# Volumes of Revolution: Disk, Washer and Shell Methods > Learn the disk, washer and shell methods for volumes of revolution, when to use which, and worked examples rotating regions about the axes and other lines. Source: https://practicecalculus.com/blog/volumes-of-revolution Updated: 2026-08-27 ## Key takeaways - Disk method: slice perpendicular to the axis, $V = \pi\int_a^b R(x)^2\,dx$. - Washer method: when there is a hole, $V = \pi\int_a^b \big(R(x)^2 - r(x)^2\big)\,dx$, outer radius squared minus inner radius squared. - Shell method: slice parallel to the axis, $V = 2\pi\int_a^b (\text{radius})(\text{height})\,dx$. - Pick the method whose slices let you write the radius and height without solving for the other variable. Spin a flat region around a line and it sweeps out a 3D solid: a vase, a bowl, a bead. Finding its volume sounds hard, but if you slice it the right way, every slice is a simple shape (a disk, a washer or a thin cylindrical shell) and the integral adds them up. The hard part is choosing the slice and getting the radius right. ## What is a volume of revolution? A **solid of revolution** is the 3D shape you get by rotating a 2D region around a line, called the **axis of rotation**. Its volume is found by cutting it into thin slices whose volumes you know, then integrating. There are three standard methods: | Method | Slice shape | Slice relative to axis | Formula | | ------- | -------------------- | ---------------------- | --------------------------------------------- | | Disk | Solid circle | Perpendicular | $\pi\int R^2$ | | Washer | Circle with a hole | Perpendicular | $\pi\int (R^2 - r^2)$ | | Shell | Hollow cylinder | Parallel | $2\pi\int (\text{radius})(\text{height})$ | ## The disk method Rotate the region under $y = f(x)$ from $a$ to $b$ around the $x$-axis. A vertical slice of thickness $dx$ becomes a thin coin with radius $R(x) = f(x)$ and volume $\pi R^2\,dx$. Adding the coins: $$V = \pi\int_a^b R(x)^2\,dx$$ ### Worked example 1: rotating y = √x about the x-axis Find the volume when the region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ is rotated about the $x$-axis. ```graph title: 'Region under y = √x from 0 to 4, rotated about the x-axis to form a solid' caption: 'Each vertical slice of this region sweeps out a disk of radius $\sqrt{x}$. The solid has volume $8\pi$.' x: [-0.5, 5] y: [-2.5, 2.5] functions: - expr: sqrt(x) label: y = √x - expr: -sqrt(x) label: mirror image after rotation dashed: true shade: {expr: sqrt(x), from: 0, to: 4} vertical: - {x: 4, label: 'x = 4'} ``` **Step 1. Radius.** The region touches the axis, so slices are disks with $R(x) = \sqrt{x}$. **Step 2. Set up.** $$V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx$$ **Step 3. Integrate.** $$V = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi(8 - 0) = 8\pi$$ > **Tip:** a quick sanity check for the disk method is a cone. Rotating $y = \frac{r}{h}x$ from $0$ to $h$ gives $\pi\int_0^h \frac{r^2}{h^2}x^2\,dx = \frac{1}{3}\pi r^2 h$, the familiar cone formula. ## The washer method If there is a gap between the region and the axis, each slice has a hole in the middle. With **outer radius** $R(x)$ and **inner radius** $r(x)$: $$V = \pi\int_a^b \big(R(x)^2 - r(x)^2\big)\,dx$$ ### Worked example 2: the region between y = x and y = x² Rotate the region between $y = x$ and $y = x^2$ about the $x$-axis. ```graph title: 'Region between y = x and y = x² from x = 0 to x = 1, used for washer and shell volumes' caption: 'The curves meet at $(0, 0)$ and $(1, 1)$. Rotated about the $x$-axis, the line gives the outer radius and the parabola the inner radius.' x: [-0.3, 1.5] y: [-0.3, 1.5] functions: - expr: x label: y = x - expr: x^2 label: y = x² points: - [0, 0, '(0, 0)'] - [1, 1, '(1, 1)'] ``` **Step 1. Limits.** $x = x^2$ gives $x = 0$ and $x = 1$ (see [area between curves](/blog/area-between-curves)). **Step 2. Radii.** On $[0, 1]$, $x \ge x^2$, so the line is farther from the axis. $R(x) = x$ and $r(x) = x^2$. **Step 3. Set up and integrate.** $$V = \pi\int_0^1 \big(x^2 - x^4\big)\,dx = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}$$ ### Worked example 3: rotating about the line y = 2 Rotate the same region about the horizontal line $y = 2$. **Step 1. Radii as distances.** Every point of the region is below $y = 2$, so the distance from a curve to the axis is $2 - y$. The parabola $y = x^2$ is lower, so it is **farther** from $y = 2$: $$R(x) = 2 - x^2, \qquad r(x) = 2 - x$$ **Step 2. Expand.** $$R^2 - r^2 = (4 - 4x^2 + x^4) - (4 - 4x + x^2) = x^4 - 5x^2 + 4x$$ **Step 3. Integrate.** $$V = \pi\int_0^1 (x^4 - 5x^2 + 4x)\,dx = \pi\left(\frac{1}{5} - \frac{5}{3} + 2\right) = \pi\cdot\frac{3 - 25 + 30}{15} = \frac{8\pi}{15}$$ Notice how the inner and outer curves swapped roles compared with example 2. Always measure from the axis. ## The shell method Now slice **parallel** to the axis. Rotating a thin vertical strip around the $y$-axis produces a hollow cylinder (a shell). Unrolled, a shell is nearly a flat sheet with length $2\pi(\text{radius})$, height $h(x)$ and thickness $dx$. So for rotation about the $y$-axis: $$V = 2\pi\int_a^b x\,h(x)\,dx$$ where $x$ is the radius of the shell and $h(x)$ is the height of the strip (top minus bottom). ### Worked example 4: the same region, about the y-axis Rotate the region between $y = x$ and $y = x^2$ about the $y$-axis. **Step 1. Radius and height.