Find the area of the region lying inside the circle x^2+y^2=8 and on or above the parabola y=(x^2/(2).

Eliminate $y$: $x^2+\frac{x^4}{4}=8$ gives $x^4+4x^2-32=0$, so $x^2=4$ and the curves meet at $(\pm2,2)$. Between $x=-2$ and $x=2$ the upper boundary is the upper semicircle $y=\sqrt{8-x^2}$ and the lower boundary is the parabola. Area $=\int_{-2}^{2}\left(\sqrt{8-x^2}-\frac{x^2}{2}\right)dx$. Using $\int\sqrt{8-x^2}\,dx=\frac x2\sqrt{8-x^2}+4\sin^{-1}\frac{x}{2\sqrt2}$, the circle part from $-2$ to $2$ equals $2\left(2+4\cdot\frac\pi4\right)=4+2\pi$. The parabola part is $\frac12\cdot\frac{16}{3}=\frac83$. So the area is $4+2\pi-\frac83=\frac43+2\pi\approx7.6165$, i.e. $7.62$. The two intersection points $(\pm2,2)$ lie on the circle since $4+4=8$, and the parabola is below the upper semicircle between them, which justifies the choice of upper and lower curves in the integral. The answer is rounded to two decimals.