Answer to Question #263790 in Real Analysis for saduni

Question

Show that the circumference of the ellipse (𝑥 ²/ 𝑎² )+( 𝑦 ²/ 𝑏²) = 1 is given by 2𝜋𝑎 [1- ∑^∞_𝑛=1 ( (2𝑛−1)!!/ (2𝑛)!! ) ²( 𝑒 ^2𝑛 /2𝑛−1 ) ]. Here 𝑒 = √(1 −( 𝑏²/ 𝑎² )), 𝑎 > 𝑏 is the eccentricity. Length of graph 𝑦 = 𝑓(𝑥) can be found by ∫ √(1 + ( 𝑑𝑦 /𝑑𝑥) ² )𝑑x.

Answer 1

An ellipse equation is given by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

If we put a set of parameters together, we get a parametrization.

$x = a \space cost, y = b \space sin t, 0 ≤ t ≤ 2π.$

Calculus's formula for determining the length of a curve is as follows:

$\int_0^{2\pi}\sqrt{(x)^2+(y)^2dt}=\int_0^{2\pi}\sqrt{a^2sin^2t+b^2cos^2tdt}$

We may assume without loss of generality that b ≥ a. The expression under the integral can be transformed as

$\sqrt{b^2-(b^2-a^2)sin^2t}=b\sqrt{1-e^2sin^2t}$ where $e=\sqrt{b^2-\frac{a^2}{b}}$ is called the eccentricity The ellipse's size and shape are described by the parameters b (the length of the bigger semi-axis) and eccentricity.

The length of the ellipse's arc in the first quadrant is sufficient because the ellipse is made up of four such arcs of equal length.

As a result, we must assess the integral.

$\int_0^{\frac{\pi}{2}}\sqrt{1-e^2sin^2tdt}$