Example 21 (2002 National High School Competition Question) 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 $V_{1}$, and 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. $V_{1}=\frac{2}{3} V_{2}$
C. $V_{1}=V_{2}$
D. $V_{1}=2 V_{2}$
Solution: Choose C. Reason: As shown in Figure 7-19, both solids of revolution 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 between the intersection and the origin is $|y|$, then the area of the resulting cross-section is $S_{1}=\pi\left(4^{2}-4|y|\right), S_{2}=\pi\left(4^{2}-y^{2}\right)-\pi\left[4-(2-|y|)^{2}\right]=\pi\left(4^{2}-4|y|\right)=S_{1}$.
By the principle of Zu Chongzhi, we know $V_{1}=V_{2}$.