1. Area in polar coordinates

Calculate in polar coordinates the area of the region outside of $P=1$, $P=2cos(\theta)$ and in inside $P=2$

This is correct:

$A=2\int_{0}^{\frac{\pi}{3}} \int_{2cos\theta}^{2} PdPd\theta + 2\int_{\frac{\pi}{3}}^{\pi} \int_{1}^{2} PdPd\theta$

and the answer is: $(\frac{8\pi}{3}-\frac{\sqrt{3}}{2})$

2. Hello, Apprentice123!

What does $P$ stand for in your equation?

Calculate in polar coordinates the area of the region
outside of $r_1=1,\;\;r_2 =2\cos\theta$, and inside $r_3=2$
I don't see the need for double integrals . . .

. . $\begin{array}{c}r_1\text{ is a circle: center (0,0), radius 1} \\ r_2 \text{ is a circle: center (1,0), radius 1} \\ r_3\text{ is a circle: center (0,0), radius 2} \end{array}$

The region is symmetric to the $x$-axis.
. . We can determine the area of the upper half and multiply by 2.

From $\theta = 0$ to $\theta = \tfrac{\pi}{3}$, the area is between $r_3\text{ and }r_2$

. . $A_1 \;=\;\tfrac{1}{2}\int^{\frac{\pi}{3}}_0\bigg[ 2^2 - (2\cos\theta)^2\bigg]\,d\theta$

From $\theta = \tfrac{\pi}{3}$ to $\theta = \pi$, the area is between $r_3\text{ and }r_1$

. . $A_2 \;=\;\tfrac{1}{2}\int^{\pi}_{\frac{\pi}{3}} \bigg[2^2 - 1^2\bigg]\,d\theta$

Add the two areas and multiply by 2 . . .