x = 4, shaded above the x-axis, with its mirror image below the axis showing the solid it sweeps out when revolved around the x-axis. One disc is drawn at x = 2 with radius 2, and the solid's volume is about 86.29.

Part 1 Think first How much wood is in the cone?

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

Predict first

A woodworker turns a solid cone on a lathe. It is 6 centimeters long, and its radius grows steadily from 0 at the tip to 3 centimeters at the base. A friend estimates the volume with a cylinder of the same length and the average radius, 1.5 centimeters: $π (1.5)^{2} (6)$ , about 42 cubic centimeters. Which range holds the cone's true volume?

In short. The disc method finds the volume of a solid formed by revolving a region around the x- or y-axis, by adding up the circular slices cut perpendicular to that axis with a definite integral.

Around the x-axis, $V = π \int_{a}^{b} {(f (x))}^{2} d x$ , where the radius $f (x)$ is the height of the region at $x$ .
Around the y-axis, the radius is a width, so write it as a function of $y$ and integrate with respect to $y$ .
The limits of integration are values on the axis of revolution, where the slices start and stop.

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

2a. A solid of revolution is a stack of discs

Spin a flat region around a line and it sweeps out a solid of revolution. The line is the axis of revolution.

Take the region under $y = f (x)$ , where $f (x) \geq 0$ , from $x = a$ to $x = b$ , and spin it around the x-axis. Cut the solid with planes perpendicular to the x-axis. Each thin slice is a disc, like a coin. Its radius is the height of the region at that $x$ , which is $f (x)$ , and its thickness is $Δ x$ , so one slice has volume about $π {(f (x))}^{2} Δ x$ .

Adding the slices is a Riemann sum, and as the slices get thinner its limit is a definite integral. That is the disc method:

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

Try it on a shape you already know. The triangle under $y = \frac{x}{2}$ from $x = 0$ to $x = 4$ , spun around the x-axis, makes a cone of radius 2 and height 4:

V = π \int_{0}^{4} \frac{x^{2}}{4} d x = π \cdot \frac{64}{12} = \frac{16 π}{3} .

The cone formula $\frac{1}{3} π r^{2} h = \frac{1}{3} π (2)^{2} (4)$ gives the same $\frac{16 π}{3}$ .

Quick check · Multiple choice

No calculator. $R$ is the region bounded by the graph of $y = 3 - x$ and the two coordinate axes. Which integral gives the volume of the solid formed when $R$ is revolved around the x-axis?

2b. Around the y-axis, slice across y

Now spin a region around the y-axis. The slices are still cut perpendicular to the axis, so this time they are horizontal, with thickness $Δ y$ . Each one is a disc whose radius is a horizontal distance: how far the region reaches out from the y-axis at height $y$ .

So the radius has to be written as a function of $y$ . If the region lies between the y-axis and the curve $x = g (y)$ , from $y = c$ to $y = d$ , then

V = π \int_{c}^{d} {(g (y))}^{2} d y .

The limits are heights, values of $y$ , because the discs stack up along the y-axis.

Take the region bounded by $y = x^{2}$ with $x \geq 0$ , the y-axis and the line $y = 3$ . Solve the curve for the radius: $x = \sqrt{y}$ . Then

V = π \int_{0}^{3} {(\sqrt{y})}^{2} d y = π \int_{0}^{3} y d y = \frac{9 π}{2} .

The orientation of the axis settles the setup. A horizontal axis means vertical slices and an integral in $x$ . A vertical axis means horizontal slices and an integral in $y$ .

Quick check · Match

No calculator. Each region is revolved around the axis named. Tap a region, then the integral that gives the volume of its solid.

2c. Getting the setup right

Most wrong disc answers come from the setup, not from the antiderivative.

Square the radius inside the integral. Each disc has area $π r^{2}$ , so the integrand is the radius squared, slice by slice. Squaring after integrating, $π {(\int_{a}^{b} f (x) d x)}^{2}$ , gives a different number. When you expand, keep the middle term: $(5 - x)^{2} = 25 - 10 x + x^{2}$ , not $25 - x^{2}$ .

The radius is a distance from the axis, along the slice. Around the x-axis it is the height of the region at $x$ . Around the y-axis it is the width of the region at $y$ . A square root often disappears here: a radius of $\sqrt{x}$ gives a disc of area $π x$ .

Keep $π$ in exact answers. Without a calculator, leave the volume as a multiple of $π$ , such as $\frac{16 π}{3}$ . With one, write the integral first and then the decimal.

A quick sanity check: a solid fits inside a cylinder as long as the solid, with its largest radius. The cone in the first idea fits inside a cylinder of volume $π (2)^{2} (4) = 16 π$ , and $\frac{16 π}{3}$ is smaller, as it must be.

Quick check · Decide and justify

No calculator. $R$ is the region bounded by the graph of $y = \sqrt{x}$ , the x-axis and the line $x = 6$ . A classmate writes the volume of the solid formed when $R$ is revolved around the x-axis as $π \int_{0}^{6} x d x = 18 π$ . Is the classmate right?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Integrating the radius, $π \int_{a}^{b} f (x) d x$ , instead of the radius squared.	Each slice is a disc, and the volume of a disc is its area $π r^{2}$ times its thickness.	Write $π {(radius)}^{2}$ for one slice first, then put the integral sign and the differential around it.
Revolving around the y-axis but integrating in $x$ with the curve's height as the radius.	Slices perpendicular to the y-axis are horizontal, so their radii are widths and they stack along the y-axis.	Solve the curve for $x$ in terms of $y$ , use that as the radius, and take the limits from the heights where the region starts and stops.
Squaring the whole integral, $π {(\int_{a}^{b} f (x) d x)}^{2}$ , instead of squaring the radius inside it.	The square of a sum is not the sum of the squares, so squaring after adding gives the wrong volume.	Square the radius inside the integral, slice by slice, and only then integrate.

Part 3 Watch it done A volume around the y-axis, step by step

Watch each step, then do the last one yourself.

$R$ is the region in the first quadrant bounded by the graph of $y = 9 - x^{2}$ and the coordinate axes. Find the volume of the solid formed when $R$ is revolved around the y-axis.

Step 1 · Pick the slices
The axis is vertical, so the slices are horizontal discs of thickness $Δ y$ . $R$ runs from $y = 0$ up to $y = 9$ , where the curve meets the y-axis.
Why this step?
Why do the limits run from 0 to 9 and not from 0 to 3?
The discs stack along the y-axis, so the limits are heights. The curve meets the x-axis at $x = 3$ , which is a width of $R$ , not a height.
Step 2 · Write the radius in terms of y
At height $y$ the radius is the $x$ -value on the curve. From $y = 9 - x^{2}$ , $x^{2} = 9 - y$ , so the radius is $x = \sqrt{9 - y}$ .
Step 3 · Set up the integral
$V = π \int_{0}^{9} {(\sqrt{9 - y})}^{2} d y = π \int_{0}^{9} (9 - y) d y$ .
Step 4 · Antidifferentiate
An antiderivative of $9 - y$ is $9 y - \frac{y^{2}}{2}$ , so $V = π {[9 y - \frac{y^{2}}{2}]}_{0}^{9}$ .
Your step · Calculate

No calculator. Finish it. The volume is $V = π {[9 y - \frac{y^{2}}{2}]}_{0}^{9}$ , which equals $k π$ . What is $k$ ? Give an exact decimal.

Your answer

Part 4 Try it A drawer knob and a cone

Try it · Calculate

Graphing calculator required. A wooden drawer knob has the shape of the solid formed when the region under $y = 1 + e^{- x^{2}}$ , for $0 \leq x \leq 3$ , is revolved around the x-axis, with $x$ and $y$ in centimeters. What is the volume of the knob, in cubic centimeters? Give your answer to three decimal places.

Your answer

Try it · Decide and justify

No calculator. $R$ is the region bounded by the graph of $y = 3 x$ , the y-axis and the line $y = 6$ . A classmate finds the volume of the solid formed when $R$ is revolved around the y-axis as $π \int_{0}^{2} (3 x)^{2} d x = 24 π$ . Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

On the exam free response is handwritten, so work on paper first. Nothing you type here leaves this page. Then compare with the model answer and score each point.

Free response is 50% of the exam score. This one is shaped like Question 5, from free-response Part B (calculator not permitted).

Free-response practice · Question 5 style

No calculator. A glassmaker shapes two paperweights from the same flat region. Let $R$ be the region in the first quadrant bounded by the graph of $y = 8 - x^{3}$ and the coordinate axes, where $x$ and $y$ are measured in centimeters.

(a) The first paperweight is the solid formed when $R$ is revolved around the x-axis. Write an integral expression for its volume, and find the volume. Indicate units of measure.

(b) The second paperweight is the solid formed when $R$ is revolved around the y-axis. Write an integral expression for its volume, and find the volume. Indicate units of measure.

(c) The glassmaker expects the two paperweights to have the same volume, since both come from $R$ . Using your answers to parts (a) and (b), explain why the volumes differ.

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

Model answer

(a) The slices are vertical discs of radius $8 - x^{3}$ , for $0 \leq x \leq 2$ . $V = π \int_{0}^{2} {(8 - x^{3})}^{2} d x = π \int_{0}^{2} (64 - 16 x^{3} + x^{6}) d x = π {[64 x - 4 x^{4} + \frac{x^{7}}{7}]}_{0}^{2} = π (128 - 64 + \frac{128}{7}) = \frac{576 π}{7}$ cubic centimeters.

(b) The slices are horizontal discs. Solving $y = 8 - x^{3}$ for $x$ gives the radius $x = (8 - y)^{1 / 3}$ , for $0 \leq y \leq 8$ . $V = π \int_{0}^{8} (8 - y)^{2 / 3} d y = π {[- \frac{3}{5} (8 - y)^{5 / 3}]}_{0}^{8} = π (0 + \frac{3}{5} \cdot 32) = \frac{96 π}{5}$ cubic centimeters.

(c) Revolving $R$ around different axes makes different solids. Around the x-axis the radii are heights of $R$ , up to 8 centimeters, on discs stacked over 2 centimeters. Around the y-axis the radii are widths of $R$ , at most 2 centimeters, on discs stacked over 8 centimeters. A disc's volume grows with the square of its radius but only in proportion to its thickness, so the first paperweight is larger: $\frac{576 π}{7} > \frac{96 π}{5}$ .

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

(a) Writes $π \int_{0}^{2} {(8 - x^{3})}^{2} d x$ , with radius $8 - x^{3}$ and limits 0 and 2.
(a) Expands, antidifferentiates and gets $\frac{576 π}{7}$ cubic centimeters.
(b) Writes the radius $(8 - y)^{1 / 3}$ and the integral $π \int_{0}^{8} (8 - y)^{2 / 3} d y$ , with limits 0 and 8.
(b) Antidifferentiates and gets $\frac{96 π}{5}$ cubic centimeters.
(c) Explains that the radii are measured from different axes (heights up to 8, widths up to 2) and that squaring the radius makes the first volume larger.

Part 6 Check yourself A candle from a mold

The two case questions use the same table. Answer them from it.

A candle is poured in a mold shaped like a solid of revolution. At a distance of $x$ centimeters from the base, measured along the candle's central axis, the cross section is a disc of radius $r (x)$ centimeters, where $r$ is continuous. The candle is 12 centimeters tall. Some values of $r$ are in the table.

Radius of the candle at distances from its base
$x$ (centimeters)	$r (x)$ (centimeters)
0	2
3	4
6	5
9	4
12	1

Case question 1 · Multiple choice

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

Case question 2 · Calculate

No calculator. Use a right Riemann sum with the four subintervals given by the table to approximate the volume of the candle. 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 · Spot the error

No calculator. $R$ is the region bounded by the graph of $y = \ln x$ , the y-axis, the x-axis and the line $y = 1$ . A student finds the volume of the solid formed when $R$ is revolved around the y-axis. Tap the first line with a mistake.

Check yourself · Fill the statement

No calculator. Tap a blank, then tap an answer.

To find the volume of a solid formed by revolving a region around the y-axis, cut the region into slices to the y-axis. Each disc's radius is then a width of the region, and the integral is taken with respect to .

Part 7 Keep it Bring back earlier ideas

Discs are one kind of cross section, and every volume ends in a definite integral. Pulling both back up is what makes them last.

Bring it back · Multiple choice

No calculator. The base of a solid is the region bounded by the graph of $y = 6 - 2 x$ and the coordinate axes. Cross sections perpendicular to the x-axis are semicircles whose diameters lie in the base. Which expression gives the volume of the solid?

From 6.10 Integrating Functions Using Long Division and Completing the Square · Calculate

No calculator. The exact value of $\int_{0}^{3} \frac{x^{2} + 5 x + 7}{x + 2} d x$ can be written as $a + \ln \frac{5}{2}$ . What is $a$ ? Give an exact decimal.

Your answer

In three lines

Revolving a region around an axis makes a solid whose slices, cut perpendicular to that axis, are discs.
Each disc has area $π r^{2}$ , so the volume is $π$ times the definite integral of the radius squared.
Around the x-axis integrate in $x$ ; around the y-axis write the radius in $y$ and integrate in $y$ .

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 setup before any arithmetic: $V = π \int_{a}^{b} {(r (x))}^{2} d x$ with the actual radius, both limits and the differential, and the units of volume, such as cubic centimeters. On a calculator question, show that integral and then the decimal to three places; on a no-calculator question, show the antiderivative and leave $π$ in the answer.

Questions students ask

Why is it called the disc method?

Because every slice of the solid, cut perpendicular to the axis, is a flat disc. The integral adds up the volumes of those discs, $π r^{2}$ times a thickness, the way a Riemann sum adds rectangles.

How do I know whether to integrate with respect to

x

y

Look at the axis. The slices are perpendicular to it, so a horizontal axis (the x-axis) gives vertical slices and an integral in $x$ , and a vertical axis (the y-axis) gives horizontal slices and an integral in $y$ .

What if the region does not reach all the way to the axis?

Then each slice has a hole in the middle where the region stops short of the axis, and it is a washer rather than a disc. That is the washer method, which comes a few topics later. The disc method in this lesson is for regions that run right up to the axis.

Do I have to expand the square before integrating?

Only if you need to. A polynomial like ${(2 + x^{2})}^{2}$ is easiest to expand and integrate term by term. Something like ${(\sqrt{4 - y})}^{2}$ simplifies to $4 - y$ at once. On a calculator question you can integrate the square as it is.

Should my answer have

π

in it?

Without a calculator, yes: an exact answer such as $\frac{16 π}{3}$ is the volume. With a calculator, give the decimal to the places the question asks for, after writing the integral that produced it.

Course alignment, for teachers

AP Calculus AB topic 8.9, Unit 8: Applications of Integration.

Skill 3.D: Apply an appropriate mathematical definition, theorem, or test.

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.