Blog · Calculus guides · 2 November 2026 · 4 min read

By the Asymptote team. Every answer on this page is checked by the same machine that grades the game.

Disk and Washer Method: Formulas & Practice Problems

An illustration of the disk and washer methods

The disk method slices a solid of revolution into thin circular disks, each of volume $π r^{2} d x$ , so $V = π \int_{a}^{b} R (x)^{2} d x$ . When the solid has a hole through the middle, each slice is a washer instead, and you subtract the hole: $V = π \int_{a}^{b} (R (x)^{2} - r (x)^{2}) d x$ , where $R$ is the outer radius and $r$ the inner one.

The formula

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

A washer is a disk that has had its middle eaten. Square each radius first, then subtract: the order is the entire trick.

When to use it

Use disks or washers when the axis of rotation is perpendicular to your strips: vertical strips spun around the $x$ -axis or any horizontal line. If the region touches the axis all the way along, every slice is a solid disk. If there is a gap between the region and the axis, the slices have holes and you need washers. When the axis is parallel to the strips instead, the shell method is usually less work.

How to do it, step by step

Sketch the region and the axis. Draw one thin strip perpendicular to the axis. Spinning it makes one disk or one washer.
Find the outer radius $R$ . The distance from the axis to the far edge of the region. About the $x$ -axis that is the curve farther from it.
Find the inner radius $r$ . The distance from the axis to the near edge. If the region touches the axis, $r = 0$ and you have a disk.
Square, subtract, integrate. $π \int (R^{2} - r^{2}) d x$ between the limits of the region. Expand before integrating.

Worked examples

Example 1: A square root around the x-axis

Easy

The region under $y = x$ from $x = 0$ to $x = 4$ is revolved about the $x$ -axis. Find the volume.

y = x, 0 \leq x \leq 4, about the x -axis

The region touches the $x$ -axis everywhere, so each slice is a solid disk of radius $R (x) = x$ .

V = π \int_{0}^{4} (x)^{2} d x = π \int_{0}^{4} x d x = π [\frac{x ^{2}}{2}]_{0}^{4} = π \cdot 8

Squaring the radius got rid of the root before any integrating happened, which is the disk method being generous for once.

Answer

V = 8 π

Example 2: A washer between two curves

Medium

The region between $y = x$ and $y = x^{2}$ is revolved about the $x$ -axis. Find the volume.

y = x and y = x^{2}, about the x -axis

The curves meet at $x = 0$ and $x = 1$ , and $x$ is above $x^{2}$ in between. The region does not touch the axis (except at the ends), so the slices are washers.

The outer radius is the curve farther from the axis, $R = x$ . The inner radius is the nearer one, $r = x^{2}$ .

V = π \int_{0}^{1} (x^{2} - (x^{2})^{2}) d x = π \int_{0}^{1} (x^{2} - x^{4}) d x = π (\frac{1}{3} - \frac{1}{5})

Picture it as a cone with a narrower, curved cone scooped out of it: the line $y = x$ alone would sweep out a cone of volume $\frac{π}{3}$ , and the hole made by $y = x^{2}$ takes $\frac{π}{5}$ of that away.

Answer

V = \frac{2 π}{15}

Example 3: Around the line y = 2

Harder

The same region, between $y = x$ and $y = x^{2}$ , is revolved about the line $y = 2$ . Find the volume.

y = x and y = x^{2}, about y = 2

Now the axis is above the region, so the radii are distances down from $y = 2$ : the line minus the curve. The curve farther from $y = 2$ is the lower one, $x^{2}$ , so it gives the outer radius.

R (x) = 2 - x^{2}, r (x) = 2 - x

That swap is the whole point of this example: the inner and outer curves trade places when the axis moves above the region.

V = π \int_{0}^{1} ((2 - x^{2})^{2} - (2 - x)^{2}) d x = π \int_{0}^{1} (x^{4} - 5 x^{2} + 4 x) d x

= π (\frac{1}{5} - \frac{5}{3} + 2) = π \cdot \frac{3 - 25 + 30}{15}

Answer

V = \frac{8 π}{15}

Disk and washer methods practice problems

Three to try, easy to harder. Type an answer and it is checked instantly by the same checker that grades the game; any correct form counts. Stuck? The answer and the first step are under each one.

Problem 1

Easy

The region under the curve from $x = 0$ to $x = 1$ is revolved about the $x$ -axis. Find the volume.

y = 2 x^{2}, 0 \leq x \leq 1

Stuck? Show the answer

Answer

\frac{4 π}{5}

How to start

Disks of radius $2 x^{2}$ , so the integrand is $π (2 x^{2})^{2} = 4 π x^{4}$ . Square the coefficient too.

Problem 2

Medium

The region between the two curves is revolved about the $x$ -axis. Find the volume.

y = 2 x and y = x^{2}

Stuck? Show the answer

Answer

\frac{64 π}{15}

How to start

They meet at $x = 0$ and $x = 2$ , and $2 x$ is farther from the axis in between. Washers: $π \int_{0}^{2} ((2 x)^{2} - (x^{2})^{2}) d x$ .

Problem 3

Harder

The region between the two curves is revolved about the $x$ -axis. Find the volume.

y = 3 - x^{2} and y = x^{2} + 1

Stuck? Show the answer

Answer

\frac{32 π}{3}

How to start

Set them equal to get $x = \pm 1$ . Outer radius $3 - x^{2}$ , inner radius $x^{2} + 1$ ; square each before subtracting, and the $x^{4}$ terms cancel.

Want more? Asymptote has 400 more disk and washer methods problems, free, with no account.

Common mistakes

Squaring the difference. $π \int (R - r)^{2} d x$ is not the washer formula. It is $π \int (R^{2} - r^{2}) d x$ : square each radius, then subtract. The two differ by $2 R r$ , which is not small.
Swapping outer and inner. A negative volume means $R$ and $r$ are the wrong way round. Outer is whichever curve is farther from the axis, which is not always the one on top.
Using the curve instead of the distance. About $y = 2$ the radius is $2 - y_{curve}$ , not $y_{curve}$ . Every radius is measured from the axis.
Forgetting the $π$ . Each slice is a circle, so $π$ is in every term. It is the most common way to be right by a factor of about three.

Practice it now

No clock, three strikes

Disk problems show up in Diverge around d8-d9: the region under $y = c x^{n}$ between two $x$ -values, spun about the $x$ -axis. They are solid disks only; there are no washer problems in the game yet. Answers are exact, in terms of $π$ .

Play Diverge See it on the formula sheet

FAQ

What is the difference between the disk and washer methods?: A disk is a solid slice; a washer is a slice with a hole. The washer formula is the disk formula for the outside minus the disk formula for the hole.
How do you know whether to use disks or washers?: Look for a gap between the region and the axis. No gap, disks. A gap, washers. A region between two curves almost always needs washers.
When should I use the shell method instead?: When the axis is parallel to the strips you would draw, or when washers would force you to solve the curve for $x$ in terms of $y$ .
Why is the disk method formula $π r^{2}$ ?: Each slice is a very thin cylinder: a circle of area $π r^{2}$ with thickness $d x$ . Adding up infinitely many of them is exactly what the integral does.
Can you use washers around the $y$ -axis?: Yes, with horizontal strips: $V = π \int_{c}^{d} (R (y)^{2} - r (y)^{2}) d y$ , with every radius written as a function of $y$ .
Where can I find disk and washer methods practice problems?: Right on this page: the practice problems section has three disk and washer methods problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 400 more disk and washer methods problems after that, free and with no account: try Diverge, which has no clock.

Disk and Washer Method: Formulas & Practice Problems

When to use it

Use disks or washers when the axis of rotation is perpendicular to your strips: vertical strips spun around the

x

-axis or any horizontal line. If the region touches the axis all the way along, every slice is a solid disk. If there is a gap between the region and the axis, the slices have holes and you need washers. When the axis is parallel to the strips instead, the shell method is usually less work.

How to do it, step by step

Sketch the region and the axis. Draw one thin strip perpendicular to the axis. Spinning it makes one disk or one washer.

Find the outer radius $R$ . The distance from the axis to the far edge of the region. About the

x

-axis that is the curve farther from it.

Find the inner radius $r$ . The distance from the axis to the near edge. If the region touches the axis,

r = 0

and you have a disk.

Square, subtract, integrate.

π \int (R^{2} - r^{2}) d x

between the limits of the region. Expand before integrating.

Worked examples

Example 1: A square root around the x-axis

Easy

The region under $y = x$ from $x = 0$ to $x = 4$ is revolved about the $x$ -axis. Find the volume.

y = x, 0 \leq x \leq 4, about the x -axis

The region touches the $x$ -axis everywhere, so each slice is a solid disk of radius $R (x) = x$ .

V = π \int_{0}^{4} (x)^{2} d x = π \int_{0}^{4} x d x = π [\frac{x ^{2}}{2}]_{0}^{4} = π \cdot 8

Squaring the radius got rid of the root before any integrating happened, which is the disk method being generous for once.

Answer

V = 8 π

Example 2: A washer between two curves

Medium

The region between $y = x$ and $y = x^{2}$ is revolved about the $x$ -axis. Find the volume.

y = x and y = x^{2}, about the x -axis

The curves meet at $x = 0$ and $x = 1$ , and $x$ is above $x^{2}$ in between. The region does not touch the axis (except at the ends), so the slices are washers.

