Unit 8 · Topic 8.10 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice, plus 1 from earlier lessons

AP Calculus AB 8.10: Volume with Disc Method: Revolving Around Other Axes

Q: How is this different from Topic 8.9?

Only the radius changes. In 8.9 the axis was the x-axis or the y-axis, so the radius was the curve's height or its distance from the y-axis. With the axis at y = k or x = h, measure from that line instead.

Q: Does it matter whether the curve is above or below the axis?

Not for the value, because the radius is squared and (k - f(x))2 = (f(x) - k)2. Still write the radius as a distance, the larger value minus the smaller, since that habit carries straight into washers in 8.11 and 8.12.

Q: When do I need a calculator for these?

When the integral has no antiderivative you can find, as with 01 (e-x2 + 2)2\,dx, or when a limit comes from an equation you cannot solve by hand. Show the integral you evaluate, then the result to three decimal places.

Q: Why rewrite the curve in terms of y for a vertical axis?

The slices are horizontal, with thickness dy, so the radius has to be a function of y before it can be integrated with respect to y. For y = 2x, that means x = y2.

Q: Can the axis run through the middle of the region?

Not for the solids in this topic. Each region lies on one side of its axis and touches it along its whole length. If the axis cut through the region, the two sides would overlap when turned, so the volume would not be a single disc integral.

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 turning a region about any horizontal or vertical line, using discs and a definite integral.

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.	The same region turned about two different lines makes two different solids, so read the line from the question before you set anything up.
Axis of revolution	The line a region is turned about. In this topic it can be any horizontal line $y = k$ or vertical line $x = h$ .	The axis can sit above, below or beside the region; mark it on a sketch before you write anything else.
Disc method	Finding the volume of a solid of revolution with a definite integral, by slicing it perpendicular to the axis into thin discs and adding their volumes.	Discs fit only when the region touches the axis along every slice. A gap between them makes each slice a ring instead.

Graph of the region between y = square root of x and the horizontal line y = -1, for x from 0 to 4, with its mirror image across y = -1 shaded more lightly. At x = 2.56 the disc the region sweeps out is drawn as an ellipse, and a segment from y = -1 up to the curve marks its radius, 2.6.

Part 1 Think first Where is the center of the circle?

Commit to a guess first. Even a wrong one makes the answer easier to keep.

Predict first

A woodturner draws the outline of a candlestick on graph paper marked in inches. Her lathe will spin the wood about the line $y = 3$ on the drawing, and the wood fills the space between the outline and that line. At $x = 2$ the outline is at height $y = 1$ , below the line. When the wood spins, what is the radius of the round slice of candlestick at $x = 2$ ?

In short. To find the volume of a solid made by turning a region about a horizontal line $y = k$ or a vertical line $x = h$ , slice it into discs perpendicular to that line and integrate $π$ times the square of each disc's radius.

A disc is centered on the axis of revolution, so its radius is the distance from that line to the curve: the larger coordinate minus the smaller.
A horizontal axis gives $V = π \int_{a}^{b} {(R (x))}^{2} d x$ , and a vertical axis gives $V = π \int_{c}^{d} {(R (y))}^{2} d y$ .
Square the whole radius, and take the limits from where the region starts and ends along the axis.

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

10a. The radius is the distance to the axis

In Topic 8.9 every solid turned about the $x$ -axis or the $y$ -axis, so a disc's radius was the height of the curve or its distance from the $y$ -axis. Now the axis can be any horizontal line $y = k$ or vertical line $x = h$ , and the radius has to be measured from that line.

Take the region between the graph of $y = f (x)$ and the line $y = k$ , and turn it about $y = k$ . Cut the solid with a plane perpendicular to the axis at some $x$ . The cut is a disc centered on the axis, and its radius runs from the axis to the curve:

R (x) = k - f (x) if the curve is below the line, R (x) = f (x) - k if it is above.

Either way the radius is the larger $y$ -value minus the smaller one, so it is never negative.

For example, take the region between $y = x^{2} + 1$ and the line $y = - 1$ for $0 \leq x \leq 2$ , turned about $y = - 1$ . The curve is above the axis, so $R (x) = (x^{2} + 1) - (- 1) = x^{2} + 2$ . At $x = 1$ the disc has radius 3, not 2. The height of the curve above the $x$ -axis is no longer the radius, because the $x$ -axis is no longer the axis the region turns about.

