Perimeter of an Ellipse

Date: 06/29/99 at 15:31:57
From: Dan Lawn
Subject: Perimeter of an oval formula question

I've tried to implement a formula from your archives for the perimeter 
of an oval.

I assumed the x and y (x^2 + y^2/b^2 = 1; b<1) were the height and 
width of the oval.

In order to check my program I tried entering the dimensions for a 
circle. Does this formula hold true for the special case of the 
circle?

My results seem to be inaccurate. I used x = .5,  y = .5. I get
b = .5773502691896257, and for perimeter I get 4.731235153812344.

It seems to me that the answer should be 1/2 pi. Can you help?

Thank you,
Dan

Date: 06/29/99 at 17:41:47
From: Doctor Anthony
Subject: Re: Perimeter of an oval formula question

Calculating the perimeter of an ellipse is a difficult calculation, as 
the following discussion will show.

The parametric coords. of an ellipse are  x = a.cos(phi)
                                          y = b.sin(phi)

e = eccentricity of the ellipse.

So for length of arc ds^2 = dx^2 + dy^2

                          = [a^2sin^2(phi) +b^2.cos^2(phi)].d(phi)

                          = [a^2 -(a^2-b^2)cos^2(phi)].d(phi)

                          = [a^2 - a^2.e^2.cos^2(phi)].d(phi)

If we take the first quarter of the ellipse, phi will vary from 0 to 
pi/2.

First quarter  s = a.INT(0 to pi/2)[(1-e^2.cos^2(phi))^(1/2).d(phi)].

This integral cannot be evaluated in any of the usual ways. For 
obvious reasons it is called an elliptic integral, and tables are 
available to give values of the integral between various limits.

An approximate value is obtained by expanding (1-e^2.cos^2(phi))^(1/2) 
by the binomial theorem and retaining only a few terms.

We get for the series:

     1 - (1/2)e^2.cos^2(phi) - (1/8)e^4.cos^4(phi) -
                                         (1/16)e^6.cos^6(phi) - ...

Integrating term by term and putting in the limits 0 to pi/2 we get:

 (1/2)pi.a[1 - (1/4)e^2 - (3/64)e^4 - (5/256)e^6 - ... ]

To get the complete perimeter of the ellipse, this value would have to 
be multiplied by 4. 

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/