Calculus I › Integrals › full formula sheet
The disk method
Spin a region around an axis — it stacks into disks. Square the radius, not the function.
Notation on this page: R(x) is the radius — the distance from the curve to the axis of rotation. Slices are perpendicular to the axis.
Before this lesson: FTC Part 2 · Area between curves
Where it comes from
The problem: you have a flat region (say, under y = x from 0 to 1) and you spin it around the x-axis. The result is a 3D solid — here, a cone. What is its volume? Slice the solid perpendicular to the axis: each slice is a thin disk — a short cylinder — with radius R(x) and thickness dx. Disk volume = (area)·(thickness) = π[R(x)]² dx. Add the disks, take the limit: an integral.
The tempting unsquared version:
Before reading on: the wrong guess writes V = π∫01 x dx, forgetting to square. But a disk of radius x has area πx2, not πx. Before reading: which power of x should be inside the integral — and what does the wrong guess actually compute?
Kill it with geometry. The solid is a cone with radius 1 and height 1, whose volume is (1/3)πr²h = π/3 — not π/2. The guess forgot that a disk’s area is πR², not πR:
Intuition: area scales with the square of radius — doubling the radius quadruples the disk. The square is not bookkeeping; it is the geometry of circles.
Derivation
Partition [a, b], approximate each slab as a cylinder, sum, and take the limit. The πR² is the area of one circular face.
Disks vs washers: this derivation assumes the region touches the axis — every slab is a solid disk. If there is a gap between the region and the axis, each slab has a hole: that is the washer method (next formula on the sheet).
How to use it
Before reading on: rotating y = x about y = −2 instead of the x-axis makes every radius 2 longer. Before reading: will the volume be bigger or smaller than the about-the-x-axis version — and does your intuition survive the actual number?
The procedure, every time:
- Identify the axis of rotation. Slices must be perpendicular to it: x-axis → vertical slices (dx); y-axis → horizontal slices (dy).
- Write R as a distance from the curve to the axis. About the x-axis: R(x) = ycurve (if the region touches the axis). About the line y = −1: R(x) = ycurve + 1 — shift by the gap.
- Square it: [R(x)]². Expand if needed.
- Compute π∫[R]²dx (FTC Part 2). Keep the π outside the integral.
- Cone/cylinder check: if the solid is a cone or cylinder, verify with (1/3)πr²h or πr²h.
Worked examples
Four problems, easiest first. In each one, read every step — the why of each move is the lesson.
Example 1 — the basic move: y = x on [0, 2], about the x-axis
- Radius: the region touches the x-axis, so R(x) = x − 0 = x.
- Square: [R(x)]² = x².
- Integrate: V = π∫02 x² dx = π[x³/3]02 = 8π/3.
- Cone check: the solid is a cone with r = 2, h = 2: (1/3)π(2)²(2) = 8π/3. Matches ✓
Your turn: y = x on [0, 3], about the x-axis. Find V.
Answer: 9π
Radius R(x) = x; square: x2. Integrate: V = π∫03 x2 dx = π[x3/3]03 = 9π. Cone check: (1/3)π(3)2(3) = 9π. ✓
Example 2 — a curve: y = √x on [0, 4], about the x-axis
- Radius: R(x) = √x.
- Square: (√x)² = x. (Why this is nice: the square undoes the root.)
- Integrate: V = π∫04 x dx = π[x²/2]04 = 8π.
- Sanity check: the solid fattens then thins? No — radius √x grows throughout, max 2 at x = 4. A cylinder of r = 2, h = 4 has volume 16π; our solid is slimmer (radius starts at 0), so 8π < 16π is plausible. ✓
Your turn: y = √x on [0, 9], about the x-axis. Find V.
