least surface of revolution
The points and have to be by an arc such that when it rotates around the -axis, the area of the surface of revolution (http://planetmath.org/SurfaceOfRevolution) formed by it is as small as possible.
The area in question, expressed by the path integral
(1) |
along , is to be minimised; i.e. we must minimise
(2) |
Since the integrand in (2) does not explicitly depend on , the Euler–Lagrange differential equation (http://planetmath.org/EulerLagrangeDifferentialEquation) of the problem, the necessary condition for (2) to give an extremal , reduces to the Beltrami identity
where is a constant of integration. After solving this equation for the derivative and separation of variables, we get
by integration of which we choose the new constant of integration such that when :
We can write two equivalent (http://planetmath.org/Equivalent3) results
i.e.
Adding these yields
(3) |
From this we see that the extremals of the problem are catenaries. It means that the least surface of revolution in the question is a catenoid.
Title | least surface of revolution |
Canonical name | LeastSurfaceOfRevolution |
Date of creation | 2013-03-22 19:12:11 |
Last modified on | 2013-03-22 19:12:11 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 6 |
Author | pahio (2872) |
Entry type | Example |
Classification | msc 49K05 |
Classification | msc 53A05 |
Classification | msc 26B15 |
Related topic | MinimalSurface |
Related topic | EquationOfCatenaryViaCalculusOfVariations |
Related topic | Catenary |
Related topic | MinimalSurface2 |
Related topic | CalculusOfVariations |
Related topic | SurfaceOfRevolution2 |