4. Given a tetrahedron $P-ABC$, $PA=a$, $PB=$ $b$, $PC=c$, the projection of $P$ on the base $ABC$ is $O$, which is the centroid of $\triangle ABC$, and $PO=h$. Then $V_{P-ABC}=$ $\qquad$ .
4. $\sqrt{3 h^{4}-2 h^{2}\left(a^{2}+b^{2}+c^{2}\right)+2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}-a^{4}-b^{4}-c^{4}} \cdot h$. As shown in Figure 1, according to the problem, we have
$$
\begin{array}{l}
A O=\sqrt{a^{2}-h^{2}}, \\
B O=\sqrt{b^{2}-h^{2}}, \\
C O=\sqrt{c^{2}-h^{2}} .
\end{array}
$$

Therefore, the lengths of the three medians of $\triangle A B C$ are
$$
\begin{array}{l}
\frac{3}{2} \sqrt{a^{2}-h^{2}}, \\
\frac{3}{2} \sqrt{b^{2}-h^{2}}, \frac{3}{2} \sqrt{c^{2}-h^{2}} .
\end{array}
$$

According to Qin Jiushao's formula, the area of the triangle formed by the three medians is
$$
\begin{array}{l}
\frac{9}{4} \sqrt{4\left(a^{2}-h^{2}\right)\left(b^{2}-h^{2}\right)-\left(a^{2}+b^{2}-c^{2}-h^{2}\right)^{2}} \\
=\frac{9}{4} \sqrt{3 h^{4}-2 h^{2}\left(a^{2}+b^{2}+c^{2}\right)+2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}-a^{4}-b^{4}-c^{4}} .
\end{array}
$$

Since the area of the triangle formed by the three medians is $\frac{3}{4}$ of the area of the original triangle and using the volume formula of the cone, we get the required volume.