What you’ll learn:

Using the disk and washer methods to find volumes of solids of revolution
How to set up integrals based on the axis of rotation

The disk method and the washer method are used to find the volumes of solids formed by rotating a region around an axis. Cross-sections in this case are always full circles (disks) or concentric circles (washers).

Disk method

This method is for solids of revolution with no hole.

Formula:

If the radius of the circular slice is $R (x)$ (rotating around a horizontal axis), then

$Volume = π \int_{a}^{b} [R (x)]^{2} d x$

Use $d y$ if rotating around a vertical axis (the radius function must be expressed in terms of $y$ ).

AP tip:

To use the disk and washer methods, the axis of rotation must be parallel to the direction of integration for the disk and washer methods. Use $d x$ if rotating about a horizontal axis and $d y$ when rotating about a vertical axis.

Horizontal axis of rotation

1. The region bounded by $y = x$ , the $x$ -axis, and $x = 2$ is rotated about the $x$ -axis. Determine the volume of the solid formed.

To find what the radius $R$ is, it’s always helpful to draw a diagram similar to the one shown below with these features:

The curve
Its reflection over the axis of rotation
A circle connecting the ends in order to label the radius $R$

At any $x$ in the interval $[0, 2]$ , the radius $R$ is the distance from the $x$ -axis to the line $y = x$ , so $R (x) = x$ .

Then the volume is

$V = π \int_{0}^{2} x^{2} d x$

$= π (\frac{x ^{3}}{3})_{0}^{2}$

$= \frac{8}{3} π$

In fact, the solid formed is a cone of radius of $2$ units and a height of $2$ units, and using the formula for the volume of a cone ( $V = \frac{1}{3} π r^{2} h$ ), we indeed arrive at $\frac{8}{3} π$ cubic units.

2. The region bounded by the $y$ -axis, the line $y = 2$ , and the curve $y = x$ is rotated about the line $y = 2$ . Determine the volume of the solid formed.

Solution

(spoiler)

R is the distance between the top and bottom functions

The radius $R$ is the distance between the top function, or the line $y = 2$ , and the bottom function $y = x$ , so

$R (x) = 2 - x$

The bounds of integration are from $x = 0$ to $x = 4$ and the volume is

$V = π \int_{0}^{4} (2 - x)^{2} d x$

$= \frac{8}{3} π$

Vertical axis of rotation

1. The region bounded by $y = x^{2} - 2 x + 1$ , the $x$ -axis, and the $y$ -axis is rotated about the $y$ -axis. Determine the volume of the solid formed.

Solution

(spoiler)

R is the distance between the right and left functions

Because the axis of rotation is vertical, the radius $R$ must be expressed in terms of $y$ . Rewriting the parabola $y = x^{2} - 2 x + 1$ in terms of $y$ :

$y = (x - 1)^{2}$

$\pm y = x - 1$

$x = \pm y + 1$

The region involved is the left half of the parabola, so we use the negative version.

$x = - y + 1$

At any point $y$ in the interval $y = 0$ to $y = 1$ , the radius of the circle $R$ is the distance between the curve and the $y$ -axis, so

$R (y) = - y + 1$

The volume is

$V = π \int_{0}^{1} (- y + 1)^{2} d y$

$= \frac{π}{6}$

2. The region bounded by $y = x - 1$ , the $x$ -axis, and the line $x = 3$ is rotated about $x = 3$ . Determine the volume of the solid formed.

Solution

(spoiler)

The radical function $y = x - 1$ meets the line $x = 3$ at $y = 2$ .

The axis of rotation is vertical so rewriting $y = x - 1$ is terms of $y$ ,

$y^{2} = x - 1$

$x = y^{2} + 1$

The radius of the circle $R$ is the distance between the axis of rotation $x = 3$ and the parabola $x = y^{2} + 1$ . This distance can be expressed as the right function (the axis) minus the left function (the parabola), or

$R (y) = 3 - (y^{2} + 1)$

$= 2 - y^{2}$

The interval to integrate over is from $y = 0$ to $y = 2$ and the volume is

$V = π \int_{0}^{2} (2 - y^{2})^{2} d y$

$= \frac{32 2}{15} π$

Washer method

Use this method when the solid is hollow in the center. Each slice is a washer - a circle with a smaller circle removed.

