4
$\begingroup$

In $\mathbb{R}^3$, suppose there is a curve on X-Y plane $y(x)$ defined on $x\in [-a,a]$ satisfying:

  1. $y(x)\geqslant 0$;
  2. $y(-a)=y(a)=0.$

Rotate $y(x)$ along x-axis in $\mathbb{R}^3$ and get a solid revolution.

Minimize surface area $A=\int_{-a}^{a}2\pi y\sqrt{1+(y')^2}\mathrm{d}x$, given fixed volume $V=\int_{-a}^{a}\pi y^2\mathrm{d}x=C$, for some constant $C$.

By standard variational method and Lagrange multiplier, ($\lambda$ is Lagrange constant), we get the Euler Lagrange equation, which is nonlinear. $$\frac{1}{(1+(y')^2)^{1/2}}+\lambda y = \frac{yy''}{(1+(y')^2)^{3/2}}$$

My question is how to do with this nonlinear equations. Is there any way to get the minimal $A$ without solving $y$ explicitly?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ The isoperimetric inequality in Euclidean space is well known, even without assuming a solid of revolution. This does not appear to be an issue of current research. $\endgroup$
    – Ben McKay
    Apr 1, 2019 at 8:28
  • $\begingroup$ I think the setting of a solid of revolution makes this question different than the isoperimetric inequality. Fixed some "ending" point (a,0,0), (-a,0,0), for most constant volume $C$, it seems that the ball (optimal in isoperimetric inequality) cannot fit into the volume constraint. $\endgroup$
    – Silentmovie
    Apr 1, 2019 at 16:55

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.