** A vertical strip at $x$ is a distance $x$ from the $y$-axis. Its height is $x - x^2$. **Step 2. Set up and integrate.** $$V = 2\pi\int_0^1 x(x - x^2)\,dx = 2\pi\int_0^1 (x^2 - x^3)\,dx = 2\pi\left(\frac{1}{3} - \frac{1}{4}\right) = \frac{\pi}{6}$$ **Check with washers.** Slicing horizontally, solve each curve for $x$: the outer radius is $x = \sqrt{y}$ and the inner is $x = y$, for $0 \le y \le 1$: $$V = \pi\int_0^1 (y - y^2)\,dy = \pi\left(\frac{1}{2} - \frac{1}{3}\right) = \frac{\pi}{6}$$ Both methods agree. ### Worked example 5: where shells are the only easy option Rotate the region under $y = 2x^2 - x^3$ from $x = 0$ to $x = 2$ about the $y$-axis. ```graph title: 'Region under y = 2x² − x³ from 0 to 2, rotated about the y-axis with the shell method' caption: 'Solving $y = 2x^2 - x^3$ for $x$ is a nightmare, so washers are impractical. Vertical strips like the one at $x = 1.2$ become cylindrical shells.' x: [-0.5, 2.5] y: [-0.3, 1.6] functions: - expr: 2x^2 - x^3 label: y = 2x² − x³ shade: {expr: 2x^2 - x^3, from: 0, to: 2} vertical: - {x: 1.2, label: 'shell at x = 1.2'} ``` Washers would need $x$ as a function of $y$, which means solving a cubic. Shells avoid that entirely. **Step 1. Radius and height.** Radius $x$, height $2x^2 - x^3$. **Step 2. Set up.** $$V = 2\pi\int_0^2 x(2x^2 - x^3)\,dx = 2\pi\int_0^2 (2x^3 - x^4)\,dx$$ **Step 3. Integrate.** $$V = 2\pi\left[\frac{x^4}{2} - \frac{x^5}{5}\right]_0^2 = 2\pi\left(8 - \frac{32}{5}\right) = 2\pi\cdot\frac{8}{5} = \frac{16\pi}{5}$$ ## Disk, washer or shell: how to choose Ask two questions: **which axis**, and **which variable are the curves written in?** | Curves given as | Axis of rotation | Easiest method | Integrate in | | --------------- | ------------------------ | ------------------------------- | ------------ | | $y = f(x)$ | Horizontal ($x$-axis, $y = k$) | Disk or washer | $x$ | | $y = f(x)$ | Vertical ($y$-axis, $x = h$) | Shell | $x$ | | $x = g(y)$ | Vertical | Disk or washer | $y$ | | $x = g(y)$ | Horizontal | Shell | $y$ | The pattern: disks and washers slice **perpendicular** to the axis, shells slice **parallel** to it. Choose whichever lets you integrate in the variable the curves are already written in. For shells around $x = h$, the radius becomes $|h - x|$ instead of $x$. ## Common mistakes with volumes of revolution 1. **Writing $(R - r)^2$ instead of $R^2 - r^2$.** Square each radius separately, then subtract. 2. **Forgetting $\pi$ (or $2\pi$).** Disks and washers use $\pi$; shells use $2\pi$. 3. **Measuring radius from the wrong place.** The radius is the distance to the **axis of rotation**, not to the $x$-axis. For $y = 2$, the radius of $y = x^2$ is $2 - x^2$. 4. **Swapping inner and outer radii.** The outer radius belongs to the curve **farther** from the axis, which is not always the top curve. 5. **Mixing variables.** A $dy$ integral needs radii and limits in terms of $y$. ## Practice volumes until the setup is automatic The integrals in these problems are usually easy. The marks are in the setup: the right slice, the right radius, the right limits. The interactive lesson below drills exactly that, with instant feedback on each radius and each integral, so a swapped inner radius gets caught the moment it happens. ## Frequently asked questions ### What is the difference between the disk and washer method? The washer method is the disk method with a hole. If the region touches the axis of rotation, each slice is a solid disk with area $\pi R^2$. If there is a gap between the region and the axis, each slice is a washer with area $\pi R^2 - \pi r^2$. ### When should I use the shell method instead of disks or washers? Use shells when disks would force you to solve the equation for the other variable, or would need several integrals. A common case is rotating a region given by $y = f(x)$ around the $y$-axis: shells let you integrate in $x$ directly with $V = 2\pi\int x f(x)\,dx$. ### Why is it R squared minus r squared and not (R minus r) squared? A washer is a big disk with a small disk removed. Its area is $\pi R^2 - \pi r^2$. The expression $\pi(R - r)^2$ would be the area of a disk with radius equal to the washer's thickness, which is a completely different shape. ### How do you find the radius when rotating about a line other than an axis? The radius is the distance from the curve to the axis of rotation. For a horizontal line $y = k$, use $|k - y|$; for a vertical line $x = h$, use $|h - x|$. Write it as (larger minus smaller) so it comes out positive. ### Do the disk and shell methods give the same answer? Yes. They slice the same solid in different directions, so if both are set up correctly the volumes match. Computing a volume both ways is an excellent check.