1
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ 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. $\endgroup$
    – Zen Harper
    Commented Jun 7, 2011 at 0:55
  • $\begingroup$ 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! $\endgroup$
    – user15626
    Commented Jun 7, 2011 at 16:00

1 Answer 1

Reset to default
2
$\begingroup$

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...

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .