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?
Best Answer
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.