4. In tetrahedron $ABCD$, $\angle ADB = \angle BDC = \angle CDA = 60^{\circ}$, the areas of $\triangle ADB$, $\triangle BDC$, and $\triangle CDA$ are $\frac{\sqrt{3}}{2}$, $2$, and $1$, respectively. Then the volume of the tetrahedron is . $\qquad$
(4) $\frac{2 \sqrt{6}}{9}$ Hint: Let $A D=a, B D=b, C D=c$, then $\frac{\sqrt{3}}{4} a b=\frac{\sqrt{3}}{2}$, $\frac{\sqrt{3}}{4} b c=2, \frac{\sqrt{3}}{4} c a=1$, obtaining $a=1, b=2, c=\frac{4}{\sqrt{3}}$. Since the volume of a regular tetrahedron with edge length 1 is $V_{0}=\frac{\sqrt{2}}{12}$, and $\frac{V}{V_{0}}=\frac{1 \times 2 \times \frac{4}{\sqrt{3}}}{1 \times 1 \times 1}=\frac{8 \sqrt{3}}{3}$, then $V=$ $\frac{8 \sqrt{3}}{3} V_{0}=\frac{2 \sqrt{6}}{9}$.
(Annotation: In the tetrahedron $P-A B C$, if $A_{1} 、 B_{1} 、 C_{1}$ are points on the rays $P A 、 P B 、 P C$ respectively, then $\frac{V_{P A_{1} B_{1} C_{1}}}{V_{P A B C}}=$ $\frac{P A_{1} \cdot P B_{1} \cdot P C_{1}}{P A \cdot P B \cdot P C}$. This is because, if $\triangle P A B$ and $\triangle P A_{1} B_{1}$ are taken as the bases, and $C H 、 C_{1} H_{1}$ as the heights, then $\frac{C_{1} H_{1}}{C H}=\frac{P C_{1}}{P C}, \frac{\triangle P A_{1} B_{1}}{\triangle P A B}=$ $\frac{P A_{1} \cdot P B_{1}}{P A \cdot P B}$, hence $\frac{V_{P A_{1} B_{1} C_{1}}}{V_{P A B C}}=\frac{P A_{1} \cdot P B_{1} \cdot P C_{1}}{P A \cdot P B \cdot P C}$. )