3. In a regular tetrahedron with edge length $a$, a cylinder is embedded such that its height is exactly half the height of the tetrahedron, and its top base is tangent to the tetrahedron's side faces. Then the surface area of this cylinder is $\qquad$ .
3. $\frac{\pi(2 \sqrt{2}+1)}{24} a^{2}$.

As shown in Figure 3, let the regular tetrahedron be $P-ABC$, and $O$ be the centroid of the base $\triangle ABC$. $E$ is the midpoint of the height $PO$, and $AD$ is the median of the equilateral $\triangle ABC$. $F$ is the midpoint of $PD$. It is easy to see that $EF \parallel AD$, and the length of $EF$ is the radius of the base of the cylinder. Given
$$
AB=AC=BC=AP=BP=PC=a
$$

we know $AD=\frac{\sqrt{3}}{2} a$, $OD=\frac{1}{3} AD=\frac{\sqrt{3}}{6} a$,
$$
EF=\frac{1}{2} OD=\frac{\sqrt{3}}{12} a, PD=\frac{\sqrt{3}}{2} a,
$$

the height $PO=\sqrt{PD^{2}-OD^{2}}=\frac{\sqrt{6}}{3} a$,
$$
OE=\frac{1}{2} PO=\frac{\sqrt{6}}{6} a.
$$

Thus, the surface area is
$$
\begin{array}{l}
2 \pi \cdot EF \cdot OE + 2 \pi \cdot EF^{2} = 2 \pi \cdot EF(OE + EF) \\
= 2 \pi \cdot \frac{\sqrt{3}}{12} \left( \frac{\sqrt{6}}{6} + \frac{\sqrt{3}}{12} \right) a^{2} = \frac{\pi(2 \sqrt{2}+1)}{24} a^{2}.
\end{array}
$$