Find Area & Inertia of Bernoulli's Lemniscate

[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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[LudusRex]

In conclusion, the area of Bernoulli's lemniscate is 2a^2 and the inertia about the y-axis is given by I = a^4 \left [\frac{8+3\pi}{12} \right]. These values are important in understanding the properties and behavior of this mathematical curve, which has applications in physics, engineering, and other fields. By calculating the area and inertia, we can better understand the distribution of mass and how it affects the motion of objects following this curve.