8.4.3 Washer method

Achievable AP Calculus AB

8. Applications of integrals

8.4. Volume

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Washer method

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What you’ll learn

Washer method: Calculate the volume of a solid of revolution with a hollow core using concentric circles as cross-sections.

The washer method, an extension of the disk method, is used when revolving a region around an axis creates a hollow 3D solid. Every cross-sectional slice perpendicular to the axis forms a washer (a large outer circle with a smaller inner circle removed).

The figure below illustrates how revolving the same region yields two different cross-sections: solid disks when rotated about the $x$ -axis, and hollow washers when rotated about the $y$ -axis.

Disk vs. washer: Revolving the same region around different axes

AP tip:

To determine which method to use, compare the region’s boundary to the axis of revolution:

Disk: The region touches the axis along the whole boundary line (no gap)
Washer: There is space between the region and the axis (gap $\to$ hole)

Horizontal axes of revolution $(d x)$

When a region is revolved around a horizontal axis and creates a hollow core, integrate with respect to $x$ :

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

where:

Outer radius $R (x)$ : The distance from the axis to the curve farther from it.
Inner radius $r (x)$ : The distance from the axis to the curve closer to it.

Watch your algebra!

Always remember that the washer cross-section is the area of the larger circle minus the area of the smaller circle: $Area = π R^{2} - π r^{2}$ .

When factoring out $π$ , the squares must stay attached to the individual radii:

Correct: $π (R^{2} - r^{2})$
Incorrect: $π (R - r)^{2}$

Example 1: Revolving around the $x$ -axis

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

Solution

(spoiler)

1. Bounds:

Find the intersection points of the two curves to set the $x$ -bounds:

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

So we integrate over $[- 1, 2]$ .

2. Find the radii $R (x)$ and $r (x)$ :

These are vertical distances from the axis of $y = 0$ :

Outer radius $R (x)$ : The top line $y = x + 3$ is farther from the axis.

$R (x) = (x + 3) - 0 = x + 3$

Inner radius $r (x)$ : The bottom parabola $y = x^{2} + 1$ is closer to the axis.

$r (x) = (x^{2} + 1) - 0 = x^{2} + 1$

3. Integrate:

$V = π \int_{a}^{b} (R^{2} - r^{2}) d x = π \int_{- 1}^{2} [(x + 3)^{2} - (x^{2} + 1)^{2}] d x = \frac{117}{5} π$

Example 2: Other horizontal axes $(y = k)$

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)

Once you’re comfortable visualizing the solid, you don’t need to sketch the washers every time. Instead, drop two straight lines from your functions to the axis of rotation:

The longer line gives you the outer radius, $R$ .
The shorter line gives you the inner radius, $r$ .

The region revolved about the line y = -1

Double-check analytically by substituting a test point $x$ within the interval of integration into both equations. The function that yields the larger value is always the outer radius $R$ , no matter what the sketch looks like.

1. Bounds:

Solving $x^{2} = x$ gives $x = 0$ and $x = 1$ so we integrate over $[0, 1]$ .

2. Find the radii $R (x)$ and $r (x)$ :

These are vertical distances from the axis of $y = - 1$ .

Outer radius $R$ : The curve $y = x$ is farther from the axis.

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

Inner radius $r$ : The parabola $y = x^{2}$ is closer to the axis.

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

3. Integrate:

$V = π \int_{0}^{1} [(x + 1)^{2} - (x^{2} + 1)^{2}] d x = \frac{29}{30} π$

Vertical axes of revolution $(d y)$

When a region is revolved around a vertical axis, integrate with respect to $y$ :

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

In similar fashion, the radii are measured as horizontal distances from the curves to the axis of revolution:

Outer radius $R (y)$ : The horizontal distance to the farther curve.
Inner radius $r (y)$ : The horizontal distance to the closer curve.

