Question

Find area of the region outside the circle r=2 and inside the lemniscate r^2=8cos2α

Answer 1

Find angles, when the circle inside the lemniscate:

$$r^2 = 8 \cos(2\alpha) \geq 2^2$$ $$\cos(2\alpha)\geq 1/2$$ $$\frac{1}{6} \left(6 \pi n-\pi \right)<\alpha <\frac{1}{6} \left(6 \pi n+\pi \right),\,\,n \in \mathbb{Z}$$

Using periodicity over $\alpha$ , lets convert usual convension over domain $0<\alpha<2\pi$ , to more convenient in this case,

$$-\pi/2<\alpha<3\pi/2$$

This correspond to n = 0 and n = 1.

Finally, domain of angles, when the circle inside the lemniscate:

$$-\frac{\pi}{6} <\alpha <\frac{\pi}{6},\,\,\,\,\text{and}\,\,\,\,\,\frac{5\pi}{6} <\alpha <\frac{7\pi}{6}$$

AREA OF A REGION BOUNDED BY A POLAR CURVE:

$$S_0 = \frac12 \int_{-\pi/6}^{\pi/6} (8cos(2α)-2^2)d\alpha = 2(\sin(2α)-\alpha)|_{-\pi/6}^{\pi/6}=2 \sqrt{3}-\frac{2 \pi }{3}$$

As integration in region $\frac{5\pi}{6} <\alpha <\frac{7\pi}{6}$ would give the same result $S_1 = S_0$ ,

therefore final result - area of the region outside the circle and inside the lemniscate equals to

$$4 \sqrt{3}-\frac{4 \pi }{3}$$