Ellipse Circumference and Elliptic Integrals

Why an Ellipse Has a Special Circumference Formula

An ellipse resembles a circle that has been stretched unequally in two perpendicular directions. For a circle, every point is the same distance from the center, and its circumference has the familiar formula $2\pi r$. An ellipse has two different semiaxes, so its arc-length integrand varies around the curve. The circumference is still an exact integral, but in general that integral does not simplify to elementary functions such as powers, trigonometric functions, exponentials, or logarithms.

This tutorial derives the circumference integral, uses the ellipse’s symmetry to reduce it to one quarter of the curve, and names the special function that represents the result. The integral itself is also a useful starting point when the circumference must be evaluated numerically.

Convention: Let $a$ be the longer semiaxis and $b$ the shorter semiaxis, with $a\ge b>0$. The circumference $C$ has units of length. We use $m$ for the parameter of the elliptic integral; some references instead use the modulus $k=\sqrt m$.

Set Up the Arc-Length Integral

An ellipse centered at the origin, with semiaxis $a$ along the horizontal axis and semiaxis $b$ along the vertical axis, has the parametrization

x(t)=a\cos t,\qquad y(t)=b\sin t,\qquad 0\le t\le 2\pi.

As $t$ runs from $0$ to $2\pi$, this parametrization traces the ellipse once. The parametric arc-length formula says that a smooth curve $(x(t),y(t))$ has length $\int \sqrt{(x'(t))^2+(y'(t))^2}\,dt$ over its parameter interval. Here,

x'(t)=-a\sin t,\qquad y'(t)=b\cos t.

Therefore the speed is

\sqrt{(x'(t))^2+(y'(t))^2} = \sqrt{a^2\sin^2 t+b^2\cos^2 t},

and the full circumference is

C=\int_0^{2\pi}\sqrt{a^2\sin^2 t+b^2\cos^2 t}\,dt.

The expression under the square root changes as $t$ changes. This variation is the central reason the circumference does not usually reduce to the circle formula.

Why One Quarter Is Enough

The ellipse is symmetric across both coordinate axes. The speed in the integral depends on $\sin^2 t$ and $\cos^2 t$, so reflecting an arc into any of the other three quadrants gives an arc with the same length. Consequently, the four quadrant lengths are equal, and the integral over the full parameter interval is four times the integral over $0\le t\le \pi/2$:

C = 4\int_0^{\pi/2} \sqrt{a^2\sin^2 t+b^2\cos^2 t}\,dt.

This is the quarter-ellipse form of the circumference integral. It is not an approximation: it follows exactly from the symmetry and the fact that the parametrization covers each quadrant once.

Key takeaway: The circumference of an ellipse is four times the length of its first-quadrant arc. The quarter-ellipse integral still has to be evaluated; symmetry reduces the interval but does not, by itself, produce an elementary antiderivative.

Eccentricity and the Elliptic Integral

To put the quarter-ellipse integral into standard form, factor out $a^2$ inside the square root. Define the eccentricity $e$ of this ellipse by

e=\sqrt{1-\frac{b^2}{a^2}}.

Since $a\ge b>0$, this gives $0\le e<1$. Using $b^2=a^2(1-e^2)$, the integrand becomes

\begin{aligned} \sqrt{a^2\sin^2 t+b^2\cos^2 t} &=\sqrt{a^2\sin^2 t+a^2(1-e^2)\cos^2 t}\\ &=a\sqrt{\sin^2 t+(1-e^2)\cos^2 t}\\ &=a\sqrt{1-e^2\cos^2 t}. \end{aligned}

Substituting $\theta=\pi/2-t$ changes $\cos^2 t$ to $\sin^2\theta$, while leaving the integration interval from $0$ to $\pi/2$. This gives

C=4a\int_0^{\pi/2}\sqrt{1-e^2\sin^2\theta}\,d\theta.

The integral on the right is a standard special function. It is called the complete elliptic integral of the second kind. “Complete” means the upper limit is $\pi/2$; “second kind” distinguishes this function from other elliptic integrals.

Definition: For $0\le m<1$, the complete elliptic integral of the second kind, with parameter $m$, is $$ E(m)=\int_0^{\pi/2}\sqrt{1-m\sin^2\theta}\,d\theta. $$

Comparing this definition with the circumference integral gives the standard result:

Theorem: An ellipse with semiaxes $a\ge b>0$, eccentricity $e=\sqrt{1-b^2/a^2}$, and parameter $m=e^2$ has circumference $$ C=4aE(e^2)=4aE\left(1-\frac{b^2}{a^2}\right). $$

Derivation of the Circumference Formula

The formula follows directly from parametric arc length, symmetry, and a change of variable. The steps also show exactly where the elliptic integral enters.

Differentiate the parametrization.
For $x=a\cos t$ and $y=b\sin t$, the derivatives are $x'=-a\sin t$ and $y'=b\cos t$. Thus the speed is $\sqrt{a^2\sin^2t+b^2\cos^2t}$.

Integrate around the ellipse.
Because $0\le t\le 2\pi$ traces the curve once, its length is the integral of that speed over the full interval.

Use the four equal quadrants.
The squared trigonometric functions repeat their values under reflection into each quadrant. Hence the full integral is four times the integral from $0$ to $\pi/2$.

Put the integral in standard form.
Factoring out $a$, using $b^2=a^2(1-e^2)$, and replacing $t$ by $\pi/2-\theta$ yields $4a\int_0^{\pi/2}\sqrt{1-e^2\sin^2\theta}\,d\theta=4aE(e^2)$.

