8.4.2 Disk method

Achievable AP Calculus AB

8. Applications of integrals

8.4. Volume

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Disk method

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What you’ll learn

Disk method: Calculate the volume of a solid of revolution with a solid core by integrating circular cross sections.
Axis of revolution: Set up the integral with respect to $x$ or $y$ based on whether the axis is horizontal or vertical.

The disk method calculates the volume of a solid formed by revolving a region around an axis. The cross-sections perpendicular to the axis of revolution are circles with an area of $A = π R^{2}$ .

Horizontal axes of revolution $(d x)$

Horizontal axes can be either the $x$ -axis or another horizontal line (like $y = 2$ or $y = - 1$ ). Because the axis is horizontal, the entire integral - bounds, functions, and radius - must be in terms of $x$ .

When a region is revolved around a horizontal axis (parallel to the $x$ -axis), integrate with respect to $x$ :

$V = π \int_{a}^{b} [R (x)]^{2} d x$

Where the radius $R (x)$ is the vertical distance from a curve $f (x)$ to the axis of revolution, or the magnitude of the difference:

$R (x) = ∣ f (x) - Axis ∣$

Example 1: Revolving around the $x$ -axis $(y = 0)$

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

Solution

(spoiler)

To get started, make a quick sketch of the region:

Pick a point on the function and draw a line segment straight to the axis of revolution.
Imagine spinning the line segment a full rotation around the axis. The circle (disk) it sweeps out is a representative cross section of the solid. Label its radius $R$ .

1. Bounds:

Integrate over $x$ -bounds: $[0, 2]$

2. Find the radius $R$ :

This is the vertical distance from the line $y = x$ to the $x$ -axis $(y = 0)$ :

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

3. Integrate:

$V = π \int_{0}^{2} (x)^{2} d x = π [\frac{x ^{3}}{3}]_{0}^{2} = \frac{8}{3} π$

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

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

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

Solution

(spoiler)

Revolving a region about the horizontal line y = 2

1. Bounds:

$x$ -bounds: $[0, 4]$

The curve $y = x$ meets the line $y = 2$ at the intersection point $x = 4$ .

2. Find the radius $R$ :

This is the vertical distance between the curve $y = x$ and the axis of revolution $y = 2$ :

$R (x) = x - 2$

Note: Even though substituting an $x$ -value within the interval bounds produces a negative radius, this can be ignored because $R$ is squared in the integral.

3. Integrate:

$V = π \int_{0}^{4} (x - 2)^{2} d x = \frac{8}{3} π$

Vertical axes of revolution $(d y)$

When a region is revolved around a vertical axis (parallel to the $y$ -axis), integrate with respect to $y$ :

$V = π \int_{c}^{d} [R (y)]^{2} d y$

Where the radius $R (y)$ is the horizontal distance from the curve $g (y)$ to the axis of revolution:

$R (y) = ∣ g (y) - Axis ∣$

AP tip:

To determine which variable to integrate with respect to when using the disk method, look at your axis of revolution: the direction of integration runs parallel to it.

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

Example 3: Revolving around the $y$ -axis $(x = 0)$

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

Solution

(spoiler)

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

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

The region is bounded by the left half of the parabola, so we choose the negative square root:

$x = - y + 1$

2. Bounds:

$y$ -bounds: $[0, 1]$
- Lower bound: The curve meets the $x$ -axis at $y = 0$ .
- Upper bound: The curve meets the $y$ -axis at $y = 1$ .

3. Find the radius $R$ :

This is the horizontal distance between the curve $x = - y + 1$ and the $y$ -axis $(x = 0)$ :

$R (y) = - y + 1$

4. Integrate:

$V = π \int_{0}^{1} (- y + 1)^{2} d y = \frac{π}{6}$

Example 4: Other vertical axes $(x = k)$

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

Solution

(spoiler)

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

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

2. Bounds:

$y$ -bounds: $[0, 2]$
- Lower bound: The region is bounded below by the $x$ -axis, where $y = 0$ .
- Upper bound: The curve $y = x - 1$ meets $x = 3$ at $y = 2$ .

3. Find the radius $R$ :

This is the horizontal distance between the curve $x = y^{2} + 1$ and the line $x = 3$ :

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

4. Integrate:

$V = π \int_{0}^{2} (y^{2} - 2)^{2} d y = \frac{32 2}{15} π$

Challenge problem

The region bounded by $y = x + 1$ , the coordinate axes, and the line $x = 4$ is revolved about $x = 4$ . Determine the volume of the solid formed.

