MHB What is the area inside the lemniscate and outside the circle?

[Fernando Revilla]

I quote a question from Yahoo! Answers

Find the area inside the lemniscate r^2=150cos(2theta) and outside the circle r=5sqrt(3).

I have given a link to the topic there so the OP can see my response.

 

[Fernando Revilla]

By symmetry arguments, the area is $A=2A_1$ where $A_1$ is the corresponding area on the right half plane. Equating modules, $$150\cos 2\theta=(4\sqrt{3})^2\Leftrightarrow 150\cos 2\theta=75 \Leftrightarrow \cos 2\theta=\frac{1}{2}\Leftrightarrow \theta=\pm \frac{\pi}{6}$$ Using a well known formula: $$A=2A_1=2\cdot\frac{1}{2}\int_{-\pi/6}^{\pi/6}\left(r_2^2-r_1^2\right)d\theta=\int_{-\pi/6}^{\pi/6}\left(150\cos 2\theta-75\right)d\theta\\=\left[75\sin 2\theta-75\theta\right]_{-\pi/6}^{\pi/6}=75\left(\sqrt{3}-\frac{\pi}{3}\right)$$