least surface of revolution

The points  $P_1=(x_1,y_1)$  and  $P_2=(x_2,y_2)$  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

$A=\mathrm{\hspace{0.33em}2}\pi {\displaystyle {\int }_{P_1}^{P_2}}y𝑑s,$(1)

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

${\displaystyle {\int }_{P_1}^{P_2}}y𝑑s={\displaystyle {\int }_{x_1}^{x_2}}\sqrt{1+y^{\prime 2}}|dx|.$(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

$$y\sqrt{1+y^{\prime 2}}-y^{\prime }\cdot \frac{y{y}^{\prime }}{\sqrt{1+y^{\prime 2}}}\equiv \frac{y}{\sqrt{1+y^{\prime 2}}}=a,$$

where $a$ is a constant of integration.  After solving this equation for the derivative $y^{\prime }$ and separation of variables, we get

$$\pm \frac{dy}{\sqrt{y^2-a^2}}=\frac{dx}{a},$$

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

$$\pm {\int }_a^y\frac{dy}{\sqrt{y^2-a^2}}={\int }_b^x\frac{dx}{a}$$

We can write two equivalent (http://planetmath.org/Equivalent3) results

$$\ln \frac{y+\sqrt{y^2-a^2}}{a}=+\frac{x-b}{a},\ln \frac{y-\sqrt{{}^2-a^2}}{a}=-\frac{x-b}{a},$$

i.e.

$$\frac{y+\sqrt{y^2-a^2}}{a}=e^{+\frac{x-b}{a}},\frac{y-\sqrt{y^2-a^2}}{a}=e^{-\frac{x-b}{a}}.$$

Adding these yields

$y={\displaystyle \frac{a}{2}}\left(e^{\frac{x-b}{a}}+e^{-\frac{x-b}{a}}\right)=a\cosh {\displaystyle \frac{x-b}{a}}.$(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.