Differential Forms on Surfaces. Show that $N\cdot (\nabla \times V)\eta=d\phi$ on $x(D)$. Let $M$ be an orientable surface in $\Bbb R^3$ with a unit normal vector field $N$ and let $x: D\to M$  be a patch. Let $\eta$ be a differential 2-form on $x(D)$ defined by $\eta(x_u,x_v)=\pm\|x_u \times x_v\|$,   where the sign is determined by the same sign as $N\cdot (x_u\times x_v)$. Let $V=(V_1,V_2,V_3)$ be a vector field on $\Bbb R^3$ and  $\phi=V_1\,dx_1+V_2\,dx_2+V_3\,dx_3$  the differential $1$-form obtained from $V$. 
   Show that $N\cdot (\nabla \times V)\eta=d\phi$  on $x(D)$.
My work.
 
I don't know how to simplify either side further to show the equality. 
I would greatly appreciate it if anyone could help me solve this problem.
 A: Interesting. That exercise got me thinking for awhile, I had no choice but to open everything in coordinates, but I nailed it. I wonder if there's a better way to do it. What you did so far is correct. Since we want to compare these forms in ${\bf x}(D)$ (and you should really use ${\bf x}$ instead of $x$ here, for reasons that'll be clear shortly), it is sufficient to check that both $2$-forms are equal in all possible pairs formed by a basis. Since: $$\eta ({\bf x}_u,{\bf x}_u) = \eta({\bf x}_v,{\bf x}_v) = {\rm d}\phi({\bf x}_u,{\bf x}_u) = {\rm d}\phi({\bf x}_v,{\bf x}_v) = 0,$$ we'll see that: $$\langle N, \nabla \times V\rangle \eta({\bf x}_u,{\bf x}_v) = {\rm d}\phi({\bf x}_u,{\bf x}_v).$$
Let's assume that $\eta({\bf x}_u,{\bf x}_v) = +\|{\bf x}_u\times{\bf x}_v\|$. This means that: $$N = \frac{{\bf x}_u\times {\bf x}_v}{\|{\bf x}_u\times{\bf x}_v\|},$$ and hence: $$\langle N, \nabla \times V\rangle \eta({\bf x}_u,{\bf x}_v) = \langle N, \nabla \times V\rangle \|{\bf x}_u\times{\bf x}_v\| = \langle {\bf x}_u\times{\bf x}_v,\nabla \times V\rangle.$$
I hope you'll forgive my laziness: I'll omit the pair $(u,v)$ from now on. Meaning: $${\bf x} = x \,\partial_1 + y\,\partial_2 + z\,\partial_3 \implies \begin{cases} {\bf x}_u = x_u \,\partial_1 + y_u\,\partial_2 + z_u\,\partial_3 \\ {\bf x}_v = x_v \,\partial_1 + y_v\,\partial_2 + z_v\,\partial_3\end{cases}$$with $\partial_1 = \frac{\partial }{\partial x_1} = (1,0,0)$, etc. Let's register here some useful info: $$\nabla \times V = \left(\frac{\partial V_3}{\partial x_2}-\frac{\partial V_2}{\partial x_3}\right)\partial_1+\left(\frac{\partial V_1}{\partial x_3}-\frac{\partial V_3}{\partial x_1}\right)\partial_2+\left(\frac{\partial V_2}{\partial x_1}-\frac{\partial V_1}{\partial x_2}\right)\partial_3.$$Also: $${\bf x}_u \times {\bf x}_v = \begin{vmatrix} \partial_1 & \partial_2 & \partial_3 \\ x_u & y_u & z_u \\ x_v & y_v & z_v\end{vmatrix} = \begin{vmatrix} y_u & z_u \\ y_v & z_v\end{vmatrix}\partial_1\color{red}{-}\begin{vmatrix} x_u & z_u \\ x_v & z_v\end{vmatrix}\partial_2+\begin{vmatrix}x_u  & y_u \\ x_v & y_v \end{vmatrix}\partial_3.$$ Now we compute ${\rm d}\phi({\bf x}_u,{\bf x}_v)$ by brute force and see that it is equal to the inner product of the above vectors and we're done. $$\begin{align} {\rm d}\phi({\bf x}_u,{\bf x}_v) &= \left(\frac{\partial V_2}{\partial x_1}-\frac{\partial V_1}{\partial x_2}\right){\rm d}x_1\wedge {\rm d}x_2({\bf x}_u,{\bf x}_v) \color{red}{-} \left(\frac{\partial V_1}{\partial x_3}-\frac{\partial V_3}{\partial x_1}\right){\rm d}x_1\wedge {\rm d}x_3({\bf x}_u,{\bf x}_v) +  \\ &\qquad\qquad + \left(\frac{\partial V_3}{\partial x_2}-\frac{\partial V_2}{\partial x_3}\right){\rm d}x_2\wedge {\rm d}x_3({\bf x}_u,{\bf x}_v) \\ &= \left(\frac{\partial V_2}{\partial x_1}-\frac{\partial V_1}{\partial x_2}\right)\begin{vmatrix} {\rm d}x_1({\bf x}_u) & {\rm d}x_1({\bf x}_v) \\  {\rm d}x_2({\bf x}_u) & {\rm d}x_2({\bf x}_v) \end{vmatrix} \color{red}{-} \left(\frac{\partial V_1}{\partial x_3}-\frac{\partial V_3}{\partial x_1}\right)\begin{vmatrix} {\rm d}x_1({\bf x}_u) & {\rm d}x_1({\bf x}_v) \\  {\rm d}x_3({\bf x}_u) & {\rm d}x_3({\bf x}_v) \end{vmatrix} +  \\ &\qquad\qquad + \left(\frac{\partial V_3}{\partial x_2}-\frac{\partial V_2}{\partial x_3}\right)\begin{vmatrix} {\rm d}x_2({\bf x}_u) & {\rm d}x_2({\bf x}_v) \\  {\rm d}x_3({\bf x}_u) & {\rm d}x_3({\bf x}_v) \end{vmatrix}  \\ &= \left(\frac{\partial V_2}{\partial x_1}-\frac{\partial V_1}{\partial x_2}\right)\begin{vmatrix} x_u & x_v \\  y_u & y_v \end{vmatrix} \color{red}{-} \left(\frac{\partial V_1}{\partial x_3}-\frac{\partial V_3}{\partial x_1}\right)\begin{vmatrix} x_u & x_v \\  z_u & z_v \end{vmatrix} +  \\ &\qquad\qquad + \left(\frac{\partial V_3}{\partial x_2}-\frac{\partial V_2}{\partial x_3}\right)\begin{vmatrix} y_u & y_v \\  z_u & z_v \end{vmatrix} \end{align}$$Now look at this crap and convince yourself that we're done. I put that minus sign in red to make it easier for you to see that this last expression is what we want.
