# Bernoulli's lemniscata

1. Jun 22, 2011

### Quinzio

Find area of Bernoulli's lemniscate: $r^2=2a^2cos2\theta$

$$A = 2 \int_{-\pi/4}^{+\pi/4} \int_{0}^{\sqrt{2a^2cos2\theta}} r \ dr\ d\theta$$

$$A = 2 \int_{-\pi/4}^{+\pi/4} \left [ \frac{r^2}{2} \right ]_{0}^{\sqrt{2a^2cos2\theta}}\ d\theta$$

$$A = 2 \int_{-\pi/4}^{+\pi/4} a^2cos2\theta \ d\theta$$

$$A = 2 a^2 \left [\frac{sen2\theta}{2} \right ]_{-\pi/4}^{+\pi/4}\$$

$$A = 2 a^2$$

Find Bernoulli's lemniscate: $r^2=2a^2cos2\theta$ inertia about $y$ axis.

$$\frac{I}{2} = \int_{-\tfrac{\pi}{4}}^{\frac{\pi}{4}} \int_{0}^{\sqrt{2a^2 cos2\theta}} r^2\ cos^2 \theta\ r \ dr\ d\theta$$

$$\frac{I}{2} = \int_{-\tfrac{\pi}{4}}^{\frac{\pi}{4}} \int_{0}^{\sqrt{2a^2 cos2\theta}} r^3\ cos^2 \theta\ dr\ d\theta$$

$$\frac{I}{2} = \int_{-\tfrac{\pi}{4}}^{\frac{\pi}{4}} \int_{0}^{\sqrt{2a^2 cos2\theta}} r^3\ cos^2 \theta\ dr\ d\theta$$

$$\frac{I}{2} = \int_{-\tfrac{\pi}{4}}^{\frac{\pi}{4}} \left [ \frac{r^4}{4} \right ]_{0}^{\sqrt{2a^2 cos2\theta}} \ cos^2 \theta\ d\theta$$

$$\frac{I}{2} = \int_{-\tfrac{\pi}{4}}^{\frac{\pi}{4}} a^4 cos^2 2\theta \ cos^2 \theta\ d\theta$$

$$\frac{I}{2} = \frac{a^4}{2}\int_{-\tfrac{\pi}{4}}^{\frac{\pi}{4}} cos^2 2\theta \ (cos 2\theta +1 )\ d\theta$$

$$\frac{I}{2} = \frac{a^4}{2}\int_{-\tfrac{\pi}{4}}^{\frac{\pi}{4}} (1-sen^2 2\theta ) \ (cos 2\theta +1 )\ d\theta$$

$$\frac{I}{2} = \frac{a^4}{2}\int_{-\tfrac{\pi}{4}}^{\frac{\pi}{4}} (cos2\theta -cos2\theta sen^2 2\theta+1-sen^2 2\theta) d\theta$$

$$\frac{I}{2} = \frac{a^4}{2}\int_{-\tfrac{\pi}{4}}^{\frac{\pi}{4}} (cos2\theta -cos2\theta sen^2 2\theta+cos^2 2\theta) d\theta$$

$$\frac{I}{2} = \frac{a^4}{2}\int_{-\tfrac{\pi}{4}}^{\frac{\pi}{4}} \left (cos2\theta -cos2\theta sen^2 2\theta+\frac{(1+cos 4\theta)}{2} \right) d\theta$$

$$\frac{I}{2} = \frac{a^4}{2}\int_{-\tfrac{\pi}{4}}^{\frac{\pi}{4}} \left (cos2\theta -cos2\theta sen^2 2\theta+\frac{cos 4\theta}{2} +\frac{1}{2} \right) d\theta$$

$$\frac{I}{2} = \frac{a^4}{2} \left [\frac{1}{2}sen2\theta -\frac{1}{6} sen^3 2\theta+\frac{sen 4\theta}{8} +\frac{\theta}{2} \right]_{-\tfrac{\pi}{4}}^{\frac{\pi}{4}}$$

$$\frac{I}{2} = \frac{a^4}{2} \left [1 -\frac{1}{3} +\frac{\pi}{4} \right]$$

$$I = a^4 \left [\frac{2}{3} +\frac{\pi}{4} \right]$$

$$I = a^4 \left [\frac{8+3\pi}{12} \right]$$