41 Two cones have the same axis of symmetry, with base diameters and heights both equal to 1. One is placed upright, and the other is inverted, with the distance between the two bases being \(x(0 \leqslant x \leqslant 2)\). Try to express the volume \(V\) of the common part of the two cones in terms of \(x\).
41 (1) $1<x \leqslant 2$. The axial section is shown in Fig. 26(a). Here, $A^{\prime} P=x-1$.
$$
\begin{array}{l}
A Q=A^{\prime} Q=\frac{1}{2}(1-(x-1))=1-\frac{x}{2} . \\
\text { The original cone volume }=\frac{1}{3} \pi \times\left(\frac{1}{2}\right)^{2}=\frac{\pi}{12} \text {. } \\
V=2 V_{\text {circular } A B C}=2 \times \frac{\pi}{12} \times\left(1-\frac{x}{2}\right)^{3} \\
=\frac{\pi}{6}\left(1-\frac{3}{2} x+\frac{3}{4} x^{2}-\frac{1}{8} x^{3}\right) \text {. } \\
\end{array}
$$
(2) $0 \leqslant x \leqslant 1$.

The axial section is shown in Fig. 26(b), where $P^{\prime} Q=\frac{x}{2}$,
$$
A P^{\prime}=1-x, A Q=1-\frac{x}{2} \text {. }
$$

Therefore,
$$
\begin{array}{l}
=2\left[\left(1-\frac{x}{2}\right)^{3} V_{0}-(1-x)^{3} V_{0}\right] \\
=\frac{\pi}{12} x\left(3-\frac{9}{2} x+\frac{7}{4} x^{2}\right) \text {. } \\
\end{array}
$$

Combining (1) and (2), we have
Fig. 26
$$
V=\left\{\begin{array}{ll}
\frac{\pi}{12} x\left(3-\frac{9}{2} x+\frac{7}{4} x^{2}\right) & (0 \leqslant x \leqslant 1), \\
\frac{\pi}{6}\left(1-\frac{3}{2} x+\frac{3}{4} x^{2}-\frac{1}{8} x^{3}\right) & (1<x \leqslant 2)
\end{array}\right.
$$