Calculus

# Disk Method: Definition, Formula & Examples

## Rachel McLean

Subject Matter Expert

In this article, we’ll discuss how to use the disk method. Then, we’ll practice with some examples. We’ll also compare the disk and washer methods and learn how to determine when to use each method.

## What Is the Disk Method?

When a two-dimensional area is rotated around a line, it creates a three-dimensional solid of revolution. For example, when a rectangle is rotated about the x-axis, it generates a disk.

In the image below, $W$ is the width of the rectangle, and $R$ is the height of the rectangle. $R$ becomes the radius of the disk when the rectangle is revolved around the x-axis.

You can think about a revolving door to help visualize this process. As a revolving door moves, the outer edge of the door follows a circular path about the axis of rotation. Once the door revolves completely about its axis, the space through which it passed is in the shape of a cylinder.

The line that a curve rotates around is called the axis of rotation. In the image above, the x-axis is our axis of rotation. The shape of the resulting solid is a disk, which is a cylinder. So, we can use the formula for the volume of a cylinder to find its volume.

$\text{Volume} = \pi(R^2)(W)$

So, to find the volume of a solid of revolution, we can cut the curve into a series of thin rectangles, which cuts the solid of revolution into a series of thin disk-shaped slices. Adding up the volume of each slice gives us an approximation of the solid’s volume.

Cutting the solid into a finite number of slices gives us an approximation of the volume. To calculate the precise volume, we need to use infinite slices. This process will require integration.

To find the volume of each infinitely small slice, we’ll reuse the formula for the volume of a cylinder. Each slice of the solid is also called a cross-section, and must be perpendicular to the axis of rotation. This time, the width is $\Delta x$ and the radius is the height of the function at $x$.

This method of slicing to determine the volume of a solid of revolution is called the disk method.

## What Is the Disc Method Formula?

There are two variations of the disk method formula. We use the first variation for a horizontal axis of rotation. We use the second variation for a vertical axis of rotation.

### Disc Method Formula with Horizontal Axis

Here’s the formula for using the disk method with a horizontal axis of rotation. In this case, we integrate with respect to $x$. If $f$ is non-negative and continuous on the interval $[a, b]$, then the volume formed by rotating the area under the curve of $f$ about a horizontal axis is the definite integral given by:

$\text{V} =\int_a^b \pi R^2\,dx$

$R$ is the distance between the top function and the axis of rotation. It is given by $R$ = (top function) - (bottom function). If the axis of rotation is the x-axis, $R$ is simply $[f(x)]$. The bounded region should be flush to the axis of rotation so that the solid of revolution has no holes.

### Disc Method Formula With Vertical Axis

Here’s the formula for using the disk method with a vertical axis of rotation. In this case, we integrate with respect to $y$. If $g$ is non-negative and continuous on the interval $[a, b]$, then the volume formed by rotating the area under the curve of $g$ about a vertical axis is the definite integral given by:

$\text{V} =\int_a^b \pi R^2\,dy$

Again, $R$ is the distance between the top function and the axis of rotation, and is given by $R$ = (top function) - (bottom function). If the axis of rotation is the y-axis, $R$ is simply $[g(y)]$.

You can think about an orange to help visualize this integration process. If you cut an orange into thin circular slices, the orange is transformed into the sum of many “disks”. Finding the volume of each disk and adding them together gives you the volume of the entire orange.

## 3 Steps on How To Use the Disk Method

Here are 3 steps to using the disk method:

1. Graph the bounded region.

2. Identify the axis of rotation, and then draw a slice perpendicular to the axis of rotation.

3. Substitute your function and its bounds into the disk method formula and integrate.

Let’s do a few disk method examples together.

### Example 1

Let $f(x) = x^2$. Consider the region bounded by $f(x)$, $x = 0$, $x = 1$, and the x-axis. Find the volume of the solid generated by revolving the described region around the x-axis.

#### Step 1 and 2

Our bounded region looks like this. The black vertical line represents one slice of the curve.

#### Step 3

Since we are rotating around the x-axis, $R = [f(x)]$. Now we can plug our function into the disk method equation. We’ll use the power rule to integrate.

