Liu Hui's Puzzle

## **Problem**

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**

Inscribe a sphere inside the bicylinder. Each vertical cross-section is a circle inscribed inside a square. The ratio of the area is

π
4

{\displaystyle \frac{\pi}{4}}

. Since this is continually true for all vertical cross-sections, the volumetric ratio of the bicylinder to the sphere is

4
:
π

{\displaystyle 4:\pi }

. 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

r

{\displaystyle r}

 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

r

2

−

h

2

{\displaystyle {\sqrt {{r}^{2}-{h}^{2}}}}

 at a given height

h

{\displaystyle h}

.

Since each vertical cross-section is a square (pictured right), the area of the cross-section of the bicylinder-octant is

r

2

−

h

2

{\displaystyle {r}^{2}-{h}^{2}}

. To analyze the vertical cross-section of the excess volume, subtract the cube-octant with the bicylinder-octant.

r

2

−
(

r

2

−

h

2

# )

h

2

{\displaystyle {r}^{2}-({r}^{2}-{h}^{2})={h}^{2}}

The cross-section of the excess volume are squares that vary continuously by the height interval

[
0
,
r
]

{\displaystyle [0,r]}

. At one end, the cross-section is a square with area

r

2

{\displaystyle {r}^{2}}

. This linearly decreases until the square reduces to a point. Thus, the excess volume is equivalent to the volume of an inverted square pyramid.

V

e
x
c
e
s
s

=

1
3

r

3

{\displaystyle {V}\_{excess}={\frac {1}{3}}{r}^{3}}

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

V

t
o
t
a
l
e
x
c
e
s
s

=

8
3

r

3

{\displaystyle {V}\_{totalexcess}={\frac {8}{3}}{r}^{3}}

Thus the volume of the bicylinder is given by

V

c
u
b
e

−

V

t
o
t
a
l
e
x
c
e
s
s

{\displaystyle {V}\_{cube}-{V}\_{totalexcess}}

.

V

b
i
c
y
l
i
n
d
e
r

=
8

r

3

−

8
3

r

3

=

16
3

r

3

{\displaystyle {V}\_{bicylinder}=8{r}^{3}-{\frac {8}{3}}{r}^{3}={\frac {16}{3}}{r}^{3}}

**Part 3**

From Part 1, it is deduced that

V

s
p
h
e
r
e

=

π
4

V

b
i
c
y
l
i
n
d
e
r

{\displaystyle {V}\_{sphere}={\frac {\pi }{4}}{V}\_{bicylinder}}

V

s
p
h
e
r
e

=

π
4

(

16
3

r

3

)

{\displaystyle {V}\_{sphere}={\frac {\pi }{4}}\left({\frac {16}{3}}{r}^{3}\right)}

V

s
p
h
e
r
e

=

4
π

3

r

3

.

{\displaystyle {V}\_{sphere}={\frac {4\pi }{3}}{r}^{3}.}

    If expressed using the diameter of the sphere,

V

s
p
h
e
r
e

=

π
6

d

3

.

{\displaystyle {V}\_{sphere}={\frac {\pi }{6}}{d}^{3}.}