1

I want to apply the formula for arc length to an ellipse in polar coordinates to find its perimeter $$s=\int_{\theta_1}^{\theta_2}\sqrt{(dr/d\theta)^2 + r^2}$$ I'm looking to numerically integrate this, so the exact answer isn't the goal. However, what I cannot understand is how this function (numerically or exactly) integrated could ever return the perimeter of an ellipse. For example, consider an ellipse with a semi-major axis of $5$, and a semi-minor axis of $3$ ($a=5$, $b=3$) See an image of the polar plot of ellipse

Now, using Ramanujan's approximation for the perimeter of the ellipse, we get $s\approx25.527$

Using a very unrefined method of numerical integration (i.e. reducing the ellipse to a diamond defined by the 4 points where the ellipse crosses $\theta=0, \pi/2, \pi$, and $3\pi/2$) and ignoring the $dr/d\theta$ term in the integral, we would get a result of $s=30$. Of course, a more refined numerical integration method would result in a larger value for $s$ and the addition of the $dr/d\theta$ term will only increase it further.

What is it that I'm missing here? How is it possible that I'm finding a value larger than the best approximation for the perimeter of the ellipse even though I'm using a method which should only under predict this value?

Alessio K
  • 10,599
TehDrunkSailor
  • 113
  • 1
    There is something contradictory in your statement. Either you are integrating $|r|$ or you are viewing the ellipse as a diamond - both cannot be true at the same time. With the diamond approximation I get $$s = 4\sqrt{34} \approx 23.323$$ – Ninad Munshi Sep 18 '21 at 06:23
  • If you're using something like a left endpoint approximation, then what you are actually integrating is the arc length of these four arcs and not a diamond. This is not necessarily an under or over approximation of the original perimeter. – Ninad Munshi Sep 18 '21 at 06:31
  • See my polar coordinate treatment for centred origin ellipse here. – Ng Chung Tak Sep 18 '21 at 06:44
  • @NinadMunshi, I don't understand why these can't be true at the same time. As I mentioned in the reply to Ares' answer, doesn't integration in polar coordinates return the area bound by the curve when the limits of integration are $0$ and $2\pi$? And if so, since the diamond that I described is fully within the ellipse, its area will be less than that of the ellipse, but this this diamond's area is still larger than the approx. perimeter of the ellipse. – TehDrunkSailor Sep 18 '21 at 10:51

1 Answers1

0

As mentioned by Ninad Munshi in the comments, you incorrectly calculated the perimeter of the diamond, as it should be $4\sqrt{34}$.

For an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with parametric representation

$$x=a\cos(\theta),\space y=b\sin(\theta),\space r\in[0,2\pi]$$

the arclength is given by

$$P=\int_{0}^{2\pi}\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta=\int_{0}^{2\pi}\sqrt{a^2\sin^2(\theta)+b^2\cos^2(\theta)} \ d\theta$$

If $a$ is not more than $3b$, then another approximation which is about $6\%$ the true value is

$$P\approx 2\pi\sqrt{\frac{a^2+b^2}{2}}$$

Ramanujan came up with the approximations

$$P\approx \pi\left(3(a+b)-\sqrt{(3a+b)(a+3b)}\right)$$

and

$$P\approx\pi(a+b)\left(1+\frac{3h}{10+\sqrt{4-3h}}\right)$$

where $h=\frac{(a-b)^2}{(a+b)^2}.$

See 1 and 2 for useful functions in finding the elliptic perimeter.

Alessio K
  • 10,599
  • I understand that the perimeter of the diamond that I illustrated is wrong, and I certainly understand where the $4\sqrt34$ comes from. Perhaps I'm misunderstanding something fundamental about integration in polar coordinates. If the ellipse is replaced by the diamond, and I am again using this simplified version of the formula where the $dr/d\theta$ term is ignored, then won't this integral return the area of that diamond, not the perimeter? Doesn't it return the area bound by the curve (when the limits of integration are $0$ and $2\pi$)? – TehDrunkSailor Sep 18 '21 at 10:45
  • For the parametrisation of the ellipse $x(\theta)=a\cos(\theta)$ and $y(\theta)=b\sin(\theta)$, the area of the ellipse can be written as $$A=\int_{0}^{2\pi}\int_{0}^{B(\theta)}rdrd\theta$$ where $B(\theta)$ is the boundary of the ellipse. So you should get the area if computed correctly. However, you are looking for an approximation for the perimeter. – Alessio K Sep 18 '21 at 10:55
  • Okay, I had forgotten that integration with polar coordinates is always a double integral. Thank you for time. Since my question stems from a fundamental misunderstanding (misremembering?) of how to integrate, do you think it would be best to delete it? – TehDrunkSailor Sep 18 '21 at 14:05
  • You're welcome! :-) It's up to you to delete it or not. If someone else encounters a similar problem in the future, then they may find the answer here. – Alessio K Sep 18 '21 at 14:59