Solution

(spoiler)

1. Axis:

Revolved about a vertical line $(x = 4) \Rightarrow$ use $d y$ .

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

$y = x + 1 ⟹ x = (y - 1)^{2}$

3. Bounds (Watch out!):

We must split the integral:

Bottom section with $y$ -bounds $[0, 1]$ :
- Region is bounded on the left by $x = 0$ ( $y$ -axis) and on the right by $x = 4$ .
- When revolved about $x = 4$ , the solid formed is a cylinder.
Top section with $y$ -bounds $[1, 3]$ :
- Region is now bounded on the left by $x = (y - 1)^{2}$ and on the right by $x = 4$ .

4. Find radii:

For $y \in [0, 1]$ : The radius is a constant distance from $x = 0$ to $x = 4$ :

$R_{1} (y) = 4 - 0 = 4$

For $y \in [1, 3]$ : The radius is the distance from the curve to $x = 4$ :

$R_{2} (y) = (y - 1)^{2} - 4$

5. Integrate:

Add the separate integrals together:

$V = π \int_{0}^{1} (4)^{2} d y + π \int_{1}^{3} [(y - 1)^{2} - 4]^{2} d y = \frac{496}{15} π$

Disk method overview

Finds volume of a solid of revolution by integrating circular cross sections
Cross-sectional area formula: $A = π R^{2}$

Horizontal axis of revolution (use $d x$ )

Axis is horizontal (e.g., $x$ -axis or $y = k$ ); integrate with respect to $x$
Formula: $V = π \int_{a}^{b} [R (x)]^{2} d x$
Radius: $R (x) = ∣ f (x) - Axis ∣$ (vertical distance from curve to axis)

Vertical axis of revolution (use $d y$ )

Axis is vertical (e.g., $y$ -axis or $x = k$ ); integrate with respect to $y$
Formula: $V = π \int_{c}^{d} [R (y)]^{2} d y$
Radius: $R (y) = ∣ g (y) - Axis ∣$ (horizontal distance from curve to axis)
- Rewrite curve as $x = g (y)$ before finding radius and bounds

Choosing integration variable (AP tip)

Direction of integration runs parallel to the axis of revolution
- Horizontal axis → $d x$
- Vertical axis → $d y$

Disk method

What you’ll learn

Disk method: Calculate the volume of a solid of revolution with a solid core by integrating circular cross sections.
Axis of revolution: Set up the integral with respect to $x$ or $y$ based on whether the axis is horizontal or vertical.

Horizontal axes of revolution $(d x)$

When a region is revolved around a horizontal axis (parallel to the $x$ -axis), integrate with respect to $x$ :

$V = π \int_{a}^{b} [R (x)]^{2} d x$

Where the radius $R (x)$ is the vertical distance from a curve $f (x)$ to the axis of revolution, or the magnitude of the difference:

$R (x) = ∣ f (x) - Axis ∣$

Example 1: Revolving around the $x$ -axis $(y = 0)$

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

Solution

(spoiler)

To get started, make a quick sketch of the region:

Pick a point on the function and draw a line segment straight to the axis of revolution.
Imagine spinning the line segment a full rotation around the axis. The circle (disk) it sweeps out is a representative cross section of the solid. Label its radius $R$ .

1. Bounds:

Integrate over $x$ -bounds: $[0, 2]$

2. Find the radius $R$ :

This is the vertical distance from the line $y = x$ to the $x$ -axis $(y = 0)$ :

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

3. Integrate:

$V = π \int_{0}^{2} (x)^{2} d x = π [\frac{x ^{3}}{3}]_{0}^{2} = \frac{8}{3} π$

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

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

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

Solution

(spoiler)

1. Bounds:

$x$ -bounds: $[0, 4]$

The curve $y = x$ meets the line $y = 2$ at the intersection point $x = 4$ .

2. Find the radius $R$ :

This is the vertical distance between the curve $y = x$ and the axis of revolution $y = 2$ :

$R (x) = x - 2$

Note: Even though substituting an $x$ -value within the interval bounds produces a negative radius, this can be ignored because $R$ is squared in the integral.

