MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

Hi all,

I'm trying to minimize the following integral : $ \int_{0}^{\pi/2} \frac{\int_{0}^{x} \sqrt{r(\theta)^2 + r'(\theta)^2}d\theta}{\sin(x)} dx $ with boundary values r(0)=1 and r(pi/2)=0. As you may have guessed, the numerator in the integrand represents arc length in polar coordinates of the curve $(\theta, r(\theta) )$. I have absolutely no idea where to start: I have tried looking into optimization and variational calculus books but wasn't lucky. Are there numerical methods which I could try? Of course, an extra requirement is that $ \underset{x\to 0^{+}}{\lim} \frac{\int_{0}^{x} \sqrt{r(\theta)^2 + r'(\theta)^2}d\theta}{\sin(x)}$ exists and is bounded.

share|cite|improve this question
Use Fubini's Theorem to reduce to a single integral, i.e. interchange the order of integration, so you are integrating a function of $\theta$ from $0$ to $\pi/2$. This requires the indefinite integral of $1/ \sin x$. Now you are in the standard realm of the Calculus of Variations; I'll leave you to fill in the integration details...! There is no guarantee that the Euler-Lagrange equations will actually have a closed form solution. But, you could at least solve them numerically, as you asked. – Zen Harper Jun 7 '11 at 0:55
Thanks for your reply Zen Harper. I'm afraid I don't quite follow: I don't see how to apply Fubini's theorem since 1/sin(x) is obstructing the way. I would greatly appreciate it if you could elaborate on this detail. Thanks! – user15626 Jun 7 '11 at 16:00

I don't understand what you mean by "getting in the way". Fubini's Thoerem is \int_A \int_B f(x,y) dy dx = \int_B \int_A f(x,y) dx dy provided \int_{AxB} |f(x,y)|d(x,y) < \infty. Since you are assuming that the last requirement is fulfilled. Then we may proceed.

Your integral \frac{\int_0^x \sqrt{r(\theta)+r'(\theta)^2}d\theta}{\sin(x)} can be written \int_0^x \frac{\sqrt{r(\theta)+r'(\theta)^2}}{\sin(x)}d\theta. From there, just follow Zen's advice...

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.