Minimal Surface of Revolution

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Calculus of variations can be used to find the curve from a point (x_1,y_1) to a point (x_2,y_2) which, when revolved around the x-axis, yields a surface of smallest surface area A (i.e., the minimal surface). This is equivalent to finding the minimal surface passing through two circular wire frames. The area element is

so the surface area is

and the quantity we are minimizing is

This equation has f_x=0, so we can use the Beltrami identity

to obtain

which is called a catenary, and the surface generated by rotating it is called a catenoid. The two constants a and b are determined from the two implicit equations

which cannot be solved analytically.

The general case is somewhat more complicated than this solution suggests. To see this, consider the minimal surface between two rings of equal radius y_0. Without loss of generality, take the origin at the midpoint of the two rings. Then the two endpoints are located at (-x_0,y_0) and (x_0,y_0), and

But cosh(-x)=cosh(x), so

Inverting each side

so b=0 (as it must by symmetry, since we have chosen the origin between the two rings), and the equation of the minimal surface reduces to

At the endpoints

but for certain values of x_0 and y_0, this equation has no solutions. The physical interpretation of this fact is that the surface breaks and forms circular disks in each ring to minimize area. Calculus of variations cannot be used to find such discontinuous solutions (known in this case as Goldschmidt solutions). The minimal surfaces for several choices of endpoints are shown above. The first two cases are catenoids, while the third case is a Goldschmidt solution.

To find the maximum value of x_0/y_0 at which catenary solutions can be obtained, let p=1/a. Then (17) gives

Now, denote the maximum value of x_0 as x_0^*. Then it will be true that dx_0/dp=0. Take d/dp of (20),

Now set dx_0/dp=0

From (20),

Take (23) ÷ (22),

Defining u=px_0^*,

This has solution u=1.1996789403.... From (22), y_0p=coshu. Divide this by (25) to obtain y_0/x_0=sinhu, so the maximum possible value of x_0/y_0 is

Therefore, only Goldschmidt ring solutions exist for x_0/y_0>0.6627....

The surface area of the minimal catenoid surface is given by

but since

Some caution is needed in solving (◇) for a. If we take x_0=1/2 and y_0=1 then (◇) becomes

which has two solutions: a_1=0.2350... ("deep"), and a_2=0.8483... ("flat"). However, upon plugging these into (◇) with x_0=1/2, we find A_1=6.8456... and A_2=5.9917.... So A_1 is not, in fact, a local minimum, and A_2 is the only true minimal solution.

The surface area of the catenoid solution equals that of the Goldschmidt solution when (◇) equals the area of two disks,

Plugging in

Defining

gives

This has a solution u=1.2113614259. The value of x_0/y_0 for which

is therefore

For x_0/y_0 in (0.52770,0.6627), the catenary solution has larger area than the two disks, so it exists only as a local minimum.

There also exist solutions with a disk (of radius r) between the rings supported by two catenoids of revolution. The area is larger than that for a simple catenoid, but it is a local minimum. The equation of the positive half of this curve is

At (0,r),

At (x_0,y_0),

The area of the two catenoids is

Now let u=x/c_1+c_3, so du=dx/c_1

The area of the central disk is

so the total area is

By Plateau's laws, the catenoids meet at an angle of 120 degrees, so

and

This means that

so

Now examine x_0/y_0,

where u=x_0/c_1. Finding the maximum ratio of x_0/y_0 gives

with c_3=sinh^(-1)(1/sqrt(3)) as given above. The solution is u=1.0799632187, so the maximum value of x_0/y_0 for two catenoids with a central disk is y_0=0.4078241702.

If we are interested instead in finding the curve from a point (x_1,y_1) to a point (x_2,y_2) which, when revolved around the y-axis (as opposed to the x-axis), yields a surface of smallest surface area A, we proceed as above. Note that the solution is physically equivalent to that for rotation about the x-axis, but takes on a different mathematical form. The area element is

and the quantity we are minimizing is

Taking the derivatives gives

so the Euler-Lagrange differential equation becomes

Solving for x then gives

which is the equation for a catenary. The surface area of the catenoid product by rotation is

Isenberg (1992, p. 80) discusses finding the minimal surface passing through two rings with axes offset from each other.

See also

Minimal Surface, Surface of Revolution

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References

Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 931-937, 1985.Goldstein, H. Classical Mechanics, 2nd ed. Reading, MA: Addison-Wesley, p. 42, 1980.Isenberg, C. The Science of Soap Films and Soap Bubbles. New York: Dover, pp. 79-80 and Appendix III, 1992.

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Minimal Surface of Revolution

Cite this as:

Weisstein, Eric W. "Minimal Surface of Revolution." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/MinimalSurfaceofRevolution.html

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