[Click here for a PDF of this post with nicer formatting]

## Motivation

Lance told me they’ve been covering the circumference of a circle in school this week. This made me think of the generalization of a circle, the ellipse, but I couldn’t recall what the circumference of an ellipse was. Sofia guessed \( \pi ( a + b ) \). Her reasoning was that this goes to \( 2 \pi r \) when the ellipse is circular, just like the area of an ellipse \( \pi a b \), goes to \( \pi a^2 \) in the circular limit. That seemed reasonable to me, but also strange since I didn’t recall any \( \pi ( a + b ) \) formula.

It turns out that there’s no closed form expression for the circumference of an ellipse, unless you count infinite series or special functions. Here I’ll calculate one expression for this circumference.

## Geometry recap

There’s two ways that I think of ellipses. One is the shape that you get when you put a couple tacks in a paper, and use a string and pencil to trace it out, as sketched in fig. 1.

The other is the basic vector parameterization of that same path

\begin{equation}\label{eqn:elipticCircumference:20}

\Br = ( a \cos\theta, b \sin\theta ).

\end{equation}

It’s been a long time since grade 11 when I would have taken it for granted that these two representations are identical. To do so, we’d have to know where the foci of the ellipse sit. Cheating a bit I find in [1] that the foci are located at

\begin{equation}\label{eqn:elipticCircumference:40}

\Bf_{\pm} = \pm \sqrt{ a^2 – b^2 } (1, 0).

\end{equation}

This and the equivalence of the pencil and tack representation of the ellipse can be verified by checking that the “length of the string” equals \( 2 a \) as expected.

That string length is

\begin{equation}\label{eqn:elipticCircumference:60}

\begin{aligned}

\Abs{ \Br – \Bf_{+} } + \Abs{ \Br – \Bf_{-} }

&=

\sqrt{ (a \cos\theta – f)^2 + b^2 \sin^2\theta }

+

\sqrt{ (a \cos\theta + f)^2 + b^2 \sin^2\theta } \\

&=

\sqrt{ a^2 \cos^2 \theta + f^2 – 2 a f \cos\theta + b^2 \lr{ 1 – \cos^2\theta } } \\

&\quad +

\sqrt{ a^2 \cos^2 \theta + f^2 + 2 a f \cos\theta + b^2 \lr{ 1 – \cos^2\theta } }.

\end{aligned}

\end{equation}

These square roots simplify nicely

\begin{equation}\label{eqn:elipticCircumference:80}

\begin{aligned}

\sqrt{ a^2 \cos^2 \theta + f^2 \pm 2 a f \cos\theta + b^2 \lr{ 1 – \cos^2\theta } }

&=

\sqrt{ (a^2 – b^2) \cos^2 \theta + a^2 – b^2 \pm 2 a f \cos\theta + b^2 } \\

&=

\sqrt{ f^2 \cos^2 \theta + a^2 \pm 2 a f \cos\theta } \\

&=

\sqrt{ (a \pm f \cos\theta)^2 } \\

&=

a \pm f \cos\theta.

\end{aligned}

\end{equation}

So the total length from one focus to a point on the ellipse, back to the other focus, is

\begin{equation}\label{eqn:elipticCircumference:100}

\Abs{ \Br – \Bf_{+} } + \Abs{ \Br – \Bf_{-} }

=

a + f \cos\theta + a – f \cos\theta = 2 a,

\end{equation}

as expected. That verifies that the trigonometric parameterization matches with the pencil and tacks representation of an ellipse (provided the foci are placed at the points \ref{eqn:elipticCircumference:40}).

## Calculating the circumference

The circumference expression can almost be written by inspection. An element of the tangent vector along the curve is

\begin{equation}\label{eqn:elipticCircumference:340}

\frac{d\Br}{d\theta} = ( -a \sin\theta, b \cos\theta ),

\end{equation}

so the circumference is just a one liner

\begin{equation}\label{eqn:elipticCircumference:280}

C = 4 \int_0^{\pi/2} \sqrt{ a^2 \sin^2\theta + b^2 \cos^2 \theta} d\theta.

\end{equation}

The problem is that this one liner isn’t easy to evaluate. The square root can be put in a slightly simpler form in terms of the eccentricity, which is defined by

\begin{equation}\label{eqn:elipticCircumference:300}

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

\end{equation}

Factoring out \( a \) and writing the sine as a cosine gives

\begin{equation}\label{eqn:elipticCircumference:320}

\begin{aligned}

C

&=

4 a \int_0^{\pi/2} \sqrt{ 1 – \cos^2\theta + \frac{b^2}{a^2} \cos^2 \theta} d\theta \\

&=

4 a \int_0^{\pi/2} \sqrt{ 1 + \lr{ \frac{b^2}{a^2} -1} \cos^2 \theta} d\theta \\

&=

4 a \int_0^{\pi/2} \sqrt{ 1 – e^2 \cos^2 \theta} d\theta.

