How ellipse area and perimeter are calculated
An ellipse's area follows directly from its two radii, the same way a circle's area follows from one radius. Its perimeter, though, has no closed-form solution in elementary functions — it requires either an infinite series, a numerical integral, or (as used here) a highly accurate approximation.
- a — semi-major axis (half the longer diameter)
- b — semi-minor axis (half the shorter diameter)
More detail
Why the perimeter needs an approximation
A circle's circumference is 2πr because every point is the same distance from the center, but an ellipse curves more sharply near its ends than its sides, so no simple multiplication captures its perimeter exactly — the true value is a complete elliptic integral of the second kind, which mathematicians only express as an infinite series or a numerical approximation. Ramanujan's first approximation (1914) used here keeps the relative error under 0.1% for aspect ratios up to about a = 11b, and under about 0.43% even for an extremely flattened ellipse.
Sanity check. When a = b, the ellipse is a circle: area should reduce to πr² and perimeter to 2πr. Try setting both axes equal above — Ramanujan's formula returns the exact circle perimeter in that case, with zero approximation error.
Frequently asked questions
What's the area and perimeter of an ellipse with a = 5 and b = 3?
Area = π × 5 × 3 = 15π ≈ 47.12 exactly. Perimeter ≈ π[3(8) − √(18×14)] = π(24 − √252) ≈ 25.53 via Ramanujan's approximation.
Why isn't the perimeter formula exact like the area formula?
Area only needs a single integral that resolves to πab in closed form. Perimeter requires integrating the ellipse's changing curvature around its full path, which produces an elliptic integral with no elementary-function solution — every perimeter formula for a non-circular ellipse is necessarily an approximation or an infinite series.
How accurate is Ramanujan's approximation?
For a circle (a = b) it's exact. As the ellipse becomes more elongated the relative error grows but stays small — under 0.1% for aspect ratios up to about a = 11b, and never exceeding about 0.43% even for an extremely flattened ellipse — which is why it's widely used in engineering and design instead of numerically integrating the exact elliptic integral.
What happens if I enter a = b (equal axes)?
The shape is a circle, and this calculator's formulas reduce correctly: area becomes πa² and Ramanujan's perimeter approximation returns exactly 2πa with no error, since 3(a+a) − √((4a)(4a)) = 6a − 4a = 2a.