Show that there is no vector field with curl $x \hat i + y \hat j + z \hat k$.

Question

I have no idea how to prove this. By assuming the field has the curl I get these 3 equations:

$$x = \frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z}$$

$$y = \frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x}$$

$$z = \frac{\partial F_{2}}{\partial z} - \frac{\partial F_{1}}{\partial y}$$

I cant't see how to get a contradiction from here.

Answer 1

Hint: If one vector field is the curl of another, what is its divergence?

Answer 2

Divergence of every curl field must be zero. In your case it is 3, so it is not a curl.