3. Integrate:

$V = π \int_{0}^{4} (x - 2)^{2} d x = \frac{8}{3} π$

Vertical axes of revolution $(d y)$

When a region is revolved around a vertical axis (parallel to the $y$ -axis), integrate with respect to $y$ :

$V = π \int_{c}^{d} [R (y)]^{2} d y$

Where the radius $R (y)$ is the horizontal distance from the curve $g (y)$ to the axis of revolution:

$R (y) = ∣ g (y) - Axis ∣$

AP tip:

To determine which variable to integrate with respect to when using the disk method, look at your axis of revolution: the direction of integration runs parallel to it.

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

Example 3: Revolving around the $y$ -axis $(x = 0)$

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

Solution

(spoiler)

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

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

The region is bounded by the left half of the parabola, so we choose the negative square root:

$x = - y + 1$

2. Bounds:

$y$ -bounds: $[0, 1]$
- Lower bound: The curve meets the $x$ -axis at $y = 0$ .
- Upper bound: The curve meets the $y$ -axis at $y = 1$ .

3. Find the radius $R$ :

This is the horizontal distance between the curve $x = - y + 1$ and the $y$ -axis $(x = 0)$ :

$R (y) = - y + 1$

4. Integrate:

$V = π \int_{0}^{1} (- y + 1)^{2} d y = \frac{π}{6}$

Example 4: Other vertical axes $(x = k)$

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

Solution

(spoiler)

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

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

2. Bounds:

$y$ -bounds: $[0, 2]$
- Lower bound: The region is bounded below by the $x$ -axis, where $y = 0$ .
- Upper bound: The curve $y = x - 1$ meets $x = 3$ at $y = 2$ .

3. Find the radius $R$ :

This is the horizontal distance between the curve $x = y^{2} + 1$ and the line $x = 3$ :

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

4. Integrate:

$V = π \int_{0}^{2} (y^{2} - 2)^{2} d y = \frac{32 2}{15} π$

Challenge problem

The region bounded by $y = x + 1$ , the coordinate axes, and the line $x = 4$ is revolved about $x = 4$ . Determine the volume of the solid formed.

Solution

(spoiler)

1. Axis:

Revolved about a vertical line $(x = 4) \Rightarrow$ use $d y$ .

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

$y = x + 1 ⟹ x = (y - 1)^{2}$

3. Bounds (Watch out!):

We must split the integral:

Bottom section with $y$ -bounds $[0, 1]$ :
- Region is bounded on the left by $x = 0$ ( $y$ -axis) and on the right by $x = 4$ .
- When revolved about $x = 4$ , the solid formed is a cylinder.
Top section with $y$ -bounds $[1, 3]$ :
- Region is now bounded on the left by $x = (y - 1)^{2}$ and on the right by $x = 4$ .

4. Find radii:

For $y \in [0, 1]$ : The radius is a constant distance from $x = 0$ to $x = 4$ :

$R_{1} (y) = 4 - 0 = 4$

For $y \in [1, 3]$ : The radius is the distance from the curve to $x = 4$ :

$R_{2} (y) = (y - 1)^{2} - 4$

5. Integrate:

Add the separate integrals together:

$V = π \int_{0}^{1} (4)^{2} d y + π \int_{1}^{3} [(y - 1)^{2} - 4]^{2} d y = \frac{496}{15} π$

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.

Disk method

8 min read

Font

Discuss

Feedback

What you’ll learn

Disk method: Calculate the volume of a solid of revolution with a solid core by integrating circular cross sections.
Axis of revolution: Set up the integral with respect to $x$ or $y$ based on whether the axis is horizontal or vertical.

Horizontal axes of revolution $(d x)$

When a region is revolved around a horizontal axis (parallel to the $x$ -axis), integrate with respect to $x$ :

$V = π \int_{a}^{b} [R (x)]^{2} d x$

Where the radius $R (x)$ is the vertical distance from a curve $f (x)$ to the axis of revolution, or the magnitude of the difference:

$R (x) = ∣ f (x) - Axis ∣$

Example 1: Revolving around the $x$ -axis $(y = 0)$

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

Solution

(spoiler)

To get started, make a quick sketch of the region:

Pick a point on the function and draw a line segment straight to the axis of revolution.
Imagine spinning the line segment a full rotation around the axis. The circle (disk) it sweeps out is a representative cross section of the solid. Label its radius $R$ .

