# Why curl of a vector field that is proportional to $1/r^2$ equal to $0$?

The curl of the vector field $${\bf F} = (-y {\bf i} + x {\bf j})/(x^2 + y^2)$$ is $$0$$.

I have an intuitive understanding of why the divergence of a radial field that is proportional to 1/r^2 is equal to 0 (because for a spherical polar volume element the face farthest from the origin has a surface area that grows proportional to r^2 but the field strength decreases proportional to 1/r^2 so the effects cancel).

My lecture notes try to use a similar argument for the curl scenario, but I thought the curl represented the line integral around an infinitesimal element. If this is the case, the length of the plane polar sidelength farthest from the origin should increase proportional to $$r$$ (as $$L = r\, d\theta$$) but the field would decrease proportional to $$1/r^2$$ and they should not cancel.

Where is my misunderstanding?

image of the vector field

## 1 Answer

My own possible thinking is that rewritten in plane polar form, F=θ^/r (since -yi+xj/(x^2+y^2)^0.5 = θ^) and therefore the field strength does die off proportional to 1/r after all, thereby cancelling with the increased side length proportional to r and producing 0 curl.

In other words, x^2+y^2 = r^2 = r*r, where one of the r's is used to normalise the theta vector, and the other r causes a field strength decrease proportional to r.

I'm not sure how correct this thinking is though.