1

I have the equation of an ellipse given in Cartesian coordinates as $\left(\frac{x}{0.6}\right)^2+\left(\frac{y}{3}\right)^2=1$ .

I need the equation for its arc length in terms of $\theta$, where $\theta=0$ corresponds to the point on the ellipse intersecting the positive x-axis, and so on.

So converting to polar coordinates with the substitutions \begin{Bmatrix} x=r\cos\theta\\y=r\sin\theta \end{Bmatrix} gives $r(\theta)=\frac{1.8}{\sqrt{9\cos^2\theta+0.36\sin^2\theta}}$ , as is illustrated here on Desmos: https://www.desmos.com/calculator/h27qsdnotm.

Then substituting into the arc length formula for polar equations: $L=\int\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} \ d\theta$ gives the rather ugly integral $L=\int\sqrt{\frac{1.8^2}{9\cos^2\theta+0.36\sin^2\theta}+\frac{1944^2\sin^2\theta\cos^2\theta}{(216\cos^2\theta+9)^3}}$ . The online https://www.integral-calculator.com/ was unable to determine an antiderivative, so my question is whether or not the above expression can be expressed in terms of elementary functions, or if there are other methods of finding the equation of the arc length of an ellipse in polar coordinates with respect to $\theta$.

H. Luo
  • 131
  • 5
  • 2
    Arclength of an ellipse generally does not have a closed form expression. Here (http://pages.pacificcoast.net/~cazelais/250a/ellipse-length.pdf) there is a simpler integral expression for arclength. – Alex Jones May 29 '18 at 15:05
  • 1
    Maybe better said: is not an elementary function, but the closed form is in terms of the standard elliptic integrals. – GEdgar May 29 '18 at 17:13
  • 1
    @GEdgar: special functions are usually not considered allowed in closed-form expression. –  Jan 08 '21 at 08:32

2 Answers2

2

It is no good idea to convert to polar coordinates, it just makes things more difficult: the polar equation of an ellipse is not simple.

Stick to the parametric equation and use

$$\int\sqrt{\dot x^2+\dot y^2}\,d\theta=\int\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}\,d\theta.$$

It turns out that this integral does not have a closed form and a special function was introduced to represent it: the... elliptic integral !

https://www.wolframalpha.com/input/?i=integrate+sqrt%28a%5E2+cos%5E2+theta%2Bb%5E2+sin%5E2+theta%29

1

\begin{align} r &= \frac{a b}{\sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta}} \\ r' &= \frac{ab(b^2-a^2)\sin \theta \cos \theta} {(a^2\sin^2 \theta+b^2\cos^2 \theta)^{3/2}} \\ ds &= \frac{ab\sqrt{a^4\sin^2 \theta+b^4\cos^2 \theta}} {(a^2\sin^2 \theta+b^2 \cos^2 \theta)^{3/2}} d\theta \\ s &= b E \left( \phi, \sqrt{1-\frac{a^2}{b^2}} \, \right) \\ \tan \phi &= \frac{a\tan \theta}{b} \\ \end{align}

where $E(\phi,k)$ is the incomplete elliptic integral of the second kind.

It is consistent with the result of $(x,y)=(a\sin \phi, b\cos \phi)$ parametrization.

Ng Chung Tak
  • 19,693