3. Given point $P \in\left\{(x, y) \left\lvert\, \frac{(x-\cos \theta)^{2}}{4}+\left(y-\frac{1}{2} \sin \theta\right)^{2}=1\right., \theta \in R\right\}$, then the area of the figure formed by points $P$ on the plane satisfying the condition is $\qquad$ (the area formula for an ellipse is $S=a b \pi, a, b$ are the semi-major axis and semi-minor axis respectively)
$3.4 \pi$
Explanation: When $\theta$ is fixed, the curve is an ellipse $C$, the center of which lies on the ellipse $x^{2}+4 y^{2}=1$. Therefore, as $\theta$ changes, the ellipse $C$ also changes, but its size remains constant. Hence, the area swept by the ellipse $C$ is the area of $\frac{x^{2}}{9}+\frac{4 y^{2}}{9}=1$ minus the area of $x^{2}+4 y^{2}=1$, which is $4 \pi$.