# Is there a simpler way of finding the circumference of an ellipse?

I found this formula for the circumference of an ellipse:

$$4aE(e)$$

where

$$e = \sqrt{1-\frac{b^2}{a^2}}$$

and

$$E(x) = \int_{0}^{\pi/2}\sqrt{1-x^2\sin^2\theta}\;d\theta$$

$a$ is the semi-major axis (or, in other words, the maximum radius), and $b$ is the semi-minor axis (minimum radius).

Now somehow, this is related to $2\pi r$ for the circumference of a circle.

Is there an easier way to find the circumference of an ellipse than to do this integral? Because I can't do integrals yet.

I can do square roots, but not integrals; and square roots in integrals are even harder than integrals themselves. Okay, so, technically, I can do integrals, but only the antiderivative kind of integral.

In other words I could do this:

$$\int x = x^2$$

But a definite integral is one of the kinds I can't do.

So, is there a way that I can more easily find the circumference of an ellipse? Would I need to know the circle it could have come from and that circumference and then scale that circle circumference by whatever factor made the circle an ellipse?

• Unfortunately, there's no close form solution (in terms of "elementary function") for the integral; that's why elliptic integrals/functions were introduced. For other approximations or efficient algorithms, see the link here Dec 2, 2016 at 23:41
• Ng is right. The area of an ellipse has a nice form: $\pi a b$. Not so for the circumference (unless you call $E(e)$ "nice," which you certainly can!)
– John
Dec 2, 2016 at 23:44
• It's not just you. Nobody can "do" this integral, that's why it is there in the first place. It is not going to be any simpler than that. Mar 13, 2017 at 8:13
• $\int x = x^2$: that's wrong in two ways! it's actually $\int x = \frac12 x^2+C.$ May 21, 2017 at 11:27

If $a$ and $b$ are positive real numbers, the perimeter of the ellipse $$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$$ can be expressed (in multiple forms) as an arc length integral, such as $$P(a, b) = \int_{0}^{2\pi} \sqrt{a^{2} \cos^{2} t + b^{2} \sin^{2} t}\, dt. \tag{1}$$ If $a = b$, the ellipse is a circle, of perimeter $2\pi a$.

If $a \neq b$, the elliptic integral (1) is not an elementary function of $a$ and $b$. Loosely, there is no closed-form algebraic expression for the perimeter of a non-circular ellipse in terms of arithmetic operations, radicals, exponentials and logs, or circular functions.

• So that means that I would have to learn how to do definite integrals before I even try to calculate the circumference of an ellipse and the closest I can get to the actual circumference of an ellipse without using integrals is an infinite series. Dec 3, 2016 at 16:13
• If you're interested in exact values, yes; if you're interested in numerical approximations, it's possible to give arbitrarily good formulas involving only radicals and circular functions. (Calculus would be required to derive such a formula, but not required to use it.) Dec 3, 2016 at 19:39
• @Caters, if you're fine with an iterative algorithm, you can use the "arithmetic geometric mean" to compute the circumference of an ellipse. May 21, 2017 at 13:01

You say you can't do integrals. Perhaps you just need the right tools. With Python and mpmath, for example, you can do

def E(x):
return quad(lambda θ: sqrt(1 - x ** 2 * sin(θ) ** 2), [0, pi / 2])

a = 10
b = 5
e = sqrt(1 - b ** 2 / a ** 2)
print(4 * a * E(e))


which prints 48.4422411027384.

• Perhaps this is not what the OP wants Mar 13, 2017 at 6:17