I'll attempt this one......
Labeling the lower left-hand corner of the square as (0,0), the upper left-hand corner as (0,1) and the lower right-had corner as (1,0), it appears that one of the curves that goes from (0,1) to (1,0) is just the quarter circle with a radius of 1 centered at the origin that has the equation x^2 + y^2 = 1. And the area of this quarter circle is just pi/4.
So, the area of the whole square less the area of this quarter circle gives us the region bounded by two sides of the square and this quarter circle and is equal to 1 - pi/4 ≈.2146. I'm going to call this "Region A." And we have a similar area (region) at the bottom left. So 2 x .2146 ≈ .4292
Now.....we need to find the areas of one of both "ends of a football" found at the top left and bottom right of the square. Both of these areas will be similar (equal) to the area bounded by the three circles whose equations are x^2 + y^2 + 1, (x-1)^2 + y^2 = 1 and x^2 + (y-1)^2 = 1. I'll label this bounded area as Region B.
To proceed, we can find the area of the small sub-region at the top left corner of the square that is bounded by the line y = 1 and the equation of the circle in the first paragraph from x = 0 to x = .5
So we have the following set up :
.5
∫ 1 - √(1 - x^2) dx
0
And....with a little help from Wolfram Alpha, this ≈ .0216943
And, using symmetry, that are four such "sub-regions" within Region A, so their total area is ≈ .086777
So, the area of Region B - one of the "ends of a football," =
[Area of Region A - Area of 4 "sub-regions"] =
[ .2146 - .086777]= 0.127823.....and we have another one of these Region B areas, as well...so....
2 x 0.127823 ≈ 0.255646
So...the central yellow region is given by [1 - 2(Area of Region A) - 2 (Area of Region B) ] =
[ 1 - .4292 - 0.255646] ≈ 0.315154 sq. units
There are probably easier methods to do this (Jacobians and change of Variables, maybe ??) but I don't know how to proceed any differently.
Also......somebody else check my reasoning and math.....I don't feel "100%" about this one....
