TITLE: Find the surface of least area spanned by a given contour
QUESTION [5 upvotes]: I'm reading a book on Calculus of Variations, and am confused by the following problem:

Find the surface of least area spanned by a given contour.

The answer is to write a functional of the form:
$$J [z] = \iint_R \sqrt {1+z_x^2+z_y^2} dx dy $$
Now, use Euler's equation:
$$F_z-\frac {\partial}{\partial x}
F_{z_x}-\frac {\partial}{\partial y} F_{z_y} = 0$$
Where $F = \sqrt {1+z_x^2+z_y^2} $
The calculation does not confuse me so much as what we are actually finding. 
I can see that the argument of the integral is highly analogous to arclength. When finding the curve of least length that connects two points, we apply the single variable Euler equation to:
$$\int_L \sqrt{1+y'^2}dx $$
However, I'm not entirely sure where the formula for $J [z] $ comes from, nor what it means for a surface to be spanned by a contour (in fact, I'm not all that certain of what a contour is)
I am enjoying the book, but I have definitely been ambitious to try and tackle this subject, as it seems to take a number of subjects I am only vaguely familiar with for granted.

REPLY [2 votes]: Setting
$$
\begin{align}
0
&=\delta\iint_R\sqrt{1+z_x^2+z_y^2}\,\mathrm{d}x\,\mathrm{d}y\\
&=\iint_R\frac{z_x\delta z_x+z_y  \delta z_y}{\sqrt{1+z_x^2+z_y^2}}\,\mathrm{d}x\,\mathrm{d}y\\
&=-\iint_R\delta z\frac{\partial}{\partial x}\frac{z_x}{\sqrt{1+z_x^2+z_y^2}}\,\mathrm{d}x\,\mathrm{d}y\\
&-\iint_R\delta z\frac{\partial}{\partial y}\frac{z_y}{\sqrt{1+z_x^2+z_y^2}}\,\mathrm{d}x\,\mathrm{d}y\\
&=-\iint_R\delta z\frac{z_{xx}\left(1+z_x^2+z_y^2\right)}{\sqrt{1+z_x^2+z_y^2}^3}\,\mathrm{d}x\,\mathrm{d}y\\
&+\iint_R\delta z\frac{z_x^2z_{xx}+z_xz_yz_{xy}}{\sqrt{1+z_x^2+z_y^2}^3}\,\mathrm{d}x\,\mathrm{d}y\\
&-\iint_R\delta z\frac{z_{yy}\left(1+z_x^2+z_y^2\right)}{\sqrt{1+z_x^2+z_y^2}^3}\,\mathrm{d}x\,\mathrm{d}y\\
&+\iint_R\delta z\frac{z_y^2z_{yy}+z_xz_yz_{xy}}{\sqrt{1+z_x^2+z_y^2}^3}\,\mathrm{d}x\,\mathrm{d}y\\
&=-\iint_R\delta z\frac{z_{xx}\left(1+z_y^2\right)+z_{yy}\left(1+z_x^2\right)-2z_xz_yz_{xy}}{\sqrt{1+z_x^2+z_y^2}^3}\,\mathrm{d}x\,\mathrm{d}y
\end{align}
$$
for all $\delta z$, requires
$$
\bbox[5px,border:2px solid #C0A000]{z_{xx}\left(1+z_y^2\right)+z_{yy}\left(1+z_x^2\right)-2z_xz_yz_{xy}=0}
$$
which is true when a surface has zero Mean Curvature.