當 $y=y(x)$ 時
$$ \int_C f(\vec r)ds=\int_{x_0}^{x_f} f(x,y(x))\sqrt{1+y'(x)^2}dx $$
當 $x=x(t),y=y(t)$ 時
$$ \int_C f(\vec r)ds = \int_{x_0}^{x_f} f(x(t),y(t))\sqrt{(\dot{x}(t))^2+(\dot{y}(t))^2}dt $$
$x=R\cos\theta, y=R\sin\theta$ 所以 $dx=-R\sin\theta d\theta, dy=R\cos\theta d\theta$ 則
$$ ds=Rd\theta $$
所以半圓的線機分為
$$ \int_C\lambda(x,y)ds = \int_0^\pi \lambda(\theta)Rd\theta \\ = \int_0^\pi\left(\lambda_0+\frac{\theta}{\pi}(\lambda_1-\lambda_0)\right)Rd\theta =\frac{1}{2}(\lambda_0+\lambda_1)\pi R $$
其中 $\lambda(\theta)=\lambda_0+\frac{\theta}{\pi}(\lambda_1-\lambda_0)$ 因為題目告訴我們線的兩端密度和線性密度的關係
Complete elliptic integral of the second kind
$$ P=\int_Cds=a\int_0^{2\pi}\sqrt{1-e^2\cos^2\theta}d\theta \\ =4a\int_0^{\pi/2}\sqrt{1-e^2\cos^2\theta}d\theta \\ \text{or,}\quad =4a\int_0^{\pi/2}\sqrt{1-e^2\sin^2\theta}d\theta $$
Eccentricity 定義為 $e=\sqrt{1-(b/a)^2}$
其中 $\sqrt{1+\delta}\approx 1 + \delta/2$ by Taylor series (https://math.stackexchange.com/questions/732540/taylor-series-of-sqrt1x-using-sigma-notation), 所以: