I found this formula for the circumference of an ellipse:



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


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

$a$ 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?

  • 3
    $\begingroup$ 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 $\endgroup$ Dec 2, 2016 at 23:41
  • 1
    $\begingroup$ 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!) $\endgroup$
    – John
    Dec 2, 2016 at 23:44
  • $\begingroup$ 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. $\endgroup$ Mar 13, 2017 at 8:13
  • $\begingroup$ $\int x = x^2$: that's wrong in two ways! it's actually $\int x = \frac12 x^2+C.$ $\endgroup$
    – TonyK
    May 21, 2017 at 11:27

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – Caters
    Dec 3, 2016 at 16:13
  • $\begingroup$ 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.) $\endgroup$ Dec 3, 2016 at 19:39
  • 2
    $\begingroup$ @Caters, if you're fine with an iterative algorithm, you can use the "arithmetic geometric mean" to compute the circumference of an ellipse. $\endgroup$ 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.

  • $\begingroup$ Perhaps this is not what the OP wants $\endgroup$
    – Shailesh
    Mar 13, 2017 at 6:17

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