How this instrument works
A cross-sectional area calculator for a ring finds the area of an annulus — the flat, washer-shaped band trapped between two concentric circles, the same shape as a pipe wall seen end-on or the metal of a washer around its hole. The formula A = π(R² − r²) is nothing more than one circle's area minus another's: sweep out the full outer disk, then cut away the smaller disk sitting inside it, and what remains is the ring.
Factor the right-hand side and the geometry behind the algebra shows itself: a difference of two squares splits into a difference times a sum, so the area becomes the wall's thickness multiplied by π times the sum of the two radii. That reading isn't a coincidence — π times the sum of the radii is also 2π times their average, the circumference of the circle running through the ring's own midline. This is Pappus's centroid theorem doing quiet work: a straight strip, swept once around an axis at its own average radius, traces out exactly the annulus, however thin or fat the band turns out to be.
Two limits pin the formula down. Shrink the inner radius to zero and the hole disappears — A collapses to πR², the ordinary area of a solid disk, because there is nothing left to subtract. Grow the inner radius until it meets the outer one and the wall thins to nothing — A falls to zero, an infinitely thin ring with no area at all. The mistake students make sits between these limits: squaring the difference of the radii, (R − r)², instead of taking the difference of the squares, R² − r². The two expressions agree only when r is zero and diverge everywhere else.
- Enter the ring's outer edge into Outer radius — the full radius out to a pipe's exterior wall or a washer's outer rim.
- Enter the radius of the hole into Inner radius — a pipe's bore, or the hole through a washer's centre.
- Read the result in Cross-sectional area (annulus); it updates the instant either radius changes.
- Set Inner radius to 0 to fall back to a solid disk's area, or match it to Outer radius to watch the ring shrink to zero.
- Keep Outer radius at or above Inner radius — the sheet flags it if the wall would need a negative thickness.
Worked example — a pipe wall, 5 cm outer and 3 cm bore
A steel pipe measures 5 cm to its outer wall and 3 cm to the inner bore — the numbers this sheet's own self-check uses. Feed R = 5 and r = 3 into the formula and the difference of squares gives 5² − 3² = 25 − 9 = 16, so A = 16π = 50.26548245743669 cm², call it 50.27 cm² — the actual steel a cross-section carries once the hollow core is subtracted out.
The factored form checks the same answer a second way: multiply the difference of the two radii by their sum, (5 − 3)(5 + 3) = 2 × 8 = 16, and 16π still lands on 50.26548245743669 cm². That 2 cm gap between the radii is the wall thickness; multiply this area by the pipe's length and a fabricator has the exact volume of metal to order, or use it directly in a bending-strength calculation for the tube.
Questions
What is the formula for the cross-sectional area of an annulus?
A = π(R² − r²), where R is the outer radius and r is the inner radius. It is the outer circle's area, πR², with the inner circle's area, πr², subtracted out. For R = 5 and r = 3 that gives 16π, about 50.27 square units.
Why is it R² − r² and not (R − r)²?
Because area scales with the square of a length, not the length itself — each circle's own area is π times its radius squared, so subtracting the circles means subtracting the squares. (R − r)² equals R² − r² only when r = 0; for any real ring the two expressions differ, and using the wrong one understates the area.
How does this connect to Pappus's centroid theorem?
Factor the formula as A = (R − r) × π(R + r): a straight strip of width R − r, revolved once around an axis at the ring's average radius (R + r)/2, sweeps out exactly this area. That is Pappus's centroid theorem for a line segment — length times the distance its centroid travels equals the swept area.
What does the calculator return if the inner radius is zero?
The plain area of a solid disk, πR², since there is no hole left to subtract. With R = 5 that is π × 25 ≈ 78.5398 square units — the formula's degenerate case rather than a special rule.
What does it mean if the inner and outer radii are equal?
The wall thickness is zero, so the area is zero too — R = 5 and r = 5 both give A = 0. It is the opposite limit from a solid disk: instead of no hole at all, there is no material left, just an infinitely thin ring.
Can this formula be used for a washer or gasket instead of a pipe?
Yes — a washer, gasket, or any flat ring is geometrically identical to a pipe's cross-section, so the same A = π(R² − r²) applies directly. Keep the outer and inner radii in one consistent unit and the area comes out in that unit squared.