Example 35 (2002 National High School Competition Question) As shown in Figure $1-12$, 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. $V_{1}=2 V_{2}$
Choose C. Reason: Observing Figure 1-12, it seems that the shaded areas of the two figures are equal, and the trend of change is from small to large, then to small again, so the answer is likely to be C. Below, we will actually derive this:

As shown in Figure 1-12, the two figures rotate around the $y$-axis to form solids of revolution, which are sandwiched between two parallel planes 8 units apart. If we cut these solids of revolution with any plane perpendicular to the $y$-axis, and let the distance of the cross-section from the origin be $|y|$, then the cross-sectional areas are
$$
S_{1}=\pi\left(4^{2}-4|y|\right), S_{2}=\pi\left(4^{2}-y^{2}\right)-\pi\left[4-\left(2-|y|^{2}\right)\right]=\pi\left(4^{2}-4|y|\right) \text {. Therefore, } S_{1}=S_{2} \text {. }
$$

By the principle of Zu Xuan, the volumes of the two geometric bodies are equal, i.e., $V_{1}=V_{2}$.