Question

Find the area of the region.
$\left\{\right(x,y):x^2+y^2\le 8,x^2\le 2y\}$.

Answer 1

REF.image

$\left\{\right(x,y):x^2+y^2\le 8;x^2\le 2y\}$

The shaded region is required region,

$\therefore$ shaded region $=2{\int }_0^2\left(\sqrt{8-x^2-\frac{x^2}{2}}\right)dx$

$=2{\int }_0^2\sqrt{8-x^{2}dx-2}{\int }_0^2\frac{x^2}{2}dx$

$=2{[\frac{\sqrt{2}}{2}\sqrt{8-x^2}+\frac{8}{2}si{n}^{-1}\frac{x}{2\sqrt{2}}]}_0^2-2\frac{x^3}{6}{\int }_0^2$

$=2[\sqrt{2}\times (2-2\sqrt{2})+4(\frac{\pi }{4})-\frac{8}{3}]$

$=2[2\sqrt{2}-4+\pi ]-\frac{8}{3}$

Area of shaded = $4\sqrt{2}(1-\sqrt{2})+2\pi -\frac{8}{3}$ sq. units