3. The polar equation of curve $C$ is $\rho=1+\cos \theta$, and the polar coordinates of point $A$ are $(2$, $0)$, If curve $C$ rotates once around $A$ in its plane, then the area of the figure it sweeps through is . $\qquad$
3. Let $p(\rho, \theta)$ be any point on the curve $C$, then $|O P|=\rho=1+\cos \theta$. In $\triangle O A P$, by the cosine rule we have
$$
\begin{aligned}
|A P|^{2} & =|O P|^{2}+|O A|^{2}-2|O P| \cdot|O A| \cos \theta \\
& =(1+\cos \theta)^{2}+4-2 \times 2(1+\cos \theta) \cos \theta \\
& =5-2 \cos \theta-3 \cos ^{2} \theta \\
& =\frac{16}{2}-3\left(\cos \theta+\frac{1}{3}\right)^{2} \leqslant \frac{16}{3} .
\end{aligned}
$$

When $\theta=\arg \cos \left(-\frac{1}{3}\right)$, the above expression reaches equality, so the maximum value of $|A P|$ is $\sqrt{\frac{16}{3}}$. It is easy to see that point $A$ lies on the curve $C$, and as $\theta$ increases from 0 to $\arccos \left(-\frac{1}{3}\right)$, $|A P|$ increases from 0 to $\sqrt{\frac{16}{3}}$.
$\therefore$ The area swept by the curve $C$ is the region enclosed by a circle with center $A$ and radius $\sqrt{\frac{16}{3}}$, which has an area of $\frac{16}{3} \pi$.