Conference rusure::math

    Re .0:
    
    The area of an ellipse is just pi*a*b, where a is the shortest radius
    and b is the longest radius.  Just imagine a circle with radius a.  It
    has area pi*a^2.  Then stretch it in one dimension by a factor of b/a. 
    That makes the radius b in that direction and multiples the area by a
    factor of b/a, so pi*a^2 * b/a = pi*a*b.
    
    
    				-- edp
    
    
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re .1, Thanks - that makes sense, I was worried that we would have to go into
the dual focal points and integrate, or something.

Now for the hard part, Imagine an elipse with the longest radius horizontal
(could be veritical but no matter), draw a line parallel to this hozizontal axis
such that it cuts the elipse - is there a method for determining the area both
above and below this line within the elipse?

Thanks, 
Tony.

Based on the ideas in .1, shrink the ellipse to a circle,
solve the problem there, and then expand:

For a circle with radius 1, and a chord  a<1 units away from
the center, the chord will cut out an arc of length
	2 * arccos(a)
and a segment of area
	2/2pi * arccos(a)
Cutting out the triangular piece between the chord and the
center leaves an area of
	2/2pi * arccos(a) - a * sin(arccos(a))
      = arccos(a)/pi - a*(1-a^2)
between the chord and the arc (does this region have a name?)

To scale back to the ellipse, just multiply by the major radius
(OK since the chord is parallel to that axis).

OK, now how about the _circumference_ of an ellipse where a and b
are the semi-major and semi-minor axes?

Allen

    Re .4:      
    
    There is no closed-form formula (a formula built out of the common
    functions of arithmetic and trigonometry) that expresses the
    circumference of a general ellipse.  The circumference can be expressed
    as an integral, but it cannot be integrated in closed-form.  Some math
    packages provide functions which will numerically evaluate the integral
    efficiently.
    
    
    				-- edp
    
    
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>    Re .4:      
>    
>    There is no closed-form formula (a formula built out of the common
>    functions of arithmetic and trigonometry) that expresses the
>    circumference of a general ellipse.  The circumference can be expressed
>    as an integral, but it cannot be integrated in closed-form.  
>    

I've seen a simple formula for it but it involves the Elliptic Function, E(k).
No doubt, E is a definite integral and there are tables of values or routines
available to compute it.

Allen

	Hi. Sorry to jump in the discussion, I just discovered this
notesfile...

RE 1841.2

	Integration to find the area of an ellipse is not difficult :
for an ellipse with radius a and b, the function to integrate is

	y = f(x) = b * SQRT ( 1 - x^2/a^2 ) derived from ellipse equation

The area of the ellipse is then 

	A = 4 * INT (0, a, f(x)dx ), where 0 and a are the lower and upper
				     bounds

		substitute x/a = cos t  ->  dx = - a sin t dt

	A = 4ab * INT (0, pi/2, (sin t)^2 dt )

		from trygonometry, (sin t)^2 = (1 - cost 2t) / 2

	A = (pi)ab - 2ab * INT (0, pi/2, cos 2t dt)

		substitute 2t = u -> dt = du/2

	A = (pi)ab - ab * INT (0, pi, cos u du)

		as sin 0 = 0 and sin pi = o, this last integral = 0

	A = (pi)ab         (CQFD)

		
			---------------------------------

	Regarding the problem where a straight line intersects the ellipse,
it can easily be solved using the same approach

	If we have a straight line with equation y = n  ( n < b ), it 
intersects the ellipse twice, one of the intersections is where y = n
and therefore 

			x = a * SQRT (1 - n^2/b^2) = a * k = ak

	The area between the straight line, the ellipse and the X axis is

A = 2 * INT (0, ak, n dx) + 2 * INT (ak, a, b * SQRT (1 - x^2/a^2))

	The first INT is trivial and = 2nak, and the second INT can
be solved as above. After calculation, we obtain

	A = 2nak + ab arccos k - ab/2 sin (2 arccos k)

	Obviously the area between the top of the ellipse and the straight
line y = n is given by

			(pi)ab/2 - A

.3> between the chord and the arc (does this region have a name?)

A segment, by my fast-fading memory.

According to my dictionary a sector is "the portion of a circle bounded by
two radii and one of the intercepted arcs."  That is what is called a segment 
in 

.3> and a segment of area
>	2/2pi * arccos(a)