1. Bounds:

Integrate over $x$ -bounds: $[0, 2]$

2. Find the radius $R$ :

This is the vertical distance from the line $y = x$ to the $x$ -axis $(y = 0)$ :

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

3. Integrate:

$V = π \int_{0}^{2} (x)^{2} d x = π [\frac{x ^{3}}{3}]_{0}^{2} = \frac{8}{3} π$

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

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

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

Solution

(spoiler)

1. Bounds:

$x$ -bounds: $[0, 4]$

The curve $y = x$ meets the line $y = 2$ at the intersection point $x = 4$ .

2. Find the radius $R$ :

This is the vertical distance between the curve $y = x$ and the axis of revolution $y = 2$ :

$R (x) = x - 2$

Note: Even though substituting an $x$ -value within the interval bounds produces a negative radius, this can be ignored because $R$ is squared in the integral.

3. Integrate:

$V = π \int_{0}^{4} (x - 2)^{2} d x = \frac{8}{3} π$

Vertical axes of revolution $(d y)$

When a region is revolved around a vertical axis (parallel to the $y$ -axis), integrate with respect to $y$ :

$V = π \int_{c}^{d} [R (y)]^{2} d y$

Where the radius $R (y)$ is the horizontal distance from the curve $g (y)$ to the axis of revolution:

$R (y) = ∣ g (y) - Axis ∣$

AP tip:

To determine which variable to integrate with respect to when using the disk method, look at your axis of revolution: the direction of integration runs parallel to it.

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

Example 3: Revolving around the $y$ -axis $(x = 0)$

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

Solution

(spoiler)

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

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

The region is bounded by the left half of the parabola, so we choose the negative square root:

$x = - y + 1$

2. Bounds:

$y$ -bounds: $[0, 1]$
- Lower bound: The curve meets the $x$ -axis at $y = 0$ .
- Upper bound: The curve meets the $y$ -axis at $y = 1$ .

3. Find the radius $R$ :

This is the horizontal distance between the curve $x = - y + 1$ and the $y$ -axis $(x = 0)$ :

$R (y) = - y + 1$

4. Integrate:

$V = π \int_{0}^{1} (- y + 1)^{2} d y = \frac{π}{6}$

Example 4: Other vertical axes $(x = k)$

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

Solution

(spoiler)

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

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

2. Bounds:

$y$ -bounds: $[0, 2]$
- Lower bound: The region is bounded below by the $x$ -axis, where $y = 0$ .
- Upper bound: The curve $y = x - 1$ meets $x = 3$ at $y = 2$ .

3. Find the radius $R$ :

This is the horizontal distance between the curve $x = y^{2} + 1$ and the line $x = 3$ :

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

4. Integrate:

$V = π \int_{0}^{2} (y^{2} - 2)^{2} d y = \frac{32 2}{15} π$

Challenge problem

The region bounded by $y = x + 1$ , the coordinate axes, and the line $x = 4$ is revolved about $x = 4$ . Determine the volume of the solid formed.

Solution

(spoiler)

1. Axis:

Revolved about a vertical line $(x = 4) \Rightarrow$ use $d y$ .

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

$y = x + 1 ⟹ x = (y - 1)^{2}$

3. Bounds (Watch out!):

We must split the integral:

Bottom section with $y$ -bounds $[0, 1]$ :
- Region is bounded on the left by $x = 0$ ( $y$ -axis) and on the right by $x = 4$ .
- When revolved about $x = 4$ , the solid formed is a cylinder.
Top section with $y$ -bounds $[1, 3]$ :
- Region is now bounded on the left by $x = (y - 1)^{2}$ and on the right by $x = 4$ .

4. Find radii:

For $y \in [0, 1]$ : The radius is a constant distance from $x = 0$ to $x = 4$ :

$R_{1} (y) = 4 - 0 = 4$

For $y \in [1, 3]$ : The radius is the distance from the curve to $x = 4$ :

$R_{2} (y) = (y - 1)^{2} - 4$

5. Integrate:

Add the separate integrals together:

$V = π \int_{0}^{1} (4)^{2} d y + π \int_{1}^{3} [(y - 1)^{2} - 4]^{2} d y = \frac{496}{15} π$

Disk method overview

Finds volume of a solid of revolution by integrating circular cross sections
Cross-sectional area formula: $A = π R^{2}$

Horizontal axis of revolution (use $d x$ )