Quick check · Match

No calculator. Each region is turned about the line named with it. Tap a region, then the radius of its disc at $x$ .

10b. A horizontal axis: discs of thickness dx

A disc perpendicular to a horizontal axis is a thin slab of thickness $d x$ whose face has area $π {(R (x))}^{2}$ . Adding the slabs from $x = a$ to $x = b$ and letting them get thinner is a definite integral, just as in 8.9:

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

Only the radius is new. For the region between $y = x^{2}$ and the line $y = 4$ , turned about $y = 4$ , the curve meets the line at $x = - 2$ and $x = 2$ , and the curve is below the line, so

V = π \int_{- 2}^{2} {(4 - x^{2})}^{2} d x = \frac{512 π}{15} .

Two habits keep this setup right. First, square the whole radius, ${(4 - x^{2})}^{2}$ . Squaring each piece, $4^{2} - {(x^{2})}^{2}$ , is a different integrand, and here it gives $51.2 π$ instead. Second, the limits are $x$ -values: where the region starts and ends, read from where the curve meets the line or from the vertical boundaries you are given. The $y$ -value of the axis is never a limit.

Quick check · Multiple choice

No calculator. Let $R$ be the region bounded by the graph of $y = \sqrt{x}$ , the line $y = 3$ and the $y$ -axis. Which integral gives the volume of the solid generated when $R$ is revolved about the line $y = 3$ ?

10c. A vertical axis: discs of thickness dy

When the axis is a vertical line $x = h$ , the slices perpendicular to it are horizontal. Each disc has thickness $d y$ , so everything in the integral is written in $y$ : the radius, the limits and the differential.

That means writing the curve as $x$ in terms of $y$ . The radius at height $y$ is the horizontal gap from the axis to the curve, the larger $x$ -value minus the smaller, and

V = π \int_{c}^{d} {(R (y))}^{2} d y,

where $c$ and $d$ are the lowest and highest $y$ -values of the region.

For example, take the region bounded by $y = 2 x$ , the line $x = 3$ and the $x$ -axis, turned about $x = 3$ . Solving $y = 2 x$ for $x$ gives $x = \frac{y}{2}$ . The region runs from $y = 0$ up to $y = 6$ , where $y = 2 x$ meets $x = 3$ . At height $y$ the curve is to the left of the axis, so $R (y) = 3 - \frac{y}{2}$ , and

V = π \int_{0}^{6} {(3 - \frac{y}{2})}^{2} d y = 18 π .

This solid is a cone of radius 3 and height 6, and the cone formula agrees: $\frac{1}{3} π (3)^{2} (6) = 18 π$ .

Quick check · Decide and justify

No calculator. Let $R$ be the region bounded by the graph of $y = x^{2}$ , the line $x = 3$ and the $x$ -axis. A classmate writes $V = π \int_{0}^{3} (3 - x)^{2} d x$ for the volume of the solid generated when $R$ is revolved about the line $x = 3$ . Is the classmate right?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Using the curve's height $f (x)$ as the radius when the axis is the line $y = k$ .	Each disc is centered on the axis of revolution, so its radius is the distance from that line to the curve, not from the $x$ -axis.	Write $R (x) = k - f (x)$ or $f (x) - k$ , the larger $y$ -value minus the smaller.
Writing $k^{2} - {(f (x))}^{2}$ in place of ${(k - f (x))}^{2}$ .	A disc's area is $π$ times its radius squared, and the radius is the whole difference.	Put the whole radius in parentheses first, then square it.
Integrating with respect to $x$ when the axis is a vertical line $x = h$ .	Slices perpendicular to a vertical axis are horizontal, so their thickness is $d y$ .	Rewrite the curve as $x$ in terms of $y$ , find the radius in $y$ , and use the region's lowest and highest $y$ -values as limits.

Part 3 Watch it done A disc about the line x = 2

Watch each step, then do the last one yourself.

Let $R$ be the region bounded by the graph of $y = x^{3}$ , the line $x = 2$ and the $x$ -axis. Find the volume of the solid generated when $R$ is revolved about the line $x = 2$ .

