Area from Polar Coordinates

The area can be trivially computed geometrically as: $A = \pi(3)^2 - \pi (1)^2= 8 \pi,$ since it is just the difference of two circle areas.

It can also be computed as a double integral, for the right-hand side. For the latter circle, the equation is:

$(r\cos \theta - 1)^2 + r^2 \sin^2 \theta = 1\implies r = 2\cos \theta.$

The double integral is then:

$\int_0^{\pi} \int_{2\cos \theta}^{3} r dr d\theta = \frac12 \int_0^{\pi} (9 - 4\cos^2 \theta) d\theta = \frac12 (9 \pi ) - 2 \int_0^{\pi} \cos^2 \theta d\theta= \frac 92 \pi- \pi .$

Since the area of the left-hand side of the region is just a semicircle of area $\frac92 \pi$ , the total area is:

$9\pi - \pi = 8\pi.$