= Solution
For an <exact differential> $M\,d\theta+N\,d\phi=dF$, equality of mixed <partial derivatives> requires $M_\phi=N_\theta$. Here the two <partial derivatives> are
$$
M_\phi=-\sinh\theta\sin\phi,\qquad N_\theta=\sinh\theta\sin\phi.
$$
They are not identically equal, so the <differential form> is not a <closed differential form> and hence \b[is not an exact differential]. Equality along special lines such as $\theta=0$ does not supply a potential on a two-dimensional open neighborhood, because every such neighborhood contains points where the two <partial derivatives> differ.
Back to article page
