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This question can be homework for elementary calculus.

The lemniscate of Bernoulli $C$ is a plane curve defined as follows.

Let $a > 0$ be a real number. Let $F_1 = (a, 0)$ and $F_2 = (-a, 0)$ be two points of $\mathbb{R}^2$.

Let $C = \{P \in \mathbb{R}^2; PF_1\cdot PF_2 = a^2\}$.

Let's get the equation of $C$ in the polar coordinates.

Let $P = (r\cos\theta, r\sin\theta)$:

$$PF_1^2 = r^2 + a^2 - 2ar\cos\theta, PF_2^2 = r^2 + a^2 + 2ar\cos\theta$$ Hence: $$(r^2 + a^2 - 2ar\cos\theta)(r^2 + a^2 + 2ar\cos\theta) = (r^2 + a^2)^2 - 4a^2r^2\cos^2\theta = a^4$$ $$r^4 + 2r^2a^2 + a^4 - 4a^2r^2\cos^2\theta = a^4$$ $$r^2 = 2a^2(2\cos^2\theta - 1) = 2a^2\cos 2\theta$$

Suppose $P \in C$ is in the first quadrant. Let $s$ be the arc length between $O = (0, 0)$ and $P$.

Then how can we express $s$ by $r$ using an integral?

This is a related question.

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I have read in detail a book where one leads off from the lemniscate to elliptic curves in Edwards form. Bernoulli, Fagnano and Euler made significant contributions to the understanding of lemniscate arcs (length addition/doubling formulas and such). My source referred to chapter one of Siegel's book Topics in complex function theory. Vol.I: Elliptic functions for a more thorough exposition. I haven't checked myself, but probably it is useful. – Jyrki Lahtonen Sep 1 '12 at 6:45
Do you know the arc length formula in terms of polar coordinates? – Raskolnikov Sep 1 '12 at 19:52
"This question can be homework for elementary calculus." - if the ones studying "elementary calculus" covered elliptic integrals, maybe... – J. M. Sep 2 '12 at 4:32
@J.M. Please see my answer. It uses only elementary calculus. You don't need any knowledge of elliptic integrals to solve this problem. – Makoto Kato Sep 2 '12 at 4:52
Yes, and I see you ended at the lemniscate integral itself, which I'm already sure you know cannot be evaluated elementarily... if the goal is to merely set up the arclength integral, then that's dandy. Otherwise... – J. M. Sep 2 '12 at 5:43

By the arc length formula in the polar coordinates,

$s = \int_{0}^{\theta}\sqrt{r^2 + (\frac{dr}{d\theta})^2} d\theta$

Since $d\theta = \frac{d\theta}{dr}dr$,

$s = \int_{0}^{r}\sqrt{r^2 + (\frac{d\theta}{dr})^{-2}} \frac{d\theta}{dr}dr$

Hence $s = \int_{0}^{r}\sqrt{1 + r^2(\frac{d\theta}{dr})^2} dr$

Since $r = a\sqrt{2\cos 2\theta}$,

$\frac{dr}{d\theta} = -\frac{2a\sin 2\theta}{\sqrt{2\cos 2\theta}}$

Hence $\frac{d\theta}{dr} = -\frac{\sqrt{2\cos 2\theta}}{2a\sin 2\theta}$

$(\frac{d\theta}{dr})^2 = \frac{\cos 2\theta}{2a^2\sin^2 2\theta}$

$\cos 2\theta = \frac{r^2}{2a^2}$

$\sin^2 2\theta = 1 - \cos^2 2\theta = \frac{4a^4 - r^4}{4a^4}$

Hence $(\frac{d\theta}{dr})^2 = \frac{r^2}{4a^4 - r^4}$

Therefore $s = \int_{0}^{r}\sqrt{\frac{4a^4}{4a^4 - r^4}}dr = \int_{0}^{r}\frac{2a^2}{\sqrt{4a^4 - r^4}}dr$

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Please do not use the same symbol as a bound of the integral and as its running variable. This basic notational (and conceptual) mistake causes much confusion when one learns integration. – Did Apr 23 '14 at 5:53

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