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= 2πP1P2y𝑑s, (1)

along c, is to be minimised; i.e. we must minimise

P1P2y𝑑s=x1x21+y2|dx|. (2)

Since the integrand in (2) does not explicitly depend on x, the Euler–Lagrange differential equationMathworldPlanetmath (http://planetmath.org/EulerLagrangeDifferentialEquation) of the problem, the necessary condition for (2) to give an extremal c, reduces to the Beltrami identity

y1+y2-yyy1+y2y1+y2=a,

where a is a constant of integration.  After solving this equation for the derivative y and separation of variablesMathworldPlanetmath, we get

±dyy2-a2=dxa,

by integration of which we choose the new constant of integration b such that  x=b  when  y=a:

±aydyy2-a2=bxdxa

We can write two equivalentPlanetmathPlanetmath (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