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) $V_{1}=2 V_{2}$
6. (C).

As shown in Figure 1, the two figures rotate around the $y$-axis to form solids of revolution that are confined between two parallel planes 8 units apart. If any plane perpendicular to the $y$-axis is used to intersect these solids of revolution, and the distance of the intersection from the origin is $|y|$, then the area of the resulting cross-section is
$$
\begin{aligned}
S_{1} & =\pi\left(4^{2}-4|y|\right), S_{2}=\pi\left(4^{2}-y^{2}\right)-\pi[4-(2- \\
\left.|y|)^{2}\right] & =\pi\left(4^{2}-4|y|\right) \text {. Therefore, } S_{1}=S_{2} .
\end{aligned}
$$

By the principle of Cavalieri, the volumes of the two solids are equal, i.e., $V_{1}=V_{2}$.