Volumes by Disks and Washers

What You’ll Learn

In this lesson you’ll learn how to find the volume of a solid formed by rotating a region around an axis. The disk method handles solid shapes, and the washer method handles shapes with a hole through the middle.

The Concept

When you rotate a 2D region around an axis, you get a 3D solid of revolution. To find its volume, imagine slicing it into thin circular cross-sections perpendicular to the axis.

Disk method (no hole)

If the region touches the axis of rotation, each slice is a solid disk with radius R(x)

$$V = \pi \int_a^b [R(x)]^2\,dx$$

Washer method (with a hole)

If the region doesn’t touch the axis, each slice is a washer (a disk with a hole). You need an outer radius and an inner radius

$$V = \pi \int_a^b \left([R_{\text{outer}}(x)]^2 - [R_{\text{inner}}(x)]^2\right)\,dx$$

How to set it up

1. Sketch the region and the axis of rotation
2. Draw a typical cross-section perpendicular to the axis
3. Identify the outer radius (farthest curve from the axis) and inner radius (closest curve, if any)
4. Square each radius, subtract inner from outer, multiply by pi, and integrate

Worked Example

Here’s what the solid of revolution looks like for Example 1. The blue surface is y = sqrt(x) rotated around the x-axis. Drag to rotate it and see the shape from different angles.

Drag to rotate · Scroll to zoom

Example 1: Disk method (rotate y = sqrt(x) around x-axis)

Find the volume of the solid formed by rotating y = sqrt(x) from x = 0 to x = 4 around the x-axis.

y = sqrt(x) region rotated R(x) = disk radius

Each cross-section is a disk with radius R(x) = sqrt(x). The orange dashed line shows the radius at a sample point.

$$V = \pi \int_0^4 (\sqrt{x})^2\,dx = \pi \int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi \cdot 8 = 8\pi$$

Example 2: Washer method (rotate region between y = x and y = x² around x-axis)

Find the volume when the region between y = x and y = x² from x = 0 to x = 1 is rotated around the x-axis.

y = x (outer) y = x² (inner) outer radius inner radius washer region

On [0, 1], y = x is farther from the x-axis than y = x², so the outer radius is x and the inner radius is x². The green dashed line shows the outer radius and the purple dashed line shows the inner radius at a sample point.

Drag to rotate · Scroll to zoom

The blue outer surface comes from rotating y = x, and the orange inner surface comes from rotating y = x². The gap between them is the washer region.

$$V = \pi \int_0^1 \left(x^2 - (x^2)^2\right)\,dx = \pi \int_0^1 (x^2 - x^4)\,dx$$ $$= \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}$$

Example 3: Disk method around the y-axis

Find the volume of the solid formed by rotating x = y² from y = 0 to y = 2 around the y-axis.

When rotating around the y-axis, integrate with respect to y. Each disk has radius R(y) = y² (the horizontal distance from the y-axis to the curve).

Drag to rotate · Scroll to zoom

This bowl-shaped paraboloid is what you get when x = y² spins around the y-axis. The green profile curve shows the generating function.

$$V = \pi \int_0^2 (y^2)^2\,dy = \pi \int_0^2 y^4\,dy = \pi\left[\frac{y^5}{5}\right]_0^2 = \pi \cdot \frac{32}{5} = \frac{32\pi}{5}$$

Real-World Application

Volumes of revolution show up anywhere you need to calculate the volume of a rotationally symmetric object.

• In manufacturing, designing bottles, cans, vases, and any container with circular cross-sections
• In engineering, calculating volumes of pistons, pipes, turbine blades, and axles
• In physics, computing moments of inertia for rotating objects
• In game development, procedurally generating 3D objects by defining a profile curve and rotating it
• In medical imaging, estimating organ volumes from cross-sectional scan data

You’ve Got This

The disk and washer methods come down to one idea: square the radius, multiply by pi, integrate. The only tricky part is identifying which curve gives the outer radius and which gives the inner radius. A quick sketch makes this obvious every time. If you can set up the radii correctly, the integration is usually straightforward.

Quiz

The disk method is used when

• A.The solid has a hole through the middle
• B.The region touches the axis of rotation (no hole)
• C.You're integrating with respect to y only
• D.The function is always negative

The volume from rotating y = sqrt(x) around the x-axis from 0 to 4 is

• A.4 pi
• B.16 pi
• C.8 pi
• D.2 pi

In the washer method, you subtract

• A.The outer radius from the inner radius
• B.The inner radius squared from the outer radius squared
• C.The two functions directly
• D.Nothing, it's the same as the disk method

When rotating the region between y = x and y = x² (from 0 to 1) around the x-axis, the outer radius is

• A.x²
• B.x
• C.1
• D.x - x²

To rotate a region around the y-axis using disks, you integrate with respect to

• A.x
• B.Neither x nor y
• C.y
• D.Both x and y simultaneously