Question

Find the area of the region $\left\{\right(x,y):x^2+y^2\le 1\le x+y\}$

Answer 1

$x^2+y^2\le 1$ represents the area with in a circle of radius $1$ with centre of origin ( shaded black )

$x+y\ge 1$ represents the region outside the line $x+y=1,$ ( shaded blue)

$\therefore$ Area of region bounded by the curves $x^2+y^2\le 1$ $\&$ $x+y\ge 1$

$=$ area of quadrent $-$ area of $△OAB$

$=\frac{\pi {\left(1\right)}^2}{4}-\frac{1}{2}\times 1\times 1$

$=\frac{\pi }{4}-\frac{1}{2}$

$=\frac{\pi -2}{4}$

$=\frac{1.142}{4}$

$=0.285uni{t}^2.$

Hence, the answer is $\mathrm{0.285.}$