# Perimeter of Ellipse

## Theorem

Let $K$ be an ellipse whose major axis is of length $2 a$ and whose minor axis is of length $2 b$.

The perimeter $\PP$ of $K$ is given by:

$\PP = 4 a \map E e$

where:

$\ds \map E e = \int_0^{\pi / 2} \sqrt{1 - e^2 \sin^2 \theta} \rd \theta$ is the complete elliptic integral of the second kind
$e = \dfrac {\sqrt {a^2 - b^2} } a$ is the eccentricity of $K$.

## Proof

Let $K$ be aligned in a cartesian plane such that:

the major axis of $K$ is aligned with the $x$-axis
the minor axis of $K$ is aligned with the $y$-axis.
$x = a \cos \theta, y = b \sin \theta$

Thus:

 $\ds \frac {\d x} {\d \theta}$ $=$ $\ds -a \sin \theta$ Derivative of Cosine Function $\ds \frac {\d y} {\d \theta}$ $=$ $\ds b \cos \theta$ Derivative of Sine Function

From Arc Length for Parametric Equations, the length of one quarter of the perimeter of $K$ is given by:

 $\ds \frac {\PP} 4$ $=$ $\ds \int_0^{\pi / 2} \sqrt {\paren {-a \sin \theta}^2 + \paren {b \cos \theta}^2} \rd \theta$ $\ds$ $=$ $\ds \int_0^{\pi / 2} \sqrt {a^2 \paren {1 - \cos^2 \theta} + b^2 \cos^2 \theta} \rd \theta$ Sum of Squares of Sine and Cosine $\ds$ $=$ $\ds \int_0^{\pi / 2} \sqrt {a^2 - \paren {a^2 - b^2} \cos^2 \theta} \rd \theta$ simplifying $\ds$ $=$ $\ds a \int_0^{\pi / 2} \sqrt {1 - \paren {1 - \frac {b^2} {a^2} } \cos^2 \theta} \rd \theta$ extracting $a$ as a factor $\text {(1)}: \quad$ $\ds$ $=$ $\ds a \int_0^{\pi / 2} \sqrt {1 - k^2 \cos^2 \theta} \rd \theta$ setting $k^2 = 1 - \dfrac {b^2} {a^2} = \dfrac {a^2 - b^2} {a^2}$

Since $\cos \theta = \map \sin {\dfrac \pi 2 - \theta}$ we can write for any real function $\map f x$:

$\ds \int_0^{\pi / 2} \map f {\cos \theta} \rd \theta = \int_0^{\pi / 2} \map f {\map \sin {\frac \pi 2 - \theta} } \rd \theta$

So substituting $t = \dfrac \pi 2 - \theta$ this can be converted to:

 $\ds \int_0^{\pi / 2} \map f {\cos \theta} \rd \theta$ $=$ $\ds -\int_{\pi / 2}^0 \map f {\sin t} \rd t$ $\ds$ $=$ $\ds \int_0^{\pi / 2} \map f {\sin t} \rd t$

justifying the fact that $\cos$ can be replaced with $\sin$ in $(1)$ above, giving:

$\ds \PP = 4 a \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \theta} \rd \theta$

$\blacksquare$