A question about calculating the perimeter of an ellipse

Question

To find the exact (ish) perimeter of a circle, we simply multiply the diameter by a ratio we have defined as being equal to the circumference / the diameter, known as $\pi$.

My question is, why do we not just do something similar for an ellipse, such that for each eccentricity there exists a unique $\pi$ value, which when multiplied by the semimajor + semiminor axes gives the circumference? Would this not be a suitable way of calculating the perimeter?

This could perhaps be done by defining a functin $f(x)$ which when the eccentricity is inputted, yields the corresponding "$\pi$" value for that eccentricity, such that the general equation for the perimeter of an ellipse is $f(x)(a+b)$.

Answer 1

Define the second-kind complete elliptic integral as $$E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2\theta}\,d\theta$$ Then the perimeter of an ellipse with semi-major axis $a$ and eccentricity $e$ is $4aE(e)$. Note that $E(0)=\pi/2$, so the circular case is a special case of this.

Indeed, the whole theory of elliptic integrals (and their inverses, elliptic functions) arose from this problem.

Answer 2

@Parcly Taxel gave the exact answer.

Using Mathematica notations, we have $$P = 4 a \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \theta} \,d \theta=4a E\left(k^2\right)$$ with $k^2=\frac {a^2-b^2}{a^2}$.

If you want to see the kind of terms you are looking form, consider the case where $b$ is "close" to $a$. By expansion around $b=a$ $$P=4a \Bigg[\frac{\pi }{2}-\frac{\pi (a-b)}{4 a}+\frac{\pi (a-b)^2}{32 a^2}++O\left((b-a)^3\right) \Bigg]$$ that is to say $$P=2\pi a-\pi(a-b)+\frac{\pi (a-b)^2}{8 a}+\cdots$$