📂Geometry

Perimeter of an Ellipse

Most materials do not provide a detailed explanation of how the elliptic integral of the second kind is derived. Even if they do, many are incorrect¹ so I wrote the ‘accurate’ and ‘detailed’ content myself. For reference, the content in Boas’ Mathematical Methods in the Physical Sciences, 3rd edition, is also incorrect.

Formula

The perimeter of an ellipse with semi-major axis $a$, semi-minor axis $b$, and eccentricity $k^{2}$ is calculated as follows.

$$ E=4a\int _{0} ^{{\textstyle \frac{\pi}{2}}} \sqrt{ 1-k^{2}\cos^{2} \theta } d\theta,\quad k^{2}=\frac{a^{2}-b^{2}}{a^{2}} $$

Derivation

First, let’s express a point $P=(x,y)$ on the ellipse through an angle. However, this angle is not the angle formed by $P$, origin $O$, and axis $x$. It’s the angle formed by the origin $O$ and the $x$ axis when a perpendicular (horizontal) line drawn from $P$ intersects the circle $x^{2}+y^{2}=a^{2}(x^{2}+y^{2}=b^{2})$. Refer to the diagram below. Ellipse ${\textstyle \frac{x^{2}}{a^{2}}}+\frac{y^{2}}{b^{2}}=1 (a>b>0)$ and two circles $x^{2}+y^{2}=a^{2}$, $x^{2}+y^{2}=b^{2}$ are drawn.

From the diagram above, it is clear that $x=a\cos \theta$. Substituting this into the equation of the ellipse gives the following.

$$ \begin{align*} &&\frac{a^{2}\cos^{2}\theta}{a^{2}}+\frac{y^{2}}{b^{2}}&=1 \\ \implies &&\cos ^{2}\theta+\frac{y^{2}}{b^{2}}&=1 \end{align*} $$

Therefore, $y=b\sin \theta$. Then, the perimeter of the ellipse $E$ can be calculated as follows.

$$ \begin{align*} E &= \int _{0} ^{2\pi} \sqrt{ \left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2}}d\theta \\ &= \int _{0} ^{2\pi} \sqrt{ \left( -a\sin \theta \right)^{2} + \left( b\cos \theta \right)^{2}}d\theta \\ &=\int _{0} ^{2\pi} \sqrt{ a^{2}\sin^{2}\theta+ b^{2}\cos^{2} \theta } d\theta \\ &=\int _{0} ^{2\pi} \sqrt{ a^{2}-a^{2}\cos^{2}\theta+ b^{2}\cos^{2} \theta } \\ &=\int _{0} ^{2\pi} a\sqrt{ 1-\frac{a^{2}-b^{2}}{a^{2}}\cos^{2} \theta } d\theta \end{align*} $$

Here, $\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}(b<a)$ represents the eccentricity of the ellipse, denoted as $\epsilon$ or $k$. By calculating the perimeter of the first quadrant and multiplying by 4, the final perimeter of the ellipse is as follows.

$$ E=4a\int _{0} ^{{\textstyle \frac{\pi}{2}}} \sqrt{ 1-k^{2}\cos^{2} \theta } d\theta,\quad k^{2}=\frac{a^{2}-b^{2}}{a^{2}} $$

If the ellipse is longer towards the $y$ axis, such that the condition is $b>a>0$, the diagram is as below.

Then, the perimeter of the ellipse is as follows.

$$ E=4b\int _{0} ^{{\textstyle \frac{\pi}{2}}} \sqrt{ 1-k^{2}\sin^{2} \theta } d\theta ,\quad k^{2}=\frac{b^{2}-a^{2} }{b^{2}} $$

Elliptic Integral of the Second Kind

This integral is specifically called the Elliptic Integral of the Second Kind, or the Complete Elliptic Integral of the Second Kind, and is denoted as follows.

$$ E(k) = \int_{0}^{{\textstyle \frac{\pi}{2}}}\sqrt{ 1-k^{2}\sin^{2} \theta } d\theta $$

Calculation

This elliptic integral cannot be represented by elementary functions and can only be calculated numerically. The value of the integral depending on $k$ is as follows.

If $b=1.1547$ and $a=1$, then $k=0.5$. Since $E(0.5)=1.351$, the perimeter of the ellipse $ x^{2}+\frac{ y^{2}}{1.1547^{2}}=1$ is as follows.

$$ 4bE(k)=4\times 1.1547 \times 1.351=6.239 $$