I didn't say it would be EASY.
Put circles on a coordinate axis.
Center the FIELD circle at the ORIGIN
x^2 + y^2 = R^2
Center the GOAT circle on the positive y-axis
x^2 + (y-R)^2 = L^2
It is important to note that R < L < R*sqrt(2), or we'll never scope out 1/2 the field. If L = R, clearly we get much less than 1/2. [You tell me where the upper bound came from.]
Our two circles intersect at x = (1/2)*(L/R)*sqrt(4*R^2 - L^2) and x = -(1/2)*(L/R)*sqrt(4*R^2 - L^2). The y-coordinate is the same for each x-value, y = (1/2)*(1/R)*(2*R^2 - L^2)
The Calculus solution to the GOAT area is now simple, being (using symmetry) twice the integral from
0 to (1/2)*(L/R)*sqrt(4*R^2 - L^2)
of the difference of the equations of the two half-circles,
sqrt(R^2 - x^2) - [R - sqrt(L^2 - x^2)]
with respect to x, of course.
However, maybe there is an easier way. The GOAT circle intersects the x-axis at x = +/- sqrt(L^2 - R^2). The calculus SOLUTION, now, and NOT just the GOAT area, is found by setting to zero the integral (again using symmetry) from 0 to sqrt(L^2 - R^2) of R - sqrt(L^2 - x^2) (with respect to x) and solving for L.
Maybe it's a little harder than I thought. I'm still thinking it is conceptually not so difficult. It doesn't seem obvious at the moment how to calculate the area of those little not-quite-triangular sections on the outside - in a geometry forum. Someone will beat me to it.
gtg