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