Volumes of Revolution: Disks and Washers
Spin a flat region around a line and you sweep out a solid of revolution. Slice it perpendicular to the axis and every cross-section is a circle — so the volume is just the integral of circular areas: \(V = \pi\int_a^b [R(x)]^2\,dx\).
By the end you'll be able to set up and evaluate disk-method volumes \(\pi\int [R(x)]^2 dx\), switch to the washer method \(\pi\int ([R(x)]^2 - [r(x)]^2)\,dx\) when the region doesn't touch the axis, and measure both radii from the correct axis.
Predict: as you drag the slice from x = 0 toward x = 4, does the disk's radius grow or shrink? Drag the slice (or use the slider) to check.
The curve is \(y = \sqrt{x}\) on \([0, 4]\), revolved about the x-axis. The vertical slice at x becomes a circular disk of radius R(x) = √x and area πR². Stack more slices and their combined volume closes in on the exact answer, 8π. Tick the checkbox to carve out an inner radius r(x) and turn each disk into a washer. Hover or focus the disk for exact readouts.
R(2.00) = √x = 1.41 · slice area πR² ≈ 6.28
8 stacked slices ≈ 21.99 · exact volume 8π ≈ 25.13
Revolving a region about an axis produces circular cross-sections, so its volume is the integral of circle areas: disks when the region touches the axis, washers when there's a hole.
When the region touches the axis of revolution, every cross-section perpendicular to the axis is a full disk of radius \(R(x)\) and area \(\pi[R(x)]^2\), so the disk method gives \(V = \pi\int_a^b [R(x)]^2\,dx\). When there's a gap between the region and the axis, each cross-section is an annulus: subtract the inner disk from the outer to get the washer method, \(V = \pi\int_a^b \left([R(x)]^2 - [r(x)]^2\right)dx\). Both \(R(x)\) and \(r(x)\) are distances measured from the axis of revolution — shift the axis and the radii change with it. Once the setup is done, you evaluate the integral exactly as the Fundamental Theorem taught you: antidifferentiate and take endpoint values.
Picture the interactive above: the flat region under \(y = \sqrt{x}\) is a blade, and spinning it around the x-axis sweeps a horn-shaped solid. Freeze one vertical slice — it sweeps a disk of radius R(x) whose area is \(\pi[R(x)]^2\). A slab of thickness \(\Delta x\) then has volume \(\pi[R(x)]^2\,\Delta x\), and stacking the slabs is a Riemann sum that becomes the integral as the slices get thinner. The washer picture is the same with a hole of radius r(x) drilled through every slab.
Anything turned on a lathe or thrown on a potter's wheel is a solid of revolution: table legs, wine glasses, rocket nose cones, machine bushings. To cost out the material in a turned part, an engineer models its profile as \(y = R(x)\) and integrates \(\pi[R(x)]^2\); if the part is hollow — a pipe or a bushing — the bore profile \(r(x)\) is subtracted washer-style, \(\pi\int(R^2 - r^2)\,dx\), giving the exact volume of metal removed or remaining.
Revolve the region under \(y = \sqrt{x}\), \(0 \le x \le 4\), about the x-axis. The region touches the axis, so use disks with \(R(x) = \sqrt{x}\): \[V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi\cdot 8 = \mathbf{8\pi}.\] Note the squaring did the simplifying: \((\sqrt{x})^2 = x\) turned a root into a polynomial.
Revolve the region under \(y = x^2\), \(0 \le x \le 1\), about the x-axis. The region touches the axis, so disks apply with \(R(x) = x^2\), and \([R(x)]^2 = x^4\). Now finish it: \(V = \pi\int_0^1 x^4\,dx = \pi\cdot\) ____
Reveal the answer
\(V = \pi\left[\dfrac{x^5}{5}\right]_0^1 = \dfrac{\pi}{5}\). The radius got squared — \((x^2)^2 = x^4\) — before integrating; integrating \(x^2\) instead would give the (wrong) answer \(\pi/3\).
More info — the washer is "top minus bottom", squared
When you found the area between two curves, the recipe was \(\int (\text{top} - \text{bottom})\,dx\). The washer method is the same idea after revolution: the outer curve plays "top", the inner curve plays "bottom" — but because each curve now sweeps a disk rather than a line segment, you subtract squared radii, \(\pi\int (R^2 - r^2)\,dx\), not the radii themselves. Subtracting first and then squaring, \(\pi\int (R - r)^2\,dx\), is the single most common error with washers. The OpenStax section in Dive deeper below builds both formulas from the general slicing principle if you want the full derivation.
More info — watch the axis and the variable
Two setups that trip people up. First, a shifted axis: revolving \(y = f(x)\) about the line \(y = 1\) makes the radius \(f(x) - 1\), because radii are measured from the axis of revolution, not from the x-axis. Second, a vertical axis: revolving about the y-axis means the circular cross-sections are stacked along y, so the radii must be written as functions of y and the integral becomes a \(dy\) integral — \(V = \pi\int_c^d [R(y)]^2\,dy\).
Check your understanding
The region under \(y = x\) for \(0 \le x \le 3\) is revolved about the x-axis. What is the volume of the resulting cone?
A washer cross-section has outer radius \(R\) and inner radius \(r\). Why is its area \(\pi(R^2 - r^2)\) and not \(\pi(R - r)^2\)?
The region between \(y = f(x)\) (with \(f(x) > 1\)) and the line \(y = 1\) is revolved about the line \(y = 1\). What is the disk radius \(R(x)\)?
The region between \(y = \sqrt{x}\) (top) and \(y = x/2\) (bottom) on \([0, 4]\) is revolved about the x-axis. Which integral gives the volume?
Recap
- A solid of revolution has circular cross-sections; its volume is the integral of their areas.
- Disk method (region touches the axis): \(V = \pi\int_a^b [R(x)]^2\,dx\), where \(R(x)\) is the distance from the axis to the outer curve.
- Washer method (region has a gap): \(V = \pi\int_a^b \left([R(x)]^2 - [r(x)]^2\right)dx\) — subtract squared radii; \((R - r)^2\) is wrong.
- Measure both radii from the axis of revolution — revolving about \(y = 1\) makes the radius \(f(x) - 1\) — and use a \(dy\) integral when the axis is vertical.
Dive deeper
- OpenStax Calculus Volume 1 — 6.2 Determining Volumes by Slicing Derive the disk and washer volume formulas from circular cross-sections
Sources
- Volumes of Revolution — Disk and Washer Methods