[GeeTwo]

Thanks, Ether! I really did put more work into making it readable than doing the math (though not nearly as much work as for a textbook, which I can vouch is a truly painstaking task if you've never done it). I don't think there's anything in there beyond second semester calculus; I know I was routinely doing more complex multidimensional integrals in second semester college physics.

Anyway, as I promised a while back, I came back to solving the square problem in radial coordinates. Once I didn't try too hard for symmetry and integrated over r first, it was pretty straightforward:

As already established, the mean segment length in Cartesian coordinates is:

Ḹ = 4∫₀¹∫₀¹ (1-x)(1-y)√(x²+y²) dx dy

Converting to polar coordinates, placing the lower left corner of the square at the origin and aligning so that the x axis is along θ=0, the y axis along θ=π/2, we replace √(x²+y²) = r, x=rcosθ, y=rsinθ, and dxdy = rdrdθ. For bounds of integration, note that the lower and left sides of the square are θ=0 and θ=π/2 for 0<=r<=1. The right side is defined by rcosθ=1, or r=secθ for 1<=r <= √2. Similarly the top side is r=cscθ for 1<=r <= √2. This yields:

Ḹ = 4∫₀^π/4∫₀^secθ(1- rcosθ)(1- rsinθ)r rdrdθ + 4∫_π/4^π/2∫₀^cscθ(1- rcosθ)(1- rsinθ)r rdrdθ

In the first integral, we substitute α=θ, and in the second we substitute α=π/2-θ, which means cscθ = secα, cosθ = sinα, and sinθ = cosα. After these substitutions and a reversal of the limits of integration on the second term, the first and second term are identical, so combine them:

Ḹ = 8∫₀^π/4∫₀^secα(1- rcosα)(1- rsinα)r rdrdα
Ḹ = 8∫₀^π/4∫₀^secα r² – (cosα+sinα)r³ + cosαsinαr⁴ drdα

The integrals over r are polynomials:

Ḹ = 8∫₀^π/4[⅓r³ – ¼(cosα+sinα)r⁴ + ⅕cosαsinαr⁵ ]₀^secα dα
Ḹ = 8∫₀^π/4[⅓sec³α – ¼(cosα+sinα)sec⁴α + ⅕cosαsinαsec⁵α]dα
Ḹ = 2/15∫₀^π/420sec³α – 15 sec³α -15tanαsec³α + 12tanαsec³α dα
Ḹ = 2/15∫₀^π/45sec³α -3tanαsec³α dα

To solve the integral of sec³α, try tanαsecα. By the product rule,

dtanαsecα/dα = sec³α + tan²αsecα = sec³α + (sec²α-1)secα = 2sec³α - secα

We can fill in with the well-known integral of secα to offset. The integral of 3tanαsec³α is seen to be sec³α:

Ḹ = 2/15[5(secαtanα + ln|secα+tanα|)/2 – sec³α]₀^π/4
Ḹ = 2/15[5(√2-0 + ln(√2+1)-ln(1))/2 - 2√2 + 1]
Ḹ = 2/15[5ln(√2+1)/2 + 5/2√2 - 2√2 + 1]
Ḹ = 2/15[5ln(√2+1)/2 + √2/2 + 1]
Ḹ = ln(√2+1)/3 + √2/15 + 2/15

Which is the same result achieved by Aren Siekmeier through a much longer process.

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The above process did not result in a population density of segment lengths (many of Ether's histograms), which was my primary motivation to do the polar integral. To get this population, leave out the "r" factor representing segment length and integrate over angle only:

Ƥ(r) = 4∫^□(1- rcosθ)(1- rsinθ)rdθ
Ƥ(r) = 4∫^□r - r²(cosθ+sinθ)+ r³cosθsinθ dθ
Ƥ(r) = 4[rθ - r²(sinθ-cosθ)+ r³sin²θ/2 ]^□

For 0<=r <=1, the limits of integration are 0 to π/2. This is the integral over the intersection of the unit circle and this unit square:

Ƥ(r)|_r<=1 = 4[rθ - r²(sinθ-cosθ) + r³sin²θ/2]₀^π/2
Ƥ(r)|_r<=1 = 4[πr/2 - 0 - r²(1 - 0 - 0 + 1)+ r³(1-0)/2]
Ƥ(r) = 2πr - 8r + 2r³

For 1<=r <= √2, the limits are cos^-1(1/r) and sin^-1(1/r) (or equivalently sec^-1r and csc^-1r); this is the integral over the part of the unit square which is outside of the unit circle. To evaluate sin(cos^-1(1/r)) and cos(sin^-1(1/r)), recall that sin²x = 1-cos²x:

Ƥ(r)|_r>=1 = 4[rθ - r²(sinθ-cosθ)+ r³sin²θ/2]_cos^-1(1/r)^sin-1(1/r)
Ƥ(r)|_r>=1 = 4[r(sin^-1(1/r)-cos^-1(1/r)) - r²(1/r-√(1-1/r²)-√(1-1/r²)+1/r)+ r³(1/r²-(1-1/r²))/2]

Ƥ(r)|_r>=1 = 4r(sin^-1(1/r)-cos^-1(1/r)) - 8r + 8r√(r²-1) + 4r - 2r³

Ƥ(r)|_r>=1 = 4r(csc^-1r - sec^-1r) - 4r + 8r√(r²-1) - 2r³

As a sanity check, I calculated the population, population * length, and the integrals of each using excel with a step size of 0.001, and plotted all four (image and excel file attached). The sum of the population was 0.99999947, good to six decimal places, and the average length came out to 0.521405427, which is good to eight decimal places.