Euler's Equation/Integrated wrt Length Element

Theorem
Let $y$ be a real mapping belonging to $C^2$ differentiability class.

Assume that:


 * $\ds J \sqbrk y = \int_a^b \map f {x, y, y'} \rd s$

where


 * $\rd s = \sqrt {1 + y'^2} \rd x$

Then Euler's Equation can be reduced to:


 * $f_y - f_x y' - f_{y'} y' y - f \dfrac {y} {\paren {1 + y'^2}^{\frac 3 2} } = 0$

Proof
Substitution of $\rd s$ into $J$ results in the following functional:


 * $\ds J \sqbrk y = \int_a^b \map f {x, y, y'} \sqrt {1 + y'^2} \rd x$

We can consider this as a functional with the following effective $F$:


 * $F = \map f {x, y, y'} \sqrt {1 + y'^2}$

Find Euler's Equation:

Due to assumptions on $y$, the prefactor does not vanish.

By Euler's Equation, the last expression vanishes.