Not really sure what the first bit means. 

The area of a square (L^2)  the area of a circle inscribed in that square (pi(L/2)^2) = (Approximately) The double integral of half of the ratio of the height with respect to the base of an equilateral triangle with sides L ( double integral ((1/2)(.86602)dL). (note that the ratio is fixed because its an equilateral triangle). 

Not really sure what the first bit means. 

Yeah I meant ∫∫sqrt3/4 dLdL. I guess that conceptually I am trying still confused about both what the significance of that area between these shapes is as well as the relationship of the triangle to this left over area may have  If there is any at all other than just a linear relationship as the size of the square increases. 

Still not sure what integral you mean. ∫∫sqrt3 dLdL gives you (sqrt(3)/2)L^{2}, but that still differs from (pi/4)L^{2} by about 10%... there's all kinds of ways you can get stuff that's kinda close to (pi/4)L^{2}; none of it has any deep meaning though. ∫∫dLdL doesn't have any obvious geometric meaning to it, for example. 

The comparison is l^2  pi(l/2)^2 to .2165l^2, the latter a result of taking the integral of sqrt(3)/4, then the integral of .433x to get .2165l^2. The difference between the two is 1.0088%. I'm not saying that because the difference is significant per se, but rather the relationship of approximating the same thing (that left over area) with two different approaches. what it comes down to I think is that the former (l^2..) because it uses pi, is more exact and therefore smaller. When you think about the four corners with the triangle shaped remaining area that isn't accounted for by the circle, you notice that obviously the base of these triangles isn't exactly equilateral, which is what the former equation does assume to be true. so ultimately what this percentage difference is  is the difference between the triangle shaped area and a crude triangle approximation of that area that is of course prone to over estimating it. 

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