Estoy tratando de hacer el cálculo que muestra que la superficie en $\Bbb R^3$ es área de minimizar si y sólo si la media de la curvatura es cero. Estoy recibiendo una señal equivocada y me estoy volviendo loco, necesito ayuda.
Notaciones: Fijar un dominio $D$. Aquí ${\bf x}$ es una parametrización, ${\bf x}^t = {\bf x}+t{\bf V}$ es una variación, ${\bf V}$ cero en $\partial D$, $\bf N$ es el vector unitario normal y $A(t)$ es el área de ${\bf x}^t$.
Hasta el momento, he $$A'(0) = \iint_D \langle {\bf N}, {\bf x}_u \times {\bf V}_v + {\bf V}_u \times {\bf x}_v \rangle \,{\rm d}u\,{\rm d}v,$$ and I'm positive that so far, so good. Then, the trick seems to use Green-Stokes with $P = \langle {\bf N}, {\bf V}\times {\bf x}_u\rangle$ and $Q = \langle {\bf N}, {\bf V}\times {\bf x}_v\rangle$.
$$\begin{align} \frac{\partial Q}{\partial u} - \frac{\partial P}{\partial v} &= \langle {\bf N}_u, {\bf V}\times {\bf x}_v\rangle+\langle {\bf N}, {\bf V}_u\times {\bf x}_v\rangle+\langle {\bf N}, {\bf V}\times {\bf x}_{uv}\rangle \\ &\qquad -\langle {\bf N}_v, {\bf V}\times {\bf x}_u\rangle-\langle {\bf N}, {\bf V}_v\times {\bf x}_u\rangle-\langle {\bf N}, {\bf V}\times {\bf x}_{uv} \rangle \\ &= \langle {\bf N}_u, {\bf V}\times {\bf x}_v\rangle-\langle {\bf N}_v, {\bf V}\times {\bf x}_u\rangle + \langle {\bf N}, {\bf x}_u \times {\bf V}_v + {\bf V}_u \times {\bf x}_v \rangle\end{align}$$
Ahora vamos a mirar sólo en: $$\begin{align} \langle {\bf N}_u, {\bf V}\times {\bf x}_v\rangle-\langle {\bf N}_v, {\bf V}\times {\bf x}_u\rangle &= \langle {\bf V}\times{\bf x}_v, {\bf N}_u\rangle - \langle {\bf V}\times{\bf x}_u,{\bf N}_v\rangle \\ &= \langle {\bf V},{\bf x}_v\times {\bf N}_u\rangle - \langle {\bf V},{\bf x}_u\times{\bf N}_v\rangle \\ &= \langle {\bf V}, {\bf N}_v\times{\bf x}_u+{\bf x}_v\times{\bf N}_u\rangle \\ &= -2H\langle {\bf V},{\bf x}_u\times{\bf x}_v\rangle,\end{align}$$ using that $${\cal S}{\bf v}\times{\bf w}+{\bf v}\times{\cal S}{\bf w} = 2H {\bf v}\times{\bf w},$$ where ${\cal S}$ stands for the shape operator, with ${\bf v} = {\bf x}_v$ and ${\bf w}={\bf x}_u$. The line integral that appears after using Green-Stokes is zero because $\bf V$ is zero in $\partial D$ and we get: $$A'(0) = 2\iint_D H \langle {\bf V},{\bf N}\rangle \,{\rm d}A.$$ But all books say that $$A'(0) = \color{red}{-}2\iint_D H \langle {\bf V},{\bf N}\rangle \,{\rm d}A.$$ ¿Dónde está el error?
