Perimeter of an Ellipse

I remembered that there was no direct formula for the perimeter of an ellipse unlike the circle where the perimeter or circumference \(C\) is given by the formula \(C=\pi d\) where \(d\) is the diameter of the circle. I asked Gemini to explain the situation. What follows is Gemini's explanation and NOT mine.

Unlike a circle (\(C = 2\pi r\)), there is no simple, exact algebraic formula for the perimeter (circumference) of an ellipse that uses only elementary functions. The exact perimeter can only be expressed using a mathematical concept known as an elliptic integral.

However, there are several highly accurate approximations. The definitive way to calculate the perimeter of an ellipse practically is to use the exact integral for theoretical precision, or Srinivasa Ramanujan's first approximation for everyday calculation.

To use these formulas, you must know two values:

• \(a\): The semi-major axis (half the length of the longest diameter).
• \(b\): The semi-minor axis (half the length of the shortest diameter).
  
  

The Best Practical Approximation (Ramanujan)

In 1914, the mathematician Srinivasa Ramanujan published a brilliant approximation that is incredibly accurate for most ellipses (unless they are extremely elongated, almost resembling a straight line). This is the formula you should use for almost any practical engineering or geometric application:

\[ P \approx \pi \Big [ 3(a+b) - \sqrt{(3a+b)(a+3b)} \,\Big ] \]

Example Calculation:
If your ellipse has a semi-major axis \(a = 5\) and a semi-minor axis \(b = 3\):

1. Calculate \(3(5+3) = 3(8) = 24\)
2. Calculate \((3(5)+3)(5+3(3)) = (15+3)(5+9) = (18)(14) = 252\)
3. Take the square root of 252: \(\sqrt{252} \approx 15.8745\)
4. Subtract: \(24 - 15.8745 = 8.1255\)
5. Multiply by \(\pi\): \(\pi \times 8.1255 \approx 25.527\)

The perimeter is approximately \(25.527 \).

The Exact Formula (For Absolute Precision)

If you need the mathematically perfect answer, you must use the complete elliptic integral of the second kind. The exact formula is:

\[ P = 4a \int_0^{\pi/2} \sqrt{1 - e^2 \sin^2 \theta} \, d\theta \]

• \(a\) is the semi-major axis.
• \(e\) is the eccentricity of the ellipse, defined as \(e = \sqrt{1 - \dfrac{b^2}{a^2}}\).

Because this integral cannot be solved to give a simple polynomial or trigonometric function, it must be evaluated using numerical methods (which computers and advanced calculators do instantly). For the \(a=5, b=3\) example above, a computer evaluating the elliptic integral gives a perimeter of \(25.5269988 \dots \) showing just how incredibly accurate Ramanujan's approximation (25.527) is.