Past exam of the mathematical methods II course of the University of Oxford / 2022 / 9 / b

For a sufficiently smooth vector field on an oriented surface $S$ with compatible boundary orientation, Stokes theorem statesApply it to $\mathbf{F}=(P,Q,0)$ in the plane. The closed line integral is the area integral of $Q_x-P_y$. Thus $P_y=Q_x$ makes it vanish whenever a spanning surface lies within the smooth domain. In a simply connected domain this implies path independence: join one path from $A$ to $B$ to the reverse of another, and their difference is a zero closed line integral.

The domain qualification is essential. On the punctured plane, $P=-y/(x^2+y^2)$, $Q=x/(x^2+y^2)$ satisfy $P_y=Q_x$, but the line integral around the unit circle is $2\pi$. A disk spanning that circle passes through the singularity. Hence the printed implication requires an appropriate domain assumption; zero curl alone is not a global path independence theorem.

Let $F^{\prime }=f$ and $G^{\prime }=g$ on the relevant intervals. The chain rule givesThus the one-form is explicitly an exact differential. Fixing its potential to vanish at $(1,1)$ determines it uniquely on the connected domain:Continuity of $f,g$ on the integration ranges suffices. For arbitrary smooth data defined throughout $x>0$, that half-plane also supplies the simply connected domain in the previous argument.

For the specialization $f(s)=s^{-2}$, the potential based at $(1,1)$ is defined in the positive quadrant; it cannot be continued through $xy=0$ using this formula. Taking $a,b>0$ keeps both endpoints in that component. Their ratios $y/x$ are both $b/a$, so the $G$ term cancels andNo condition on $g$ beyond existence on the traversed ratios enters this difference.