CalConverters

Annulus Calculator: Ring Area, Radii, Circumference

Find the area, radii, and circumference of an annulus (ring) — the region between two concentric circles — from any two known measures, with exact steps.

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The annulus area formula

An annulus is the flat ring between two circles that share the same center — the outer circle with radius r1r_1 and the inner circle with radius r2r_2. Its area is simply the outer circle's area minus the inner circle's area:

A1=πr12A2=πr22A0=A1−A2=π(r12−r22)A_1 = \pi r_1^2 \qquad A_2 = \pi r_2^2 \qquad A_0 = A_1 - A_2 = \pi(r_1^2 - r_2^2)

The circumferences of the two boundary circles follow the same formula as any circle: C1=2πr1C_1 = 2\pi r_1 and C2=2πr2C_2 = 2\pi r_2.

Tip: Square EACH radius first, then subtract, then multiply by π\pi — π(r12−r22)\pi(r_1^2 - r_2^2), never π(r1−r2)2\pi(r_1 - r_2)^2. Those two expressions are only equal when r2=0r_2 = 0.

Solving from any two known measures

This calculator works from three different pairs of known measures, solving for the radii first and then deriving everything else:

KnownSolve for radiiThen
Outer radius r1r_1, inner radius r2r_2— (already known)A0=π(r12−r22)A_0 = \pi(r_1^2 - r_2^2), C1=2πr1C_1 = 2\pi r_1, C2=2πr2C_2 = 2\pi r_2
Outer radius r1r_1, annulus area A0A_0r2=r12−A0/πr_2 = \sqrt{r_1^2 - A_0/\pi}C1=2πr1C_1 = 2\pi r_1, C2=2πr2C_2 = 2\pi r_2
Outer circumference C1C_1, inner circumference C2C_2r1=C1/(2π)r_1 = C_1/(2\pi), r2=C2/(2π)r_2 = C_2/(2\pi)A0=π(r12−r22)A_0 = \pi(r_1^2 - r_2^2)

Worked example: r₁ = 10, r₂ = 6

A1=π(100)≈314.1593A2=π(36)≈113.0973A_1 = \pi(100) \approx 314.1593 \qquad A_2 = \pi(36) \approx 113.0973 A0=π(100−36)=64π≈201.0619A_0 = \pi(100 - 36) = 64\pi \approx 201.0619 C1=2π(10)≈62.8319C2=2π(6)≈37.6991C_1 = 2\pi(10) \approx 62.8319 \qquad C_2 = 2\pi(6) \approx 37.6991

Because both radii are exactly 10 and 6 here, the area also has an EXACT symbolic form (64π64\pi) in addition to the decimal approximation.

Worked example: outer radius r₁ = 10, area A₀ = 201.0619

Rearranging the area formula gives r2=102−201.0619/π=100−64=36=6r_2 = \sqrt{10^2 - 201.0619/\pi} = \sqrt{100 - 64} = \sqrt{36} = 6. From there the circumferences follow the usual way: C1≈62.8319C_1 \approx 62.8319, C2≈37.6991C_2 \approx 37.6991.

Tip: If r12−A0/π≤0r_1^2 - A_0/\pi \le 0, no real inner radius produces that much annulus area inside that outer circle — the calculator reports this as a degenerate input rather than a nonsense negative-square-root result.

Worked example: circumferences C₁ = 62.8319, C₂ = 37.6991

Dividing each by 2π2\pi gives r1≈10r_1 \approx 10 and r2≈6r_2 \approx 6, matching the first example — the area comes out to the same ≈201.0619\approx 201.0619.

Common mistakes

  • Squaring the difference instead of the sum of squares. π(r1−r2)2\pi(r_1 - r_2)^2 is a different, smaller formula. For r₁=10, r₂=6 it gives π(4)2=16π≈50.27\pi(4)^2 = 16\pi \approx 50.27 — less than a quarter of the correct 64π≈201.0664\pi \approx 201.06.
  • Swapping inner and outer radii. The outer radius must be the larger one; if r2>r1r_2 > r_1 the "ring" would have negative area, which isn't geometrically meaningful.
  • Forgetting the inner circle entirely. The annulus area is NOT just πr12\pi r_1^2 — that's the area of the full disk, ring and hole both included.

Where this shows up

  • Washers, gaskets, and O-rings: the annulus area determines how much material sits between the bolt hole and the outer edge.
  • Pipes and tubing: the cross-sectional wall area (outer minus inner radius) is exactly an annulus, used to estimate material volume per unit length.
  • CDs, DVDs, and running tracks: any flat surface that's a ring rather than a full disk uses this same area formula.

The annulus is a direct extension of the circle: master A=πr2A = \pi r^2 for one circle, and the ring between two concentric circles is just that formula applied twice and subtracted.

Frequently asked questions

What is an annulus?
An annulus is the ring-shaped region between two concentric circles (circles sharing the same center) — like a washer, a CD, or a donut viewed from above. Its area is the outer circle's area minus the inner circle's area.
What is the formula for the area of an annulus?
A₀ = π(r₁² − r₂²), where r₁ is the outer radius and r₂ is the inner radius. You square each radius FIRST, then subtract, then multiply by π — not the other way around.
Why isn't the area π(r₁ − r₂)²?
Squaring the difference of the radii is a completely different (and smaller) number than the difference of the squares. For r₁ = 10, r₂ = 6: the correct area is π(100 − 36) = 64π ≈ 201.06, while the wrong formula gives π(4)² = 16π ≈ 50.27 — less than half the true value.
How do I find the inner radius if I only know the outer radius and the area?
Rearrange the area formula: r₂ = √(r₁² − A₀/π). For example, an outer radius of 10 and an area of 201.0619 gives r₂ = √(100 − 64) = √36 = 6.
How do I find the radii if I only know the two circumferences?
Divide each circumference by 2π: r₁ = C₁/(2π) and r₂ = C₂/(2π). Once both radii are known, the area follows from π(r₁² − r₂²).
What happens if the inner radius is larger than the outer radius?
That is not a valid annulus — the inner circle must fit entirely inside the outer one, so r₁ must be strictly greater than r₂. This calculator flags swapped or equal radii as an error rather than silently returning a negative area.
What is the most common mistake with the annulus area formula?
Squaring the difference of the radii instead of taking the difference of their squares — writing π(r₁ − r₂)² instead of π(r₁² − r₂²). The two formulas only agree when one of the radii is zero.
Where does the annulus shape show up in real life?
Washers and gaskets, pipe cross-sections (wall thickness), CDs and DVDs, running tracks, and any flat ring-shaped surface where you need to know how much material fills the gap between an inner and outer boundary.

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