8.4.2 Disk & washer methods

Achievable AP Calculus AB

8. Applications of integrals

8.4. Volume

Our AP Calculus AB course is currently in development and is a work-in-progress.

Disk & washer methods

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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 help you find the volume of a solid formed by rotating a region around an axis. In both methods, the cross-sections perpendicular to the axis of rotation are circles:

A disk is a full circle (no hole).
A washer is a circle with a smaller circle removed (a hole).

Disk method

Use the disk method when the solid of revolution has 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. Use $d x$ if rotating about a horizontal axis and $d y$ when rotating about a vertical axis.

Horizontal axis of rotation

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 identify the radius $R$ , it helps to sketch a diagram like the one below and include:

The curve
Its reflection over the axis of rotation
A circle connecting the ends so you can label the radius $R$

For any $x$ in $[0, 2]$ , the radius is the vertical distance from the $x$ -axis to the line $y = x$ . That distance is

$R (x) = x .$

Now apply the disk formula:

$V = π \int_{0}^{2} x^{2} d x$

$= π (\frac{x ^{3}}{3})_{0}^{2}$

$= \frac{8}{3} π$

This solid is a cone with radius $2$ and height $2$ . Using the cone volume formula $V = \frac{1}{3} π r^{2} h$ also gives $\frac{8}{3} π$ cubic units.

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

Here the axis of rotation is the horizontal line $y = 2$ . The radius is the vertical distance from $y = 2$ down to the curve $y = x$ :

$R (x) = 2 - x .$

The region runs from $x = 0$ (the $y$ -axis) to where $x = 2$ , which is $x = 4$ . So the volume is

$V = π \int_{0}^{4} (2 - x)^{2} d x$

$= \frac{8}{3} π$

Vertical axis of rotation

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 must be written as a function of $y$ , and we integrate with respect to $y$ .

Start by rewriting the parabola in terms of $y$ :

$y = x^{2} - 2 x + 1$

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

$\pm y = x - 1$

$x = \pm y + 1$

The region uses the left half of the parabola, so we take the negative branch:

$x = - y + 1.$

For $y$ from $0$ to $1$ , the radius is the horizontal distance from the $y$ -axis ( $x = 0$ ) to the curve:

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

Now compute the volume:

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

$= \frac{π}{6}$

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 curve meets the line $x = 3$ at

$y = 3 - 1 = 2 .$

Because the axis of rotation is vertical, rewrite the function in terms of $y$ :

$y = x - 1$

$y^{2} = x - 1$

$x = y^{2} + 1.$

The radius is the horizontal distance from the axis $x = 3$ to the curve $x = y^{2} + 1$ :

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

$= 2 - y^{2} .$

The region runs from $y = 0$ up to $y = 2$ , so

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

$= \frac{32 2}{15} π$

Washer method

Use the washer method when the solid is hollow in the center. Each cross-section is a washer: an outer circle minus an inner circle.

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

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 axis of rotation is the $x$ -axis ( $y = 0$ ), so radii are vertical distances to that axis.

Outer radius (farther from the axis): from $y = x + 3$ down to $y = 0$

$R (x) = x + 3$

Inner radius (closer to the axis): from $y = x^{2} + 1$ down to $y = 0$

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

Find the intersection points to set the bounds:

$x^{2} + 1 = x + 3 \Rightarrow x^{2} - x - 2 = 0 \Rightarrow (x - 2) (x + 1) = 0.$

So $x = - 1$ and $x = 2$ . The volume is

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

$= \frac{117}{5} π$

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 axis of rotation is $y = - 1$ , so each radius is a vertical distance up from $y = - 1$ .

Outer radius (farther from $y = - 1$ ): from $y = x$ to $y = - 1$

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

$= x + 1$

Inner radius (closer to $y = - 1$ ): from $y = x^{2}$ to $y = - 1$

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

$= x^{2} + 1$

The curves intersect where $x^{2} = x$ , which occurs at $x = 0$ and $x = 1$ . So

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

$= \frac{29}{30} π$

Vertical axis of rotation

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, write both curves as $x$ in terms of $y$ .

