Show that $\int_C d\mathbf{r} \times \mathbf{F} = \int_S (d\mathbf{S} \times \nabla) \times \mathbf{F}$

Question

Question: Prove that $$\int_C d\mathbf{r} \times \mathbf{F} = \int_S (d\mathbf{S} \times \nabla) \times \mathbf{F}$$ where $\mathbf{F}$ is a vector field, $S$ is an open surface and $C$ is the boundary of the surface.

This question states that we need to apply Stokes's theorem to the vector field $\mathbf{a} \times \mathbf{F}$ where $\mathbf{a}$ is an arbitrary constant vector. I have 2 questions regarding the linked question and answer.

They write

Stokes's theorem gives $\int_{S}(\nabla\times (\mathbf{a} \times \mathbf{F}))\cdot d\mathbf{S}=\int_{C}(\mathbf{a} \times \mathbf{F})\cdot d\mathbf{r}$. The right hand side is $\mathbf{a} \cdot \int_{C} d\mathbf{r} \times \mathbf{F}(\mathbf{r})$ [...]

Question: Should we not get $\int_{C}(\mathbf{a}\times \mathbf{F})\cdot d\mathbf{r} = \int_{C}(\mathbf{F} \times d\mathbf{r}) \cdot \mathbf{a}$ instead?

For the term on the left, the question and answer state that:

[....] The left hand side is $\int_{S}(\mathbf{a} (\nabla\cdot \mathbf{F})-(\mathbf{a} \cdot \nabla)\mathbf{F})\cdot d\mathbf{S}=\mathbf{a} \cdot \int_{S}(\nabla\cdot \mathbf{F})\cdot d\mathbf{S}-\int_{S}d\mathbf{S} \cdot (\mathbf{a} \cdot \nabla)\mathbf{F}$

[Perhaps with a sign error] \begin{align} &\mathbf{a}\cdot\int_S (d\mathbf{S}(\nabla\cdot \mathbf{F})-d\mathbf{S}\cdot(\nabla \mathbf{F})) \\&=\mathbf{a}\cdot\int_S (d\mathbf{S}\times \nabla)\times \mathbf{F} \end{align}

I can't really see how this equality holds. I have by the BAC-CAB formula that $$(d\mathbf{S} \times \nabla) \times \mathbf{F} = -d\mathbf{S}(\nabla \cdot \mathbf{F}) + \nabla(d\mathbf{S} \cdot \mathbf{F}) = -d\mathbf{S}(\nabla \cdot \mathbf{F}) + \nabla(d\mathbf{S}) \cdot \mathbf{F} + \nabla \mathbf{F} \cdot d\mathbf{S} $$

Is the above correct?

Question: How can I salvage my efforts to obtain the required result?

Answer 1

To answer your first question, yes, there should be an extra negative sign.

As for your second question, I quote the answer to the linked question and add the extra negative sign:

\begin{align} \dots &= \displaystyle a_j \int_S (\mathrm dS_i \partial_j F_i -\mathrm d S_j \partial_i F_i) \tag{1} \\ &= a \cdot \int_S (\mathrm d S\cdot ( \nabla F) - \mathrm d S ( \nabla \cdot F)) \tag{2} \\ &= a \cdot \int_S(\mathrm dS \times \nabla) \times F. \tag{3} \end{align}

I'm not sure about the validity of line $(2)$, but relating lines $(1)$ and $(3)$, you can quite easily show that $$ [(\mathrm d S \times \nabla) \times F]_j = \mathrm dS_i \partial_j F_i -\mathrm d S_j \partial_i F_i $$ using suffix notation (here are some notes on the topic, all the identities I use here are proven there). Starting with the left-hand side, \begin{align} [(\mathrm d S \times \nabla) \times F]_j &= \epsilon_{jki} (\mathrm dS \times \nabla)_k F_i = \epsilon_{jki}\epsilon_{klm}\mathrm dS_l \partial_mF_i \\ &= \epsilon_{kij}\epsilon_{klm}\mathrm dS_l \partial_mF_i = (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}) \mathrm dS_l \partial_mF_i\\ &= \mathrm dS_i \partial_j F_i -\mathrm d S_j \partial_i F_i, \end{align} as required. The second term $\mathrm dS_j \partial _i F_i$ is indeed $\mathrm dS(\nabla \cdot F)$, as you have obtained. Now, as far as I know, there isn't a super nice way of expressing the first term $\mathrm dS_i \partial_j F_i$ neatly; expanded out in full vectorial form it looks something like this: \begin{align} &\left( \mathrm dS_x \frac{\partial}{\partial x}F_x + \mathrm dS_y \frac{\partial}{\partial x}F_y+ \mathrm dS_z \frac{\partial}{\partial x}F_z \right) \hat x + \left( \mathrm dS_x \frac{\partial}{\partial y}F_x + \mathrm dS_y \frac{\partial}{\partial y}F_y+ \mathrm dS_z \frac{\partial}{\partial y}F_z \right) \hat y \\ &+ \left( \mathrm dS_x \frac{\partial}{\partial z}F_x + \mathrm dS_y \frac{\partial}{\partial z}F_y+ \mathrm dS_z \frac{\partial}{\partial z}F_z \right) \hat z, \end{align} which emphatically is not the same as $\nabla(\mathrm dS \cdot F)$ , since $$ [\nabla(\mathrm dS \cdot F)]_j = \partial_j(\mathrm dS_i F_i) = \mathrm dS_i \partial_j F_i + F_i\partial_j\mathrm dS_i \neq \mathrm dS_i \partial_j F_i. $$ Your mistake was treating $\nabla$ like a regular vector with numerical components and applying the BAC-CAB rule blindly, when it is in fact an operator which does not always commute with functions.

Hope this helped, and let me know if you spot any mistakes.