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) $V_{1}=2 V_{2}$
(C)
6. 【Analysis and Solution】As shown in the figure, the two figures rotate around the $y$-axis to form solids of revolution, which are sandwiched between two parallel planes 8 units apart. If any plane perpendicular to the $y$-axis is used to intersect these two solids of revolution, and the distance from the intersection to the origin is $|y|$, then the area of the intersection is
$$
\begin{array}{l}
S_{1}=\pi\left(4^{2}-4|y|\right), \\
S_{2}=\pi\left(4^{2}-y^{2}\right)-\pi\left[4-\left(2-|y|^{2}\right)\right]=\pi\left(4^{2}-4|y|\right) . \\
\therefore S_{1}=S_{2} .
\end{array}
$$

By the principle of Cavalieri, the volumes of the two solids are equal,
$$
\therefore V_{1}=V_{2} \text {, hence the answer is C. }
$$