AP tip:

Just like the disk method, the orientation of the axis of revolution dictates the variable of integration:

Horizontal axis $\Rightarrow$ use $d x$
Vertical axis $\Rightarrow$ use $d y$

Example 3: Revolving about a vertical axis $(x = k)$

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)

A representative washer cross-section when the region is revolved about the vertical line x = 1

1. Rewrite the equations in terms of $y$ :

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

2. Bounds:

$y$ -bounds: $[0, 1]$

3. Find radii $R (y)$ and $r (y)$ :

Radii are horizontal distances from the axis of revolution $x = 1$ .

Outer radius $R (y)$ : The line $x = y + 1$ is farther from the axis.

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

Inner radius $r (y)$ : The curve $x = y^{3} + 1$ is closer to the axis.

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

4. Integrate:

$V = π \int_{c}^{d} [R (y)^{2} - r (y)^{2}] d y = π \int_{0}^{1} [y^{2} - (y^{3})^{2}] d y = \frac{4}{21} π$

Example 4: Constant inner radius

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)

A constant inner radius (r) and variable outer radius (R) measured horizontally from the axis of revolution.

1. Rewrite the equation in terms of $y$ :

$y = x + 3 ⟹ x = y^{2} - 3$

2. Bounds:

$y$ -bounds: $[0, 1]$ (where the curve meets the right boundary $x = - 2$ )

3. Find radii $R (y)$ and $r (y)$ :

Radii are horizontal distances from the axis of revolution $x = - 1$ .

Outer radius $R (y)$ : The curve $x = y^{2} - 3$ is farther from the axis.

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

Note: The order of terms does not matter since $R$ will be squared. $R = 2 - y^{2}$ is also valid.

Inner radius $r (y)$ : The line $x = - 2$ is a constant distance away from the axis.

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

4. Integrate:

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

$= \frac{28}{15} π$

Washer Method Overview

Used for hollow solids of revolution (disk with a hole)
Formula: $V = π \int_{a}^{b} [R (x)^{2} - r (x)^{2}] d x$
Use $d y$ and express in terms of $y$ when rotating about a vertical axis

Identifying Radii

Outer radius $R$ : distance from axis to the farther function
Inner radius $r$ : distance from axis to the closer function
When axis is not at origin, add/subtract axis value to each radius

Horizontal Axis of Rotation

Radii are vertical distances measured from the axis
Find bounds by setting the two functions equal and solving
Axis shift example: rotating about $y = - 1$ adds 1 to each radius

Vertical Axis of Rotation

Rewrite both curves as $x$ in terms of $y$
Radii are horizontal distances from the vertical axis
Bounds come from $y$ -values at intersection points

Washer method

What you’ll learn

Washer method: Calculate the volume of a solid of revolution with a hollow core using concentric circles as cross-sections.

The figure below illustrates how revolving the same region yields two different cross-sections: solid disks when rotated about the $x$ -axis, and hollow washers when rotated about the $y$ -axis.

AP tip:

To determine which method to use, compare the region’s boundary to the axis of revolution:

Disk: The region touches the axis along the whole boundary line (no gap)
Washer: There is space between the region and the axis (gap $\to$ hole)

Horizontal axes of revolution $(d x)$

When a region is revolved around a horizontal axis and creates a hollow core, integrate with respect to $x$ :

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

where:

Outer radius $R (x)$ : The distance from the axis to the curve farther from it.
Inner radius $r (x)$ : The distance from the axis to the curve closer to it.

Watch your algebra!

Always remember that the washer cross-section is the area of the larger circle minus the area of the smaller circle: $Area = π R^{2} - π r^{2}$ .

When factoring out $π$ , the squares must stay attached to the individual radii:

Correct: $π (R^{2} - r^{2})$
Incorrect: $π (R - r)^{2}$

Example 1: Revolving around the $x$ -axis

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

Solution

(spoiler)

1. Bounds:

Find the intersection points of the two curves to set the $x$ -bounds:

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

So we integrate over $[- 1, 2]$ .

2. Find the radii $R (x)$ and $r (x)$ :

These are vertical distances from the axis of $y = 0$ :

Outer radius $R (x)$ : The top line $y = x + 3$ is farther from the axis.

$R (x) = (x + 3) - 0 = x + 3$

Inner radius $r (x)$ : The bottom parabola $y = x^{2} + 1$ is closer to the axis.

