How do I know whether to use the disc method or the washer method?

Look at whether the region touches the axis of revolution. If the region sits flush against the axis with no gap, each slice is a solid disc and you use r^2. If there is empty space between the region and the axis, each slice is a ring with a hole, so you use the washer formula (R^2 - r^2).

Why do I square the radius before integrating instead of after?

The area of a circular cross section is r^2, and r is a function of the variable of integration. The integral adds up these areas times thickness. Squaring happens inside the area formula, so it must be applied to the radius function before you integrate. Squaring after integrating would compute a completely different quantity.

When should I integrate with respect to y instead of x?

Match the variable to the axis. For revolution about a horizontal line like the x-axis, slices are vertical with thickness dx, so integrate in x. For revolution about a vertical line like the y-axis, slices are horizontal with thickness dy, so integrate in y — which usually means solving each curve for x in terms of y.

What is the most common mistake on these problems?

Writing (R - r)^2 instead of R^2 - r^2 in the washer method. These are not equal. You must square the outer and inner radii separately and then subtract. Other frequent errors are forgetting the factor of and using the wrong limits of integration.

AP-CALCBC-8.9-8.11

U8.9 Volumes of Revolution: Disc and Washer Methods

Master the disc and washer methods for AP Calculus BC volumes of revolution: set up π∫r² dx and π∫(R²−r²) dx, and revolve around horizontal or vertical axes.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U8.9 Volumes of Revolution: Disc and Washer Methods, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U8.9 Volumes of Revolution: Disc and Washer Methods free

What this lesson covers

When you spin a region around a line, it sweeps out a three-dimensional solid. The disc and washer methods let you find the exact volume of that solid by slicing it into thin circular cross sections and adding up their volumes with an integral. This is one of the most reliable point-earners in Unit 8, and it shows up on both the multiple-choice and free-response sections.

In this lesson you will learn how to identify the radius of each slice, decide whether you need a disc or a washer, choose the correct variable of integration, and set up the integral carefully around any horizontal or vertical axis. Get the setup right and the arithmetic follows.

The Disc Method: Solid Slices

Imagine revolving a region that sits flush against the axis of rotation. Each thin slice perpendicular to the axis becomes a solid circle — a disc — of radius

r

and thickness

dx

(or

dy

). The area of one disc is

\pi r^2

, and its volume is

\pi r^2\,dx

. Summing over the interval gives

V = \pi\int_a^b r^2\,dx.

The key skill is writing the radius as a function. When you revolve about the

x

-axis, the radius of a slice at position

x

is simply the height of the curve there, so

r = f(x)

and

V = \pi\int_a^b [f(x)]^2\,dx.

When you revolve about the

y

-axis, slices are horizontal, thickness is

dy

, and the radius is the horizontal distance, so you must solve for

x

as a function of

y

and integrate in

y

.

The most common misconception is squaring after integrating, or forgetting the square entirely. Remember: you square the radius first, then integrate. Also, the disc method only works when the region touches the axis with no gap. If there is a gap between the region and the axis, you need a washer.

The Washer Method: Slices With Holes

When the region being revolved does not touch the axis, each cross section is a ring — a washer — with an outer radius

R

and an inner radius

r

. The area of the ring is the big circle minus the hole:

\pi R^2 - \pi r^2

. Integrating gives

V = \pi\int_a^b \left(R^2 - r^2\right)dx.

Here

R

is the distance from the axis to the outer (farther) curve, and

r

is the distance from the axis to the inner (nearer) curve. A frequent error is writing

(R-r)^2

instead of

R^2 - r^2

— these are not equal, so keep the squares separate.

Feature	Disc	Washer
Cross section	Solid circle	Ring
Formula	$\pi\int r^2$	$\pi\int (R^2-r^2)$
When to use	Region touches axis	Gap between region and axis
Radii needed	One	Outer and inner

To find each radius, always measure distance from the axis of rotation, not from the origin. Sketching the region and a representative slice makes this automatic and prevents sign or subtraction mistakes.

Finding Radii and Choosing the Variable

The single most important habit is drawing a representative rectangle perpendicular to the axis of rotation. If the axis is horizontal, the rectangle is vertical, its thickness is

dx

, and you integrate in

x

. If the axis is vertical, the rectangle is horizontal, its thickness is

dy

, and you integrate in

y

.

The radius is the length of that rectangle measured from the axis to the curve. For revolution about the

x

-axis, a slice from the curve

y=f(x)

down to the axis has radius

f(x)

. For a washer between two curves

f(x)\ge g(x)\ge 0

, the outer radius is

R=f(x)

and inner radius is

r=g(x)

.

To find the limits of integration, locate the boundaries of the region. For

x

-integration these are the leftmost and rightmost

x

-values, often the intersection points found by setting

f(x)=g(x)

. For

y

-integration, use the bottom and top

y

-values.

Exam tip: the AP exam frequently gives a region bounded by two curves and asks for the volume when revolved about the

x

-axis or a horizontal line. Recognize instantly whether the region hugs the axis (disc) or leaves a gap (washer). Revolving about lines other than the axes is previewed in U8.12, but the same distance principle applies: radius equals the distance from the slice to the line.

How the AP Exam Tests This

On multiple-choice, you are usually asked to identify the correct integral setup rather than evaluate it, so setup precision is everything. Watch for answer choices that swap

R^2-r^2

for

(R-r)^2

, forget

\pi

, or use the wrong variable of integration.

On free-response, a calculator-active problem often gives two curves and asks you to set up and evaluate a volume. You earn points for the correct integrand with squared radii, the correct limits, and the correct numerical answer. Set up the integral fully before pressing any calculator buttons, and always include

\pi

.

