Past exam of the mathematical methods II course of the University of Oxford / 2014 resit / 5 / Solution

The proposed change of variables gives $u=x-3y$, $v=2x+y$, with inverse $x=(u+3v)/7$, $y=(v-2u)/7$. The Jacobian determinant isThe two lines of slope $-2$ become $v=1,4$, while the two lines of slope $1/3$ become $u=0,7$. Thus the parallelogram becomes $0\le u\le 7$, $1\le v\le 4$. Since $v$ stays positive, the transformed integrand has no singularity there. The double integral is