$r (x) = (x^{2} + 1) - 0 = x^{2} + 1$

3. Integrate:

$V = π \int_{a}^{b} (R^{2} - r^{2}) d x = π \int_{- 1}^{2} [(x + 3)^{2} - (x^{2} + 1)^{2}] d x = \frac{117}{5} π$

Example 2: Other horizontal axes $(y = k)$

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)

Once you’re comfortable visualizing the solid, you don’t need to sketch the washers every time. Instead, drop two straight lines from your functions to the axis of rotation:

The longer line gives you the outer radius, $R$ .
The shorter line gives you the inner radius, $r$ .

1. Bounds:

Solving $x^{2} = x$ gives $x = 0$ and $x = 1$ so we integrate over $[0, 1]$ .

2. Find the radii $R (x)$ and $r (x)$ :

These are vertical distances from the axis of $y = - 1$ .

Outer radius $R$ : The curve $y = x$ is farther from the axis.

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

Inner radius $r$ : The parabola $y = x^{2}$ is closer to the axis.

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

3. Integrate:

$V = π \int_{0}^{1} [(x + 1)^{2} - (x^{2} + 1)^{2}] d x = \frac{29}{30} π$

Vertical axes of revolution $(d y)$

When a region is revolved around a vertical axis, integrate with respect to $y$ :

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

In similar fashion, the radii are measured as horizontal distances from the curves to the axis of revolution:

Outer radius $R (y)$ : The horizontal distance to the farther curve.
Inner radius $r (y)$ : The horizontal distance to the closer curve.

AP tip:

Just like the disk method, the orientation of the axis of revolution dictates the variable of integration:

Horizontal axis $\Rightarrow$ use $d x$
Vertical axis $\Rightarrow$ use $d y$

Example 3: Revolving about a vertical axis $(x = k)$

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)

1. Rewrite the equations in terms of $y$ :

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

2. Bounds:

$y$ -bounds: $[0, 1]$

3. Find radii $R (y)$ and $r (y)$ :

Radii are horizontal distances from the axis of revolution $x = 1$ .

Outer radius $R (y)$ : The line $x = y + 1$ is farther from the axis.

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

Inner radius $r (y)$ : The curve $x = y^{3} + 1$ is closer to the axis.

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

4. Integrate:

$V = π \int_{c}^{d} [R (y)^{2} - r (y)^{2}] d y = π \int_{0}^{1} [y^{2} - (y^{3})^{2}] d y = \frac{4}{21} π$

Example 4: Constant inner radius

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)

1. Rewrite the equation in terms of $y$ :

$y = x + 3 ⟹ x = y^{2} - 3$

2. Bounds:

$y$ -bounds: $[0, 1]$ (where the curve meets the right boundary $x = - 2$ )

3. Find radii $R (y)$ and $r (y)$ :

Radii are horizontal distances from the axis of revolution $x = - 1$ .

Outer radius $R (y)$ : The curve $x = y^{2} - 3$ is farther from the axis.

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

Note: The order of terms does not matter since $R$ will be squared. $R = 2 - y^{2}$ is also valid.

Inner radius $r (y)$ : The line $x = - 2$ is a constant distance away from the axis.

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

4. Integrate:

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

$= \frac{28}{15} π$

Achievable AP Calculus AB

8. Applications of integrals

8.4. Volume

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Washer method

8 min read

Font

Discuss

Feedback

What you’ll learn

Washer method: Calculate the volume of a solid of revolution with a hollow core using concentric circles as cross-sections.

The figure below illustrates how revolving the same region yields two different cross-sections: solid disks when rotated about the $x$ -axis, and hollow washers when rotated about the $y$ -axis.

AP tip:

To determine which method to use, compare the region’s boundary to the axis of revolution:

Disk: The region touches the axis along the whole boundary line (no gap)
Washer: There is space between the region and the axis (gap $\to$ hole)

Horizontal axes of revolution $(d x)$

When a region is revolved around a horizontal axis and creates a hollow core, integrate with respect to $x$ :

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

where:

Outer radius $R (x)$ : The distance from the axis to the curve farther from it.
Inner radius $r (x)$ : The distance from the axis to the curve closer to it.

