From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math.num-analysis
Subject: Re: Lissajous
Date: 13 Feb 1998 22:12:53 GMT

In article <34e27ba1.0@news.potato.ne.jp>, Hary <haryh@usa.net> wrote:

>Please help me find a mathematical way how to calculate the
>area of the Lissajous curve.

I'm not entirely sure what you mean by the area of a self-intersecting
curve, but the curve in question is usually given parametrically, and
areas bounded by parameterized curves can be evaluated by simple integrals:
the area is by Green's theorem the line integral of  x dy  around the
boundary (taken counterclockwise). So if the curve is parameterized
by functions  p_i(t) = (x_i(t),y_i(t))  on various intervals  [t_(i-1), t_i],
then one needs only sum the integrals
	Int( x_i(t) * y_i ' (t)  dt,  t_(i-1) <= t <= t_i ).

This is fairly straightforward for the Lissajou figures given by
parameterizations such as  p(t) = ( cos(n*t), sin(m*t) ); here the 
antiderivative is (for  m <> n) 

                        /sin((n - m) t)   sin((n + m) t)\
            F(t) =  m/2 |-------------- + --------------| .
                        \     n - m            n + m    /

The difficulty is in determining a counterclockwise parameterization of
the appropriate boundary, which (depending on what you meant to 
accomplish) will probably require using the above parameterization or
its inverse ( p(-t) ) each integrated over a large number of intervals,
whose endpoints  t_i  must be determined as solutions to trig equations.
So you end up with sums and differences of quite a few values of  F(t_i).

(Actually it can be done algebraically but that's certainly not easier!)

There is also the matter of defining precisely just what class of curves
you mean to include as "Lissajou figures", since by allowing phase
differences and scaling you can include many additional figures. Certainly
by taking large  m  and  n  you can have your curve surround most of a
rectangle in the plane, so the limiting areas are easier to find.

dave