Quite a simple question:
$\textbf{u} = (a\cos(x)\cos(y)\cos(z), b\sin(x)\sin(y)\cos(z), c\sin(z))$ is a curl vector field. Which of the following may correctly be deduced?
A: None of the Above
B: $c = 0$, but $a$ and $b$ are arbitrary;
C: $c = 0$, and $a = b$;
D: $a$, $b$ and $c$ are all arbitrary
E: $a = 1$, $b = 1$ and $c = 0$;
I believed the correct answer to be D, however, I know now that the correct answer is C. But I don't know how this can be deduced?
You need to use the properties of curl vector fields. Let $\textbf{u} = \nabla \times\textbf{f}$
Now, we know that
$$\nabla \cdot\textbf{u} = \nabla\cdot(\nabla \times\textbf{f}) = 0$$
$$\implies -a\sin x \cos y\cos z + b\sin x \cos y \cos z + c\cos z = 0$$
Since this is an identity, it means that this should be true for all $x,y,z$
$$\implies \cos z \left(c + (b-a) \sin x \cos y\right) = 0$$
Now for this to be true for all $x,y,z$, we have
$$c = 0, a=b$$