Perimeter of Ellipse

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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 $\mathcal P$ of $K$ is given by:

$\displaystyle \mathcal P = 4 a \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \theta} \rd \theta$

where:

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


The definite integral:

$\displaystyle \mathcal P = \int_0^{\pi / 2} \sqrt{1 - k^2 \sin^2 \theta} \rd \theta$

is the complete elliptic integral of the second kind.


Proof

Let $K$ be aligned in a cartesian coordinate 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:

\(\displaystyle \frac {\d x} {\d \theta}\) \(=\) \(\displaystyle -a \sin \theta\) Derivative of Cosine Function
\(\displaystyle \frac {\d y} {\d \theta}\) \(=\) \(\displaystyle b \cos \theta\) Derivative of Sine Function


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

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


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

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


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

\(\displaystyle \int_0^{\pi / 2} \map f {\cos \theta} \rd \theta\) \(=\) \(\displaystyle -\int_{\pi / 2}^0 \map f {\sin t} \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle \int_0^{\pi / 2} \map f {\sin t} \rd t\)


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

$\displaystyle \mathcal P = 4 a \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \theta} \rd \theta$

$\blacksquare$


Sources