Does there exist a vector field $\vec F$ such that $curl \vec F=x \vec i+y\vec j+z \vec k$?

Question

Does there exist a vector field $\vec F$ such that curl of $\vec F$ is $x \vec i+y\vec j+z \vec k$ ?

UPDATE : I did $div(curl \vec F)=0$ as the answers did ; but that assumes a lot i.e. it assumes that components of $F$ have second partial derivatives and continuous mixed partial derivatives ; whereas for curl to be defined , we only need components of $F$ to have first order partial derivatives . Is the answer still no with this less assumption ? Please help . Thanks in advance

Answer 1

No: a vector field $F$ can only be the curl of something if $\operatorname{div}{F}=0$, because $$\operatorname{div}\operatorname{curl} G=0$$ for any twice-differentiable $G$ by antisymmetry. The divergence of your vector field, on the other hand, is $3$.

Answer 2

Using the theorem $div(curl(\vec F))= 0 $ we can show that the vector $F=x \vec i+y\vec j+z \vec k$ cannot be the curl of any field because $F=x \vec i+y\vec j+z \vec k$ has a divergence of 3.