Find a vector field $F$ such that $$\operatorname{curl} F = xi + 2yj + 3zk,$$
or, explain why such a vector field does not exist
What i tried
Since $\operatorname{curl} F$ is given, in order to find the original vector field, can i say that since $div (curl F)$ not equals to 0, the second order partial derivative does not exists and hence its potential function (vector field) does not exists.
$\endgroup$ 31 Answer
$\begingroup$The divergence of the curl of a vector field must be zero. Here, we see that \begin{equation} \text{div} \, \text{curl} \, F = \nabla \cdot \left(\begin{array}{c} x \\ 2y \\ 3z \end{array}\right) = \frac{\partial}{\partial x}x + \frac{\partial}{\partial y}2y + \frac{\partial}{\partial z}3z = 1 + 2 + 3 = 6 \neq 0. \end{equation}
Then what we're calling "$\text{curl} \, F$" isn't really the curl of any vector field and such a vector field does not exist, precisely because the condition $\text{div} \, \text{curl} \, F = 0$ condition is not met.
$\endgroup$ 2
