Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 3 / 9A (Vector Calculus)

Parametrize the straight segment by $x=1-t$, $y=t$, $0\le t\le 1$. Then $dx=-dt$, $dy=dt$, and $x^2+xy=1-t$. The line integral is

On the first leg $y=0$ and $x$ decreases from one to zero; on the second $x=0$ and $y$ increases from zero to one. Adding their line integrals givesThe negative sign in the first term follows from the direction of traversal.

For the anticlockwise arc use $x=\cos \theta$, $y=\sin \theta$, $0\le \theta \le \pi /2$. Since $dx=-\sin \theta d\theta$ and $dy=\cos \theta d\theta$, the line integral becomesThe two integrals multiplying $-\alpha$ each equal $1/3$, by substituting $\cos \theta$ and $\sin \theta$ respectively. The remaining integral equals one. Thus

When $\alpha =2\beta$, all three values reduce to $-\beta /3$. To prove path independence for every piecewise smooth path, construct a global potential rather than infer it from agreement on only three paths. Integrating the coefficient of $dx$ and then matching the coefficient of $dy$ givesIndeed ${\Phi }_x=2\beta (x^2+xy)=\alpha (x^2+xy)$ and ${\Phi }_y=\beta (x^2+y^2)$. The integrand is the exact differential $d\Phi$ throughout ${\mathbb{R}}^2$. The fundamental theorem for line integrals therefore gives, for every such path from $A$ to $B$,