Answer: 81π/2
Radius √x; square: (√x)2 = x. Integrate: V = π∫09 x dx = π[x2/2]09 = 81π/2. Sanity: radius grows 0→3; a full cylinder (r = 3, h = 9) is 81π, and our slimmer solid is half of that. ✓
Example 3 — shifted axis: y = x on [0, 3], about the line y = −2
- Radius: distance from the curve y = x down to y = −2: R(x) = x − (−2) = x + 2. (Why +2? The axis sits 2 units below the x-axis, so every radius is 2 longer.)
- Square: (x+2)² = x² + 4x + 4.
- Integrate: V = π∫03 (x²+4x+4) dx = π[x³/3 + 2x² + 4x]03 = π(9 + 18 + 12) = 39π.
- Sanity check: radii run 2→5, so the solid is chunkier than Example 1’s cone (8π/3) — 39π reflects that. ✓
Your turn: y = x on [0, 2], about the line y = −1. Find V.
Answer: 26π/3
Radius: distance from y = x down to y = −1: R(x) = x + 1. Square: (x+1)2 = x2+2x+1. Integrate: V = π[x3/3 + x2 + x]02 = π(8/3 + 4 + 2) = 26π/3. Sanity: radii 1→3, chunkier than the cone 8π/3; 26π/3 ≈ 27.2 reflects that. ✓
Example 4 — dy version: region between x = y and the y-axis, y in [0, 3], about the y-axis
- Slice horizontally (perpendicular to the y-axis): disks of thickness dy.
- Radius: R(y) = y − 0 = y (distance from the line x = y to the y-axis).
- Integrate: V = π∫03 y² dy = π[y³/3]03 = 9π.
- Cone check: the solid is a cone with r = 3, h = 3: (1/3)π(3)²(3) = 9π. Matches ✓
Your turn: Region between x = 2y and the y-axis, y in [0, 2], about the y-axis. Find V.
Answer: 32π/3
Slice horizontally (dy), perpendicular to the y-axis: R(y) = 2y. Integrate: V = π∫02 (2y)2 dy = π∫02 4y2 dy = π[4y3/3]02 = 32π/3. Cone check: r = 4, h = 2: (1/3)π(4)2(2) = 32π/3. ✓
Memorization tips
- Say it aloud: “pi times the integral of R-squared dee-x.” The square is in the sentence — say it every time.
- The unit test: π∫R dx has area units; π∫R² dx has volume units. If your setup does not smell like volume, the square is missing.
- Radius is a distance: R = (curve) − (axis), with the axis’s position accounted for. About y = −2, R = ycurve + 2.
- Cone check: whenever the solid is a cone, (1/3)πr²h verifies your integral in seconds. Memorize y = x → cone.
- Perpendicular slices: x-axis → dx; y-axis → dy. Say “slices cross the axis” while setting up.
- Gap → washers: if the region does not touch the axis, disks are wrong — each slab has a hole. That is the next formula.
Final challenge
Five mixed questions — the square trap, shifted axes, and a hemisphere. Score 5/5 and disks are yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What is the disk method?
Revolve a region around an axis: each slice perpendicular to the axis becomes a disk of radius R(x) and thickness dx, with volume π[R(x)]² dx. Integrate: V = π∫ab[R(x)]² dx.
Why do you square the radius?
Because each slice is a disk with area πR² — area scales with the square of the radius. Forgetting the square (π∫R dx) gives the wrong units and the wrong number.
When do I use disks vs washers?
Disks when the region touches the axis of rotation (solid, no hole). Washers when there’s a gap between the region and the axis (a hole to subtract).
What is the volume when y = x on [0, 2] revolves around the x-axis?
V = π∫02x² dx = 8π/3. Check: it’s a cone with r = 2, h = 2, and (1/3)πr²h = 8π/3. ✓
How do I set up disks around the y-axis?
Slice horizontally (dy): V = π∫[R(y)]² dy, with R expressed as a function of y. The slice must be perpendicular to the axis of rotation.
More from the codex
Support the codex
This page is free, with no account and no ads. If it helped you learn, consider supporting the indie dev behind it.
Questions or a bug to report? Email [email protected].