- 1. Introduction
- 2. The use of Cavalieris theorem by Liu Hui and Zu Gengzhi.
- 3. Summary of the proof
- 4. The attribution of Liu Huis commentary
- 5. Note on the translation
- 6. Translation
- [6.1. Problems 23 and 24 of the Jiu zhang suanshu]
- [6.2.] Method for the extraction of the spherical root:
- [6.3. Liu Huis comment]
- [6.3.1. The incorrect derivation of the Jiu zhang suanshu]
- [6.3.2. This derivation is incorrect]
- [6.3.3. The formula is nevertheless not too bad]
- [6.3.4. The geometers frustration]
- [6.4. Zhang Heng on the sphere]
- [6.5. Li Chunfengs comment]
- 7. References

[The name Zu Gengzhi is now often transcribed, probably more correctly, Zu Xuanzhi.]

I translate here a remarkable proof of the formula for the volume of a
sphere, written by Zu Gengzhi 祖暅之 in the late fifth century A.D. The proof is quoted from some lost book in Li
Chunfeng's 李淳風 (602-670) commentary on the *Jiu zhang suanshu* 九章算術 (Arithmetic in nine chapters). The *Jiu zhang suanshu* appears to have
reached its present form in the first century A.D. (Qian Baocong 1964: 32-33).
It has two problems in which the volume of a sphere is given and the diameter
is to be calculated. The algorithm given is equivalent to

Since the true formula is

i.e.

it might be thought that the *Jiu zhang suanshu* here used the value π ≈ 3^{3}/_{8} (cf. Vogel 1968: 137; Juschkewitsch 1964: 57). This is unlikely, since
everywhere else in the *Jiu zhang suanshu* the value π≈3 is used.

A commentary attributed to Liu Hui 劉徽 (active third century A.D.)[2] suggests instead that the formula should be interpreted as

or, equivalently,

and gives the following reasoning. The volume of a cylinder inscribed in a cube is π/4 times the volume of the cube. If one then makes the incorrect assumption that the volume of an inscribed sphere is π/4 times the volume of the inscribed cylinder, the formula above is obtained.

Liu Hui proves that the assumption is incorrect by showing that this relation
in fact holds between the volume of a sphere and that of another object,
smaller than the cylinder. This object he calls a "box-lid" (*he gai* 合[= 盒]蓋):
it is the intersection of two cylinders inscribed in the cube with
perpendicular axes.[3] Having proved that the
given formula is incorrect, Liu Hui is unable to find the correct formula, and
he concludes with a little poem which I have titled "The geometer's
frustration" (Section 6.3.4 below).

Two centuries later Zu Gengzhi takes up the problem, finds and proves the correct formula, and concludes with a poem which I have titled "The geometer's triumph" (Section 6.5.2 below).

The relevant passages from the *Jiu zhang suanshu* and the commentaries of
Liu Hui and Li Chunfeng are translated in Section 6 below. In the next two
Sections I discuss the methods employed in the proofs, and in Section 4 I
discuss a textual problem.

Any treatment of the volume of a sphere must use infinitesimal considerations in one form or another. Zu Gengzhi explicitly uses an assumption which is equivalent to Cavalieri's Theorem. Cavalieri's Theorem, as applied to volumes, may be stated as follows:

If two objects can be so oriented that there exists a plane such that any plane parallel to it intersects equal areas in both objects, then the volumes of the two objects are equal.[4]Zu Gengzhi's statement of his basic assumption (Section 6.5.1 below) is rhymed and rather cryptic:

疊棋成立積Though this statement might be understood in any number of ways, it is clear from the context that it is in fact a statement of Cavalieri's Theorem. The commentator Li Huang 李潢 (d. A.D. 1812) emended the second character,

緣幂勢既同

則積不容異

If blocks are piled up to form volumes,

And corresponding areas are equal,

Then the volumes cannot be unequal.

Zu Gengzhi's explicit use of Cavalieri's Theorem has a forerunner in an implicit assumption used several times by Liu Hui. It may be stated as follows:

If an object with circular cross-section is inscribed in an object with square cross-section, and every circular cross-section is inscribed in the corresponding square cross-section, then the ratio of the volumes of the objects is equal to the ratio of the areas of a circle and a circumscribed square, i.e. π:4.An example of Liu Hui's use of this assumption can be seen in his treatment of the volumes of the box-lid and sphere, translated in Section 6.3.2 below.[6]

The origin of Zu Gengzhi's use of Cavalieri's Theorem may therefore be reconstructed as follows. The assumption used by Liu Hui was a kind of inspired common sense, useful in working with such common solids as the cylinder and the cone. Zu Gengzhi, or someone before him, was led to think about the reason why it should be true, and hit upon the more general statement which he used.