This derivation justifies the formula for all $a\ge b>0$. It does not require a separate assumption that the ellipse is a graph over the entire horizontal or vertical axis; the parametrization covers the complete closed curve.

Worked Examples

Worked Example: A Circle as a Special Ellipse

Find the circumference when both semiaxes are $3$ units.

Here $a=b=3$, so the eccentricity and elliptic-integral parameter are

e=\sqrt{1-\frac{3^2}{3^2}}=\sqrt{1-1}=0,\qquad m=e^2=0.

From the definition,

E(0)=\int_0^{\pi/2}\sqrt{1-0\sin^2\theta}\,d\theta =\int_0^{\pi/2}1\,d\theta =\frac{\pi}{2}.

Therefore,

C=4aE(0)=4(3)\left(\frac{\pi}{2}\right)=6\pi\text{ units}.

As a check, a circle of radius $3$ has circumference $2\pi(3)=6\pi$ units, matching the result.

Worked Example: An Ellipse with Semiaxes 5 and 3

Set up the circumference and express it using a complete elliptic integral. Then give its numerical value to eight decimal places.

The longer semiaxis is $a=5$ units and the shorter is $b=3$ units. Thus

e=\sqrt{1-\frac{3^2}{5^2}} =\sqrt{1-\frac{9}{25}} =\sqrt{\frac{16}{25}} =\frac45, \qquad m=e^2=\frac{16}{25}=0.64.

The exact special-function expression is

C=4(5)E(0.64)=20E(0.64)\text{ units}.

Numerical evaluation of the defining complete elliptic integral gives $E(0.64)\approx1.27634994$. Therefore

C\approx20(1.27634994)=25.52699880\text{ units}.

Using the unrounded value $E(0.64)\approx1.276349943169906$ gives $C\approx25.52699886339812$, so the circumference to eight decimal places is $25.52699886$ units. As a check, the quarter-ellipse integrand is between $b=3$ and $a=5$, so $2\pi b\le C\le2\pi a$. Numerically, $18.84956\le25.52699886\le31.41593$, as required.

Worked Example: Setting Up a More Elongated Ellipse

Find an exact elliptic-integral expression and an approximate circumference for semiaxes $4$ units and $2$ units.

First calculate the eccentricity and parameter:

e=\sqrt{1-\frac{2^2}{4^2}} =\sqrt{1-\frac{4}{16}} =\sqrt{\frac34} =\frac{\sqrt3}{2}, \qquad m=e^2=\frac34.

Then the exact expression is

C=4(4)E\left(\frac34\right)=16E\left(\frac34\right)\text{ units}.

Numerical evaluation gives $E(3/4)\approx1.21105603$, so

C\approx16(1.21105603)=19.37689648\text{ units}.

Using $E(3/4)\approx1.211056027568459$, the circumference is approximately $19.37689644$ units to eight decimal places. The bound $2\pi b\le C\le2\pi a$ becomes $4\pi\le C\le8\pi$, or approximately $12.56637\le19.37689644\le25.13274$. The computed value lies within these bounds.

Why the General Answer Is Not Elementary

For a circle, $e=0$, and the integrand defining $E(0)$ is simply $1$. For a noncircular ellipse, $e>0$, and the square root $\sqrt{1-e^2\sin^2\theta}$ varies through the interval of integration. Its antiderivative does not generally have an expression in elementary functions. The elliptic integral gives a precise standard name to this definite integral, rather than implying that the circumference is unknown or cannot be computed.

For an exact symbolic answer, $4aE(e^2)$ is the natural form. For a decimal answer, the value of $E(e^2)$ can be evaluated numerically. The later study of numerical integration develops methods for approximating such definite integrals; here, the main point is to identify the correct integral and its exact special-function representation.

A useful check: On the quarter interval, the speed $\sqrt{a^2\sin^2t+b^2\cos^2t}$ lies between $b$ and $a$. Integrating these bounds over an interval of length $\pi/2$, then multiplying by four, gives $$ 2\pi b\le C\le2\pi a. $$ For a circle, both bounds are equal. For a noncircular ellipse, the circumference is strictly between them.

Common Mistake: Treating an Ellipse as a Stretched Circle Formula

A tempting but incorrect approach is to use $2\pi a$ as the ellipse’s circumference simply because $a$ is the larger semiaxis. For the ellipse with semiaxes $5$ and $3$, this gives

2\pi a=2\pi(5)=10\pi\approx31.41593\text{ units}.

This is not the circumference: it is the upper bound $2\pi a$. The error is treating the maximum semiaxis as though it were the radius at every point. The speed is actually $\sqrt{25\sin^2t+9\cos^2t}$, which varies with $t$. The correct setup is

C=4\int_0^{\pi/2}\sqrt{25\sin^2t+9\cos^2t}\,dt =20E(0.64)\approx25.52699886\text{ units}.

The two answers differ because the ellipse is not a circle. The circle formula applies exactly only when $a=b$.

Check Your Understanding

Use the definitions and circumference formula to answer these questions.

For an ellipse with $a=6$ and $b=4$, calculate the eccentricity $e$ and the parameter $m=e^2$.
Why does the integral from $0$ to $\pi/2$ need to be multiplied by four to obtain the full circumference?
Write the circumference of an ellipse with semiaxes $a\ge b>0$ in terms of the complete elliptic integral of the second kind.
What does the elliptic integral reduce to when $a=b$, and what familiar circumference formula follows?
For semiaxes $a=7$ and $b=5$, state a lower and an upper bound for the circumference using $2\pi b\le C\le2\pi a$.