Example 1
- R = 2 · r = 1
≈ 9.424778
What is the annulus area at R = 2 and r = 1? ≈ 9.424778.
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.
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.
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.
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).
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.
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.
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⁺.
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.
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.
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 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.
Annulus area is π(R² − r²). With R = 2 and r = 1 the result is ≈ 9.424778. Domain: R ≥ r ≥ 0.
≈ 9.424778
What is the annulus area at R = 2 and r = 1? ≈ 9.424778.
≈ 50.265482
What is the area at R = 5 and r = 3? ≈ 50.265482.
≈ 28.274334
What is the area at R = 3 and r = 0? ≈ 28.274334.
≈ 201.061930
What is the area at R = 10 and r = 6? ≈ 201.061930.
≈ 150.796447
What is the area at R = 8 and r = 4? ≈ 150.796447.
≈ 2.356194
What is the area at R = 1 and r = 0.5? ≈ 2.356194.
≈ 141.371669
What is the area at R = 7 and r = 2? ≈ 141.371669.
≈ 197.920337
What is the area at R = 12 and r = 9? ≈ 197.920337.
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).
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”.
The result is 16π ≈ 50.265482, because π(25 − 9) = 16π. Equivalently π(5 − 3)(5 + 3) = π·2·8 = 16π.
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.
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.
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.
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.
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.
Yes. 2.5 and 2,5 are the same radius. The calculator accepts a comma or a period as the decimal separator.
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.
The result is 64π ≈ 201.061930, because π(100 − 36) = 64π. The width is 4 and the mean radius is 8.
Not in this calculator. When r > R, you get a domain error instead of a negative number. The fields are not swapped automatically.
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).
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.
No. Here only a flat concentric annulus. Torus volume and a ring with different centers need other models — they are not on this card.
75π ≈ 235.619449. Computation: π(100 − 25) = 75π. Enter R = 10 and r = 5; the width is 5.
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 annulus definition and area formula below come from established mathematical sources.
Page Updated November 2, 2026.