= Solution
The original integration region is the triangle with vertices $(0,0)$, $(0,1)$ and $(1/2,1/2)$. Under $r=x+y$, $s=y-x$, its inverse map is $x=(r-s)/2$, $y=(r+s)/2$, with absolute inverse <Jacobian determinant> $1/2$. The boundaries $x=0$, $y=x$, $x+y=1$ become $s=r$, $s=0$, $r=1$. Thus the transformed triangle is $0\leq s\leq r\leq1$, and
$$
I=\frac12\int_0^1\int_0^r\frac{s}{r^2\sqrt{1-r^2}}\,ds\,dr=\frac14\int_0^1\frac{dr}{\sqrt{1-r^2}}=\boxed{\frac\pi8}.
$$
The endpoint singularities are integrable after the inner integral: $r=0$ is only a collapsed vertex, while $r=1$ has the usual inverse-square-root behavior. The diagrams indicate the complete triangles, rather than rectangular bounds that would include points absent from the original region.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2014-triangular-change-of-variables.png]
{title=Original and sum-difference-coordinate triangles for the transformed double integral}
Back to article page
