6. The volume of the solid of revolution obtained by rotating the figure enclosed 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. $\mathrm{C}$

As shown in the figure below, the two figures, when rotated around the $y$-axis, form rotational solids 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 rotational solids, and the distance of the intersection from the origin is $|y|$, then the area of the resulting cross-section is $S_{1}=\pi\left(4^{2}-4|y|\right), S_{2}=\pi\left(4^{2}-y^{2}\right)-\pi\left[4 \cdots(2-|y|)^{2}\right\}$ $=\pi\left(4^{2}-4|y|\right)$, thus $S_{1}=S_{2}$.

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