Formula:

If $R (x)$ is the outer radius and $r (x)$ is the inner radius when rotating about a horizontal axis,

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

As with the disk method, express functions in terms of $y$ and use $d y$ when rotating about a vertical axis.

AP tip:

The outer radius $R (x)$ is always the distance/difference between the axis of revolution and the function that is farther from it, while the inner radius $r (x)$ is the difference between the axis and the function closer to it.

Horizontal axis of rotation

1. The region bounded by $y = x^{2} + 1$ and $y = x + 3$ is rotated about the $x$ -axis. Determine the volume of the solid formed.

The radius of the bigger circle, $R$ , is the distance from the top function $y = x + 3$ to the axis of rotation ( $y = 0$ ), so

$R (x) = x + 3$

The radius of the smaller circle, $r$ , is the distance from the bottom function $y = x^{2} + 1$ to the axis of rotation ( $y = 0$ ), so

$r (x) = x^{2} + 1$

The two functions intersect at $(- 1, 2)$ and $(2, 5)$ so the bounds of integration are from $x = - 1$ to $x = 2$ and the volume is

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

$= \frac{117}{5} π$

2. The region bounded by $y = x^{2}$ and $y = x$ is rotated about the line $y = - 1$ . Determine the volume of the solid formed.

Solution

(spoiler)

The radius of the bigger circle, $R$ , is the distance from the top function $y = x$ to the axis of rotation ( $y = - 1$ ), so

$R (x) = x - (- 1)$

$= x + 1$

The radius of the smaller circle, $r$ , is the distance from the bottom function $y = x^{2}$ to the axis of rotation ( $y = - 1$ ), so

$r (x) = x^{2} + 1$

The two functions intersect at $(0, 0)$ and $(1, 1)$ so the bounds of integration are from $x = 0$ to $x = 1$ and the volume is

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

$= \frac{29}{30} π$

Vertical axis of rotation

1. The region bounded above by $y = (x - 1)^{1/3}$ and below by $y = x - 1$ is rotated about the line $x = 1$ . Determine the volume of the solid formed.

Solution

(spoiler)

Washer method; vertical axis of rotation

Because the axis of revolution is vertical, our functions must be in terms of $y$ :

$y = (x - 1)^{1/3}$ becomes

$x = y^{3} + 1$

$y = x - 1$ becomes

$x = y + 1$

Upon rotating, the radius of the bigger circle, $R$ , is the distance from the function $x = y + 1$ to the axis of rotation $x = 1$ , so

$R (y) = (y + 1) - 1$

$= y$

The radius of the smaller circle, $r$ , is the distance from the function $x = y^{3} + 1$ to the axis of rotation $x = 1$ , so

$r (y) = (y^{3} + 1) - 1$

$= y^{3}$

The two curves meet at $(1, 0)$ and $(2, 1)$ so the bounds of integration are from $y = 0$ to $y = 1$ and the volume is

$V = π \int_{0}^{1} [y^{2} - (y^{3})^{2}] d y$

$= \frac{4}{21} π$

2. The region bounded by $y = x + 3$ , the line $x = - 2$ , and the $x$ -axis is rotated about $x = - 1$ . Determine the volume of the solid formed.

Solution

(spoiler)

The axis of rotation is vertical so we express the function $y = x + 3$ in terms of $y$ . It’s the top half of the parabola

$x = y^{2} - 3$

The radius of the bigger circle, $R$ , is the distance from the axis of rotation $x = - 1$ to the parabola $x = y^{2} - 3$ . This distance can be expressed as the right function (the axis) minus the left function (the parabola), so

$R (y) = - 1 - (y^{2} - 3)$

$= 2 - y^{2}$

The radius of the smaller circle, $r$ , is a constant distance from the axis of rotation $x = - 1$ to the line $x = - 2$ , so

$r (y) = - 1 - (- 2)$

$= 1$

The upper half of the curve $x = y^{2} - 3$ and the line $x = - 2$ meet at $(- 2, 1)$ so the bounds of integration are from $y = 0$ to $y = 1$ and the volume is

$V = π \int_{0}^{1} [(2 - y^{2})^{2} - 1^{2}] d y$

$= \frac{28}{15} π$

What you’ll learn:

Using the disk and washer methods to find volumes of solids of revolution
How to set up integrals based on the axis of rotation

Disk method

This method is for solids of revolution with no hole.

