The Curl of  a  vector  field $X=P\partial_x+Q\partial_y+R\partial_z$ is  equal  to  $$Curl X= (R_y-Q_z)\partial_x +(P_z-R_x)\partial_y+ (Q_x-P_y)\partial_z$$

For the moment we replace $\partial_x,\partial_y,\partial_z$ with $dy\wedge dz, dx\wedge dz, dx\wedge dy$ respectively.(In fact we apply the  Hodge  star operator to the  dual of the base   $\partial_x,\partial_y,\partial_z$.

So  actually the component of  $Curl X$ is  identical  to the  compenents of  the  $2$  form $$\alpha=(R_y-Q_z)dy\wedge dz +(P_z-R_x)dx\wedge dz +(Q_x-P_y)dx\wedge dy$$

On the other hand the  vector  field  $X$, being a  section of  the  tangent bundle  $T\mathbb{R}^3$, can  be  considered as a  map  $$X:\mathbb{R}^3 \to \mathbb{R}^3\times \mathbb{R}^3\\X(x,y,z)=(x,y,z, P(x,y,z),Q(x,y,z),R(x,y,z))$$.
Note that the following equality hold:

$$\alpha =X^* \omega$$  where  $\omega$ is the  natural symplectic  structure  of  $\mathbb{R}^3 \times \mathbb{R}^3$  with $\omega= dx\wedge dp +dy\wedge dq+dz \wedge dr$.

The  situation described  above  is  a  motivation  to  consider the following  generalization of the  concept of the  Curl of  a vector  field  on an  arbitrary  Riemannian  manifold.

>**A  generalized Curl:**  Let  $(M,g)$  be  a  Riemannian manifold. The metric  $g$  gives an isomorphism(hence  diffeomorphism) between  $TM$ and  $T^* M$. So the  standard intrinsic symplectic  structure of the  cotangant  bundle is carried to a  symplectic  structure  $\omega$  on $TM$.  Now  assume  that  $X:M \to TM$  is  a  vector  field. We  define the  Curle  of  $X$ as a $2$-form with the  following  formula $$Curl  X=X^* \omega$$

This  is  already  mentioned  at this  MO question.

https://mathoverflow.net/questions/291099/a-generalization-of-gradient-vector-fields-and-curl-of-vector-fields