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

$x = y^{3} + 1$

From $y = x - 1$ :

$x = y + 1$

The axis of rotation is $x = 1$ , so radii are horizontal distances from $x = 1$ .

Outer radius (farther from the axis): to $x = y + 1$

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

$= y$

Inner radius (closer to the axis): to $x = y^{3} + 1$

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

$= y^{3}$

The curves meet at $(1, 0)$ and $(2, 1)$ , so $y$ runs from $0$ to $1$ . The volume is

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

$= \frac{4}{21} π$

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 rewrite $y = x + 3$ in terms of $y$ . Squaring gives the (right-opening) parabola:

$y^{2} = x + 3$

$x = y^{2} - 3$

Now measure horizontal distances from the axis $x = - 1$ .

Outer radius: from $x = - 1$ to the curve $x = y^{2} - 3$

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

$= 2 - y^{2}$

Inner radius: from $x = - 1$ to the line $x = - 2$ (a constant distance)

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

$= 1$

The curve $x = y^{2} - 3$ meets $x = - 2$ when $y^{2} - 3 = - 2$ , so $y = 1$ (using the upper half shown). The region runs from $y = 0$ to $y = 1$ , so

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

$= \frac{28}{15} π$

Disk method

Used when solid of revolution has no hole (cross-sections are full circles)
Formula: $V = π \int_{a}^{b} [R (x)]^{2} d x$ or $V = π \int_{a}^{b} [R (y)]^{2} d y$
Axis of rotation must be parallel to direction of integration ( $d x$ for horizontal, $d y$ for vertical)

Horizontal axis of rotation (disk)

Radius is vertical distance from axis to function: $R (x)$
Bounds determined by region endpoints in $x$
Example: For $y = x$ rotated about $x$ -axis, $R (x) = x$ , $V = π \int_{0}^{2} x^{2} d x = \frac{8}{3} π$

Vertical axis of rotation (disk)

Radius is horizontal distance from axis to function: $R (y)$
Express function as $x$ in terms of $y$ ; integrate with respect to $y$
Example: For $y = x^{2} - 2 x + 1$ rotated about $y$ -axis, $R (y) = - y + 1$ , $V = π \int_{0}^{1} (- y + 1)^{2} d y = \frac{π}{6}$

Washer method

Used when solid is hollow in the center (cross-sections are washers)
Formula: $V = π \int_{a}^{b} ([R (x)]^{2} - [r (x)]^{2}) d x$ or $V = π \int_{a}^{b} ([R (y)]^{2} - [r (y)]^{2}) d y$
$R (x)$ = outer radius (farther from axis), $r (x)$ = inner radius (closer to axis)

Horizontal axis of rotation (washer)

Radii are vertical distances from axis to each function
Bounds found by intersection points in $x$
Example: For $y = x^{2} + 1$ and $y = x + 3$ about $x$ -axis, $R (x) = x + 3$ , $r (x) = x^{2} + 1$ , $V = π \int_{- 1}^{2} [(x + 3)^{2} - (x^{2} + 1)^{2}] d x$

Vertical axis of rotation (washer)

Radii are horizontal distances from axis to each function, written as $x$ in terms of $y$
Bounds found by intersection points in $y$
Example: For $y = (x - 1)^{1/3}$ and $y = x - 1$ about $x = 1$ , $R (y) = y$ , $r (y) = y^{3}$ , $V = π \int_{0}^{1} [y^{2} - (y^{3})^{2}] d y$

General setup tips

Always sketch region and axis of rotation to identify radii
Outer radius = distance from axis to farther curve; inner radius = distance to nearer curve
Match variable of integration ( $d x$ or $d y$ ) to axis orientation (horizontal or vertical)

Disk & washer methods

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

A disk is a full circle (no hole).
A washer is a circle with a smaller circle removed (a hole).

Disk method

Use the disk method when the solid of revolution has 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

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 identify the radius $R$ , it helps to sketch a diagram like the one below and include:

The curve
Its reflection over the axis of rotation
A circle connecting the ends so you can label the radius $R$

For any $x$ in $[0, 2]$ , the radius is the vertical distance from the $x$ -axis to the line $y = x$ . That distance is

$R (x) = x .$

Now apply the disk formula:

$V = π \int_{0}^{2} x^{2} d x$

$= π (\frac{x ^{3}}{3})_{0}^{2}$

$= \frac{8}{3} π$

This solid is a cone with radius $2$ and height $2$ . Using the cone volume formula $V = \frac{1}{3} π r^{2} h$ also gives $\frac{8}{3} π$ cubic units.