Step 1 · Choose the slices
The axis $x = 2$ is vertical, so the slices are horizontal discs of thickness $d y$ . Solve $y = x^{3}$ for $x$ : $x = y^{1 / 3}$ . The region runs from $y = 0$ up to $y = 8$ , where $y = x^{3}$ meets $x = 2$ .
Step 2 · Find the radius
At height $y$ the curve is at $x = y^{1 / 3}$ , to the left of the axis, so $R (y) = 2 - y^{1 / 3}$ .
Why this step?
Why is the radius $2 - y^{1 / 3}$ and not $y^{1 / 3}$ ?
Each disc is centered on the line it turns about, $x = 2$ , so its radius is the horizontal distance from that line to the curve. $y^{1 / 3}$ is the distance from the $y$ -axis, which is not the axis here.
Step 3 · Set up and expand
$V = π \int_{0}^{8} {(2 - y^{1 / 3})}^{2} d y = π \int_{0}^{8} (4 - 4 y^{1 / 3} + y^{2 / 3}) d y$ .
Step 4 · Antidifferentiate
An antiderivative of $4 - 4 y^{1 / 3} + y^{2 / 3}$ is $4 y - 3 y^{4 / 3} + \frac{3}{5} y^{5 / 3}$ , so $V = π {[4 y - 3 y^{4 / 3} + \frac{3}{5} y^{5 / 3}]}_{0}^{8}$ .
Your step · Calculate

No calculator. Finish it. For the solid above, $V = π {[4 y - 3 y^{4 / 3} + \frac{3}{5} y^{5 / 3}]}_{0}^{8}$ , and that value is $k π$ . What is $k$ ? Give an exact decimal.

Your answer

Part 4 Try it A new axis, and a claim to test

Try it · Calculate

Graphing calculator required. Let $R$ be the region bounded by the graph of $y = \sin (x^{2})$ , the line $y = - 1$ , and the vertical lines $x = 0$ and $x = 2$ . Find the volume of the solid generated when $R$ is revolved about the line $y = - 1$ . Use radian mode, and give your answer to three decimal places.

Your answer

Try it · Decide and justify

No calculator. Region $A$ lies between the graph of $y = \sqrt{x}$ and the line $y = 2$ for $0 \leq x \leq 4$ , and is revolved about the line $y = 2$ . Region $B$ lies between the graph of $y = 2 - \sqrt{x}$ and the $x$ -axis for $0 \leq x \leq 4$ , and is revolved about the $x$ -axis. A classmate says the two solids have the same volume. Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Answer it on paper, then score it

Free response is handwritten on the exam, so write your answer on paper first, with your calculator beside you. Nothing you type here leaves this page. Then open the model answer and score yourself on each point.

Free response is 50% of the exam score. This one is shaped like Question 2, from free-response Part A (graphing calculator required).

Free-response practice · Question 2 style

Graphing calculator required. A glass bead is modeled by the region $R$ bounded by the graph of $f (x) = x + 4 e^{- x}$ and the line $y = 3$ , revolved about the line $y = 3$ . Both axes are measured in centimeters. Give decimal answers to three places.

(a) The graph of $f$ meets the line $y = 3$ at $x = A$ and $x = B$ , where $A < B$ . Find $A$ and $B$ .

(b) Find the volume of the bead. Show the integral you evaluate on your calculator. Indicate units of measure.

(c) A plane perpendicular to the line $y = 3$ at $x = c$ cuts the bead into two pieces of equal volume. Write, but do not solve, an equation involving one or more integrals whose solution is $c$ .

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

Model answer

(a) $A$ and $B$ are the solutions of $x + 4 e^{- x} = 3$ . On the calculator, $A = 0.450$ and $B = 2.742$ , stored for part (b).

(b) Between $A$ and $B$ the graph of $f$ is below $y = 3$ , so each disc has radius $3 - f (x)$ . $V = π \int_{A}^{B} {(3 - f (x))}^{2} d x = π \int_{0.450}^{2.742} {(3 - x - 4 e^{- x})}^{2} d x = 1.419$ cubic centimeters (1.418 truncated).

(c) $π \int_{A}^{c} {(3 - f (x))}^{2} d x = \frac{1}{2} π \int_{A}^{B} {(3 - f (x))}^{2} d x$ , with $A$ and $B$ the values stored in part (a).

