Calculus I › Integrals › full formula sheet
The washer method
Disks with a hole — subtract the inner disk. Square first, then subtract.
Notation on this page: R(x) is the outer radius (farther from the axis), r(x) the inner radius (nearer). Slices are perpendicular to the axis.
Before this lesson: Disk method
Where it comes from
The problem: the disk method assumed the region touches the axis. But revolve the region between y = x and y = x² around the x-axis, and there is a gap — the region never reaches the axis. Each perpendicular slice is not a disk but a washer: a disk with a circular hole punched out. Washer volume = outer disk minus inner disk = πR² dx − πr² dx.
The tempting subtract-first:
Before reading on: expand (R−r)2. Before reading: which extra term appears that π(R2−r2) does not have — and for the region between y = x and y = x2, does that term inflate or shrink the answer?
Kill it with constants: R = 3, r = 1. A washer with outer radius 3 and inner radius 1 has area π(9−1) = 8π. The guess gives π(3−1)² = 4π — half the truth. The algebra explains why:
Intuition: you cannot shortcut “area of the ring” into “area of a disk with radius (R−r)” — a ring of outer radius 3 and inner radius 1 is much fatter than a disk of radius 2. The hole’s position matters, not just its width.
Derivation
Same slab argument as disks, but each slab is a washer: outer cylinder minus inner cylinder. The −2Rr trap is why the order matters.
Which is R, which is r? R is the outer radius — the curve farther from the axis. r is the inner radius — nearer the axis. R² − r² must stay non-negative; if your integrand goes negative, you swapped them.
How to use it
Before reading on: the region sits above y = 1 but the axis is the x-axis, so the solid has a hollow core. Before reading: what is the inner radius — 0, 1, or something else?
The procedure, every time:
- Confirm the gap: the region does not touch the axis — there is empty space between the region’s near edge and the axis.
- Name R and r as distances to the axis. R = farther curve minus axis; r = nearer curve minus axis. Account for shifted axes (about y = −1, add 1 to both).
- Square each, then subtract: [R(x)]² − [r(x)]². Expand before integrating.
- Compute π∫(R²−r²)dx. Keep π outside.
- Check the sign: R² − r² ≥ 0 everywhere. Negative → swapped R and r.
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: region between y = x and y = x² on [0, 1], about the x-axis
- Outer vs inner: on [0,1], x ≥ x², so the farther curve from the x-axis is y = x: R(x) = x, r(x) = x². (Why? R is the outer radius — larger distance.)
- Square first, then subtract: R² − r² = x² − x⁴.
- Integrate: V = π∫01 (x²−x⁴) dx = π[x³/3 − x⁵/5]01 = π(1/3 − 1/5) = 2π/15.
- Sanity check: the region is thin (max gap 1/4 at x = 1/2), so the volume should be small: 2π/15 ≈ 0.42. The solid disk version (π/3 ≈ 1.05 for y = x alone) is bigger — the hole removed most of it. ✓
Your turn: Region between y = x and y = x3 on [0, 1], about the x-axis. Find V.
Answer: 4π/21
On [0,1], x ≥ x3: R = x, r = x3. Square first, then subtract: x2 − x6. Integrate: V = π[x3/3 − x7/7]01 = π(1/3 − 1/7) = 4π/21. Sanity: the gap here is wider than the x-vs-x2 region’s, so 4π/21 ≈ 0.60 > 2π/15 ≈ 0.42. ✓
Example 2 — shifted axis: y = √x on [0, 4], about the line y = −1
- Radii as distances to y = −1: R(x) = √x − (−1) = √x + 1 (outer); r(x) = 0 − (−1) = 1 (inner — the gap from the x-axis down to y = −1).
- Square first, then subtract: (√x+1)² − 1² = (x + 2√x + 1) − 1 = x + 2√x.
- Integrate: V = π∫04 (x + 2√x) dx = π[x²/2 + (4/3)x3/2]04 = π(8 + 32/3) = 56π/3.
