The arc length of a graph of a function $f(x)$ is

$$\mathcal{l} = \int_a^b \sqrt {1 + f'(x)^2} dx$$

If you choose $f$ to be a parametrization of the ellipse with $a=1, b=\frac{1}{2}$, i.e. $f(x) = \frac{1}{2} \sqrt{1-x^2}$, the arc length of the graph of this function is

$$\mathcal{l} = \int_0^1 \sqrt\frac{1-\frac{3x^2}{4} }{1-x^2}dx$$

If you enter int_0^1 sqrt((1-(3x^2)/4)/(1-x^2))dx at wolframalpha.com you will get the result:

enter image description here

What I would have expected here as the argument of $E$ (the complete elliptic integral of the second kind) was the eccentricity $e = \sqrt{1-(b/a)^2} = \sqrt{\frac{3}{4}}$ but it's $e^2$.

Is this a Wolframalpha bug or did I make or understand something wrong?

  • $\begingroup$ Show exactly how you manipulated the upper and lower bounds of integration to get your integral $\endgroup$ – Ethan Dec 20 '12 at 5:51
  • $\begingroup$ Is it wrong the way I did it: "int_0^1..."? $\endgroup$ – Hans-Peter Stricker Dec 20 '12 at 8:19

Wolfram has an unconventional definition of the elliptic integrals. If you look at the manual you see that the argument of EllipticE is $m$ which is related to the usual argument $k$ of $E(k)$ via $m=k^2$.

  • $\begingroup$ It's not "unconventional"; some people prefer the modulus, and some people prefer the parameter... $\endgroup$ – J. M. is a poor mathematician Mar 23 '13 at 19:04

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