A limited elliptical arc length

Question

I'm trying to calculate the length of an elliptical arc using Excel and later C++.

The range is not $[0, \pi/2]$, and not $[0, \pi]$, and not $[0 ,2\pi]$, but $[0, \pi/3]$.

Nowhere can I find a clear explanation of a way to calculate this. Not even an infinite series using two angles as arguments. Everywhere the question is answered by giving the circumference or $1/2$ or $1/4$ of it, not the arc length as defined by an argument.

I'd just like to find an expansion with which I can find a numerical solution to the incomplete elliptic integral of the second kind, a series which doesn't contain 'new' functions e.g. gamma, K(), B(), C(), D(), etc.

Thank you.

Answer 1

The arc length of any curve in Cartesian coordinate system is given by $\displaystyle\int_{x_1}^{x_2}\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$. You can check the proof here.

Now, for a standard ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$ we have to find the length of the arc $P_1P_2$ subtending an angle $\theta_2-\theta_1$ at the center of ellipse. Here, $OP_1$ and $OP_2$ makes an angle $\theta_1$ and $\theta_2$, respectively, with $x$-axis.

Its parametric coordinates are $(a\cos\alpha, b\sin\alpha)$. Differentiate the curve, we'll get $$\frac{dy}{dx}=-\frac{b\cos\alpha}{a\sin\alpha}$$

Substitute this, we get $$\text{Arc Length}=\int_{\alpha_1}^{\alpha_2}\sqrt{(a\sin\alpha)^2+(b\cos\alpha)^2}\ d\alpha$$

Note: $(\alpha_1, \alpha_2)\ne(\theta_1, \theta_2)$ since $\alpha$ is eccentric angle and not the central angle of ellipse. $$\alpha=\tan^{-1}\left(\frac{a}{b}\tan\theta\right)$$

Answer 2

Fourier series approach:

The canonical form for the elliptic integral of the second kind is $$\sqrt{1-k^2\sin^2\theta}.$$

You can expand this in terms of the powers of $k^2\sin^2\theta$, using the Taylor development. Then you get a sum of even powers of the sine, which you can reduce to sums of cosines of the even multiples of the argument. https://mathworld.wolfram.com/TrigonometricPowerFormulas.html

You get a double summation, and by regrouping the terms, the Fourier series of the integrand. Term-wise integration is straightforward.

I would not recommend this method, as a general formula is complex and I guess that convergence will be poor.