Which of the following is neither a gradient field nor a rotation field?
$$\mathbf{f}(\mathbf{r})= ||\mathbf{r}||^2\mathbf{r}$$
or $$\mathbf{f}(\mathbf{r})= \mathbf{r} \times \mathbf{i}$$
I know that if the curl is $0$ then it is a gradient field however I don't understand how to work that out.
Hint:
$f(r) = ||r||^2 \,\vec{r} = ||r||^2 (x\hat{i}+y\hat{j}+z\hat{k})$
$curl F = \nabla \times F = ||r||^2 \,\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x & y & z\end {vmatrix} = ?$
In second case,
$f(r) = \vec{r} \times \hat{i} = \begin{vmatrix} i & j & k \\ x & y & z \\ 1 & 0 & 0\end {vmatrix} = ? $
Once you get $f(r)$, find its curl just like in the first case.