Axis is horizontal (e.g., $x$ -axis or $y = k$ ); integrate with respect to $x$
Formula: $V = π \int_{a}^{b} [R (x)]^{2} d x$
Radius: $R (x) = ∣ f (x) - Axis ∣$ (vertical distance from curve to axis)

Vertical axis of revolution (use $d y$ )

Axis is vertical (e.g., $y$ -axis or $x = k$ ); integrate with respect to $y$
Formula: $V = π \int_{c}^{d} [R (y)]^{2} d y$
Radius: $R (y) = ∣ g (y) - Axis ∣$ (horizontal distance from curve to axis)
- Rewrite curve as $x = g (y)$ before finding radius and bounds

Choosing integration variable (AP tip)

Direction of integration runs parallel to the axis of revolution
- Horizontal axis → $d x$
- Vertical axis → $d y$

Disk method

What you’ll learn

Disk method: Calculate the volume of a solid of revolution with a solid core by integrating circular cross sections.
Axis of revolution: Set up the integral with respect to $x$ or $y$ based on whether the axis is horizontal or vertical.

Horizontal axes of revolution $(d x)$

When a region is revolved around a horizontal axis (parallel to the $x$ -axis), integrate with respect to $x$ :

$V = π \int_{a}^{b} [R (x)]^{2} d x$

Where the radius $R (x)$ is the vertical distance from a curve $f (x)$ to the axis of revolution, or the magnitude of the difference:

$R (x) = ∣ f (x) - Axis ∣$

Example 1: Revolving around the $x$ -axis $(y = 0)$

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

Solution

(spoiler)

To get started, make a quick sketch of the region:

Pick a point on the function and draw a line segment straight to the axis of revolution.
Imagine spinning the line segment a full rotation around the axis. The circle (disk) it sweeps out is a representative cross section of the solid. Label its radius $R$ .

1. Bounds:

Integrate over $x$ -bounds: $[0, 2]$

2. Find the radius $R$ :

This is the vertical distance from the line $y = x$ to the $x$ -axis $(y = 0)$ :

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

3. Integrate:

$V = π \int_{0}^{2} (x)^{2} d x = π [\frac{x ^{3}}{3}]_{0}^{2} = \frac{8}{3} π$

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

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

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

Solution

(spoiler)

1. Bounds:

$x$ -bounds: $[0, 4]$

The curve $y = x$ meets the line $y = 2$ at the intersection point $x = 4$ .

2. Find the radius $R$ :

This is the vertical distance between the curve $y = x$ and the axis of revolution $y = 2$ :

$R (x) = x - 2$

Note: Even though substituting an $x$ -value within the interval bounds produces a negative radius, this can be ignored because $R$ is squared in the integral.

3. Integrate:

$V = π \int_{0}^{4} (x - 2)^{2} d x = \frac{8}{3} π$

Vertical axes of revolution $(d y)$

When a region is revolved around a vertical axis (parallel to the $y$ -axis), integrate with respect to $y$ :

$V = π \int_{c}^{d} [R (y)]^{2} d y$

Where the radius $R (y)$ is the horizontal distance from the curve $g (y)$ to the axis of revolution:

$R (y) = ∣ g (y) - Axis ∣$

AP tip:

To determine which variable to integrate with respect to when using the disk method, look at your axis of revolution: the direction of integration runs parallel to it.

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

Example 3: Revolving around the $y$ -axis $(x = 0)$

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

Solution

(spoiler)

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

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

The region is bounded by the left half of the parabola, so we choose the negative square root:

$x = - y + 1$

2. Bounds:

$y$ -bounds: $[0, 1]$
- Lower bound: The curve meets the $x$ -axis at $y = 0$ .
- Upper bound: The curve meets the $y$ -axis at $y = 1$ .

3. Find the radius $R$ :

This is the horizontal distance between the curve $x = - y + 1$ and the $y$ -axis $(x = 0)$ :

$R (y) = - y + 1$

4. Integrate:

$V = π \int_{0}^{1} (- y + 1)^{2} d y = \frac{π}{6}$

Example 4: Other vertical axes $(x = k)$

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

Solution

(spoiler)

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

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

2. Bounds:

$y$ -bounds: $[0, 2]$
- Lower bound: The region is bounded below by the $x$ -axis, where $y = 0$ .
- Upper bound: The curve $y = x - 1$ meets $x = 3$ at $y = 2$ .

