Minimal Surface of Revolution
Calculus of variations can be used to find the curve from a point to a point which, when revolved around the x-axis, yields a surface of smallest surface area (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 , so we can use the Beltrami identity
which is called a catenary, and the surface generated by rotating it is called a catenoid. The two constants and 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 . Without loss of generality, take the origin at the midpoint of the two rings. Then the two endpoints are located at and , and
But , so
Inverting each side
so (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 and , 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 at which catenary solutions can be obtained, let . Then (17) gives
Now, denote the maximum value of as . Then it will be true that . Take of (20),
Take (23) (22),
This has solution . From (22), . Divide this by (25) to obtain , so the maximum possible value of is
Therefore, only Goldschmidt ring solutions exist for .
The surface area of the minimal catenoidsurface is given by
Some caution is needed in solving (◇) for . If we take and then (◇) becomes
which has two solutions: ("deep"), and ("flat"). However, upon plugging these into (◇) with , we find and . So is not, in fact, a local minimum, and 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,
This has a solution . The value of for which
For , 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 ) 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
The area of the two catenoidsis
Now let , so
The area of the central disk is
so the total area is
By Plateau's laws, the catenoids meet at an angle of , so
This means that
Now examine ,
where . Finding the maximum ratio of gives
with as given above. The solution is , so the maximum value of for two catenoids with a central disk is .
If we are interested instead in finding the curve from a point to a point which, when revolved around the y-axis (as opposed to the x-axis), yields a surface of smallest surface area , 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 differentialequation becomes
Solving for then gives
which is the equation for a catenary. The surfacearea of the catenoid product by rotation is
Isenberg (1992, p. 80) discusses finding the minimalsurface passing through two rings with axes offset from each other.