**Problem**

The *Mou He Fang Gai* (牟合方蓋) is a solid formed by the intersection of two perpendicular cylinders of equal diameter. This shape was named by the Chinese mathematician Liu Hui (c. 225 - 295 AD) during the Three Kingdoms Era. The volume was found by Zu Geng during the 5th century AD, the son of Zu Chongzhi (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)**

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 .

**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 .

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)**

From part (1), it is deduced that

- .

If expressed using the diameter of the sphere,

- .