- Sanity check: outer radii run 1→3 (vs the disk version’s 0→2 with volume 8π). Bigger radii throughout → much bigger volume: 56π/3 ≈ 58.6 > 8π ≈ 25.1. ✓
Your turn: y = √x on [0, 1], about the line y = −1. Find V.
Answer: 11π/6
Radii to y = −1: R = √x + 1 (outer), r = 1 (inner, the gap down to the axis). Square first: (√x+1)2 − 1 = x + 2√x. Integrate: V = π[x2/2 + (4/3)x3/2]01 = π(1/2 + 4/3) = 11π/6. ✓
Example 3 — a band: region under y = 2−x² and above y = 1 on [−1, 1], about the x-axis
- Radii: R(x) = 2−x² (outer), r(x) = 1 (inner — the strip from y = 0 to y = 1 is empty).
- Square first, then subtract: (2−x²)² − 1 = (4 − 4x² + x⁴) − 1 = 3 − 4x² + x⁴.
- Integrate (even function — double [0,1]): V = π∫−11 (3−4x²+x⁴) dx = 2π[3x − 4x³/3 + x⁵/5]01 = 2π(3 − 4/3 + 1/5) = 2π(28/15) = 56π/15.
Your turn: Region under y = 3−x2 and above y = 2 on [−1, 1], about the x-axis. Find V.
Answer: 32π/5
R = 3−x2 (outer), r = 2 (inner — the strip 0–2 is empty). Square first: (3−x2)2 − 4 = 5 − 6x2 + x4. Integrate (even, double [0,1]): V = 2π[5x − 2x3 + x5/5]01 = 2π(16/5) = 32π/5. ✓
Example 4 — washers collapse to disks: y = x² on [0, 1], about the x-axis
- Radii: the region touches the x-axis, so r(x) = 0 and R(x) = x².
- The formula gives: V = π∫01 ((x²)² − 0) dx = π∫01 x⁴ dx = π/5 — exactly the disk method.
- The lesson: disks are washers with r = 0. One formula covers both — washers are the general case.
Your turn: y = x3 on [0, 1], about the x-axis. Find V (washers collapse).
Answer: π/7
The region touches the axis: r = 0, R = x3. The washer formula gives V = π∫01 x6 dx = π/7 — exactly the disk method. The lesson: disks are washers with r = 0. ✓
Memorization tips
- Say it aloud: “big squared minus little squared, times pi.” Outer (big) first, inner (little) second.
- The (R−r)² alarm: expand it mentally — R² − 2Rr + r². If you see a −2Rr in your work, you subtracted too early.
- R is farther: point at the axis, then at each curve — the longer arrow is R. Five seconds, no swaps.
- Non-negative check: R² − r² ≥ 0 everywhere. Negative → you swapped R and r.
- r = 0 → disks: no gap means the washer formula becomes the disk formula. One framework, two names.
- Shifted axes shift both radii: about y = −1, add 1 to both R and r. Forgetting the inner shift is the classic half-error.
Final challenge
Five mixed questions — the subtract-first trap, shifted axes, and the collapse check. Score 5/5 and washers 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 washer method?
Disks with a hole: when the region doesn’t touch the axis, each slice is a washer with volume π[R(x)]² dx − π[r(x)]² dx. Integrate: V = π∫ab([R(x)]² − [r(x)]²) dx.
Why square first, then subtract?
Because (R−r)² = R² − 2Rr + r² ≠ R² − r². The washer’s area is outer disk minus inner disk: πR² − πr². Subtracting before squaring invents a −2Rr term.
When do I use washers instead of disks?
When there’s a gap between the region and the axis of rotation. No gap (r = 0) collapses washers back to disks — disks are the special case.
Which is the outer radius?
The one farther from the axis of rotation. R is always the larger distance; r the smaller. If you mix them up, R² − r² goes negative.
How do I remember the order?
Say “big squared minus little squared.” Outer (big) first, inner (little) second — then the integrand stays non-negative.
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