Formula:

If the radius of the circular slice is $R (x)$ (rotating around a horizontal axis), then

$Volume = π \int_{a}^{b} [R (x)]^{2} d x$

Use $d y$ if rotating around a vertical axis (the radius function must be expressed in terms of $y$ ).

AP tip:

Horizontal axis of rotation

1. The region bounded by $y = x$ , the $x$ -axis, and $x = 2$ is rotated about the $x$ -axis. Determine the volume of the solid formed.

To find what the radius $R$ is, it’s always helpful to draw a diagram similar to the one shown below with these features:

The curve
Its reflection over the axis of rotation
A circle connecting the ends in order to label the radius $R$

At any $x$ in the interval $[0, 2]$ , the radius $R$ is the distance from the $x$ -axis to the line $y = x$ , so $R (x) = x$ .

Then the volume is

$V = π \int_{0}^{2} x^{2} d x$

$= π (\frac{x ^{3}}{3})_{0}^{2}$

$= \frac{8}{3} π$

2. The region bounded by the $y$ -axis, the line $y = 2$ , and the curve $y = x$ is rotated about the line $y = 2$ . Determine the volume of the solid formed.

Solution

(spoiler)

The radius $R$ is the distance between the top function, or the line $y = 2$ , and the bottom function $y = x$ , so

$R (x) = 2 - x$

The bounds of integration are from $x = 0$ to $x = 4$ and the volume is

$V = π \int_{0}^{4} (2 - x)^{2} d x$

$= \frac{8}{3} π$

Vertical axis of rotation

1. The region bounded by $y = x^{2} - 2 x + 1$ , the $x$ -axis, and the $y$ -axis is rotated about the $y$ -axis. Determine the volume of the solid formed.

Solution

(spoiler)

Because the axis of rotation is vertical, the radius $R$ must be expressed in terms of $y$ . Rewriting the parabola $y = x^{2} - 2 x + 1$ in terms of $y$ :

$y = (x - 1)^{2}$

$\pm y = x - 1$

$x = \pm y + 1$

The region involved is the left half of the parabola, so we use the negative version.

$x = - y + 1$

At any point $y$ in the interval $y = 0$ to $y = 1$ , the radius of the circle $R$ is the distance between the curve and the $y$ -axis, so

$R (y) = - y + 1$

The volume is

$V = π \int_{0}^{1} (- y + 1)^{2} d y$

$= \frac{π}{6}$

2. The region bounded by $y = x - 1$ , the $x$ -axis, and the line $x = 3$ is rotated about $x = 3$ . Determine the volume of the solid formed.

Solution

(spoiler)

The radical function $y = x - 1$ meets the line $x = 3$ at $y = 2$ .

The axis of rotation is vertical so rewriting $y = x - 1$ is terms of $y$ ,

$y^{2} = x - 1$

$x = y^{2} + 1$

$R (y) = 3 - (y^{2} + 1)$

$= 2 - y^{2}$

The interval to integrate over is from $y = 0$ to $y = 2$ and the volume is

$V = π \int_{0}^{2} (2 - y^{2})^{2} d y$

$= \frac{32 2}{15} π$

Washer method

Use this method when the solid is hollow in the center. Each slice is a washer - a circle with a smaller circle removed.

Formula:

If $R (x)$ is the outer radius and $r (x)$ is the inner radius when rotating about a horizontal axis,

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

As with the disk method, express functions in terms of $y$ and use $d y$ when rotating about a vertical axis.

