Calculus of variations can be used to find the curve from a point to a point which, when revolved around the xaxis, 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
(1)

so the surface area is
(2)

and the quantity we are minimizing is
(3)

This equation has , so we can use the Beltrami identity
(4)

to obtain
(5)

(6)

(7)

(8)

(9)

(10)

(11)

(12)

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
(13)
 
(14)

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
(15)

But , so
(16)

Inverting each side
(17)

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
(18)

At the endpoints
(19)

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
(20)

Now, denote the maximum value of as . Then it will be true that . Take of (20),
(21)

Now set
(22)

From (20),
(23)

(24)

Defining ,
(25)

This has solution . From (22), . Divide this by (25) to obtain , so the maximum possible value of is
(26)

Therefore, only Goldschmidt ring solutions exist for .
The surface area of the minimal catenoid surface is given by
(27)

but since
(28)
 
(29)

(30)
 
(31)
 
(32)
 
(33)
 
(34)
 
(35)
 
(36)

Some caution is needed in solving (◇) for . If we take and then (◇) becomes
(37)

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,
(38)

(39)

(40)

Plugging in
(41)

(42)

Defining
(43)

gives
(44)

This has a solution . The value of for which
(45)

is therefore
(46)

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
(47)

At ,
(48)

At ,
(49)

The area of the two catenoids is
(50)
 
(51)
 
(52)

Now let , so
(53)
 
(54)
 
(55)
 
(56)
 
(57)

The area of the central disk is
(58)

so the total area is
(59)

By Plateau's laws, the catenoids meet at an angle of , so
(60)
 
(61)
 
(62)

and
(63)

This means that
(64)
 
(65)
 
(66)

so
(67)

Now examine ,
(68)

where . Finding the maximum ratio of gives
(69)

(70)

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 yaxis (as opposed to the xaxis), yields a surface of smallest surface area , we proceed as above. Note that the solution is physically equivalent to that for rotation about the xaxis, but takes on a different mathematical form. The area element is
(71)

(72)

and the quantity we are minimizing is
(73)

Taking the derivatives gives
(74)
 
(75)

so the EulerLagrange differential equation becomes
(76)

(77)

(78)

(79)

(80)

(81)

Solving for then gives
(82)

which is the equation for a catenary. The surface area of the catenoid product by rotation is
(83)
 
(84)
 
(85)
 
(86)
 
(87)

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