How do I prove the circumference of an ellipse

Question

I'm trying to find the circumference of an ellipse with a horizontal radius of $h$ and a vertical radius of $k$. The equation for such an ellipse centered at the origin would by $(x/h)^2 + (y/k)^2 = 1$. I've tried the arc-length formula in both cartesian form and parametric form, and got stuck on both of them.

First, I set $f(x) = \dfrac{k}{h}\sqrt{(h^2 - x^2)}$. I tried to solve $2\displaystyle\int_{-h}^h \sqrt{f'(x))^2 + 1} \:dx$, and managed to get it down to the following:

$\displaystyle\dfrac{2k}{h}\int_{-h}^h\sqrt{\dfrac{x^2}{(h^2 - x^2) + (h/k)^2}}dx$

But I didn't know how to go from there either.

Next, I set $f(t) = h\cos(t)$ and $g(t) = k\sin(t)$. I tried to solve $\displaystyle\int_{-\pi}^{\pi}(f'(t))^2 + (g'(t))^2 dx$, and managed to get it down to the following.

$\displaystyle\int_{-\pi}^{\pi} \sqrt{h\sin^2(t) + k\cos^2(t)}dt$

Once again, I didn't know how to go on.

Answer 1

The perimeter of an ellipse cannot be in general calculated analytically, you can solve it numerically. Here you can see the wikipedia article for elliptic integrals. What you are concretly looking for is the complete elliptic integral forms.

I said in general because we know analytically the perimeter of a circunference, which is a particular case of an ellipse.

Answer 2

You can get the answer in the form of an infinite series. For an ellipse $E$ with semi major axis of length $a$ and semi minor axis of length $b$ with $a>b>0$, then we have the circumference $C$ having the formula $$C=2\pi a\left[ 1-\sum\limits_{i=1}^{\infty }{{{\left( \prod\limits_{j=1}^{i}{\frac{2j-1}{2j}} \right)}^{2}}\frac{{{e}^{2i}}}{2i-1}} \right],$$ where $e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}$ denotes the eccentricity of $E$.

Proof

We first parametrise the ellipse. That is $x=a\cos\theta$ and $y=b\sin\theta$. Then, by considering the eccentricity, we have ${{b}^{2}}={{a}^{2}}\left( 1-{{e}^{2}} \right)$.

Now, we compute $C$.

$$\begin{align} & \int_{0}^{2\pi }{\sqrt{{{\left( \frac{\text{d}x}{\text{d}\theta } \right)}^{2}}+{{\left( \frac{\text{d}y}{\text{d}\theta } \right)}^{2}}}\text{ d}\theta }=\int_{0}^{2\pi }{\sqrt{{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta }\text{ d}\theta } \\ & =4\int_{0}^{\frac{\pi }{2}}{\sqrt{{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta }\text{ d}\theta } \\ & =4\int_{0}^{\frac{\pi }{2}}{\sqrt{{{a}^{2}}{{\sin }^{2}}\theta +{{a}^{2}}\left( 1-{{e}^{2}} \right){{\cos }^{2}}\theta }\text{ d}\theta } \\ & =4\int_{0}^{\frac{\pi }{2}}{\sqrt{{{a}^{2}}-{{a}^{2}}{{e}^{2}}{{\cos }^{2}}\theta }\text{ d}\theta } \\ & =4a\int_{0}^{\frac{\pi }{2}}{\sqrt{1-{{e}^{2}}{{\cos }^{2}}\theta }\text{ d}\theta } \\ & =4a\int_{0}^{\frac{\pi }{2}}{\sqrt{1-{{e}^{2}}{{\sin }^{2}}\left( \theta +\frac{\pi }{2} \right)}\text{ d}\theta } \\ & =4a\int_{\frac{\pi }{2}}^{\pi }{\sqrt{1-{{e}^{2}}{{\sin }^{2}}t}\text{ d}t} \\ & =4a\int_{0}^{\frac{\pi }{2}}{\sqrt{1-{{e}^{2}}{{\sin }^{2}}\theta }\text{ d}\theta } \end{align}$$

By the Reduction Formula which involves integration by parts, one can show that$$\int_{0}^{\frac{\pi }{2}}{{{\sin }^{2i}}\theta \text{ d}\theta }=\frac{\pi }{2}\prod\limits_{j=1}^{i}{\frac{2j-1}{2j}}.$$

Using the Binomial Theorem and Fubini's Theorem, $$\begin{align} & 4a\int_{0}^{\frac{\pi }{2}}{\sqrt{1-{{e}^{2}}{{\sin }^{2}}\theta }\text{ d}\theta }=4a\int_{0}^{\frac{\pi }{2}}{\sum\limits_{i=0}^{\infty }{\left( \begin{matrix} {}^{1}/{}_{2} \\ i \\ \end{matrix} \right)}{{\left( -{{e}^{2}}{{\sin }^{2}}\theta \right)}^{i}}\text{ d}\theta } \\ & =4a\sum\limits_{i=0}^{\infty }{\left( {{e}^{2i}}\prod\limits_{j=1}^{i}{\frac{{}^{3}/{}_{2}-j}{j}}\int_{0}^{\frac{\pi }{2}}{{{\sin }^{2i}}\theta \text{ d}\theta } \right)} \\ & =2\pi a\sum\limits_{i=0}^{\infty }{\left( {{e}^{2i}}\prod\limits_{j=1}^{i}{\frac{{}^{3}/{}_{2}-j}{j}}\prod\limits_{j=1}^{i}{\frac{2j-1}{2j}} \right)} \\ & =2\pi a\sum\limits_{i=0}^{\infty }{\left( {{e}^{2i}}\prod\limits_{j=1}^{i}{\frac{3-2j}{2j}}\prod\limits_{j=1}^{i}{\frac{2j-1}{2j}} \right)} \\ & =2\pi a\sum\limits_{i=0}^{\infty }{\left( {{e}^{2i}}{{\left( -1 \right)}^{i}}\prod\limits_{j=1}^{i}{\frac{2j-3}{2j}}\prod\limits_{j=1}^{i}{\frac{2j-1}{2j}} \right)} \\ & =2\pi a\sum\limits_{i=0}^{\infty }{\left[ {{\left( \prod\limits_{j=1}^{i}{\frac{2j-1}{2j}} \right)}^{2}}\frac{{{e}^{2i}}{{\left( -1 \right)}^{i}}}{1-2i} \right]} \\ & =2\pi a\sum\limits_{i=0}^{\infty }{\left[ {{\left( \prod\limits_{j=1}^{i}{\frac{2j-1}{2j}} \right)}^{2}}\frac{{{e}^{2i}}{{\left( -1 \right)}^{i}}}{1-2i} \right]} \\ & =2\pi a\left[ 1-\sum\limits_{i=1}^{\infty }{{{\left( \prod\limits_{j=1}^{i}{\frac{2j-1}{2j}} \right)}^{2}}\frac{{{e}^{2i}}}{2i-1}} \right]. \end{align}$$

Remark: $$\prod\limits_{j=1}^{i}{\frac{2j-1}{2j}}=\frac{\left( 2i \right)!}{{{2}^{2i}}{{\left( i! \right)}^{2}}}$$