Akhil Challenger Question

If we look just at the first quadrant $(x>0,y>0)$ , we can write two following inequalities:

$\sqrt{x^2+y^2}<1-x$

$\sqrt{x^2+y^2}<1-y$

By rearranging them a bit we can get:

$y^2<1-2x$

$x^2<1-2y$

Now, we need to find area of region where both inequalities stand, easiest way to do that is by integration. I will do it by looking at the functions $f(x)=\sqrt{1-2x}$ and $g(x)=\frac{1-x^2}{2}$ . First we need to find for witch $x$ $f(x)=g(x)$ , by solving this equation we can find that the only real solution is $x=\sqrt{2}-1$ . I didn't know how to find area directly so i wrote it like this:

$\displaystyle A=\int_{0}^{\frac{1}{2}}f(x) dx - \int_{0}^{\sqrt{2}-1}(f(x)-g(x)) dx=\frac{1}{3}-(2-\frac{4\sqrt{2}}{3})=\frac{4\sqrt{2}}{3}-\frac{5}{3}$

$A$ is area just in first quadrant, since other three are equal $\displaystyle S=4A=4(\frac{4\sqrt{2}}{3}-\frac{5}{3})\approx0.8758$