Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 3 / 12C (Vector Calculus)

The inequalities $y\le x\le 2y$ force $y\ge 0$, and the reciprocal bounds exclude zero. Thus the whole region lies in the positive quadrant. Introduce ratio-product coordinatesThe region becomes the rectangle $1\le u\le 2$, $1\le v\le 2$. The inverse map and its Jacobian determinant areThe change of variables formula then gives

Take the usual positive, counterclockwise boundary orientation; the printed question does not specify one, and reversing it negates the answer. Write the successive corner points as $P=(1,1)$, $Q=(\sqrt{2},1/\sqrt{2})$, $R=(2,1)$, $S=(\sqrt{2},\sqrt{2})$. Traverse $P\to Q\to R\to S\to P$, keeping the region on the left.

On $PQ$, use $y=1/x$ with $1\le x\le \sqrt{2}$. Here $dy=-dx/x^2$, so the line integral isOn $QR$, use $y=x/2$ with $\sqrt{2}\le x\le 2$:On $RS$, use $y=2/x$ with $x$ decreasing from $2$ to $\sqrt{2}$:Finally, on $SP$, use $y=x$ with $x$ decreasing from $\sqrt{2}$ to $1$:Adding all four line integrals gives ${\oint }_C(x^2/(2y))dy-dx=1/2$ for the positive orientation, or $-1/2$ for the reverse orientation.

The equality with part (i) follows from Green's theorem. For the coefficients $P=-1$ and $Q=x^2/(2y)$,They are smooth on a neighborhood of this region, since $y>0$. Thus for the positively oriented boundary,The term $-dx$ itself has zero line integral around the closed curve, although it contributes to the four individual edge integrals.