Past exam of the mathematical methods II course of the University of Oxford / 2015 resit / 4 / Solution

The proposed linear change of variables makes the two pairs of boundary lines constant-coordinate lines: $v=1,4$ and $u=0,7$. Thus the image of the parallelogram is the rectangle $0\le u\le 7$, $1\le v\le 4$.

The forward Jacobian matrix and its determinant areInverting gives $x=(u+3v)/7$, $y=(v-2u)/7$, so the area element is $dxdy=\frac{1}{7}dudv$. The absolute inverse Jacobian determinant, rather than the forward determinant, belongs in the transformed area integral. Consequentlywhich yields $\boxed{\frac{7}{2}\log 4=7\log 2}$. The transformed denominator is bounded below by one throughout the region, so no singular or improper integral is involved.