Suppose $V\subset \mathbb{R}^3$ be non-empty and at least twice differentiable (Smooth) and let $S$ be the surface that encloses $V$ (for example a sphere). Let $\textbf{F}\in \mathbb{R}^3$ be a smooth vector field for all space. Let $\textbf{n}$ denote the normal to the surface $S$. Does the surface integral over $S$ preserve the curl operation with respect to the vector field $\textbf{F}$. In other words, Does the surface integral of $\textbf{n}\times\textbf{F}$ commute with the curl operation $$\textbf{curl}\biggl(\oint\_{S}^{}{\textbf{n}\times\dfrac{\textbf{F}(\textbf{r}^\prime)}{|\textbf{r}-\textbf{r}^\prime|} ~ds}\biggr) = \oint\_{S}^{}{\textbf{n}\times \dfrac{ \textbf{curl} (\textbf{F}) (\textbf{r}^\prime)}{|\textbf{r}-\textbf{r}^\prime|}~ds}?$$ Here the surface intergrals are evaluated with respect to the position $\textbf{r}^\prime$ and produce vector fields.



Does surface integral preserve the curl operation?
$\renewcommand\r{\mathbf r}\newcommand\n{\mathbf n}\newcommand\F{\mathbf F}\newcommand\0{\mathbf 0}\newcommand\curl{\operatorname{\mathbf{curl}}}$No. E.g., let $S$ be the unit sphere and let $\F:=(1,0,0)$, so that $\curl\F=\0$ and hence the right-hand side of the identity in question is $\0$.


On the other hand, the left-hand side of the identity in question is
\begin{equation}
\begin{aligned} 
\curl\oint\_{S}ds(\r')\,\n(\r')\times\dfrac{\F(\r')}{|\r-\r'|}
&=\oint\_{S}ds(\r')\,\curl\Big(\n(\r')\times\dfrac{\F(\r')}{|\r-\r'|}\Big), 
\end{aligned}
\end{equation}
where the latter $\curl$ is of course with respect to $\r$.


For $\r=(1,0,0)$, the first coordinate of the latter (vector) integral (rewritten in the spherical coordinates) is
\begin{equation}
\int\_0^\pi d\phi\int\_0^{2\pi}d\theta\, g(\phi,\theta), 
\end{equation}
where
\begin{equation}
 g(\phi,\theta):=-\frac{\cos^2\phi+\sin^2\phi\;\sin^2\theta}
 {(2-2 \cos \theta \sin\phi)^{3/2}} \, \sin\phi, 
\end{equation}
which is manifestly $<0$ for $(\phi,\theta)\in(0,\pi)\times(0,2\pi)$.


So, the left-hand side of the identity in question is not $\0$, and thus  

the identity does not hold in general. $\quad\Box$


