Three, a rectangle with a length of $\sqrt{2}$ and a width of 1, is rotated once around the line containing one of its diagonals. Find the volume of the resulting solid of revolution.
$$
\frac{23 \sqrt{3}}{72} \pi
$$

Three, 【Analysis and Solution】As shown in the figure, point $D'$ is the symmetric point of point $D$ with respect to $AC$, and $CD'$ intersects $AB$ at $H$. Construct $BH \perp AC$ at point $F$, $E$, and $HF \perp AC$ at point $F$,
$$
AC = \sqrt{AB^2 + BC^2} = \sqrt{3}, \quad BE = \frac{AB \cdot BC}{AC} = \frac{\sqrt{2}}{\sqrt{3}}, \quad HF = \frac{BC \cdot AF}{AB} = \frac{\sqrt{3}}{\sqrt{2}}
$$

Let the volumes of the solids of revolution formed by rotating $\triangle ABC$, $\triangle ADC$, and $\triangle AHC$ around the line $AC$ be $V_1$, $V_2$, and $V_3$, respectively. Then we have
$$
\begin{array}{l}
V_1 = V_2 = \frac{1}{3} \pi \cdot AC \cdot BE^2 = \frac{2 \sqrt{3}}{9} \pi, \\
V_3 = \frac{1}{3} \pi \cdot AC \cdot HF^2 = \frac{\sqrt{3}}{8} \pi .
\end{array}
$$

The required volume $V = V_1 + V_1 - V_3$
$$
= 2 \times \frac{2 \sqrt{3}}{9} \pi - \frac{\sqrt{3}}{8} \pi = \frac{23 \sqrt{3}}{72} \pi .
$$