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)

Here the axis of rotation is the horizontal line $y = 2$ . The radius is the vertical distance from $y = 2$ down to the curve $y = x$ :

$R (x) = 2 - x .$

The region runs from $x = 0$ (the $y$ -axis) to where $x = 2$ , which is $x = 4$ . So the volume is

$V = π \int_{0}^{4} (2 - x)^{2} d x$

$= \frac{8}{3} π$

Vertical axis of rotation

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 must be written as a function of $y$ , and we integrate with respect to $y$ .

Start by rewriting the parabola in terms of $y$ :

$y = x^{2} - 2 x + 1$

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

$\pm y = x - 1$

$x = \pm y + 1$

The region uses the left half of the parabola, so we take the negative branch:

$x = - y + 1.$

For $y$ from $0$ to $1$ , the radius is the horizontal distance from the $y$ -axis ( $x = 0$ ) to the curve:

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

Now compute the volume:

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

$= \frac{π}{6}$

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 curve meets the line $x = 3$ at

$y = 3 - 1 = 2 .$

Because the axis of rotation is vertical, rewrite the function in terms of $y$ :

$y = x - 1$

$y^{2} = x - 1$

$x = y^{2} + 1.$

The radius is the horizontal distance from the axis $x = 3$ to the curve $x = y^{2} + 1$ :

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

$= 2 - y^{2} .$

The region runs from $y = 0$ up to $y = 2$ , so

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

$= \frac{32 2}{15} π$

Washer method

Use the washer method when the solid is hollow in the center. Each cross-section is a washer: an outer circle minus an inner circle.

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

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 axis of rotation is the $x$ -axis ( $y = 0$ ), so radii are vertical distances to that axis.

Outer radius (farther from the axis): from $y = x + 3$ down to $y = 0$

$R (x) = x + 3$

Inner radius (closer to the axis): from $y = x^{2} + 1$ down to $y = 0$

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

Find the intersection points to set the bounds:

$x^{2} + 1 = x + 3 \Rightarrow x^{2} - x - 2 = 0 \Rightarrow (x - 2) (x + 1) = 0.$

So $x = - 1$ and $x = 2$ . The volume is

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

$= \frac{117}{5} π$

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 axis of rotation is $y = - 1$ , so each radius is a vertical distance up from $y = - 1$ .

Outer radius (farther from $y = - 1$ ): from $y = x$ to $y = - 1$

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

$= x + 1$

Inner radius (closer to $y = - 1$ ): from $y = x^{2}$ to $y = - 1$

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

$= x^{2} + 1$

The curves intersect where $x^{2} = x$ , which occurs at $x = 0$ and $x = 1$ . So

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

$= \frac{29}{30} π$

Vertical axis of rotation

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, write both curves as $x$ in terms of $y$ .

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

$x = y^{3} + 1$

From $y = x - 1$ :

$x = y + 1$

The axis of rotation is $x = 1$ , so radii are horizontal distances from $x = 1$ .

Outer radius (farther from the axis): to $x = y + 1$

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

$= y$

Inner radius (closer to the axis): to $x = y^{3} + 1$

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

$= y^{3}$

The curves meet at $(1, 0)$ and $(2, 1)$ , so $y$ runs from $0$ to $1$ . The volume is

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

$= \frac{4}{21} π$

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 rewrite $y = x + 3$ in terms of $y$ . Squaring gives the (right-opening) parabola:

$y^{2} = x + 3$

$x = y^{2} - 3$

Now measure horizontal distances from the axis $x = - 1$ .

Outer radius: from $x = - 1$ to the curve $x = y^{2} - 3$

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

$= 2 - y^{2}$

Inner radius: from $x = - 1$ to the line $x = - 2$ (a constant distance)

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

$= 1$

The curve $x = y^{2} - 3$ meets $x = - 2$ when $y^{2} - 3 = - 2$ , so $y = 1$ (using the upper half shown). The region runs from $y = 0$ to $y = 1$ , so

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

$= \frac{28}{15} π$

Achievable AP Calculus AB

8. Applications of integrals

8.4. Volume

Our AP Calculus AB course is currently in development and is a work-in-progress.

