Area of Ellipse

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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 $\mathcal A$ of $K$ is given by:

$\mathcal A = \pi a b$


Proof 1

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


Then from Equation of Ellipse in Reduced Form:

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

Thus:

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

From the geometric interpretation of the definite integral:

\(\displaystyle \mathcal A\) \(=\) \(\displaystyle b \int_{-a}^a \paren {\sqrt {1 - \dfrac {x^2} {a^2} } - \paren {-\sqrt{1 - \dfrac {x^2} {a^2} } } } \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle b \int_{-a}^a 2 \sqrt {1 - \dfrac {x^2} {a^2} } \rd x\)

Let $x = a \sin \theta$ (note that we can do this because $-a \le x \le a$).

Thus:

$\theta = \arcsin \paren {\dfrac x a}$

and:

$\d x = a \cos \theta \rd \theta$

Then:

\(\displaystyle \mathcal A\) \(=\) \(\displaystyle b \int_{\arcsin \paren {\frac {-a} a} }^{\arcsin \paren {\frac a a} } 2 a \sqrt {1 - \frac {\paren {a \sin \theta}^2} {a^2} } \cos \theta \rd \theta\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle b \int_{-\frac {\pi} 2}^{\frac {\pi} 2} 2 a \sqrt {1 - \sin^2 \theta} \cos \theta \rd \theta\)
\(\displaystyle \) \(=\) \(\displaystyle b \int_{-\frac {\pi} 2}^{\frac {\pi} 2} 2 a \sqrt {\cos^2 \theta} \cos \theta \rd \theta\) Sum of Squares of Sine and Cosine
\(\displaystyle \) \(=\) \(\displaystyle a b \int_{-\frac {\pi} 2}^{\frac {\pi} 2} 2 \cos^2\theta \rd \theta\)
\(\displaystyle \) \(=\) \(\displaystyle a b \int_{-\frac {\pi} 2}^{\frac {\pi} 2} \paren {1 + \cos \paren {2 \theta} } \rd \theta\) Double Angle Formula for Cosine: Corollary 1
\(\displaystyle \) \(=\) \(\displaystyle a b \sqbrk {\theta + \frac 1 2 \sin \paren {2 \theta} }_{-\frac {\pi} 2}^{\frac {\pi} 2}\) Integration of Constant and Primitive of Cosine Function
\(\displaystyle \) \(=\) \(\displaystyle a b \paren {\frac {\pi} 2 + \frac 1 2 \sin \paren {2 \cdot \frac {-\pi} 2} - \frac {-\pi} 2 - \frac 1 2 \sin \paren {2 \cdot \frac {\pi} 2} }\)
\(\displaystyle \) \(=\) \(\displaystyle a b \paren {2 \cdot \frac {\pi} 2 + 2 \cdot \frac 1 2 \cdot 0}\)
\(\displaystyle \) \(=\) \(\displaystyle \pi a b\)

$\blacksquare$


Proof 2

Let $K$ be an ellipse aligned in a cartesian coordinate 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: Cartesian: Corollary 2, 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:

\(\displaystyle \mathcal A_E\) \(=\) \(\displaystyle \dfrac b a \mathcal A_C\) where $\mathcal A_E$ and $\mathcal A_C$ are the areas of the ellipse and the circle respectively
\(\displaystyle \) \(=\) \(\displaystyle \dfrac b a \paren {\pi a^2}\) Area of Circle
\(\displaystyle \) \(=\) \(\displaystyle \pi a b\)


$\blacksquare$


Historical Note

The area of an ellipse was proved by Archimedes in his On Conoids and Spheroids.


Sources