Watch your algebra!

Always remember that the washer cross-section is the area of the larger circle minus the area of the smaller circle: $Area = π R^{2} - π r^{2}$ .

When factoring out $π$ , the squares must stay attached to the individual radii:

Correct: $π (R^{2} - r^{2})$
Incorrect: $π (R - r)^{2}$

Example 1: Revolving around the $x$ -axis

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

Solution

(spoiler)

1. Bounds:

Find the intersection points of the two curves to set the $x$ -bounds:

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

So we integrate over $[- 1, 2]$ .

2. Find the radii $R (x)$ and $r (x)$ :

These are vertical distances from the axis of $y = 0$ :

Outer radius $R (x)$ : The top line $y = x + 3$ is farther from the axis.

$R (x) = (x + 3) - 0 = x + 3$

Inner radius $r (x)$ : The bottom parabola $y = x^{2} + 1$ is closer to the axis.

$r (x) = (x^{2} + 1) - 0 = x^{2} + 1$

3. Integrate:

$V = π \int_{a}^{b} (R^{2} - r^{2}) d x = π \int_{- 1}^{2} [(x + 3)^{2} - (x^{2} + 1)^{2}] d x = \frac{117}{5} π$

Example 2: Other horizontal axes $(y = k)$

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)

Once you’re comfortable visualizing the solid, you don’t need to sketch the washers every time. Instead, drop two straight lines from your functions to the axis of rotation:

The longer line gives you the outer radius, $R$ .
The shorter line gives you the inner radius, $r$ .

1. Bounds:

Solving $x^{2} = x$ gives $x = 0$ and $x = 1$ so we integrate over $[0, 1]$ .

2. Find the radii $R (x)$ and $r (x)$ :

These are vertical distances from the axis of $y = - 1$ .

Outer radius $R$ : The curve $y = x$ is farther from the axis.

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

Inner radius $r$ : The parabola $y = x^{2}$ is closer to the axis.

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

3. Integrate:

$V = π \int_{0}^{1} [(x + 1)^{2} - (x^{2} + 1)^{2}] d x = \frac{29}{30} π$

Vertical axes of revolution $(d y)$

When a region is revolved around a vertical axis, integrate with respect to $y$ :

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

In similar fashion, the radii are measured as horizontal distances from the curves to the axis of revolution:

Outer radius $R (y)$ : The horizontal distance to the farther curve.
Inner radius $r (y)$ : The horizontal distance to the closer curve.

AP tip:

Just like the disk method, the orientation of the axis of revolution dictates the variable of integration:

Horizontal axis $\Rightarrow$ use $d x$
Vertical axis $\Rightarrow$ use $d y$

Example 3: Revolving about a vertical axis $(x = k)$

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)

1. Rewrite the equations in terms of $y$ :

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

2. Bounds:

$y$ -bounds: $[0, 1]$

3. Find radii $R (y)$ and $r (y)$ :

Radii are horizontal distances from the axis of revolution $x = 1$ .

Outer radius $R (y)$ : The line $x = y + 1$ is farther from the axis.

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

Inner radius $r (y)$ : The curve $x = y^{3} + 1$ is closer to the axis.

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

4. Integrate:

$V = π \int_{c}^{d} [R (y)^{2} - r (y)^{2}] d y = π \int_{0}^{1} [y^{2} - (y^{3})^{2}] d y = \frac{4}{21} π$

Example 4: Constant inner radius

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)

1. Rewrite the equation in terms of $y$ :

$y = x + 3 ⟹ x = y^{2} - 3$

2. Bounds:

$y$ -bounds: $[0, 1]$ (where the curve meets the right boundary $x = - 2$ )

3. Find radii $R (y)$ and $r (y)$ :

Radii are horizontal distances from the axis of revolution $x = - 1$ .

Outer radius $R (y)$ : The curve $x = y^{2} - 3$ is farther from the axis.

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

Note: The order of terms does not matter since $R$ will be squared. $R = 2 - y^{2}$ is also valid.

