# Incomplete Elliptic Integrals of the second kind -- evaluation

Discussion in 'Differentiation and Integration' started by Sequoia, Apr 6, 2020.

1. ### Sequoia

Joined:
Apr 6, 2020
Messages:
1
0
I am programming as a hobby and trying something with elliptical orbits of satellites.
I need to understand how to measure the length of an arc on an ellipse, e.g. from 0 to pi/3.
Now I've come upon Elliptic Integrals and an associated brick wall.

For the "Complete Elliptic Integral of the second kind" the length of an arc from 0 to pi/2, I've found a series the program can use to approximate the value of this kind of arc.

However, for the INCOMPLETE sort, there is nowhere a description of how the value is calculated, no series, no formula. Nothing.

They tell you in most places either :
1. Look it up
or
2. use a website to calculate it, such as https://keisan.casio.com/

The casio.com website can indeed calculate the value where k = 0.6 and the limits are 0 to pi/3 bringing the value:
0.98922159...... so there must be an infinite series one utilizes to calculate this.

It is most certainly not an elementary function. So, what is the series?

Sequoia, Apr 6, 2020

2. ### Country Boy

Joined:
Dec 15, 2021
Messages:
157
38
I am not clear why you say "there must be an infinite series one utilizes to calculate this." I would be inclined to assume that they used a numerical integration.

The "incomplete elliptic integral of the second kind" for k= 0.6 and limits 0 and $\pi/3$ is

Country Boy, Jan 19, 2022

3. ### Country Boy

Joined:
Dec 15, 2021
Messages:
157
38
I accidently terminated my previous post:

I am not clear why you say "there must be an infinite series one utilizes to calculate this." I would be inclined to assume that they used a numerical integration.
You want to find int_0^(pi/3) sqrt(1- 0.36 sin^2(x)) dx.
The "incomplete elliptic integral of the second kind" for k= 0.6 and limits 0 and pi/3 is int_(0 to pi/3) sqrt(1- 0.6 sin^2(x))dx. The "trapezoid rule" divides the interval into n equal sections, raises vertical lines at each endpoint from the x-axis to the curve, then replaces the curve above each interval by the straight line between the two points. So the area under the curve is approximated by n trapezoids. Here, the ith trapezoid has "height", on the x-axis, from (pi/3n)(i-1) to (pi/3n)i, a length of pi/3n. the "base" on the left is sqrt(1- 0.36 sin^2((pi/3n)(i-1))) and the "base" on the right is sqrt(1- 0.36 sin^2((pi/3n)i).

The area of each such trapezoid is the average of the two "bases", [sqrt(1- 0.36 sin^2((pi/3n)(i-1)))+ sqrt(1- 0.36 sin^2((pi/3n)i)]/2, times the "height", pi/3n, so (pi/3n)[sqrt(1- 0.36 sin^2((pi/3n)(i-1)))+ sqrt(1- 0.36 sin^2((pi/3n)i)]/2.

The integral is the sum of those for i= 1 to n.

Country Boy, Jan 19, 2022