## 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  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 } } \\
\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

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