How this instrument works
An ellipse's area is A = πab, where a and b are the two semi-axis lengths — half the long diameter and half the short one. The formula falls out of a single idea: stretch a unit circle by a factor of a along one axis and by b along the perpendicular axis, and the result is exactly this ellipse. A linear stretch multiplies area by the product of its two scale factors, so the circle's area π becomes π · a · b — no calculus required, just how areas behave under scaling.
That exactness is unusual for an ellipse. Nearly everything else about the shape resists a clean formula — its perimeter has no elementary closed form at all and is normally estimated with Ramanujan's approximation — yet the area comes out as a plain product of the two semi-axes times π, with zero approximation anywhere in the result. A circle is simply the special case a = b = r, where the formula collapses to the familiar πr².
The most common slip is feeding in the full axis lengths instead of the half-lengths: doubling both a and b quadruples the reported area, since (2a)(2b) = 4ab. At the other extreme, letting either semi-axis shrink toward zero flattens the ellipse into a line segment, and the area falls to zero right along with it — a sanity check worth remembering whenever one input is much smaller than the other.
- Enter the longer half-axis length into the Semi-major axis, a field — half the ellipse's full length, not the whole span.
- Enter the shorter half-axis length into the Semi-minor axis, b field, using the same unit as a.
- Read the enclosed surface off the Area field — the sheet multiplies π by both semi-axes as soon as you finish typing.
- Swap the two entries to sanity-check: A = πab does not care which axis is called major or minor, so the area stays identical either way.
Worked example — semi-axes of 5 and 3
Take an ellipse with a semi-major axis of a = 5 and a semi-minor axis of b = 3 — a flattened oval rather than a circle, since the two lengths differ. The area is A = π × 5 × 3 = 15π, which this sheet reports as 47.12388980384689, matching fifteen times π carried to full double precision.
Compare that to circles built from either semi-axis alone: a circle of radius 5 encloses π × 25 ≈ 78.540, and a circle of radius 3 encloses π × 9 ≈ 28.274. The ellipse's 47.124 sits between those two figures, as it must — ab is always the geometric mean territory between a² and b² whenever a and b differ, so every non-circular ellipse's area lands strictly between its two circular extremes.
Questions
What is the formula for the area of an ellipse?
A = πab, where a and b are the semi-major and semi-minor axes — half the ellipse's longest and shortest diameters. It is the direct generalization of a circle's πr²: set a = b = r and the two formulas agree exactly, since a circle is just an ellipse whose two axes happen to match.
Why does the ellipse area formula use π even though only a circle involves π directly?
Because an ellipse is a circle viewed through a stretch. Scaling a unit circle by a in one direction and by b in the perpendicular direction turns its area, π, into π·a·b — the stretch multiplies area by the product of the two scale factors, and π survives untouched since it came along from the original circle.
Do I enter the full axis lengths or the semi-axes?
Semi-axes — half the length across the ellipse in each direction, the same way a circle's formula wants a radius rather than a diameter. Entering the full major and minor axis lengths instead of their halves inflates the reported area by a factor of four, since both the a term and the b term would be doubled.
Why doesn't the ellipse's perimeter have an equally simple formula?
Area behaves simply under stretching because it scales by a fixed factor regardless of direction, but a boundary's length does not — stretching an ellipse changes how sharply it curves at different points, so no single algebraic expression captures the perimeter exactly. It is instead written as an elliptic integral and normally estimated, for instance with Ramanujan's approximation.
What happens to the area as one semi-axis shrinks toward zero?
The ellipse flattens into a line segment and the area shrinks toward zero right along with it, since A = πab is directly proportional to each semi-axis on its own. Halve b while holding a fixed and the area exactly halves too — there is no floor or asymptote to worry about.
Is the area formula exact, or an approximation like some ellipse formulas are?
Exact. Unlike the ellipse's perimeter, which has no elementary closed form, A = πab involves no series, no iteration, and no truncation — the only rounding in the answer comes from how many digits of π and the inputs you choose to carry, not from the formula itself.