# Finding the Curl of a vector field. (Vector calculus)

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.

• If you compute the curl of what you obtained, do you get the original field? Commented Nov 2, 2014 at 14:25
• Does div(curl F) = 0? Commented Nov 2, 2014 at 14:25
• No it dosent give 0. Commented Nov 2, 2014 at 14:26

The divergence of the curl of a vector field must be zero. Here, we see that $$\text{div} \, \text{curl} \, F = \nabla \cdot \left(\begin{array}{c} x \\ 2y \\ 3z \end{array}\right) = \frac{\partial}{\partial x}x + \frac{\partial}{\partial y}2y + \frac{\partial}{\partial z}3z = 1 + 2 + 3 = 6 \neq 0.$$
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.
• So can i say that since $div curlF$ not equals to 0, the second order partial derivative does not exists and hence its potential function (vector field) does not exists. Commented Nov 2, 2014 at 14:35
• That gets at the right idea here. The mixed second partial derivatives of at least one component of $F$ do not agree with each other. For example, if we say $F = P\hat{i} + Q\hat{j} + R\hat{k}$, where $P$, $Q$, and $R$ are all functions of $x$, $y$, and $z$, then $\text{div} \, \text{curl} \, F \neq 0$ means that $R_{xy} \neq R_{yx}$ or that a similar condition is violated by $P$, $Q$, or $R$. Commented Nov 2, 2014 at 14:41