= Solution
The boundary ray has $y/x=3/5$, and its intersection with the <hyperbola> is $(5/4,3/4)$. Under the given <change of variables>, $x^2-y^2=r^2$ and $y/x=\tanh\theta$. Thus the region becomes
$$
0\leq r\leq1,\qquad 0\leq\theta\leq\operatorname{artanh}(3/5)=\log2.
$$
The <hyperbolic functions> satisfy $\cosh^2\theta-\sinh^2\theta=1$, so the <Jacobian determinant> is
$$
\det\frac{\partial(x,y)}{\partial(r,\theta)}
=\det\begin{pmatrix}\cosh\theta&r\sinh\theta\\\sinh\theta&r\cosh\theta\end{pmatrix}=r.
$$
It is positive in the interior. The coordinate degeneracy at $r=0$ is a boundary set of area zero and does not affect the <change of variables formula>. Therefore
$$
\int_A y\,dx\,dy
=\int_0^{\log2}\int_0^1 r^2\sinh\theta\,dr\,d\theta
=\frac13(\cosh(\log2)-1)
=\boxed{\frac1{12}}.
$$
Here $\cosh(\log2)=5/4$. The PDF specifies two line segments and one hyperbolic arc; the duplicated $y=0$ line in the TeX is not an extra boundary condition.
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