least surface of revolution
The points P1=(x1,y1) and P2=(x2,y2) have to be by an arc c such that when it rotates around the x-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
A= 2π∫P2P1y𝑑s, | (1) |
along c, is to be minimised; i.e. we must minimise
∫P2P1y𝑑s=∫x2x1√1+y′2|dx|. | (2) |
Since the integrand in (2) does not explicitly depend on x, the Euler–Lagrange differential equation (http://planetmath.org/EulerLagrangeDifferentialEquation) of the problem, the necessary condition for (2) to give an extremal c, reduces to the Beltrami identity
y√1+y′2-y′⋅yy′√1+y′2≡y√1+y′2=a, |
where a is a constant of integration. After solving this equation for the derivative y′ and separation of variables, we get
±dy√y2-a2=dxa, |
by integration of which we choose the new constant of integration b such that x=b when y=a:
±∫yady√y2-a2=∫xbdxa |
We can write two equivalent (http://planetmath.org/Equivalent3) results
lny+√y2-a2a=+x-ba,lny-√-2a2a=-x-ba, |
i.e.
y+√y2-a2a=e+x-ba,y-√y2-a2a=e-x-ba. |
Adding these yields
y=a2(ex-ba+e-x-ba)=acoshx-ba. | (3) |
From this we see that the extremals c 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 |