Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given the hyperbolic metric $ds^2=\frac{dx^2+dy^2}{x^2}$ on the half plane $x > 0$, find the length of the arc of the circle $x^2+y^2=1$ from $(\cos\alpha,\sin\alpha)$ to $(\cos \beta, \sin\beta)$

I found that $ds^2=\displaystyle\frac{d\theta^2}{\cos^2\theta}$ but when I try to plug in $\pi/3, -\pi/3$, which should give me the arc length of $2\pi/3$,

I get $4\pi/3=\sqrt{\displaystyle\frac{{(\pi/3-(-\pi/3))}^2}{cos^2{(\pi/3})}}$

I feel like I'm making a simple mistake but I cant place it

share|cite|improve this question
What do you mean: "...plug in $\pi/3$, $-\pi/3$...? Do you mean that you want to integrate over $-\pi \le 3\theta \le \pi$? – Fly by Night Sep 28 '12 at 19:09
up vote 3 down vote accepted

The circle $x^2 + y^2 = 1$ can be parametrised by $(\cos \theta, \sin \theta)$. If $x(\theta) = \cos \theta$ and $y(\theta) = \sin \theta$ then

$$ds^2 = \frac{dx^2+dy^2}{x^2} = \frac{(\sin^2\theta+\cos^2\theta) \, d\theta^2}{\cos^2\theta} = \sec^2\theta \, d\theta^2.$$

The arc-length that you are interested in is given by:

$$s = \int \sqrt{ds^2} = \int_{\alpha}^{\beta} |\sec \theta| \, d\theta \, . $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.