\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
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\begin{document}

{\bf Question}

A surface $z=u(x,y)$ passes through a given closed curve $C$ in
$x,y,z$ space whose projection on the $x,y$ plane is $\Gamma$.
Show that the area of that part of the surface enclosed by $C$ is

$$A=\ds\int \!\!\! \ds\int_S (1+u_x^2+u^2_y)^{\frac{1}{2}} dxdy$$

where $S$ is the area in the plane bounded by $\Gamma$.

Show that when this area is a minimum, $u$ satisfies the partial
differential equation

$$(1+u_y^2)u_{xx}-2u_xu_yu_{xy}+(1+u_x^2)u_{yy}=0.$$

(This is known as Plateau's problem).


\medskip

{\bf Answer}

For a surface $f(x,y,z)=0$ a normal to the surface is given by
$\un{\bigtriangledown}f$, i.e., $(f_x,f_y,f_z)$

Hence for $u(x,y)-z=0$ a normal vector is $(u_x,u_y,-1)$ and a
\un{unit} normal is $(1+u_x^2+u_y^2)^{-\frac{1}{2}}(u_x,u_y,-1)$.

it follows that is $ds$ is an elements of area of the surface,
then by projection onto the $x-y$ plane,

$$(1+u_x^2+u_y^2)^{-\frac{1}{2}} ds=dxdy$$

and so $$\ds\int ds=\ds\int \!\!\! \ds\int_S
(1+u_x^2+u_y^2)^{\frac{1}{2}}dxdy$$

The E-L equation is then

$\ds\frac{\pl F}{\pl u}-\ds\frac{\pl}{\pl x}\left(\ds\frac{\pl
F}{\pl u_x}\right)-\ds\frac{\pl}{\pl y}\left(\ds\frac{\pl F}{\pl
u_y}\right)=0$ with $F=(1+u_x^2+u_y^2)^{\frac{1}{2}}$.

This gives:

$$\ds\frac{\pl}{\pl
x}\left(\ds\frac{u_x}{(1+u_x^2+u_y^2)^{\frac{1}{2}}}\right)+\ds\frac{\pl}{\pl
y}\left(\ds\frac{u_y}{(1+u_x^2+u_y^2)^{\frac{1}{2}}}\right)=0$$

which after boring algebra gives the result in the question.

\end{document}