Why is the area of an ellipse simpler than its perimeter?

[losang]

Why does the area of an ellipse have a closed form while the perimeter does not? Obviously if the area is finite so is the boundary so it seems the perimeter should be calculable in a closed form.

 

[marcusl]

The perimeter is

P=4aE(\pi/2,e)

where a is the length of the semi-major axis, e is eccentricity and E is the complete elliptic integral of the second kind, a well-known, tabulated function.

 

[arildno]

Why does the area of an ellipse have a closed form while the perimeter does not?

Well, it just doesn't.

Obviously if the area is finite so is the boundary

Wrong implication. You can perfectly well have regions with finite areas, yet infinite boundaries.

so it seems the perimeter should be calculable in a closed form.

Why should "closed form" have anything to do with finiteness?

 

[losang]

The perimeter is

P=4aE(\pi/2,e)

where a is the length of the semi-major axis, e is eccentricity and E is the complete elliptic integral of the second kind, a well-known, tabulated function.

read the question.

 

[slider142]

We could lay a string on the perimeter and in the limit get the exact length by measuring the string. In other words for a circle there is a closed form but as the axis vary wrt one another there is no cloded form. Why?

Do you consider multiplying by a transcendental number (Ie., the closed form for the circle being a parameter multiplied by the limit we named pi) to be "more closed" than using a transcendental function (whose properties weren't taught in grade school)? E(a,b) isn't much stranger than sin(a). Can you express sin(3) in closed form without using another transcendental function?

 

[D H]

show me.

If you have access to Science, you can read this seminal article by Benoit Mandelbrot, "How Long Is the Coast of Britain? Statistical Self-Similarity and Fractional Dimension." http://www.sciencemag.org/cgi/content/abstract/156/3775/636

 

[arildno]

show me.

Sure enough:

First, consider a unit square. Then, place beside the unit square a rectangle of length 1, and height 1/2.
Then, continue by placing beside the rectangle a new rectangle with length 1, and height 1/4

And so on.

Consider now the figure as determined by this arrangement:

If you have n rectangles (the first being the square), its perimeter is simply 2n+2.

As for the area A of the figure, it equals:
A=1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}++++\frac{1}{2^{n-1}}=2*(1-(\frac{1}{2})^{n})

Thus, as n is allowed to proceed to infinity, the resulting figure will have a perimeter of infinite length, whereas its area equals 2, a finite number.

 

[marcusl]

Why does the area of an ellipse have a closed form while the perimeter does not? Obviously if the area is finite so is the boundary so it seems the perimeter should be calculable in a closed form.

Since the radius varies strongly as a function of polar angle theta, it is not surprising that the perimeter expression is complicated. The real question is why the area expression is simple.

It arises from the following coincidental property of the ellipse: if you draw an ellipse of semi-axes a and b and also draw a circle of radius \sqrt{ab}, both centered on the origin, then the areas of the sectors formed by the x axis, a radius at angle \theta, and each curve are equal. The ellipse extends outside the circle near the x-axis but is inside the circle towards the y-axis in such a way that the area differences exactly balance out. EDIT (for clarity): A circle is an ellipse, too, so it is reasonable that one whose radius is the geometric mean of the semi-major and -minor axes of the ellipse has the same area.