Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics V / 1

Use the change of variables $u=x^2-y^2$, $v=xy$. Its Jacobian determinant in the first quadrant is $2(x^2+y^2)>0$, and the inverse is uniquely determined byBoth are positive for $v>0$, so the transformation is one-to-one on the region. It becomes the rectangle $-2\le u\le 4$, $1\le v\le 3$. The integrand is $uv$, and $x^2+y^2=\sqrt{u^2+4v^2}$. Thus

Green's theorem for a positively oriented simple closed planar curve statesHere $Q_x-P_y=1-(-1)=2$. The trapezium has parallel sides of lengths two and one, separated by height one, so its area is $3/2$. ThereforeEvery congruent trapezium gives the same answer when its boundary is traversed anticlockwise, since the scalar curl is constant and geometric congruence preserves area. Reversing the traversal would instead reverse the sign.

A parametrization of the cylindrical patch isThe last range uses the planes $x=0$ and $x=y$. The surface area element is $|{\mathbf{r}}_x\times {\mathbf{r}}_{\theta }|dxd\theta =2dxd\theta$. HenceIncluding or excluding the boundary curves does not change the surface area.