4. Given that the radius of $\odot O$ is 1. Then the area of the region formed by the orthocenters of all inscribed triangles in $\odot O$ is $\qquad$ .
4. $9 \pi$.

If point $H$ is inside $\odot O$, and a chord $AD$ is drawn through $H$ such that $AH \leqslant HD$, a chord $BC$ is drawn perpendicular to and bisecting $HD$. Connect $BD$, $BH$, and extend $BH$ to intersect $AC$ at point $E$, as shown in Figure 5.
$$
\begin{array}{c}
\text { Then } \angle C=\angle D \\
=\angle 1=\angle 2 . \\
\text { Therefore, } \angle BEC=\angle CAD+\angle 2 \\
=\angle CAD+\angle C=90^{\circ} .
\end{array}
$$

Thus, $H$ is the orthocenter of the inscribed $\triangle ABC$ in $\odot O$.
If point $H$ is on $\odot O$, then $H$ is the orthocenter of the inscribed right triangle in $\odot O$ with $H$ as the right angle vertex.

If point $H$ is outside $\odot O$ and $OH<3$, then the line $OH$ intersects $\odot O$ at two points $A$ and $D$, with $HA < HD$.

Draw a chord $BC$ of $\odot O$ perpendicular to and bisecting $HD$ (if $OH \geqslant 3$, then the chord $BC$ does not exist), and let $CA$ intersect $BH$ at point $F$, as shown in Figure 6. Thus,
$$
\angle C=\angle D=\angle H \text {. }
$$

Then $\angle CFH=90^{\circ}$.
Therefore, $H$ is the orthocenter of $\triangle ABC$.
Hence, the region formed by the orthocenters is the interior of a circle with center $O$ and radius 3, whose area is $9 \pi$.