Disk & washer methods

11 min read

Font

Discuss

Feedback

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

A disk is a full circle (no hole).
A washer is a circle with a smaller circle removed (a hole).

Disk method

Use the disk method when the solid of revolution has 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

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 identify the radius $R$ , it helps to sketch a diagram like the one below and include:

The curve
Its reflection over the axis of rotation
A circle connecting the ends so you can label the radius $R$

For any $x$ in $[0, 2]$ , the radius is the vertical distance from the $x$ -axis to the line $y = x$ . That distance is

$R (x) = x .$

Now apply the disk formula:

$V = π \int_{0}^{2} x^{2} d x$

$= π (\frac{x ^{3}}{3})_{0}^{2}$

$= \frac{8}{3} π$

This solid is a cone with radius $2$ and height $2$ . Using the cone volume formula $V = \frac{1}{3} π r^{2} h$ also gives $\frac{8}{3} π$ cubic units.

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)

Here the axis of rotation is the horizontal line $y = 2$ . The radius is the vertical distance from $y = 2$ down to the curve $y = x$ :

$R (x) = 2 - x .$

The region runs from $x = 0$ (the $y$ -axis) to where $x = 2$ , which is $x = 4$ . So the volume is

$V = π \int_{0}^{4} (2 - x)^{2} d x$

$= \frac{8}{3} π$

Vertical axis of rotation

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 must be written as a function of $y$ , and we integrate with respect to $y$ .

Start by rewriting the parabola in terms of $y$ :

$y = x^{2} - 2 x + 1$

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

$\pm y = x - 1$

$x = \pm y + 1$

The region uses the left half of the parabola, so we take the negative branch:

$x = - y + 1.$

For $y$ from $0$ to $1$ , the radius is the horizontal distance from the $y$ -axis ( $x = 0$ ) to the curve:

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

Now compute the volume:

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

$= \frac{π}{6}$

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 curve meets the line $x = 3$ at

$y = 3 - 1 = 2 .$

Because the axis of rotation is vertical, rewrite the function in terms of $y$ :

$y = x - 1$

$y^{2} = x - 1$

$x = y^{2} + 1.$

The radius is the horizontal distance from the axis $x = 3$ to the curve $x = y^{2} + 1$ :

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

$= 2 - y^{2} .$

The region runs from $y = 0$ up to $y = 2$ , so

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

$= \frac{32 2}{15} π$

Washer method

Use the washer method when the solid is hollow in the center. Each cross-section is a washer: an outer circle minus an inner circle.

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

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 axis of rotation is the $x$ -axis ( $y = 0$ ), so radii are vertical distances to that axis.

Outer radius (farther from the axis): from $y = x + 3$ down to $y = 0$

$R (x) = x + 3$

Inner radius (closer to the axis): from $y = x^{2} + 1$ down to $y = 0$

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

Find the intersection points to set the bounds:

$x^{2} + 1 = x + 3 \Rightarrow x^{2} - x - 2 = 0 \Rightarrow (x - 2) (x + 1) = 0.$

So $x = - 1$ and $x = 2$ . The volume is

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

$= \frac{117}{5} π$

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 axis of rotation is $y = - 1$ , so each radius is a vertical distance up from $y = - 1$ .

Outer radius (farther from $y = - 1$ ): from $y = x$ to $y = - 1$

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

$= x + 1$

Inner radius (closer to $y = - 1$ ): from $y = x^{2}$ to $y = - 1$

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

$= x^{2} + 1$

The curves intersect where $x^{2} = x$ , which occurs at $x = 0$ and $x = 1$ . So

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

$= \frac{29}{30} π$

Vertical axis of rotation

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, write both curves as $x$ in terms of $y$ .

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

$x = y^{3} + 1$

From $y = x - 1$ :

$x = y + 1$

The axis of rotation is $x = 1$ , so radii are horizontal distances from $x = 1$ .

