3. The area of the closed figure enclosed by the curve $x^{2}+2 y^{2} \leqslant 4|y|$ is $\qquad$
3. $2 \sqrt{2} \pi$.

By the symmetry of the graph of the given curve equation, we only need to consider the case where $y \geqslant 0$.
When $y \geqslant 0$, the curve equation becomes
$$
\frac{x^{2}}{2}+(y-1)^{2} \leqslant 1 \text {, }
$$

which represents an elliptical surface. The area of this part is
$$
S=\pi \times \sqrt{2} \times 1=\sqrt{2} \pi .
$$

Therefore, the required area is $2 \sqrt{2} \pi$.