---
title: "Volumes of Revolution: Disks, Washers and Shells"
description: "Set up a volume as an integral of cross-sections, and choose between slicing perpendicular to the axis and using cylindrical shells based on which integral is easier."
canonical: https://lightmysky.com/learn/mathematics/volumes-of-revolution-disks-washers-and-shells-mt_mzZ22MkWD4
source: https://lightmysky.com/learn/mathematics/volumes-of-revolution-disks-washers-and-shells-mt_mzZ22MkWD4.md
retrieved: 2026-09-12
---

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# Volumes of Revolution: Disks, Washers and Shells

Set up a volume as an integral of cross-sections, and choose between slicing perpendicular to the axis and using cylindrical shells based on which integral is easier.

Subject: Mathematics · Area: Calculus & Analysis · Ages 19 to 20
Page: https://lightmysky.com/learn/mathematics/volumes-of-revolution-disks-washers-and-shells-mt_mzZ22MkWD4

## Ready when they can

- Write the integral for a solid generated by rotating a region about the x-axis
- Use washers when the solid has a hole and identify both radii
- Say when shells give a simpler integral than disks for the same solid

## Lesson: Turning flat regions into volumes

Picture a region under a curve, cut into thin vertical strips. Spin one strip around the x-axis and you get a thin disk. Its radius equals the function value, so its volume is pi times radius squared times dx. For y equals the square root of x from 0 to 4, each disk has radius square root of x, and the square is x, so the volume is pi times the integral from 0 to 4 of x dx. A flat check: under y equals 3 from 0 to 2, every disk has radius 3, which is a cylinder of radius 3.

**Example.** Maya rotates the region between y equals x on top and y equals x squared below, from 0 to 1, about the x-axis. The gap between the curves leaves a hole, so each slice is a washer, not a full disk. The outer radius follows the top curve, x, and the inner radius follows the bottom curve, x squared. Her volume is pi times the integral from 0 to 1 of x squared minus x to the fourth dx. With numbers alone, outer radius 5 and inner radius 3 give 25 minus 9, which is 16, so the washer area is 16 pi.

Shells wrap the region in nested tubes instead of stacking slices. One shell at radius r with height h gives 2 pi times r times h times dx. Take y equals x squared from 0 to 1 rotated about the y-axis: a vertical strip at x has height x squared, so each shell gives 2 pi times x times x squared, which is 2 pi times x cubed dx. Shells win here because disks would force you to rewrite the curve as x equals the square root of y.

**Tip.** Name your method before you integrate, then fill radius and height from the picture. Run strips across the axis for disks and washers, and along the axis for shells. Square each radius first and then subtract, since squaring the difference gives the wrong area.

**Recap.** Disks fill the solid, washers subtract the hole, and shells wrap it in tubes, so match the method to the picture.

## Practice

14 questions on this page, each with its working shown.

## Needs first

- [Improper Integrals and Their Convergence](https://lightmysky.com/learn/mathematics/improper-integrals-and-their-convergence-mt_gvtZPFRA6Z)
- [Areas Between Two Curves](https://lightmysky.com/learn/mathematics/areas-between-two-curves-mt_lDBZmnLK1b)
- [Volume and Surface Area of Cones, Spheres and Pyramids](https://lightmysky.com/learn/mathematics/volume-and-surface-area-of-cones-spheres-and-pyramids-mt_s4K5YMBSzG)

## Opens up

- [Arc Length and the Area of a Surface of Revolution](https://lightmysky.com/learn/mathematics/arc-length-and-the-area-of-a-surface-of-revolution-mt_W4uujT9Z1t)
