Why is curl considered the differential operator in 3-space?

Question

Why is the curl considered the differential operator in 3-space instead of the gradient? It would seem that the gradient is the corollary to the derivative in 2-space when extending to 3-space. This is mostly w/r/t Stokes' theorem and how the fundamental theorem of calculus seems to extend to 3-space in a not so intuitive way to me.

Answer 1

There exists a grand generalization and universal framework encompassing all of vector analysis in all dimensions $n\geq1$. In this realm there is indeed one single differential operator, which is denoted by $d$. The various integral theorems you have met, e.g., $$f(b)-f(a)=\int_a^b f'(t)\>dt\ ,$$ then appear as corollaries to a single formula, namely $$\int_{\partial M}\omega =\int_Md\omega\ .$$ Now this general theory is less intuitive than the vector analysis we learn in advanced calculus or physics classes. In the environment taught in these classes we find all sorts of fields: scalar fields, force fields, flow fields, densities, and for all of these we have differential operators which we can interpret geometrically or physically in an intuitive way. In this view of things the operators $\nabla$, ${\rm curl}$, ${\rm div}$, as well as $\Delta$, are all standing on an equal footing, and there is no question of ${\rm curl}$ being "the" differential operator in 3-space.