Annulus area calculator

Enter the outer radius R and the inner radius r (R ≥ r ≥ 0). The result is the circular annulus area A = π(R² − r²) = π(R − r)(R + r) — the difference of two concentric disks. With R = 2 and r = 1 you get 3π ≈ 9.424778; with R = 5 and r = 3 — 16π ≈ 50.265482. This is not torus volume and not an eccentric ring with different centers.

Input

Result

Enter R and r (R ≥ r ≥ 0).

The first field is R (outer), the second is r (inner). The calculator does not swap them. When r > R, you get a domain error instead of a negative area. When r = 0 you recover the full disk πR².

How it works

How the area is computed at R = 2 and r = 1
R² − r²4 − 1
× π3π
A≈ 9.424778

A circular annulus is the plane region between two circles that share a center. Its area is the difference of two disk areas: the larger disk of radius R minus the smaller disk of radius r. That yields A = πR² − πr², usually written A = π(R² − r²). This is classical Euclidean plane geometry — the same model that appears in geometry textbooks and in calculus with polar coordinates.

The domain is R ≥ r ≥ 0. Radii cannot be negative. When r > R, the difference of squares would be negative; the calculator rejects that input instead of showing a negative “area”. When R = r, the thickness is zero and A = 0. When r = 0, the hole vanishes and A = πR² — the same circle area that the dedicated circle page computes from one radius.

The algebraic factorization R² − r² = (R − r)(R + r) gives the equivalent form A = π(R − r)(R + r). With R = 2 and r = 1: π(4 − 1) = 3π ≈ 9.424778, and also π·1·3 = 3π. With R = 5 and r = 3: 16π ≈ 50.265482. With R = 3 and r = 0: 9π ≈ 28.274334.

Geometric reading: ring width is w = R − r and mean radius is ρ̄ = (R + r)/2. Then A = 2πρ̄ · w — the circumference at the mean radius times the thickness. In calculus, A = ∫_r^R 2πρ dρ = π(R² − r²). The calculator uses the R and r values you enter; the header unit switch only changes the cm²/in² label and does not rescale the entered numbers.

Applications include a pipe wall cross-section, gaskets, simplified flat models of planetary rings, and regions between two distances from a point. This calculator does not compute an angular sector or a cylinder side area — only the flat annulus area from two radii.

Check examples: R = 10 and r = 6 give 64π ≈ 201.061930; R = 8 and r = 4 give 48π ≈ 150.796447; R = 1 and r = 0.5 give 0.75π ≈ 2.356194.

Formula

A = π(R² − r²) = π(R − r)(R + r), where R ≥ r ≥ 0.

Symbol R is the outer radius (first field). Symbol r is the inner radius (second field). The constant π comes from the disk-area definition. Computation: π · (R² − r²); the panel also shows the factored form as a check.

Mathematical model: annulus area as the difference of two concentric disk areas (standard plane geometry / polar calculus).

Theory, factorization, and integral

Definition

A circular annulus is the set of plane points lying between two concentric circles of radii R and r, with R > r > 0 (the calculator also accepts the limiting cases R = r, area 0, and r = 0, a full disk). Its area is A = π(R² − r²) = π(R − r)(R + r). That is the difference of the areas of two disks: the larger of radius R and the smaller of radius r. This calculator accepts both radii (or the page’s equivalent inputs) and returns A.

The condition R > r guarantees positive area; when R = r the annulus degenerates to a circle of planar area zero. The centres coincide — an eccentric ring (distinct centres) needs integration or lens formulas and is not this calculator’s model.

Historical and conceptual context

Disk area A = πρ² is Archimedean (Measurement of a Circle): the disk equals a triangle with legs ρ and the circumference. Subtracting two such areas yields A_annulus = πR² − πr² at once. Euclid’s Elements prepare the result via similarity and exhaustion without modern algebra.

In applications the annulus models a pipe cross-section, a washer, a bearing race, or a zone between two isolines on a map. The formula computes planar area, not torus volume and not mass — a third-dimension thickness requires a separate multiplication by height.

Formal properties

Domain of a proper annulus: R > r > 0; the form accepts R ≥ r ≥ 0. The map A(R, r) increases in R and decreases in r. Homogeneity: under length scaling by λ > 0, area scales as λ². The identity A = π(R − r)(R + r) is convenient when the ring width w = R − r and the sum of radii are known.

Mean radius (R + r)/2 and width w give A = 2π·((R+r)/2)·w = π(R+r)w — exact algebraically for a concentric annulus, interpretable as a “rectangle” of length equal to the mid-circle circumference and width w.

