Annulus Area Calculator

Formula

A = π × (R² − r²)   where R > r.

Area of a circular annulus

A circular annulus, also called a circular ring, is whatever sits between two circles that share the same centre: an outer one of radius R and an inner one of radius r, where R has to be bigger than r. To get its area you take one disc and subtract the other, which is exactly A = π(R² - r²). Factor that difference of squares and you get A = π(R + r)(R - r). That version comes in handy with thin rings, where the thickness (R - r) is the number you actually have on hand. Take R = 5 m and r = 3 m: A = π(25 - 9) = 16π ≈ 50.27 m².

Working from the outer diameter D and inner diameter d instead? Use R = D/2 and r = d/2, and the area becomes A = π(D² - d²)/4. Add up the two circular edges and the boundary's perimeter works out to P = 2π(R + r).

Applications: washers, pipes and athletics tracks

Anything round with a hole punched through the middle takes this shape. Think of a flat washer, an O-ring, a pizza ring with the centre cut away, the lanes on a running track, or the cross-section of a hollow pipe — for that last one, A_steel = π(R² - r²) is what you use to figure out mass per metre. You also see it in the band a clock hand sweeps between two distances from the centre, or in a climate zone caught between two parallels on Earth's surface. Statisticians lean on it too, using the annulus to define bins for radial histograms in 2D distributions.

FAQ

What if R and r are equal? You're left with a circle of zero thickness, so the area comes out to zero. That's why the calculator insists on R > r.

Why is it not just π·R·thickness? That shortcut, A ≈ 2π·R̄·t (with R̄ the mean radius and t = R - r), only holds up as an approximation when the ring is very thin. For the exact answer you stick with π(R² - r²), since the outer and inner circumferences aren't the same length.

How do I compute the area of a ring sector (not the full ring)? Just scale by the fraction of the angle you're keeping. With θ in radians that's A = (θ/2)(R² - r²), or in degrees A = (θ°/360)·π(R² - r²).

And if the two circles are not concentric? Then you no longer have an annulus. As long as the smaller circle sits fully inside the larger, the area is still the difference of the disc areas, but the shape itself (a "lune", or eccentric ring) varies in width from one side to the other.