Elliptic Integral

**amcavoy** · Oct3-05, 03:25 PM

Taken from http://en.wikipedia.org/wiki/Elliptic_integral:

In integral calculus, elliptic integrals originally arose in connection with the problem of giving the arc length of an ellipse and were first studied by Fagnano and Leonhard Euler.

Is it just me, or does it seem like there is an easier way to find the arc length of an ellipse? I thought elliptic integrals arose in giving the arc length of elliptic curves, which as far as I know are a lot different than ellipses.

**Tide** · Oct4-05, 12:05 AM

I thought elliptic integrals arose in giving the arc length of elliptic curves, which as far as I know are a lot different than ellipses.

But an ellipse is an elliptic curve! :)

**HallsofIvy** · Oct4-05, 07:43 AM

I suspect apmcavoy was thinking of "elliptic functions", as in number theory, which are quite different. In any case, he is wrong. Elliptic integrals did, indeed, arise from trying to find the arc length of an ellipse which is NOT as simple as he seems to think. The arclength of an ellipse cannot be written in any simple formula.

(The area is very simple. The area of the ellipse $LaTeX Code: \\frac{x^2}{a^2}+ \\frac{y^2}{b^2}= 1$ is just $LaTeX Code: \\pi ab$ . The distance around (circumference?) an ellipse is not.)

selfAdjoint · Oct4-05, 12:06 PM

Originally Posted by HallsofIvy

I suspect apmcavoy was thinking of "elliptic functions", as in number theory, which are quite different. In any case, he is wrong. Elliptic integrals did, indeed, arise from trying to find the arc length of an ellipse which is NOT as simple as he seems to think. The arclength of an ellipse cannot be written in any simple formula.

(The area is very simple. The area of the ellipse $LaTeX Code: \\frac{x^2}{a^2}+ \\frac{y^2}{b^2}= 1$ is just $LaTeX Code: \\pi ab$ . The distance around (circumference?) an ellipse is not.)

Elliptic functions are the inverse functions to elliptic integrals.

$LaTeX Code: sn^{-1}(x) = \\int^x_0 \\frac{dt}{\\sqrt{(1-t^2)}\\sqrt{(1-k^2t^2)}}$ , etc.