= Solution
The region lies in the first quadrant, inside the unit <circle> and above the positive branch of $xy=1/4$. In <polar coordinates>, $xy=(r^2/2)\sin2\theta$, so its radial bounds are $(2\sin2\theta)^{-1/2}\leq r\leq1$. Such an interval exists when $\sin2\theta\geq1/2$, hence
$$
\boxed{\frac\pi{12}\leq\theta\leq\frac{5\pi}{12},\qquad\frac1{\sqrt{2\sin2\theta}}\leq r\leq1.}
$$
At the intersections with the <circle>, $r=1$ and $\sin2\theta=1/2$, giving exactly the two stated polar angles. Their Cartesian coordinates are $(\cos(\pi/12),\sin(\pi/12))$ and the swapped pair.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-1-circle-hyperbola-region.png]
{title=The first-quadrant region between a hyperbola and the unit-circle arc}
{description=Original shaded integration region. The dashed rays pass through the two intersections, at polar angles $\pi/12$ and $5\pi/12$.}
The <Jacobian determinant> contributes $dx\,dy=r\,dr\,d\theta$, while $xy(x^2+y^2)=r^4\sin2\theta/2$. Thus the <double integral> is
$$
I=\int_{\pi/12}^{5\pi/12}\int_{(2\sin2\theta)^{-1/2}}^1\frac{r^5}{2}\sin2\theta\,dr\,d\theta=\int_{\pi/12}^{5\pi/12}\left(\frac{\sin2\theta}{12}-\frac{\csc^2(2\theta)}{96}\right)d\theta.
$$
Its <antiderivative> is $-\cos2\theta/24+\cot2\theta/192$. Evaluating the bounds gives
$$
\boxed{I=\frac{\sqrt3}{24}-\frac{\sqrt3}{96}=\frac{\sqrt3}{32}.}
$$
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