As r → 0⁺ with fixed R, A → πR² (full disk). As R → r⁺, A → 0⁺.

Full derivation

Step 1. Area of a disk of radius R: A_R = πR². Area of a disk of radius r: A_r = πr².

Step 2. The annulus is the set difference of the disks (common centre), so A = A_R − A_r = π(R² − r²).

Step 3 (polar integral). ∫₀²π ∫_r^R ρ dρ dθ = 2π · (1/2)(R² − r²) = π(R² − r²).

Step 4 (factorisation). R² − r² = (R − r)(R + r), hence A = π(R − r)(R + r). This form helps when wall thickness w = R − r is the measured datum.

Special cases

For R = 2, r = 1: A = π(4 − 1) = 3π. For R = 10 cm, r = 8 cm (2 cm wall): A = π(100 − 64) = 36π cm² ≈ 113.1 cm². When r = R/2: A = (3/4)πR² — three quarters of the outer disk.

If diameters D and d are supplied instead of radii, A = (π/4)(D² − d²) for D > d > 0. Confusing diameter with radius scales each πρ² term by four.

Numerical and computational notes

Evaluating π·(R² − r²) is stable when R and r are not nearly equal at floating-point range. When R ≈ r at large magnitude, the factorisation (R−r)·(R+r) reduces cancellation loss from subtracting close squares.

Input with r ≥ R or non-positive values is rejected. Comma and period are equivalent. The calculator does not model an elliptical ring or an annular sector.

In mechanics of materials an annular cross-section appears for pipes, hollow shafts, and sealing rings. The second moment of area of an annulus about its centre is (π/4)(R⁴ − r⁴) — that is not the area, but a quantity obtained by integrating ρ³. This calculator returns only A = π(R² − r²).

Ring width w = R − r and mean circumference 2π·((R+r)/2) multiply to the area: A = w · π(R+r). That identity helps when a micrometer measures wall thickness and a tape measures a mean or outer girth. One must still convert consistently to a pair (R, r) before substituting.

As a figure with fixed radii: once R and r are fixed, the concentric annulus has no further shape freedom. For fixed A many pairs (R, r) satisfy R² − r² = A/π — the calculator does not choose a pair; it only computes A from the radii supplied.

Historically the Latin annulus (“small ring”) is standard in English texts; school Polish usually says “pierścień kołowy”, with “wycinek pierścienia” for an annular sector. MathWorld and Wikipedia (Annulus) give the same π(R² − r²) formula and the difference-of-disks reading.

Polar coordinates make the derivation transparent: the area element ρ dρ dθ integrates in ρ from r to R and in θ over a full turn. Restricting θ to an interval of length α yields an annular sector of area (α/(2π))·π(R² − r²) = (α/2)(R² − r²). That sector model is separate from the full-annulus calculator.

Related concepts and authorities

Related models: circle area, circumference, sector area, and pipe volume (annulus × height). Annulus area is not the surface area or volume of a torus.

Authorities: Archimedes, Measurement of a Circle; Euclid, Elements; MathWorld — Annulus; Wikipedia — Annulus (mathematics); NIST DLMF (π). The calculator is educational.

How to use

  1. Enter the outer radius R, for example 2.
  2. Enter the inner radius r, for example 1 (or 0 for a full disk).
  3. Click Calculate or wait for the live result. 2 and 1 give ≈ 9.424778. 5 and 3 give ≈ 50.265482.
  4. Check the panel: πR², πr², width, mean radius, and π(R−r)(R+r). The diagram draws two concentric circles.
  5. When r > R, correct the inputs — the calculator does not swap fields. A full disk from one r also lives on the circle-area page.

Symbol glossary

Annulus area is π(R² − r²). With R = 2 and r = 1 the result is ≈ 9.424778. Domain: R ≥ r ≥ 0.

A
Annulus area. R = 2, r = 1 → ≈ 9.424778. R = 5, r = 3 → ≈ 50.265482.
R
Outer radius (first field). Must be ≥ r. With R = 3 and r = 0 you get 9π ≈ 28.274334.
r
Inner radius (second field). 0 means no hole. When r > R, the result is not computed.
R−r
Ring width. At 2 and 1 it is 1; at 5 and 3 it is 2.
(R+r)/2
Mean radius. At 2 and 1 it is 1.5; then A = 2π · 1.5 · 1 = 3π.

Examples

Example 1

  • R = 2 · r = 1

≈ 9.424778

What is the annulus area at R = 2 and r = 1? ≈ 9.424778.

