Math & Physics Problems Wikia
Register
Advertisement

Problem[]

Bicylinder

Figure 1. Two intersecting cylinders and the mou he fang fai (牟合方蓋)

The mou he fang gai (牟合方蓋) is a solid formed by the intersection of two perpendicular cylinders of equal diameter (Figure 1). This shape was named by the Chinese mathematician Liu Hui (劉徽, c. 225 - 295 AD) during the Three Kingdoms Era (220 - 280 AD). The volume was later found by Zu Geng (祖暅) during the 5th century AD, the son of Zu Chongzhi (祖沖之, 429 - 500 AD).  Nowadays this solid is called a bicylinder.

Part 1: Determine the volumetric ratio of the bicylinder to the sphere.

Part 2: Obtain the formula for the volume of the bicylinder.

Part 3: Using the volume of the bicylinder obtain the formula for the volume of the sphere.

Solution[]

Part 1

Screen-shot-2014-04-06-at-8-03-02-pm

Figure 2. Mou he fang gai and the inscribed sphere

Inscribe a sphere inside the bicylinder. Each vertical cross-section is a circle inscribed inside a square. The ratio of the area is . Since this is continually true for all vertical cross-sections, the volumetric ratio of the bicylinder to the sphere is . This deduction exploits the Liu-Zu principle (劉-祖原理), more commonly known as the Cavalieri principle or the "method of indivisibles".

Part 2

Inscribe the bicylinder in a cube. Partition the solids into octants. Let be the width of the cube-octant. The horizontal cross-section of the bicylinder-octant is a circular quadrant; hence, the width of the vertical cross-section is given by at a given height .

41

Figure 3, Zu Geng's geometric reasoning 1

Since each vertical cross-section is a square (pictured right), the area of the cross-section of the bicylinder-octant is . To analyze the vertical cross-section of the excess volume, subtract the cube-octant with the bicylinder-octant.

The cross-section of the excess volume are squares that vary continuously by the height interval . At one end, the cross-section is a square with area . This linearly decreases until the square reduces to a point. Thus, the excess volume is equivalent to the volume of an inverted square pyramid.

Multiply the excess volume by 8 to obtain the volume of the excess in all octants.

Thus the volume of the bicylinder is given by .


Part 3

Sphere volume derivation using bicylinder

Figure 4, Zu Geng's geometric reasoning 2

From Part 1, it is deduced that

Bicylinder Steinmetz solid

Figure 5. Animation of the formation of the mou he fang gai

If expressed using the diameter of the sphere,


Advertisement