相交圆柱体体积

答案是 $\frac{16}{3}$。

对高度 $z$ 的水平截面，交集截面为边长 $2\sqrt{1-z^2}$ 的正方形，面积为 $4(1-z^2)$。

积分：

$$V=\int_{-1}^{1}4(1-z^2)\,dz=\frac{16}{3}.$$

---

Original Explanation

16/3

Solution

If you cut the intersection by a horizontal plane at distance z from center, the cut will be a square with side-length 2 * sqrt( 1-z^2). Integrate to get volume 16/3.

Another way is to imagine the largest possible sphere inscribed at the center of intersection. The sphere should have a radius of 1. At each cut perpendicular to the z-axis, the circle from the sphere is inscribed in the square from the intersection as well, So Area of cut-circle = (Pi/4) * Area of cut-square. This is true for all z, hence Volume of sphere = (Pi/4) * Volume of Intersection, this also gives 16/3