$\text{V} =\int_a^b \pi R^2\,dx$
$\text{V} =\int_0^1 \pi [f(x)]^2 \,dx$
$\text{V} =\pi \int_0^1 (x^2)^2 \,dx$
$\text{V} =\pi \int_0^1 x^4 \,dx$
$\text{V} = \pi \frac{x^5}{5} \Big|_0^1$
$\text{V} = \frac{\pi}{5}$

The volume of the solid of revolution is $\frac{\pi}{5}$. The generated solid looks like a curved funnel.

Dr. Tim Chartier—an award-winning professor of mathematics and computer science at Davidson College—explains more about the power rule:

### Example 2

Let $f(x) = x^2$. Consider the region bounded by $f(x)$, $x = 0$, $x = 1$, and the y-axis. Find the volume of the solid generated by revolving the described region around the y-axis. Note that our axis of revolution has changed to the y-axis.

#### Step 1 and 2

Since our axis of rotation is the y-axis, we need to rewrite our equation to solve for $x$ as a function of $y$, instead of $y$ as a function of $x$. Taking the square root of each side, this gives us $x = \sqrt{y}$.

Now our bounded region looks like this, where the white horizontal line represents one slice of the curve.

#### Step 3

Since we are rotating around the y-axis, $R = [f(y)]$. Now, we can plug our function into the disk method equation. Again, we’ll use the power rule to integrate.

$\text{V} =\int_a^b \pi R^2\,dy$
$\text{V} =\int_0^1 \pi [f(y)]^2 \,dy$
$\text{V} =\pi \int_0^1 (\sqrt{y})^2 \,dy$
$\text{V} =\pi \int_0^1 y \,dy$
$\text{V} = \pi \frac{y^2}{2} \Big|_0^1$
$\text{V} = \frac{\pi}{2}$

The volume of the solid of revolution is $\frac{\pi}{2}$. Note that both the volume and shape of the generated solid are different than the volume and shape of the same function revolved around the x-axis.

## Disk vs Washer Method

When we can’t use the disk method, the washer method is another way to calculate the volume of a solid of revolution.

Like the disk method, the washer method requires that the slice is perpendicular to the axis of revolution. However, in this method, the bounded region is not flush to the axis of revolution. Therefore, the solid generated by each slice looks like a washer—a disk with a hole in it. The corresponding solid of revolution also has a hole.

Here’s the formula for using the washer method with a horizontal axis of rotation. In this case, we integrate with respect to $x$. If $f$ and $g$ are non-negative and continuous on the interval $[a, b]$, and $f\geq g$ for all $x$ in $[a,b]$, then the volume formed by rotating the area bounded by $f$, $g$, $x = a$ and $x = b$ about a horizontal axis is given by:

$\text{V} =\int_a^b \pi (R^2 - r^2)\,dx$

### 3 Steps To Use Washer Method

Here are 3 steps to using the washer method:

1. Graph the bounded region.

2. Identify the axis of rotation by drawing a slice perpendicular to the axis of rotation.

3. Calculate $R$ and $r$. Substitute these values and the bounds into the washer method formula and integrate.

## Washer Method Example

Let’s look at an example to showcase the difference between the disk vs washer method. Consider the region bounded by $y = x^2$ and $y = x$ on the interval $[0, 1]$. Find the volume of the region when it is rotated about the line $y = 2$.

### Step 1 and 2

Our bounded region looks like this. The small white vertical black line represents one slice of the curve.

### Step 3

$R$ is the distance from the axis of rotation to the outer curve. In this case, our axis of rotation is $y = 2$ and the outer curve is $y = x^2$. So, $R = 2 - x^2$.

The lowercase $r$ is the distance from the axis of rotation to the inner curve. Our axis of rotation is $y = 2$ and the inner curve is $y = x$. So, $r = 2 - x$. Now, we can plug these values into the washer method formula and solve.

$\text{V} = \int_a^b \pi (R^2 - r^2)\,dx$
$\text{V} = \int_0^1 \pi ([2-x^2]^2 - [2-x]^2)\,dx$
$\text{V} = \pi \int_0^1 (x^4 - 5x^2 + 4x)\,dx$
$\text{V} = \pi(\frac{x^5}{5} - \frac{5x^3}{3}+\frac{4x^2}{2}) \Big|_0^1$
$\text{V} =\frac{8\pi}{15}$

Using the washer method, we’ve found that the volume of the solid of revolution is $\frac{8\pi}{15}$.

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