Outer radius (farther from the axis): to $x = y + 1$

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

$= y$

Inner radius (closer to the axis): to $x = y^{3} + 1$

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

$= y^{3}$

The curves meet at $(1, 0)$ and $(2, 1)$ , so $y$ runs from $0$ to $1$ . The volume is

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

$= \frac{4}{21} π$

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 rewrite $y = x + 3$ in terms of $y$ . Squaring gives the (right-opening) parabola:

$y^{2} = x + 3$

$x = y^{2} - 3$

Now measure horizontal distances from the axis $x = - 1$ .

Outer radius: from $x = - 1$ to the curve $x = y^{2} - 3$

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

$= 2 - y^{2}$

Inner radius: from $x = - 1$ to the line $x = - 2$ (a constant distance)

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

$= 1$

The curve $x = y^{2} - 3$ meets $x = - 2$ when $y^{2} - 3 = - 2$ , so $y = 1$ (using the upper half shown). The region runs from $y = 0$ to $y = 1$ , so

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

$= \frac{28}{15} π$

Disk method

Used when solid of revolution has no hole (cross-sections are full circles)
Formula: $V = π \int_{a}^{b} [R (x)]^{2} d x$ or $V = π \int_{a}^{b} [R (y)]^{2} d y$
Axis of rotation must be parallel to direction of integration ( $d x$ for horizontal, $d y$ for vertical)

Horizontal axis of rotation (disk)

Radius is vertical distance from axis to function: $R (x)$
Bounds determined by region endpoints in $x$
Example: For $y = x$ rotated about $x$ -axis, $R (x) = x$ , $V = π \int_{0}^{2} x^{2} d x = \frac{8}{3} π$

Vertical axis of rotation (disk)

Radius is horizontal distance from axis to function: $R (y)$
Express function as $x$ in terms of $y$ ; integrate with respect to $y$
Example: For $y = x^{2} - 2 x + 1$ rotated about $y$ -axis, $R (y) = - y + 1$ , $V = π \int_{0}^{1} (- y + 1)^{2} d y = \frac{π}{6}$

Washer method

Used when solid is hollow in the center (cross-sections are washers)
Formula: $V = π \int_{a}^{b} ([R (x)]^{2} - [r (x)]^{2}) d x$ or $V = π \int_{a}^{b} ([R (y)]^{2} - [r (y)]^{2}) d y$
$R (x)$ = outer radius (farther from axis), $r (x)$ = inner radius (closer to axis)

Horizontal axis of rotation (washer)

Radii are vertical distances from axis to each function
Bounds found by intersection points in $x$
Example: For $y = x^{2} + 1$ and $y = x + 3$ about $x$ -axis, $R (x) = x + 3$ , $r (x) = x^{2} + 1$ , $V = π \int_{- 1}^{2} [(x + 3)^{2} - (x^{2} + 1)^{2}] d x$

Vertical axis of rotation (washer)

Radii are horizontal distances from axis to each function, written as $x$ in terms of $y$
Bounds found by intersection points in $y$
Example: For $y = (x - 1)^{1/3}$ and $y = x - 1$ about $x = 1$ , $R (y) = y$ , $r (y) = y^{3}$ , $V = π \int_{0}^{1} [y^{2} - (y^{3})^{2}] d y$

General setup tips

Always sketch region and axis of rotation to identify radii
Outer radius = distance from axis to farther curve; inner radius = distance to nearer curve
Match variable of integration ( $d x$ or $d y$ ) to axis orientation (horizontal or vertical)

Disk & washer methods

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

A disk is a full circle (no hole).
A washer is a circle with a smaller circle removed (a hole).

Disk method

Use the disk method when the solid of revolution has 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

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 identify the radius $R$ , it helps to sketch a diagram like the one below and include:

The curve
Its reflection over the axis of rotation
A circle connecting the ends so you can label the radius $R$

For any $x$ in $[0, 2]$ , the radius is the vertical distance from the $x$ -axis to the line $y = x$ . That distance is

$R (x) = x .$

Now apply the disk formula:

$V = π \int_{0}^{2} x^{2} d x$

$= π (\frac{x ^{3}}{3})_{0}^{2}$

$= \frac{8}{3} π$

This solid is a cone with radius $2$ and height $2$ . Using the cone volume formula $V = \frac{1}{3} π r^{2} h$ also gives $\frac{8}{3} π$ cubic units.