Example 2

  • R = 5 · r = 3

≈ 50.265482

What is the area at R = 5 and r = 3? ≈ 50.265482.

Example 3

  • R = 3 · r = 0

≈ 28.274334

What is the area at R = 3 and r = 0? ≈ 28.274334.

Example 4

  • R = 10 · r = 6

≈ 201.061930

What is the area at R = 10 and r = 6? ≈ 201.061930.

Example 5

  • R = 8 · r = 4

≈ 150.796447

What is the area at R = 8 and r = 4? ≈ 150.796447.

Example 6

  • R = 1 · r = 0.5

≈ 2.356194

What is the area at R = 1 and r = 0.5? ≈ 2.356194.

Example 7

  • R = 7 · r = 2

≈ 141.371669

What is the area at R = 7 and r = 2? ≈ 141.371669.

Example 8

  • R = 12 · r = 9

≈ 197.920337

What is the area at R = 12 and r = 9? ≈ 197.920337.

Related calculators

Common questions

What is the domain of the annulus area formula?

You need R ≥ r ≥ 0. R is the outer radius, r the inner. When r > R, the calculator shows an error and does not display a negative area. When R = r the result is 0 (zero thickness).

What is the annulus area at R = 2 and r = 1?

The result is 3π ≈ 9.424778, because π(4 − 1) = 3π. That is the first worked example on this page and the same fill from “Fill this example”.

What is the area at R = 5 and r = 3?

The result is 16π ≈ 50.265482, because π(25 − 9) = 16π. Equivalently π(5 − 3)(5 + 3) = π·2·8 = 16π.

What happens when r = 0?

The hole vanishes. With R = 3 and r = 0 the result is 9π ≈ 28.274334, the ordinary disk area πR². The circle-area page computes the same quantity from one radius.

Where does A = π(R − r)(R + r) come from?

From the difference of squares: R² − r² = (R − r)(R + r). Multiplying by π gives A = π(R − r)(R + r). It is the same number as π(R² − r²); the factored form is often easier by hand.

What do ring width and mean radius mean?

Width is R − r. Mean radius is (R + r)/2. Then A = 2π · ((R + r)/2) · (R − r): the circumference at the mean radius times the ring thickness.

How does an annulus differ from a sector and from circle area?

Circle area uses one radius. A sector uses an angle and one radius. An annulus uses two concentric radii and subtracts the inner disk from the outer disk.

Do the currency and unit settings in the page header affect the result?

No. The unit switch only changes the cm² or in² label. Entered 2 and 1 still give 9.424778 in the chosen square scale. R and r must share the same length unit.

Does a comma in 2.5 work?

Yes. 2.5 and 2,5 are the same radius. The calculator accepts a comma or a period as the decimal separator.

How does the integral derivation look?

In polar coordinates the annulus area is ∫_{ρ=r}^{R} ∫_{θ=0}^{2π} ρ dθ dρ = ∫_r^R 2πρ dρ = π(R² − r²). That is the same formula the calculator uses.

What is the area at R = 10 and r = 6?

The result is 64π ≈ 201.061930, because π(100 − 36) = 64π. The width is 4 and the mean radius is 8.

Can the result be negative?

Not in this calculator. When r > R, you get a domain error instead of a negative number. The fields are not swapped automatically.

How do I calculate the area of a circular annulus?

Use A = π(R² − r²), where R is the outer radius and r the inner. With R = 2 and r = 1: π(4 − 1) = 3π ≈ 9.424778. Equivalently A = π(R − r)(R + r).

May I enter diameters instead of radii?

First divide diameters by 2: R = D/2, r = d/2. With diameters the formula is A = (π/4)(D² − d²). Entering D and d as R and r makes the result four times too large.

Does the calculator compute a torus or an eccentric ring?

No. Here only a flat concentric annulus. Torus volume and a ring with different centers need other models — they are not on this card.

What is the area at R = 10 and r = 5?

75π ≈ 235.619449. Computation: π(100 − 25) = 75π. Enter R = 10 and r = 5; the width is 5.

How does π(R − r)² differ from π(R² − r²)?

π(R − r)² is the classic mistake: subtract first, then square. Correctly square each radius separately: π(R² − r²). With R = 10 and r = 6 the wrong π·16 gives 16π, while the true area is 64π.

The result is A = π(R² − r²): with R = 2 and r = 1 the area is 3π ≈ 9.424778. The check form π(R−r)(R+r) matches. Separate circle area lives on the circle area page.

Knowledge sources

The annulus definition and area formula below come from established mathematical sources.

Page Updated November 2, 2026.