Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 12 / b

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 $a^2=(\sqrt{5}-1)/2$. In particular, $\sqrt{1-a^2}=a^2$. The region isIt is symmetric about the $y$ axis, with endpoints $(\pm a,a^2)$; at $x=0$ it extends from $y=0$ to $y=1$.

First perform the inner integral, which gives the vertical width of the region:Using $\int \sqrt{1-x^2}dx=\frac{1}{2}[x\sqrt{1-x^2}+\arcsin x]$ givesAt the intersection, $\sqrt{1-a^2}=a^2$, so