Multivariable Calculus Exam Mistake? This question was from an exam taken in January 2022 on a course on introductory multivariable calculus and was worded exactly as follows:
"For a general surface $S$ bounded by a closed curve $C$ show using Stokes theorem that for a vector field, $\mathbf{F}(\mathbf{r})$
$$\int_S\nabla\times(\mathbf{F}\times\mathrm{d}\mathbf{S})=\alpha\int_C\mathbf{F}\times\mathrm{d}\mathbf{r}$$
and identify the constant $\alpha$."
The "$\mathrm{d}\mathbf{S}$" is used to denote a surface integral and the "$\mathrm{d}\mathbf{r}$" to denote a line integral.
I have asked a couple of mathematicians who work in applied mathematics and they have not been able to prove this either. The LHS is apparently the area of concern - having "$\nabla\times(\mathbf{F}\times\mathrm{d}\mathbf{S})$" seems to be what's throwing people off.
I have been told that the answer should result in a vector even though, in the course, vector-valued integrals were never defined and so the notions of "$\times\mathrm{d}\mathbf{S}$" and "$\times\mathrm{d}\mathbf{r}$" were also not defined, so while I personally believe this question was unfair, I'm asking more about whether or not it's possible.
If this is a mistake, could you explain why? And if it isn't, can you prove it?
 A: It resembles the following vector variant of Stokes' theorem
$$\iint_S(\mathbf{n}\times\nabla)\times \mathbf{F}\,dS=-\int_C\mathbf{F}\times\mathrm{d}\mathbf{r}.$$
Proof. Note that
$$
(\mathbf{n}\times\nabla)\times \mathbf{F}=
\left| \begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
n_2\frac{\partial}{\partial z}-n_3\frac{\partial}{\partial y} 
&  
n_3\frac{\partial}{\partial x}-n_1\frac{\partial}{\partial z} 
& 
n_1\frac{\partial}{\partial y}-n_2\frac{\partial}{\partial x} \\
F_1 & F_2 &  F_3 
\end{array} \right|
$$
and therefore, by comparing the $x$ components of both sides we find
$$
\begin{align}\left(\iint_S(\mathbf{n}\times\nabla)\times \mathbf{F}\,dS\right)_x
&=\iint_S \left(n_3\frac{\partial F_3}{\partial x}-n_1\frac{\partial F_3}{\partial z}
-n_1\frac{\partial F_2}{\partial y}+n_2\frac{\partial F_2}{\partial x}\right)dS\\
&=\iint_S \left(-\frac{\partial F_3}{\partial z}-\frac{\partial F_2}{\partial y},\frac{\partial F_2}{\partial x},\frac{\partial F_3}{\partial x}\right)\cdot d\mathbf{S}\\
&=\iint_S \left(\nabla \times\left(0,F_3,-F_2\right)\right)\cdot d\mathbf{S}\quad \text{(Stokes' Theorem)}\\
&=\int_C \left(0,F_3,-F_2\right)\cdot d\mathbf{r}\\
&=\left(-\int_C\mathbf{F}\times\mathrm{d}\mathbf{r}\right)_x.
\end{align}$$
Similarly we can verfy that the equality holds also for $y$ and $z$ components.
P.S. I wonder if one can play (at least in a formal way) with the cross-product properties in order to obtain $\nabla\times(\mathbf{F}\times d\mathbf{S})$:
$$\nabla\times(\mathbf{F}\times d\mathbf{S})=
(\nabla\times \mathbf{F})\times d\mathbf{S}+(d\mathbf{S}\times \nabla)\times 
\mathbf{F}=
(\nabla\times \mathbf{F})\times d\mathbf{S}+(\mathbf{n}\times \nabla)\times 
\mathbf{F}dS.$$
Note that the term $(\nabla\times \mathbf{F})\times d\mathbf{S}$ is related to a variant of the divergence theorem.