3. Find the radius $R$ :

This is the horizontal distance between the curve $x = y^{2} + 1$ and the line $x = 3$ :

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

4. Integrate:

$V = π \int_{0}^{2} (y^{2} - 2)^{2} d y = \frac{32 2}{15} π$

Challenge problem

The region bounded by $y = x + 1$ , the coordinate axes, and the line $x = 4$ is revolved about $x = 4$ . Determine the volume of the solid formed.

Solution

(spoiler)

1. Axis:

Revolved about a vertical line $(x = 4) \Rightarrow$ use $d y$ .

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

$y = x + 1 ⟹ x = (y - 1)^{2}$

3. Bounds (Watch out!):

We must split the integral:

Bottom section with $y$ -bounds $[0, 1]$ :
- Region is bounded on the left by $x = 0$ ( $y$ -axis) and on the right by $x = 4$ .
- When revolved about $x = 4$ , the solid formed is a cylinder.
Top section with $y$ -bounds $[1, 3]$ :
- Region is now bounded on the left by $x = (y - 1)^{2}$ and on the right by $x = 4$ .

4. Find radii:

For $y \in [0, 1]$ : The radius is a constant distance from $x = 0$ to $x = 4$ :

$R_{1} (y) = 4 - 0 = 4$

For $y \in [1, 3]$ : The radius is the distance from the curve to $x = 4$ :

$R_{2} (y) = (y - 1)^{2} - 4$

5. Integrate:

Add the separate integrals together:

$V = π \int_{0}^{1} (4)^{2} d y + π \int_{1}^{3} [(y - 1)^{2} - 4]^{2} d y = \frac{496}{15} π$

Disk method

What you’ll learn

Horizontal axes of revolution (dx)

Example 1: Revolving around the x-axis (y=0)

Solution

Example 2: Other horizontal axes (y=k)

Solution

Vertical axes of revolution (dy)

Example 3: Revolving around the y-axis (x=0)

Solution

Example 4: Other vertical axes (x=k)

Solution

Challenge problem

Solution

Disk method

What you’ll learn

Horizontal axes of revolution (dx)

Example 1: Revolving around the x-axis (y=0)

Solution

Example 2: Other horizontal axes (y=k)

Solution

Vertical axes of revolution (dy)

Example 3: Revolving around the y-axis (x=0)

Solution

Example 4: Other vertical axes (x=k)

Solution

Challenge problem

Solution

Disk method

What you’ll learn

Horizontal axes of revolution (dx)

Example 1: Revolving around the x-axis (y=0)

Solution

Example 2: Other horizontal axes (y=k)

Solution

Vertical axes of revolution (dy)

Example 3: Revolving around the y-axis (x=0)

Solution

Example 4: Other vertical axes (x=k)

Solution

Challenge problem

Solution

Disk method

What you’ll learn

Horizontal axes of revolution (dx)

Example 1: Revolving around the x-axis (y=0)

Solution

Example 2: Other horizontal axes (y=k)

Solution

Vertical axes of revolution (dy)

Example 3: Revolving around the y-axis (x=0)

Solution

Example 4: Other vertical axes (x=k)

Solution

Challenge problem

Solution

Horizontal axes of revolution $(d x)$

Example 1: Revolving around the $x$ -axis $(y = 0)$

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

Vertical axes of revolution $(d y)$

Example 3: Revolving around the $y$ -axis $(x = 0)$

Example 4: Other vertical axes $(x = k)$

Horizontal axes of revolution $(d x)$

Example 1: Revolving around the $x$ -axis $(y = 0)$

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

Vertical axes of revolution $(d y)$

Example 3: Revolving around the $y$ -axis $(x = 0)$

Example 4: Other vertical axes $(x = k)$

Horizontal axes of revolution $(d x)$

Example 1: Revolving around the $x$ -axis $(y = 0)$

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

Vertical axes of revolution $(d y)$

Example 3: Revolving around the $y$ -axis $(x = 0)$

Example 4: Other vertical axes $(x = k)$

Horizontal axes of revolution $(d x)$

Example 1: Revolving around the $x$ -axis $(y = 0)$

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

Vertical axes of revolution $(d y)$

Example 3: Revolving around the $y$ -axis $(x = 0)$

Example 4: Other vertical axes $(x = k)$