AP tip:

Horizontal axis of rotation

1. The region bounded by $y = x^{2} + 1$ and $y = x + 3$ is rotated about the $x$ -axis. Determine the volume of the solid formed.

The radius of the bigger circle, $R$ , is the distance from the top function $y = x + 3$ to the axis of rotation ( $y = 0$ ), so

$R (x) = x + 3$

The radius of the smaller circle, $r$ , is the distance from the bottom function $y = x^{2} + 1$ to the axis of rotation ( $y = 0$ ), so

$r (x) = x^{2} + 1$

The two functions intersect at $(- 1, 2)$ and $(2, 5)$ so the bounds of integration are from $x = - 1$ to $x = 2$ and the volume is

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

$= \frac{117}{5} π$

2. The region bounded by $y = x^{2}$ and $y = x$ is rotated about the line $y = - 1$ . Determine the volume of the solid formed.

Solution

(spoiler)

The radius of the bigger circle, $R$ , is the distance from the top function $y = x$ to the axis of rotation ( $y = - 1$ ), so

$R (x) = x - (- 1)$

$= x + 1$

The radius of the smaller circle, $r$ , is the distance from the bottom function $y = x^{2}$ to the axis of rotation ( $y = - 1$ ), so

$r (x) = x^{2} + 1$

The two functions intersect at $(0, 0)$ and $(1, 1)$ so the bounds of integration are from $x = 0$ to $x = 1$ and the volume is

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

$= \frac{29}{30} π$

Vertical axis of rotation

1. The region bounded above by $y = (x - 1)^{1/3}$ and below by $y = x - 1$ is rotated about the line $x = 1$ . Determine the volume of the solid formed.

Solution

(spoiler)

Because the axis of revolution is vertical, our functions must be in terms of $y$ :

$y = (x - 1)^{1/3}$ becomes

$x = y^{3} + 1$

$y = x - 1$ becomes

$x = y + 1$

Upon rotating, the radius of the bigger circle, $R$ , is the distance from the function $x = y + 1$ to the axis of rotation $x = 1$ , so

$R (y) = (y + 1) - 1$

$= y$

The radius of the smaller circle, $r$ , is the distance from the function $x = y^{3} + 1$ to the axis of rotation $x = 1$ , so

$r (y) = (y^{3} + 1) - 1$

$= y^{3}$

The two curves meet at $(1, 0)$ and $(2, 1)$ so the bounds of integration are from $y = 0$ to $y = 1$ and the volume is

$V = π \int_{0}^{1} [y^{2} - (y^{3})^{2}] d y$

$= \frac{4}{21} π$

2. The region bounded by $y = x + 3$ , the line $x = - 2$ , and the $x$ -axis is rotated about $x = - 1$ . Determine the volume of the solid formed.

Solution

(spoiler)

The axis of rotation is vertical so we express the function $y = x + 3$ in terms of $y$ . It’s the top half of the parabola

$x = y^{2} - 3$

$R (y) = - 1 - (y^{2} - 3)$

$= 2 - y^{2}$

The radius of the smaller circle, $r$ , is a constant distance from the axis of rotation $x = - 1$ to the line $x = - 2$ , so

$r (y) = - 1 - (- 2)$

$= 1$

The upper half of the curve $x = y^{2} - 3$ and the line $x = - 2$ meet at $(- 2, 1)$ so the bounds of integration are from $y = 0$ to $y = 1$ and the volume is

$V = π \int_{0}^{1} [(2 - y^{2})^{2} - 1^{2}] d y$

$= \frac{28}{15} π$

Disk & washer methods

What you’ll learn:

Disk method

Horizontal axis of rotation

Solution

Vertical axis of rotation

Solution

Solution

Washer method

Horizontal axis of rotation

Solution

Vertical axis of rotation

Solution

Solution

Disk & washer methods

What you’ll learn:

Disk method

Horizontal axis of rotation

Solution

Vertical axis of rotation

Solution

Solution

Washer method

Horizontal axis of rotation

Solution

Vertical axis of rotation

Solution

Solution