Inner radius $r (y)$ : The line $x = - 2$ is a constant distance away from the axis.

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

4. Integrate:

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

$= \frac{28}{15} π$

Washer Method Overview

Used for hollow solids of revolution (disk with a hole)
Formula: $V = π \int_{a}^{b} [R (x)^{2} - r (x)^{2}] d x$
Use $d y$ and express in terms of $y$ when rotating about a vertical axis

Identifying Radii

Outer radius $R$ : distance from axis to the farther function
Inner radius $r$ : distance from axis to the closer function
When axis is not at origin, add/subtract axis value to each radius

Horizontal Axis of Rotation

Radii are vertical distances measured from the axis
Find bounds by setting the two functions equal and solving
Axis shift example: rotating about $y = - 1$ adds 1 to each radius

Vertical Axis of Rotation

Rewrite both curves as $x$ in terms of $y$
Radii are horizontal distances from the vertical axis
Bounds come from $y$ -values at intersection points

Washer method

What you’ll learn

Washer method: Calculate the volume of a solid of revolution with a hollow core using concentric circles as cross-sections.

The figure below illustrates how revolving the same region yields two different cross-sections: solid disks when rotated about the $x$ -axis, and hollow washers when rotated about the $y$ -axis.

AP tip:

To determine which method to use, compare the region’s boundary to the axis of revolution:

Disk: The region touches the axis along the whole boundary line (no gap)
Washer: There is space between the region and the axis (gap $\to$ hole)

Horizontal axes of revolution $(d x)$

When a region is revolved around a horizontal axis and creates a hollow core, integrate with respect to $x$ :

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

where:

Outer radius $R (x)$ : The distance from the axis to the curve farther from it.
Inner radius $r (x)$ : The distance from the axis to the curve closer to it.

Watch your algebra!

Always remember that the washer cross-section is the area of the larger circle minus the area of the smaller circle: $Area = π R^{2} - π r^{2}$ .

When factoring out $π$ , the squares must stay attached to the individual radii:

Correct: $π (R^{2} - r^{2})$
Incorrect: $π (R - r)^{2}$

Example 1: Revolving around the $x$ -axis

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

Solution

(spoiler)

1. Bounds:

Find the intersection points of the two curves to set the $x$ -bounds:

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

So we integrate over $[- 1, 2]$ .

2. Find the radii $R (x)$ and $r (x)$ :

These are vertical distances from the axis of $y = 0$ :

Outer radius $R (x)$ : The top line $y = x + 3$ is farther from the axis.

$R (x) = (x + 3) - 0 = x + 3$

Inner radius $r (x)$ : The bottom parabola $y = x^{2} + 1$ is closer to the axis.

$r (x) = (x^{2} + 1) - 0 = x^{2} + 1$

3. Integrate:

$V = π \int_{a}^{b} (R^{2} - r^{2}) d x = π \int_{- 1}^{2} [(x + 3)^{2} - (x^{2} + 1)^{2}] d x = \frac{117}{5} π$

Example 2: Other horizontal axes $(y = k)$

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)

Once you’re comfortable visualizing the solid, you don’t need to sketch the washers every time. Instead, drop two straight lines from your functions to the axis of rotation:

The longer line gives you the outer radius, $R$ .
The shorter line gives you the inner radius, $r$ .

1. Bounds:

Solving $x^{2} = x$ gives $x = 0$ and $x = 1$ so we integrate over $[0, 1]$ .

2. Find the radii $R (x)$ and $r (x)$ :

These are vertical distances from the axis of $y = - 1$ .

Outer radius $R$ : The curve $y = x$ is farther from the axis.

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

Inner radius $r$ : The parabola $y = x^{2}$ is closer to the axis.

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

3. Integrate:

$V = π \int_{0}^{1} [(x + 1)^{2} - (x^{2} + 1)^{2}] d x = \frac{29}{30} π$

Vertical axes of revolution $(d y)$

When a region is revolved around a vertical axis, integrate with respect to $y$ :

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

In similar fashion, the radii are measured as horizontal distances from the curves to the axis of revolution:

Outer radius $R (y)$ : The horizontal distance to the farther curve.
Inner radius $r (y)$ : The horizontal distance to the closer curve.

