(Area of a square is a2, where is the side length; area of a circle is πa2, where a is its radius, so area of quadrant is ¼πa2.)
No actual numeric values have been given so I'll assume the side of square ABCD has length a. The enclosed area AECF will be in terms of a. Because of symmetry, a is also the length of the radius of both quadrants AFCB and AECD. The areas of the two quadrants are ¼πa2. The diagonal AC splits the square into two right isosceles triangles with equal area=½a2. Therefore, because of symmetry, the two halves of the football-shaped area AECF (E and F on the arcs are not shown) common to the two quadrants each have an area of ¼πa2-½a2. Doubling this gives us the area of the football: 2(¼πa2-½a2)=½πa2-a2=½(π-2)a2. This evaluates to about 0.57a2 (that is, 57% of the area of the square).
SOLUTION USING CALCULUS
Ler D be the origin (0,0), AC is a segment of line y1=a-x. AFC is the quadrant arc of the circle centre B(a,a) radius a: (y-a)2+(x-a)2=a2.
Therefore y=a±√(a2-(x-a)2). However, the lower qudrant restricts this to:
y2=a-√(a2-(x-a)2) which produces only points within the square ABCD, that is, y values not exceeding a.
The area enclosed by arc AFC and diagonal AC is:
x=0∫x=a(y1-y2)dx=0∫a(a-x-(a-√(a2-(x-a)2))dx, which simplifies to:
0∫a(-x+√(a2-(x-a)2))dx.
Let x-a=asinθ, then dx=acosθdθ and, when x=0, -a=asinθ, so θ=-½π; when x=a, θ=0.
The integral splits to become:
-0∫axdx+a2θ=-½π∫θ=0cos2θdθ=[-½x2]0a+a2-½π∫0½(1+cos(2θ))dθ=
-½a2+½a2[θ+½sin(2θ)]-½π0=-½a2+½a2(0-(-½π))=-½a2+¼πa2.
The arc AEC (circle centre at D) and the diagonal AC enclose the same sized area, so the football area is -a2+½πa2, or ½(π-2)a2, as before.
The same solution can be obtained by calculus using the equations of the two circles (the circle with centre D(0,0) has the equation x2+y2=a2) and ignoring the diagonal. This 2-circle method is actually easier to evaluate.
It's obvious that the geometric solution is easier to work with than calculus!