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 satisfying $x^{2}+y^{2} \leqslant 16, x^{2}+(y-2)^{2} \geqslant 4, x^{2}+(y+2)^{2} \geqslant 4$ around the $y$-axis is $V_{2}$, then
A. $V_{1}=\frac{1}{2} V_{2}$
B. $V_{1}=\frac{2}{2} V_{2}$
C. $V_{1}=V_{2}$
D. $V_{1}=2 V_{2}$
Answer
As shown in the figure, by symmetry, we only consider the area of the shapes cut by the plane $y=t(t>0)$ with the two geometric bodies.
For the left figure, the outer radius is 4, and the inner radius is $2 \sqrt{t}$, the area of the annulus cut is $S_{1}=\pi(16-4 t)$; for the right figure, the outer radius is $\sqrt{16-t^{2}}$, and the inner radius is $\sqrt{4 t-t^{2}}$, the area of the annulus cut is $S_{2}=\pi\left[\left(16-t^{2}\right)-\left(4 t-t^{2}\right)\right]=\pi(16-4 t)$. Thus, by the principle of Zu Chongzhi, $S_{1}=S_{2} \Rightarrow V_{1}=V_{2}$. Therefore, the answer is $C$.