Find the area of the region `[x,y): x^2+y^2 le 1 le x+ y }`

Explanation

Let R `={(x,y):x^2+y^2 le 1 le x+y}`
`={(x,y):x^2+y^2 le 1 } cap {(x,y):x+y ge 1}`
`=R_1 cap R_2 `
Clearly,`R_1` is the interior of the circle `x^2+y^2=1` with its centre at O(0,0) and radius =1 uint
and `R_2 ` is the region lying above the lune x+y=1
Consider the equations
`x^2y^=1`
and x+y=1 ...(ii)
Putting y =(1-x) from (ii) in (i) ,
we get
`x^2 +(1-x)^2 =1 rArr 2x^2-2x =0`
`rArr 2x(x-1)=0`
`rArr x=0 or x=1 `
Now `(x=0 rArr y =1) " and " (x=1 rArr y=0 )`
Thus , the points of intersection of (i) and (ii) are A(0,1) and B(1,0)
So , the required ara is the shaded region
Required area= area BCAB
=(area AOBCA )-(area OBAO)
`underset(0)overset(1)int sqrt(1-x^2)dx - underset(0)overset(1)int (1-x)dx`
`=[1/2sin^(-1)+x/2sqrt(1-x^2)]_0^1-[x-x^2/2]_0^1`
`=(1/2sin^(-1))-1/2=(1/2xxpi/2)-1/2=(pi/4-1/2)`sq units
Hence , the required area is `(pi/4-1/2)`sq units