The outer radius is the curve farther from the axis, $R = x$ . The inner radius is the nearer one, $r = x^{2}$ .

V = π \int_{0}^{1} (x^{2} - (x^{2})^{2}) d x = π \int_{0}^{1} (x^{2} - x^{4}) d x = π (\frac{1}{3} - \frac{1}{5})

Answer

V = \frac{2 π}{15}

Example 3: Around the line y = 2

Harder

The same region, between $y = x$ and $y = x^{2}$ , is revolved about the line $y = 2$ . Find the volume.

y = x and y = x^{2}, about y = 2

Now the axis is above the region, so the radii are distances down from $y = 2$ : the line minus the curve. The curve farther from $y = 2$ is the lower one, $x^{2}$ , so it gives the outer radius.

R (x) = 2 - x^{2}, r (x) = 2 - x

That swap is the whole point of this example: the inner and outer curves trade places when the axis moves above the region.

V = π \int_{0}^{1} ((2 - x^{2})^{2} - (2 - x)^{2}) d x = π \int_{0}^{1} (x^{4} - 5 x^{2} + 4 x) d x

= π (\frac{1}{5} - \frac{5}{3} + 2) = π \cdot \frac{3 - 25 + 30}{15}

Answer

V = \frac{8 π}{15}

Disk and washer methods practice problems

Three to try, easy to harder. Type an answer and it is checked instantly by the same checker that grades the game; any correct form counts. Stuck? The answer and the first step are under each one.

Problem 1

Easy

The region under the curve from $x = 0$ to $x = 1$ is revolved about the $x$ -axis. Find the volume.

y = 2 x^{2}, 0 \leq x \leq 1

Stuck? Show the answer

Answer

\frac{4 π}{5}

How to start

Disks of radius $2 x^{2}$ , so the integrand is $π (2 x^{2})^{2} = 4 π x^{4}$ . Square the coefficient too.

Problem 2

Medium

The region between the two curves is revolved about the $x$ -axis. Find the volume.

y = 2 x and y = x^{2}

Stuck? Show the answer

Answer

\frac{64 π}{15}

How to start

They meet at $x = 0$ and $x = 2$ , and $2 x$ is farther from the axis in between. Washers: $π \int_{0}^{2} ((2 x)^{2} - (x^{2})^{2}) d x$ .

Problem 3

Harder

The region between the two curves is revolved about the $x$ -axis. Find the volume.

y = 3 - x^{2} and y = x^{2} + 1

Stuck? Show the answer

Answer

\frac{32 π}{3}

How to start

Set them equal to get $x = \pm 1$ . Outer radius $3 - x^{2}$ , inner radius $x^{2} + 1$ ; square each before subtracting, and the $x^{4}$ terms cancel.

Common mistakes

Squaring the difference.

π \int (R - r)^{2} d x

is not the washer formula. It is

π \int (R^{2} - r^{2}) d x

: square each radius, then subtract. The two differ by

2 R r

, which is not small.

Swapping outer and inner. A negative volume means

R

and

r

are the wrong way round. Outer is whichever curve is farther from the axis, which is not always the one on top.

Using the curve instead of the distance. About

y = 2

the radius is

2 - y_{curve}

, not

y_{curve}

. Every radius is measured from the axis.

Forgetting the $π$ . Each slice is a circle, so

π

is in every term. It is the most common way to be right by a factor of about three.

Practice it now

No clock, three strikes

Play Diverge See it on the formula sheet

FAQ

What is the difference between the disk and washer methods?

A disk is a solid slice; a washer is a slice with a hole. The washer formula is the disk formula for the outside minus the disk formula for the hole.

How do you know whether to use disks or washers?

Look for a gap between the region and the axis. No gap, disks. A gap, washers. A region between two curves almost always needs washers.

When should I use the shell method instead?

When the axis is parallel to the strips you would draw, or when washers would force you to solve the curve for

x

in terms of

y

Why is the disk method formula

π r^{2}

Each slice is a very thin cylinder: a circle of area

π r^{2}

with thickness

d x

. Adding up infinitely many of them is exactly what the integral does.

Can you use washers around the

y

-axis?

Yes, with horizontal strips:

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

, with every radius written as a function of

y

Where can I find disk and washer methods practice problems?

Right on this page: the practice problems section has three disk and washer methods problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 400 more disk and washer methods problems after that, free and with no account: try Diverge, which has no clock.

Disk and Washer Method: Formulas & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A square root around the x-axis

Example 2: A washer between two curves

Example 3: Around the line y = 2

Disk and washer methods practice problems

Common mistakes

FAQ

Keep going

Disk and Washer Method: Formulas & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A square root around the x-axis

Example 2: A washer between two curves

Example 3: Around the line y = 2

Disk and washer methods practice problems

Common mistakes

FAQ

Keep going