# Area of Ellipse/Proof 2

## Theorem

Let $K$ be an ellipse whose major axis is of length $2 a$ and whose minor axis is of length $2 b$.

The area $\AA$ of $K$ is given by:

- $\AA = \pi a b$

## Proof

Let $K$ be an ellipse aligned in a cartesian plane in reduced form.

Then from Equation of Ellipse in Reduced Form:

- $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$

Thus:

- $y = \pm \dfrac b a \sqrt {a^2 - x^2}$

Consider a circle of radius $a$ whose center is at the origin.

From Equation of Circle center Origin, its equation is given by:

- $x^2 + y^2 = a^2$

and so:

- $y = \pm \sqrt {a^2 - x^2}$

The formulas show that each ordinate of the ellipse is $\dfrac b a$ the ordinate of the circle.

Since the same thing is true of the vertical chords:

\(\ds \AA_E\) | \(=\) | \(\ds \dfrac b a \AA_C\) | where $\AA_E$ and $\AA_C$ are the areas of the ellipse and the circle respectively | |||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac b a \paren {\pi a^2}\) | Area of Circle | |||||||||||

\(\ds \) | \(=\) | \(\ds \pi a b\) |

$\blacksquare$

## Historical Note

This was the proof given by the method of Bonaventura Francesco Cavalieri.

It is essentially the same method as Johannes Kepler, except that while Cavalieri used chords to divide the area up, Kepler used thin rectangles.

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.14$: Cavalieri ($\text {1598}$ – $\text {1647}$)