AP tip:

Just like the disk method, the orientation of the axis of revolution dictates the variable of integration:

Horizontal axis $\Rightarrow$ use $d x$
Vertical axis $\Rightarrow$ use $d y$

Example 3: Revolving about a vertical axis $(x = k)$

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)

1. Rewrite the equations in terms of $y$ :

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

2. Bounds:

$y$ -bounds: $[0, 1]$

3. Find radii $R (y)$ and $r (y)$ :

Radii are horizontal distances from the axis of revolution $x = 1$ .

Outer radius $R (y)$ : The line $x = y + 1$ is farther from the axis.

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

Inner radius $r (y)$ : The curve $x = y^{3} + 1$ is closer to the axis.

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

4. Integrate:

$V = π \int_{c}^{d} [R (y)^{2} - r (y)^{2}] d y = π \int_{0}^{1} [y^{2} - (y^{3})^{2}] d y = \frac{4}{21} π$

Example 4: Constant inner radius

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)

1. Rewrite the equation in terms of $y$ :

$y = x + 3 ⟹ x = y^{2} - 3$

2. Bounds:

$y$ -bounds: $[0, 1]$ (where the curve meets the right boundary $x = - 2$ )

3. Find radii $R (y)$ and $r (y)$ :

Radii are horizontal distances from the axis of revolution $x = - 1$ .

Outer radius $R (y)$ : The curve $x = y^{2} - 3$ is farther from the axis.

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

Note: The order of terms does not matter since $R$ will be squared. $R = 2 - y^{2}$ is also valid.

Inner radius $r (y)$ : The line $x = - 2$ is a constant distance away from the axis.

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

4. Integrate:

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

$= \frac{28}{15} π$

Washer method

What you’ll learn

Horizontal axes of revolution (dx)

Example 1: Revolving around the x-axis

Solution

Example 2: Other horizontal axes (y=k)

Solution

Vertical axes of revolution (dy)

Example 3: Revolving about a vertical axis (x=k)

Solution

Example 4: Constant inner radius

Solution

Washer method

What you’ll learn

Horizontal axes of revolution (dx)

Example 1: Revolving around the x-axis

Solution

Example 2: Other horizontal axes (y=k)

Solution

Vertical axes of revolution (dy)

Example 3: Revolving about a vertical axis (x=k)

Solution

Example 4: Constant inner radius

Solution

Washer method

What you’ll learn

Horizontal axes of revolution (dx)

Example 1: Revolving around the x-axis

Solution

Example 2: Other horizontal axes (y=k)

Solution

Vertical axes of revolution (dy)

Example 3: Revolving about a vertical axis (x=k)

Solution

Example 4: Constant inner radius

Solution

Washer method

What you’ll learn

Horizontal axes of revolution (dx)

Example 1: Revolving around the x-axis

Solution

Example 2: Other horizontal axes (y=k)

Solution

Vertical axes of revolution (dy)

Example 3: Revolving about a vertical axis (x=k)

Solution

Example 4: Constant inner radius

Solution

Horizontal axes of revolution $(d x)$

Example 1: Revolving around the $x$ -axis

Example 2: Other horizontal axes $(y = k)$

Vertical axes of revolution $(d y)$

Example 3: Revolving about a vertical axis $(x = k)$

Horizontal axes of revolution $(d x)$

Example 1: Revolving around the $x$ -axis

Example 2: Other horizontal axes $(y = k)$

Vertical axes of revolution $(d y)$

Example 3: Revolving about a vertical axis $(x = k)$

Horizontal axes of revolution $(d x)$

Example 1: Revolving around the $x$ -axis

Example 2: Other horizontal axes $(y = k)$

Vertical axes of revolution $(d y)$

Example 3: Revolving about a vertical axis $(x = k)$

Horizontal axes of revolution $(d x)$

Example 1: Revolving around the $x$ -axis

Example 2: Other horizontal axes $(y = k)$

Vertical axes of revolution $(d y)$

Example 3: Revolving about a vertical axis $(x = k)$