The solid angle subtended by the surface S at a point P is: $$ \Omega=\iint_{S} \frac{\hat{r} \cdot \hat{n}}{r^{2}} d S $$ where $\hat{r}$ and $\hat{n}$ are unit vectors and $r =|\vec {r}|$ is the distance from dS to point P.
Is it possible to write $\frac{\hat{r}}{r^{2}}$ as the curl of a vector $\vec{A}$, in order to calculate the solid angle as the line integral along the boundary of S? $$ \Omega=\iint_{S} (\nabla \times \vec{A}) \cdot \hat{n} \, d S = \int_{\partial S}\vec{A} \cdot d\hat{l}. $$
It would be useful to calculate the solid angle enclosed by a line forming loops, e.g. a spiral or a solenoid, similar to what is done in Electromagnetism to calculate the magnetic flux surface integral (https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction).
Motivation
If $\vec {r}=[x,y,z]$ then we can see that:
$ \nabla \cdot \frac{\hat{r}}{r^{2}} = \nabla \cdot \frac{[x,y,z]}{(x^2+y^2+z^2)^{3/2}} = \frac{(x^2+y^2+z^2)^{3/2}- x \frac{3}{2}2x (x^2+y^2+z^2)^{1/2}}{(x^2+y^2+z^2)^{5/2}} + \frac{\partial (\cdot)}{\partial y} + \frac{\partial (\cdot)}{\partial z} = \\ = \frac{3(x^2+y^2+z^2)^{3/2}-3(x^2+y^2+z^2)(x^2+y^2+z^2)^{1/2}}{(x^2+y^2+z^2)^{5/2}} = 0; \quad (\text{for }\vec {r}\neq[0,0,0]). $
Being $\frac{\hat{r}}{r^{2}}$ divergence-free, it should be possible to express it as the curl of a vector: $$ \nabla \cdot \frac{\hat{r}}{r^{2}} = \nabla \cdot (\nabla \times \vec{A}) =0 $$
It is possible to calculate the solid angle subtended by a surface by performing a line integral along the boundary of the surface.
This is particularly useful for surfaces in the 3D space or when the surface shape is unclear but you have a well-defined boundary. Think for example at the surface enclosed by a 3D spiral coil.
The method finds applications in Electromagnetism, e.g., to calculate mutual inductances or scalar potentials.
The solid angle subtended by the surface $\Sigma$ at a point $\mathbf{r}$ is given by: \begin{equation} \label{eq4:solidAngle} \Omega=\iint_{\Sigma} \frac{\mathbf{r} \cdot \mathbf{n}}{R^{3}} d \Sigma \end{equation} where $\mathbf{n}$ is the unit vector normal to $d\Sigma$ and $R=\left|\mathbf{r}\right|$.
As $\frac{\mathbf{r}}{R^{3}}$ is divergence-free, it is possible to express it as the curl of a new vector, $\mathbf{G}$, and the equation can be rephrased using Stokes' theorem: \begin{equation} \label{eq4:solidAngle_G} \Omega=\iint_{\Sigma} (\nabla \times \mathbf{G}) \cdot \mathbf{n} \, d \Sigma = \oint_{\Gamma} \mathbf{G}\cdot \mathbf{t} \, d \Gamma, \end{equation} where $\Gamma$ is the boundary of the area $\Sigma$ and $\mathbf{t}$ is the unit vector tangent to $d\Gamma$.
One possible solution for $\mathbf{G}$ is: \begin{equation} \mathbf{G} = \left( -\frac{y z}{\left(x^{2}+z^{2}\right) \sqrt{x^{2}+y^{2}+z^{2}}}, 0, \frac{x y}{\left(x^{2}+z^{2}\right) \sqrt{x^{2}+y^{2}+z^{2}}} \right). \end{equation} You can verify that $\nabla \times \mathbf{G} =\frac{\mathbf{r}}{R^{3}} $.
For example, for a rectangular plate on the XY plane, the integral vanishes for the lines in the Y direction and the contribution of the two edges in the X direction is given by: \begin{equation} \begin{aligned} \Omega_i &= \int_{P_1}^{P_2}-\frac{y z}{\left(x^{2}+z^{2}\right) \sqrt{x^{2}+y^{2}+z^{2}}} d x \\[.75em] &= \left. -\tan ^{-1}\left(\frac{x y}{z \sqrt{x^{2}+y^{2}+z^{2}}}\right) \right|_{[x_1,y_1,z_1]}^{[x_2,y_2=y_1,z_2=z_1]} \end{aligned} \end{equation} where the only variable quantity is $x$, while $y$ and $z$ are fixed for each X-oriented line, and $P_1$ and $P_2$ are the $i^{th}$ horizontal edge end-points. This formula is the same as equation (A2) in "Solid Angle of a Rectangular Plate", Richard J. Mathar, 2014, https://wiki.kern.phys.au.dk/mathar20051002.pdf. However, this method is more general and applies to any 2D/3D closed-line that encloses a surface.
For more details, see "Appendix A. Solid Angle of the Area Enclosed by a Spiral" in "Electromagnetic Tracking of Elongated Sensors for Endoscopic Navigation" by M. Cavaliere and P.Cantillon-Murphy, https://www.mdpi.com/2673-8724/2/3/20/htm