Disk method | AP Calc AB Volume

Disk method

What you’ll learn

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

The disk method is used to find the volume of a solid formed by rotating a region around an axis. The cross-sections perpendicular to the axis of rotation are circles.

Disk method

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 method, 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 $R$

Next, identify 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

$R (x) = x$

Applying 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.

Regions can be rotated about any axis:

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.

(spoiler)

R is the distance between the top and bottom functions

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

$R (x) = 2 - x$

The bounds for the integral are from $x = 0$ (the $y$ -axis) to where $y = x$ meets $y = 2$ , which is at $x = 4$ . 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.

(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.

(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

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

$= \frac{32 2}{15} π$

Disk Method Formula

Rotating a region creates circular cross-sections; volume = $π \int_{a}^{b} [R (x)]^{2} d x$
$R (x)$ = radius = distance from axis of rotation to the curve
Use $d y$ (and express $R$ in terms of $y$ ) when rotating about a vertical axis

Horizontal Axis of Rotation

Integrate with respect to $x$ ; radius is a vertical distance
Rotating about a non- $x$ -axis: $R (x) = (axis value) - (curve value)$
- e.g., rotating about $y = 2$ : $R (x) = 2 - x$

Vertical Axis of Rotation

Integrate with respect to $y$ ; rewrite curve as $x = f (y)$
Radius is a horizontal distance from the axis to the curve
- e.g., rotating about $x = 3$ : $R (y) = 3 - (y^{2} + 1)$
Bounds must be converted to $y$ -values

AP Tip

Axis of rotation must be parallel to direction of integration
Horizontal axis → $d x$ ; vertical axis → $d y$

Disk method

What you’ll learn

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

The disk method is used to find the volume of a solid formed by rotating a region around an axis. The cross-sections perpendicular to the axis of rotation are circles.

Disk method

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 method, 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 $R$

Next, identify 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

$R (x) = x$

Applying 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.

Regions can be rotated about any axis:

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.

(spoiler)

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

$R (x) = 2 - x$

The bounds for the integral are from $x = 0$ (the $y$ -axis) to where $y = x$ meets $y = 2$ , which is at $x = 4$ . 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.

(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.

(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

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

$= \frac{32 2}{15} π$