# Perimeter of an ellipse

1. Nov 10, 2008

### 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.

2. Nov 10, 2008

### 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.

3. Nov 10, 2008

### arildno

Well, it just doesn't.
Wrong implication. You can perfectly well have regions with finite areas, yet infinite boundaries.
Why should "closed form" have anything to do with finiteness?

4. Nov 10, 2008

### losang

5. Nov 10, 2008

### slider142

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?

6. Nov 10, 2008

### Staff: Mentor

7. Nov 11, 2008

### arildno

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.

8. Nov 11, 2008

### marcusl

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.

Last edited: Nov 12, 2008