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)

Here the axis of rotation is the horizontal line $y = 2$ . The radius is the vertical distance from $y = 2$ down to the curve $y = x$ :

$R (x) = 2 - x .$

The region runs from $x = 0$ (the $y$ -axis) to where $x = 2$ , which is $x = 4$ . So the volume is

$V = π \int_{0}^{4} (2 - x)^{2} d x$

$= \frac{8}{3} π$

Vertical axis of rotation

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 must be written as a function of $y$ , and we integrate with respect to $y$ .

Start by rewriting the parabola in terms of $y$ :

$y = x^{2} - 2 x + 1$

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

$\pm y = x - 1$

$x = \pm y + 1$

The region uses the left half of the parabola, so we take the negative branch:

$x = - y + 1.$

For $y$ from $0$ to $1$ , the radius is the horizontal distance from the $y$ -axis ( $x = 0$ ) to the curve:

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

Now compute the volume:

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

$= \frac{π}{6}$

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 curve meets the line $x = 3$ at

$y = 3 - 1 = 2 .$

Because the axis of rotation is vertical, rewrite the function in terms of $y$ :

$y = x - 1$

$y^{2} = x - 1$

$x = y^{2} + 1.$

The radius is the horizontal distance from the axis $x = 3$ to the curve $x = y^{2} + 1$ :

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

$= 2 - y^{2} .$

The region runs from $y = 0$ up to $y = 2$ , so

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

$= \frac{32 2}{15} π$

Washer method

Use the washer method when the solid is hollow in the center. Each cross-section is a washer: an outer circle minus an inner circle.

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

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 axis of rotation is the $x$ -axis ( $y = 0$ ), so radii are vertical distances to that axis.

Outer radius (farther from the axis): from $y = x + 3$ down to $y = 0$

$R (x) = x + 3$

Inner radius (closer to the axis): from $y = x^{2} + 1$ down to $y = 0$

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

Find the intersection points to set the bounds:

$x^{2} + 1 = x + 3 \Rightarrow x^{2} - x - 2 = 0 \Rightarrow (x - 2) (x + 1) = 0.$

So $x = - 1$ and $x = 2$ . The volume is

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

$= \frac{117}{5} π$

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 axis of rotation is $y = - 1$ , so each radius is a vertical distance up from $y = - 1$ .

Outer radius (farther from $y = - 1$ ): from $y = x$ to $y = - 1$

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

$= x + 1$

Inner radius (closer to $y = - 1$ ): from $y = x^{2}$ to $y = - 1$

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

$= x^{2} + 1$

The curves intersect where $x^{2} = x$ , which occurs at $x = 0$ and $x = 1$ . So

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

$= \frac{29}{30} π$

Vertical axis of rotation

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, write both curves as $x$ in terms of $y$ .

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

$x = y^{3} + 1$

From $y = x - 1$ :

$x = y + 1$

The axis of rotation is $x = 1$ , so radii are horizontal distances from $x = 1$ .

Outer radius (farther from the axis): to $x = y + 1$

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

$= y$

Inner radius (closer to the axis): to $x = y^{3} + 1$

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

$= y^{3}$

The curves meet at $(1, 0)$ and $(2, 1)$ , so $y$ runs from $0$ to $1$ . The volume is

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

$= \frac{4}{21} π$

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 rewrite $y = x + 3$ in terms of $y$ . Squaring gives the (right-opening) parabola:

$y^{2} = x + 3$

$x = y^{2} - 3$

Now measure horizontal distances from the axis $x = - 1$ .

Outer radius: from $x = - 1$ to the curve $x = y^{2} - 3$

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

$= 2 - y^{2}$

Inner radius: from $x = - 1$ to the line $x = - 2$ (a constant distance)

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

$= 1$

The curve $x = y^{2} - 3$ meets $x = - 2$ when $y^{2} - 3 = - 2$ , so $y = 1$ (using the upper half shown). The region runs from $y = 0$ to $y = 1$ , so

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

$= \frac{28}{15} π$