# Area of Lobe of Lemniscate of Bernoulli

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## Theorem

Consider the lemniscate of Bernoulli $M$ embedded in a Cartesian plane such that its foci are at $\tuple {a, 0}$ and $\tuple {-a, 0}$ respectively.

Let $O$ denote the origin.

The area of one lobe of $M$ is $a^2$.

## Proof

By the definition of the lemniscate of Bernoulli, we have that the polar equation of $M$ is:

- $r^2 = 2 a^2 \cos 2 \theta$

Let $\mathcal A$ denote the area of one lobe of $M$.

The boundary of the right hand lobe of $M$ is traced out where $-\dfrac \pi 2 \le 2 \theta \le \dfrac \pi 2$.

Thus:

\(\displaystyle \mathcal A\) | \(=\) | \(\displaystyle \int_{-\pi / 4}^{\pi / 4} \dfrac {\map {r^2} \theta} 2 \rd \theta\) | Area between Radii and Curve in Polar Coordinates | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int_{-\pi / 4}^{\pi / 4} \dfrac {2 a^2 \cos 2 \theta} 2 \rd \theta\) | Definition of Lemniscate of Bernoulli | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a^2 \int_{-\pi / 4}^{\pi / 4} \cos 2 \theta \rd \theta\) | simplifying | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a^2 \intlimits {\dfrac {\sin 2 \theta} 2} {-\pi / 4} {\pi / 4}\) | Primitive of $\cos a x$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {a^2} 2 \paren {\sin \dfrac \pi 2 - \map \sin {-\dfrac \pi 2} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {a^2} 2 \paren {1 - \paren {-1} }\) | Sine of Right Angle etc. | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a^2\) |

$\blacksquare$

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 11$: Special Plane Curves: Lemniscate: $11.4$