\end{aligned}

\end{equation}

For the square root, it’s not hard to show that the fractional binomial expansion is

\begin{equation}\label{eqn:elipticCircumference:360}

\sqrt{1 + a}

=

1 – \sum_{k=1}^\infty \frac{(-a)^k}{2k – 1}\frac{ (2k – 1)!!}{(2k)!!},

\end{equation}

so the circumference is

\begin{equation}\label{eqn:elipticCircumference:380}

C

=

4 a \int_0^{\pi/2} d\theta

\lr{ 1 – \sum_{k=1}^\infty \frac{(e \cos\theta)^{2k}}{2k – 1}\frac{ (2k – 1)!!}{(2k)!!}}.

\end{equation}

Using \ref{eqn:elipticCircumference:260}, this is

\begin{equation}\label{eqn:elipticCircumference:400}

C

=

2 \pi a

–

4 a

\sum_{k=1}^\infty \frac{e^{2k}}{2k – 1}\frac{ (2k – 1)!!}{(2k)!!}

\frac{(2k – 1)!!}{(2k)!!} \frac{\pi}{2}

\end{equation}

\begin{equation}\label{eqn:elipticCircumference:420}

\boxed{

C

=

2 \pi a \lr{ 1 –

\sum_{k=1}^\infty \frac{e^{2k}}{2k – 1} \lr{ \frac{ (2k – 1)!!}{(2k)!!} }^2

}.

}

\end{equation}

Observe that this does reduce to \( 2 \pi r \) for the circle (where \( e = 0 \)), and certainly isn’t as nice as \( \pi (a + b) \).

## Appendix. Integral of even cosine powers.

The integral

\begin{equation}\label{eqn:elipticCircumference:120}

\int_0^{\pi/2} \cos^{2k} \theta d\theta

\end{equation}

can be evaluated using integration by parts.

\begin{equation}\label{eqn:elipticCircumference:140}

\begin{aligned}

\int_0^{\pi/2} \cos^{2k} \theta d\theta

&=

\int_0^{\pi/2} \cos^{2k-1} \theta \frac{d \sin\theta}{d\theta} d\theta \\

&=

\evalrange{

\cos^{2k-1} \theta \sin\theta

}{0}{\pi/2}

–

(2 k -1)

\int_0^{\pi/2} \cos^{2k-2} \theta (-\sin\theta) \sin\theta d\theta \\

&=

(2 k -1)

\int_0^{\pi/2} \cos^{2k-2} \theta (1 – \cos^2\theta) d\theta \\

&=

(2 k -1)

\int_0^{\pi/2} \cos^{2k-2} \theta d\theta

–

(2 k -1)

\int_0^{\pi/2} \cos^{2k} \theta d\theta.

\end{aligned}

\end{equation}

Bringing the \( 2k \) power integral to the other side and solving for the original integral gives a recurrence relation

\begin{equation}\label{eqn:elipticCircumference:160}

\begin{aligned}

\int_0^{\pi/2} \cos^{2k} \theta d\theta

&= \frac{2 k – 1}{2 k}

\int_0^{\pi/2} \cos^{2k -2} \theta d\theta \\

&=

\frac{2 k – 1}{2 k}

\frac{2 k – 3}{2 k – 2}

\int_0^{\pi/2} \cos^{2k -4} \theta d\theta \\

&=

\frac{2 k – 1}{2 k}

\frac{2 k – 3}{2 k – 2}

\cdots

\frac{3}{4}

\int_0^{\pi/2} \cos^{2} \theta d\theta.

\end{aligned}

\end{equation}

This last can also be solved using integration by parts

\begin{equation}\label{eqn:elipticCircumference:180}

\begin{aligned}

\int_0^{\pi/2} \cos^{2} \theta d\theta

&=

\int_0^{\pi/2} \cos \theta \frac{d \sin\theta}{d\theta} d\theta \\

&=

–

\int_0^{\pi/2} (-\sin \theta) \sin\theta d\theta \\

&=

\int_0^{\pi/2} \lr{ 1 – \cos^2 \theta } d\theta,

\end{aligned}

\end{equation}

or

\begin{equation}\label{eqn:elipticCircumference:200}

\begin{aligned}

\Abs{ \Br – \Bf_{+} } + \Abs{ \Br – \Bf_{-} }

&=

\int_0^{\pi/2} \cos^{2} \theta d\theta \\

&=

\inv{2} \frac{\pi}{2}.

\end{aligned}

\end{equation}

This gives

\begin{equation}\label{eqn:elipticCircumference:220}

\int_0^{\pi/2} \cos^{2k} \theta d\theta

=

\frac{2 k – 1}{2 k}

\frac{2 k – 3}{2 k – 2}

\cdots

\frac{3}{4}

\frac{1}{2}

\frac{\pi}{2}.

\end{equation}

Using the double factorial notation (factorial that skips every other value), this is

\begin{equation}\label{eqn:elipticCircumference:260}

\boxed{

\int_0^{\pi/2} \cos^{2k} \theta d\theta

=

\frac{(2k – 1)!!}{(2k)!!} \frac{\pi}{2}

}

\end{equation}

# References

[1] Wikipedia. Ellipse — wikipedia, the free encyclopedia, 2015. URL https://en.wikipedia.org/w/index.php?title=Ellipse&oldid=650116160. [Online; accessed 9-March-2015].