Consider the scalar field defined below: $$f:\mathbb{R}^3 \rightarrow \mathbb{R}^3, \hspace{2mm} F(x,y,z)=(x^2y^3,xy,xz^4)$$ Find the curl of $f$ at each point where it exists.
I am a bit confused on where it says 'at each point where it exists'. Do we just find the curl normally and that's it? I got it as $(0,-z^4,y-3xy^3)$. Do we have to state anything else to get full marks?
On
The idea is that $curl$ is applied for continuously differentiable functions. So, first you have to check if you function is continuously differentiable, and this is the case as each component of your function is a polynomial in $x,y$ and $z$. Note that, in your answer, maybe you are missing a square on $x$, so $curl$ is $ (0,-z^4, y-3x^2y^2)$.
"At each point where it exists" is a bit superfluous here, since $\text{curl }f$ is defined everywhere in this case. Probably it's to remind you that because the components of $\text{curl}$ are differences of derivatives, the curl of a vector field isn't defined where any of those derivatives fails to exist.
For example, consider the vector field $${\bf g}(x, y, z) := (0, |x|, 0).$$ The last component of the curl is, by definition, $$\partial_x |x| - \partial_y |0| = \partial_x |x| = \left\{\begin{array}{cc}-1, & x < 0 \\ 1, & x > 0 \end{array} \right.,$$ but $\partial_x |x|$ is undefined where $x = 0$, hence so is this component. The other components of the curl are identically zero, so $\text{curl }{\bf g}$ is defined precisely on $\{(x, y, z) \in \mathbb{R}^3 : x \neq 0\}$, which is particular is not the full domain of ${\bf g}$ itself.