Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 3 / 10A (Vector Calculus)

For a one-to-one continuously differentiable change of variables with nonzero Jacobian determinant, the change of variables formula rule iswhere $E$ is the transformed region and the inverse map covers $D$ once. The absolute value is required even if the coordinate map reverses orientation.

Here $x,y$ are positive. For $u=y/x$, $v=xy$, the inverse is $x=\sqrt{v/u}$ and $y=\sqrt{uv}$. On the first portion of $D$, the bounds give $v\ge 1$, $u\le 4$ and $v\le u$ because $x\le 1$. Together these give $1\le v\le u\le 4$; conversely these inequalities imply $1/2\le x\le 1$ and the original bounds. On the second portion, they give $u\ge 1$, $v\le 4$ and $v\ge u$, hence $1\le u\le v\le 4$, with the converse following similarly. The two triangles combine into the rectangle $E=\{1\le u\le 4,1\le v\le 4\}$, sharing only the diagonal corresponding to $x=1$.

The Jacobian isAlso $x{y}^3=u{v}^2$ and $x^2+y^2=v(1+u^2)/u$. The transformed integral consequently factors:Therefore