6 The volume of the solid of revolution obtained by rotating the figure bounded by the curves $x^{2}=4 y, x^{2}=-4 y, x=4, x=-4$ around the $y$-axis is $V_{1}$; the volume of the solid of revolution obtained by rotating the figure composed of points $(x, y)$ that satisfy $x^{2}+y^{2} \leqslant 16$, $x^{2}+(y-2)^{2} \geqslant 4, x^{2}+(y+2)^{2} \geqslant 4$ around the $y$-axis is $V_{2}$, then ( ).
(A) $V_{1}=\frac{1}{2} V_{2}$
(B) $V_{1}=\frac{2}{3} V_{2}$
(C) $V_{1}=V_{2}$
(D) $V_{1}=2 V_{2}$
Solve As shown in the figure, the two figures rotate around the $y$-axis to form solids of revolution that are confined between two parallel planes separated by a distance of 8. If any plane perpendicular to the $y$-axis is used to intersect these two solids of revolution, and the distance of the intersection from the origin is $|y|(|y| \leqslant 4)$, then the area of the resulting cross-section is
$$
\begin{array}{l}
S_{1}=\pi\left(4^{2}-4|y|\right), \\
S_{2}=\pi\left(4^{2}-y^{2}\right)-\pi\left[4-(2-|y|)^{2}\right]=\pi\left(4^{2}-4|y|\right),
\end{array}
$$

Therefore, $S_{1}=S_{2}$. By the principle of Cavalieri, the volumes of the two solids are equal, i.e., $V_{1}=V_{2}$, so the answer is (C).