= Solution
For a one-to-one continuously differentiable change of variables with nonzero <Jacobian determinant> in the interior, the <change of variables formula> is
$$
\int_R f(x,y)\,dx\,dy=\int_{\widetilde R}f(x(u,v),y(u,v))\left|\det\frac{\partial(x,y)}{\partial(u,v)}\right|du\,dv.
$$
The absolute value is necessary even if the transformation reverses orientation.
The original region is the first-quadrant area between the hyperbola and circle, above the horizontal-axis segment. More explicitly,
$$
0\leq y\leq\frac1{\sqrt2},\qquad \sqrt{1+y^2}\leq x\leq\sqrt{2-y^2}.
$$
Its vertices are $(1,0)$, $(\sqrt2,0)$ and $(\sqrt{3/2},1/\sqrt2)$. Under the <squared-radius difference substitution>, the inverse is
$$
x=\sqrt{\frac{u+v}{2}},\qquad y=\sqrt{\frac{u-v}{2}},
$$
and the image is the triangle $1\leq v\leq u\leq2$, with vertices $(1,1),(2,1),(2,2)$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-3-change-of-variables.png]
{title=Circular-hyperbolic integration region and its triangular image}
The forward <Jacobian determinant> is
$$
\det\begin{pmatrix}2x&2y\\2x&-2y\end{pmatrix}=-8xy.
$$
It is nonzero in the interior; the zero on $y=0$ lies on a measure-zero boundary, so the integral formula applies by taking interior limits. Since $x^5y-xy^5=xyuv$ and $4x^2y^2=u^2-v^2$, the inverse-Jacobian factor cancels $xy$, giving
$$
I=\frac18\int_1^2\int_1^u uv\,e^{u^2-v^2}\,dv\,du.
$$
Integrating in $v$ first and then $u$,
$$
I=\frac1{16}\int_1^2u\bigl(e^{u^2-1}-1\bigr)\,du=\frac1{32}(e^3-1)-\frac3{32}.
$$
Therefore
$$
\boxed{I=\frac{e^3-4}{32}.}
$$
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