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

where:
 * $k = \dfrac {\sqrt {a^2 - b^2} } a$

The definite integral:
 * $\displaystyle \PP = \int_0^{\pi / 2} \sqrt{1 - k^2 \sin^2 \theta} \rd \theta$

is the complete elliptic integral of the second kind.

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.

Then from Equation of Ellipse in Reduced Form: parametric form:
 * $x = a \cos \theta, y = b \sin \theta$

Thus:

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

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:

justifying the fact that $\cos$ can be replaced with $\sin$ in $(1)$ above, giving:
 * $\ds \PP = 4 a \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \theta} \rd \theta$