Two pi or not two pi

We will determine the path length of the quarter in the first quadrant, with $x,y \geq 0$ . For the sake of symmetry, I parametrize the curve as $u \in [-1,+1]$ , and set $x(u),y(u) = 1 - \left(\frac{1\pm u}2\right)^2.$ (The positive sign gives the $x$ -coordinate, the negative sign the $y$ -coordinate.)

Then $\frac{dx}{du}, \frac{dy}{du} = -\frac{u \pm 1}2;$ the infinitesimal path length when traveling along the curve over an increment $du$ is $ds = \sqrt{(dx)^2 + (dy)^2} = \frac 12 \sqrt{(u + 1)^2 + (u - 1)^2}du = \sqrt{\frac{u^2 +1} 2}.$ Integrating this over an interval $u \in [a,b]$ , we get $s_{a\to b} = \frac 1 {\sqrt 2} \int_a^b \sqrt{u^2+1} = \frac 1 {2 \sqrt 2} \left[u\sqrt{1+u^2}+\ln (u + \sqrt{1+u^2})\right]_a^b.$ Substituting $[a,b] = [-1,1]$ , we get for one quarter curve (note that $\ln (\sqrt 2 - 1) = -\ln (\sqrt 2 + 1)$ ) $s = 1 + \frac {\ln (1 + \sqrt 2)}{\sqrt 2},$ . and for the entire circumference $C = 4s = 4 + 2\sqrt 2 \ln (1 + \sqrt 2) \approx 6.492901.$ Comparing to the circumference of the unit circle, $C^\circ = 2\pi$ , $\frac C{C^\circ} \approx \frac{6.492901}{6.283185} \approx 1.033377,$ showing that the circumference is $\boxed{3.337}\%$ greater than that of the unit circle.