Ellipse Area

An ellipse is a circle that has been stretched by different amounts along two perpendicular directions, and that fact is the derivation. Stretching any figure by a factor $k$ in one direction multiplies its area by $k$ — every thin strip keeps its width and gains length in proportion. So start with a circle of radius $b$, area $\pi b^2$, and stretch it along one axis by $a/b$. The area becomes $\pi b^2 \times (a/b) = \pi ab$. Where the circle has $r \times r$, the ellipse has $a \times b$, and setting $a = b = r$ returns the circle exactly, as any honest generalisation must.

A worked instance: oval duct sized 600 mm by 300 mm has semi-axes of 0.30 m and 0.15 m, so its cross-section is $\pi(0.30)(0.15) = 0.141$ m² — against 0.180 m² for a rectangular duct of the same outside dimensions, which is the trade you make for a shape that fits between joists and resists collapse. Backwards, $b = A/(\pi a)$: 0.141 m² with a fixed 0.30 m semi-major axis needs 0.15 m the other way.

The ellipse is also where a famous asymmetry lives. Its area is this effortless product, but its perimeter has no elementary closed form at all — the exact answer is a complete elliptic integral of the second kind, which cannot be written with ordinary functions, and every practical formula for it is an approximation. Kepler's second law is the other reason this area matters: a planet sweeps equal areas in equal times, so the total $\pi ab$ divided by the orbital period gives the constant rate at which the radius vector sweeps.

One mistake dominates, and it is the diameter-for-radius trap with two chances to happen. $a$ and $b$ are semi-axes — half the full width, measured from the centre out — not the overall dimensions. Entering 0.60 and 0.30 for that duct returns 0.565 m², four times the truth, and the number is plausible enough to survive a glance. Halve both before you start. The same sanity check as for the circle applies: an ellipse fills $\pi/4$, a bit under 80%, of the rectangle that boxes it, so 600 by 300 must land a little below 0.18 m². Two lesser notes: the product is symmetric, so it makes no difference which semi-axis you call $a$, whatever the convention says about the major one being larger; and do not estimate the perimeter as $\pi(a+b)$, which is only the crudest first approximation and runs low by up to 10% on an elongated ellipse.