= Solution
The region lies above the <parabola> $y=x^2$ and below the upper semicircle $y=\sqrt{1-x^2}$. Their intersections satisfy $x^4+x^2-1=0$, hence $x=\pm a$ with $a^2=(\sqrt5-1)/2$. In particular, $\sqrt{1-a^2}=a^2$. The region is
$$
\boxed{D=\{(x,y):-a\leq x\leq a,\quad x^2\leq y\leq\sqrt{1-x^2}\}}.
$$
It is symmetric about the $y$ axis, with endpoints $(\pm a,a^2)$; at $x=0$ it extends from $y=0$ to $y=1$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2-q12-integration-region.png]
{title=Integration region between a parabola and the upper unit semicircle}
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