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:
\(\ds A \sqbrk y\) | \(=\) | \(\ds 2 \pi \int_{x_0}^{x_1} y \sqrt {1 + y'^2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \pi \int_{x_0}^{x_1} C \, \map {\cosh^2} {\frac {x + C_1} C} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x_1 - x_0} C \pi + \frac {\pi C^2} 2 \paren {\map \sinh {\frac {2 \paren {x_1 + C_1} } C} - \map \sinh {\frac {2 \paren {x_0 + C_1} } C} }\) |
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Sources
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 1.4$: The Simplest Variational Problem. Euler's Equation