# 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

 $\displaystyle A\;=\;2\pi\int_{P_{1}}^{P_{2}}\!y\,ds,$ (1)

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

 $\displaystyle\int_{P_{1}}^{P_{2}}\!y\,ds\;=\;\int_{x_{1}}^{x_{2}}\sqrt{1\!+\!y% ^{\prime 2}}\,|dx|.$ (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

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

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 which we 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

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

i.e.

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

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