3. The polar equation of curve $C$ is $\rho=1+\cos \theta$, and the polar coordinates of point $A$ are $(2,0)$. Curve $C$ rotates once around $A$ in its plane, then the area of the figure it sweeps through is
$$
f(\theta)=1+\cos \theta.
$$

Since $2=1+\cos 0$, point $A$ lies on the curve $C$. To find the area swept over, the key is to calculate the maximum distance from a point on $C$ to $A$. For a point $B(1+\cos \theta, \theta)$ on $C$,
$$
\begin{aligned}
|A B|^{2} & =(1+\cos \theta)^{2}+4-2 \times 2(1+\cos \theta) \cos \theta \\
& =5-2 \cos \theta-3 \cos ^{2} \theta \\
& =\frac{16}{3}-3\left(\cos \theta+\frac{1}{3}\right)^{2} \leqslant \frac{16}{3} .
\end{aligned}
$$

Equality holds when $\cos \theta=-\frac{1}{3}$, so the maximum value of $|A B|$ is $\sqrt{\frac{16}{3}}$. Therefore, the area swept over is the area of a circle with center $A$ and radius $\sqrt{\frac{16}{3}}$, which is $\frac{16}{3} \pi$.