A classic trap: if the region is bounded above by

f

and below by

g

but both lie above the

x

-axis, revolving about the

x

-axis produces a washer, not a disc — the lower curve

g

carves out the hole. Students who forget the inner radius overcount the volume.

Another check: your radii must be nonnegative on the interval. If a curve dips below the axis, the distance is

|f(x)|

, but since you square it,

[f(x)]^2

handles the sign automatically. Just be sure you are still measuring from the axis, not between two curves that straddle it — in that case split the region.

Key terms

Solid of revolution.: A three-dimensional solid formed by rotating a plane region about a fixed line called the axis of revolution.
Disc method.: A technique computing volume as $\pi\int r^2\,dx$ when cross sections perpendicular to the axis are solid circles.
Washer method.: A technique computing volume as $\pi\int (R^2 - r^2)\,dx$ when cross sections are rings with an outer radius $R$ and inner radius $r$ .
Outer radius.: The distance from the axis of revolution to the farther boundary curve of the region.
Inner radius.: The distance from the axis of revolution to the nearer boundary curve; it defines the hole in a washer.
Representative slice.: A thin rectangle drawn perpendicular to the axis of rotation whose revolution generates one disc or washer.
Axis of revolution.: The fixed line about which a region is rotated; all radii are measured as distances from this line.

Worked example

Let

R

be the region bounded by

y = \sqrt{x}

y = 0

, and

x = 4

. Find the volume of the solid formed when

R

is revolved about the

x

-axis.

First sketch the region:

y=\sqrt{x}

rises from the origin, bounded below by the

x

-axis and on the right by

x=4

. The region touches the axis of rotation (the

x

-axis) along its bottom edge, so cross sections are solid discs — use the disc method.

Draw a vertical representative slice at position

x

. Its thickness is

dx

and its radius is the height of the curve,

r = \sqrt{x}

.

The area of one disc is

\pi r^2 = \pi(\sqrt{x})^2 = \pi x

.

The region runs from

x=0

x=4

, so

V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx.

Evaluate the integral:

\pi\left[\frac{x^2}{2}\right]_0^4 = \pi\left(\frac{16}{2} - 0\right) = 8\pi.

The volume is

8\pi

cubic units. Notice how squaring the radius first turned the awkward

\sqrt{x}

into a simple

x

, which is exactly why you square before integrating.

Practice questions

The region bounded by

y = x^2

and

y = 4

is revolved about the

x

-axis. Which integral gives the volume of the resulting solid?

$\pi\int_{-2}^{2} (4 - x^2)^2\,dx$
$\pi\int_{-2}^{2} \left(4^2 - (x^2)^2\right)dx$
$\pi\int_{0}^{4} (4 - x^2)\,dx$
$\pi\int_{-2}^{2} (4 - x^2)\,dx$

Answer: $\pi\int_{-2}^{2} \left(4^2 - (x^2)^2\right)dx$

The curves meet where

x^2 = 4

, so

x = \pm 2

; these are the limits. Both boundaries lie above the

x

-axis, leaving a gap, so this is a washer. The outer radius is the farther curve

y=4

, giving

R=4

, and the inner radius is

r=x^2

. The integrand is

R^2 - r^2 = 4^2 - (x^2)^2

. The choice using

(4-x^2)^2

wrongly subtracts before squaring.

The region bounded by

y = e^x

y = 0

x = 0

, and

x = 1

is revolved about the

x

-axis. Set up and evaluate the volume of the solid.

Answer: $V = \dfrac{\pi(e^2 - 1)}{2}$

The region sits on the

x

-axis, so use discs with radius

r = e^x

. Then

V = \pi\int_0^1 (e^x)^2\,dx = \pi\int_0^1 e^{2x}\,dx

. Antidifferentiate:

\pi\left[\tfrac{1}{2}e^{2x}\right]_0^1 = \tfrac{\pi}{2}(e^2 - e^0) = \tfrac{\pi}{2}(e^2 - 1)

The region bounded by

x = y^2

and

x = 4

is revolved about the

y

-axis. Write the integral for the volume.

Answer: $V = \pi\int_{-2}^{2} \left(4^2 - (y^2)^2\right)dy$

Revolving about the

y

-axis means horizontal slices with thickness

dy

, so integrate in

y

. The curves meet where

y^2 = 4

, i.e.

y = \pm 2

. The outer radius reaches to

x=4

, so

R=4

; the inner radius is the parabola

x=y^2

, so

r=y^2

. The washer integrand is

4^2 - (y^2)^2

, giving

V = \pi\int_{-2}^{2}\left(16 - y^4\right)dy

FAQ

How do I know whether to use the disc method or the washer method?: Look at whether the region touches the axis of revolution. If the region sits flush against the axis with no gap, each slice is a solid disc and you use $\pi\int r^2$ . If there is empty space between the region and the axis, each slice is a ring with a hole, so you use the washer formula $\pi\int (R^2 - r^2)$ .
Why do I square the radius before integrating instead of after?: The area of a circular cross section is $\pi r^2$ , and $r$ is a function of the variable of integration. The integral adds up these areas times thickness. Squaring happens inside the area formula, so it must be applied to the radius function before you integrate. Squaring after integrating would compute a completely different quantity.
When should I integrate with respect to $y$ instead of $x$ ?: Match the variable to the axis. For revolution about a horizontal line like the $x$ -axis, slices are vertical with thickness $dx$ , so integrate in $x$ . For revolution about a vertical line like the $y$ -axis, slices are horizontal with thickness $dy$ , so integrate in $y$ — which usually means solving each curve for $x$ in terms of $y$ .
What is the most common mistake on these problems?: Writing $(R - r)^2$ instead of $R^2 - r^2$ in the washer method. These are not equal. You must square the outer and inner radii separately and then subtract. Other frequent errors are forgetting the factor of $\pi$ and using the wrong limits of integration.

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