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Minimal Surface

Prove that the minimum surface of revolution formed by a curve joining two points \({x}_{1},{y}_{1}\) and \({x}_{1},{y}_{1}\) is described by revolving the catenary.


You should familiarize yourself with this identity first.

Given the arclength formula \(dS = \sqrt{1+{y'}^{2}} dx\), the surface area is thus \[2 \pi \int _{ { x }_{ 1 } }^{ { x }_{ 2 } }{ y\sqrt { 1+{ y' }^{ 2 } } } dx.\]

Since the integrand is independent of \(x\), we can apply the above Euler-Lagrange identity and show that \[y\sqrt { 1+{ y' }^{ 2 } } - y' \left[\frac{y}{2} {(1+{y'}^{2})}^{-1/2} 2y' \right] = A\] for some arbitrary constant \(A\).

With a little algebra, \[y' = \frac{\sqrt{{y}^{2} - {A}^{2}}}{A}.\] Solving this differential equation we find \[y= A \cosh {\left(\frac{x+k}{A}\right)} \] where \(k\) is another arbitrary constant. This curve is the famous catenary.

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Note by Steven Zheng
2 years, 3 months ago

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