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 $\paren {x_0, y_0}$ and $\paren {x_1, y_1}$.

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:

$\displaystyle y = C \cosh \sqbrk{\frac {x + C_1} {C} }$.

Furthermore, its area is:

$\displaystyle A = \paren{x_1 - x_0} C \pi + \frac {\pi C^2} 2 \paren {\sinh \sqbrk {\frac {2 \paren {x_1 + C_1} } C} - \sinh \sqbrk {\frac {2 \paren {x_0 + C_1} } C} }$


Proof

The area of the surface of revolution is:

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

Integrand does not depend on $x$.

By Euler's Equation:

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

i.e.

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

which is equivalent to:

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

Solving for $y'$ yields:

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

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

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

Integrate this once:

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

Solving for $y$ results in:

$\displaystyle y = C \cosh {\frac {x + C_1} {C} }$

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

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



$\blacksquare$

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