= Solution
{c}
Let $a$ be an arbitrary constant <vector>. By the cyclic identity for the <scalar triple product>,
$$
a\cdot\oint_{\partial\Sigma}d\mathbf r\times\mathbf F
=\oint_{\partial\Sigma}(\mathbf F\times a)\cdot d\mathbf r.
$$
Apply <Stokes theorem> to $\mathbf F\times a$. Since $a$ is constant, the <curl> product identity gives
$$
\nabla\times(\mathbf F\times a)
=(a\cdot\nabla)\mathbf F-a(\nabla\cdot\mathbf F).
$$
Consequently
$$
\begin{aligned}
a\cdot\oint_{\partial\Sigma}d\mathbf r\times\mathbf F
&=\iint_\Sigma\left[\mathbf n\cdot(a\cdot\nabla)\mathbf F
-(\mathbf n\cdot a)(\nabla\cdot\mathbf F)\right]dS\\
&=a\cdot\iint_\Sigma(\mathbf n\times\nabla)\times\mathbf F\,dS.
\end{aligned}
$$
For precision, the operator in the last line has components
$$
\bigl[(\mathbf n\times\nabla)\times\mathbf F\bigr]_i
=\sum_{j=1}^3n_j\partial_iF_j-n_i\sum_{j=1}^3\partial_jF_j.
$$
The <partial derivatives> act on $\mathbf F$, with $\mathbf n$ multiplying their values; they do not differentiate $\mathbf n$. Since the <dot products> agree for every constant $a$, the <cross-product form of Stokes theorem> follows:
$$
\boxed{\oint_{\partial\Sigma}d\mathbf r\times\mathbf F
=\iint_\Sigma(\mathbf n\times\nabla)\times\mathbf F\,dS.}
$$
Here $\wedge$ in the paper denotes the <cross product>, and the printed surface integral has its <area element> $dS$ understood.
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