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

Sets up $x + 4 e^{- x} = 3$ and gets $A = 0.450$ and $B = 2.742$ .
Writes $π \int_{A}^{B} {(3 - f (x))}^{2} d x$ : the radius $3 - f (x)$ squared, both limits and $d x$ .
Gets 1.419 (or 1.418) cubic centimeters, with units.
Writes an equation that sets the integral from $A$ to $c$ equal to half the volume, or equal to the integral from $c$ to $B$ .

Part 6 Check yourself A machine part from a table

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

A machine part is made by turning a region about the line $y = 6$ . The region lies between the graph of a function $f$ and the line $y = 6$ for $0 \leq x \leq 8$ , where $f (x) < 6$ . Both axes are measured in centimeters, and $f$ is known only from the table.

Values of f for the machine part
$x$ (centimeters)	$f (x)$ (centimeters)
0	5
2	3
4	1
6	2
8	4

Case question 1 · Calculate

No calculator. What is the radius, in centimeters, of the disc the region sweeps out at $x = 4$ ?

Your answer

Case question 2 · Multiple choice

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

Case question 3 · Calculate

No calculator. Approximate the volume of the machine part with a right Riemann sum, using the four subintervals the table gives. The estimate 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. Let $R$ be the region bounded by the graph of $y = \ln x$ , the $x$ -axis and the line $x = e$ . A student finds the volume of the solid generated when $R$ is revolved about the line $x = e$ . Tap the first line with a mistake.

Check yourself · Sort it

No calculator. Sort each solid by the slices the disc method uses for it.

Discs of thickness $d x$

Discs of thickness $d y$

Part 7 Keep it Bring back earlier ideas

This lesson stands on discs about the coordinate axes and on evaluating a definite integral. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. Let $R$ be the region bounded by the graph of $y = x^{2}$ , the $x$ -axis and the line $x = 2$ . What is the volume of the solid generated when $R$ is revolved about the $x$ -axis?

From 6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation · Calculate

Graphing calculator required. The function $\sqrt{1 + x^{3}}$ has no closed-form antiderivative, so $\int_{0}^{2} \sqrt{1 + x^{3}} d x$ cannot be found from an antiderivative by hand. Use your calculator to evaluate it. Give your answer to three decimal places.

Your answer

In three lines

A disc is centered on the axis of revolution, so its radius is measured from that line to the curve.
A horizontal axis means discs of thickness $d x$ ; a vertical axis means discs of thickness $d y$ , with the curve written as $x$ in terms of $y$ .
The volume is $π$ times the integral of the whole radius squared, between the region's limits along the 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.

On a calculator question, write the whole setup before the number, for example $π \int_{a}^{b} {(k - f (x))}^{2} d x$ with both limits and $d x$ , then give the volume to three decimal places with cubic units. If a limit comes from where a graph meets the axis line, say which equation you solved, and store the result rather than retyping a rounded value.

Questions students ask

How is this different from Topic 8.9?

Only the radius changes. In 8.9 the axis was the $x$ -axis or the $y$ -axis, so the radius was the curve's height or its distance from the $y$ -axis. With the axis at $y = k$ or $x = h$ , measure from that line instead.

Does it matter whether the curve is above or below the axis?

Not for the value, because the radius is squared and ${(k - f (x))}^{2} = {(f (x) - k)}^{2}$ . Still write the radius as a distance, the larger value minus the smaller, since that habit carries straight into washers in 8.11 and 8.12.

When do I need a calculator for these?

When the integral has no antiderivative you can find, as with $π \int_{0}^{1} {(e^{- x^{2}} + 2)}^{2} d x$ , or when a limit comes from an equation you cannot solve by hand. Show the integral you evaluate, then the result to three decimal places.

Why rewrite the curve in terms of

y

for a vertical axis?

The slices are horizontal, with thickness $d y$ , so the radius has to be a function of $y$ before it can be integrated with respect to $y$ . For $y = 2 x$ , that means $x = \frac{y}{2}$ .

Can the axis run through the middle of the region?

Not for the solids in this topic. Each region lies on one side of its axis and touches it along its whole length. If the axis cut through the region, the two sides would overlap when turned, so the volume would not be a single disc integral.

Course alignment, for teachers

AP Calculus AB topic 8.10, 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.