Minimal Smooth Surface of Revolution

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Theorem

Let $\map y x$ be a real mapping in 2-dimensional real Euclidean space.

Let $y$ pass through the points $\tuple {x_0, y_0}$ and $\tuple {x_1, y_1}$.

Consider a surface of revolution constructed by rotating $y$ around the $x$-axis.

Suppose this surface is smooth for any $x$ between $x_0$ and $x_1$.


Then its surface area is minimized by the following curve, known as a catenoid:

$y = C \map \cosh {\dfrac {x + C_1} C}$

Furthermore, its area is:

$A = \paren {x_1 - x_0} C \pi + \dfrac {\pi C^2} 2 \paren {\map \sinh {\dfrac {2 \paren {x_1 + C_1} } C} - \map \sinh {\dfrac {2 \paren {x_0 + C_1} } C} }$


Proof

The area functional of the surface of revolution is:

$\ds A \sqbrk y = 2 \pi \int_{x_0}^{x_1} y \sqrt {1 + y'^2} \rd x$

The integrand does not depend on $x$.

By Euler's Equation:

$F - y' F_{y'} = C$

that is:

$y \sqrt {1 + y'^2} - \dfrac {y y'^2} {\sqrt {1 + y'^2} } = C$

which is equivalent to:

$y = C \sqrt {1 + y'^2}$

Solving for $y'$ yields:

$y' = \sqrt {\dfrac {y^2 - C^2} {C^2} }$

Consider this as a differential equation for $\map x y$:


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$\dfrac {\d x} {\d y} = \dfrac C {\sqrt {y^2 - C^2} }$

Integrate this once:

$x + C_1 = C \ln {\dfrac {y + \sqrt {y^2 - C^2} } C}$


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Solving for $y$ results in:

$y = C \cosh {\dfrac {x + C_1} C}$

To find the area, substitute this function into the formula for the area:

\(\ds A \sqbrk y\) \(=\) \(\ds 2 \pi \int_{x_0}^{x_1} y \sqrt {1 + y'^2} \rd x\)
\(\ds \) \(=\) \(\ds 2 \pi \int_{x_0}^{x_1} C \, \map {\cosh^2} {\frac {x + C_1} C} \rd x\)
\(\ds \) \(=\) \(\ds \paren {x_1 - x_0} C \pi + \frac {\pi C^2} 2 \paren {\map \sinh {\frac {2 \paren {x_1 + C_1} } C} - \map \sinh {\frac {2 \paren {x_0 + C_1} } C} }\)


This needs considerable tedious hard slog to complete it.
In particular: Connect this to the fact that this does not always minimizes the area
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Sources

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