It seems that we have no other Chinese mathematical writings before modern
times which have any relation to Cavalieri's Theorem. The mathematical writings
of Zu Gengzhi and his father, Zu Chongzhi 祖冲之 (A.D. 429–500), are completely lost except for some odd fragments like the one
translated here. Thus we have no way of knowing whether Cavalieri's Theorem was
used in other problems, or whether Zu Gengzhi wrote anything more precise about
it. We may note, however, that a mathematical book, now lost, titled *Zhui
shu* 綴術,
was written by either Zu Gengzhi or his father. The book is generally believed
to have been concerned with successive approximations and other infinitesimal
considerations; for example, Zu Chongzhi's calculation of π is thought to
have been in this book.[7]

It seems possible, then, that Zu Gengzhi could have worked out some sort of derivation of Cavalieri's Theorem. The "blocks" in his statement of the theorem might then be cross-sections of non-zero thickness which, in his derivation, would be reduced in thickness to a limit.

Liu Hui's comment, translated in Section 6.3.2 below, gives the
following construction. Circumscribe a cube on the sphere. Inscribe two
cylinders with perpendicular axes in the cube. Then the intersection of these
two cylinders is referred to as a "box-lid". Figure 5 shows one corner of this
object. The volume of the sphere is stated to be _{ }times
the volume of the box-lid. This is correct, as can be shown using Cavalieri's
Theorem: it is only necessary to observe that any plane parallel to the plane
of the axes of the cylinders intersects the box-lid in a square, and intersects
the sphere in a circle inscribed in this square. The area of the circle is π/4 times
the area of the square.

The problem that remains is to find the relationship between the volume of the box-lid and the volume of the cube. The treatise by Zu Gengzhi quoted in Li Chunfeng's comment solves the problem as follows.

Consider a smaller cube cut out of the original cube as in Figure 1. Its edge
is *r* =* d */2 = one-half the edge of the original cube. Figure 3 shows the situation in this
smaller cube. Figures 5-8 show the pieces cut out of the smaller cube by the
inscribed cylinders.

Consider the intersection of this smaller cube with an arbitrary plane NPQR
parallel to the base ABCD of the cube as shown in Figure 4. Let *a* = AN
be the altitude of this plane above the base, and let *b* = PW = NV = ST.
Then the following areas are known:

NPQR=r^{2}SPWT=b^{2}

Since *NPQR* = *SPWT* + *NSTV* + *RVTX* + *TWQX* we have,

NSTV+RVTX+TWQX=r^{2}–b^{2}

But considering the right triangle ANV, by the Pythagorean Theorem,

r^{2}–b^{2}=a^{2}

NSTV+RVTX+TWQX=a^{2}

This is the area of the intersection of the plane NPQR with the volume bounded by the box-lid and the cube (the sum of the shaded areas in Figures 6–8).

Now consider a pyramid (Figure 9) with square base and one lateral edge
parallel to the base, with the altitude and the side of the base equal to *r* = the edge of the small cube. Liu Hui has already proved[8] that the volume of this pyramid is *r*^{3}/3.
The intersection of this pyramid with a plane parallel to its base at a
distance *a* from the apex has area *a*^{2}. Thus by Cavalieri's Theorem the volume bounded by the box-lid and the cube is
equal to the volume of the pyramid. Therefore the volume of the corner of the
box-lid (ABCDH, see Figure 5) is 2*r*^{3}/3,
and the volume of the whole box-lid is

8 × 2r^{3}/3 = 8 × 2(d/2)^{3}/3 =2d^{3}/3

Taking the results of Liu Hui and Zu Gengzhi together, we have the volume of the sphere:

V = π/4 × volume of box-lid = π/4 × 2

d^{3}/3= π

d^{3}/6

I have elsewhere expressed doubts as to whether the commentary on the *Jiu zhang suanshu* attributed to Liu Hui is in fact entirely by him
(Wagner 1978). Some parts are certainly his; others seem to be later
interpolations. It would seem that the commentary is a conflation of two or
more commentaries, probably all written before the time of Li Chunfeng.

One of the passages in which the problem of attribution arises is the one with which we are concerned here. The comment attributed to Liu Hui clearly states and proves that it is incorrect to take the ratio of the volumes of the sphere and the cylinder as π : 4 (Section 6.3.2 below); yet Zu Gengzhi states that Liu Hui did just that (Sections 6.5 and 6.5.2). But he must have seen the passage attributed to Liu Hui, for "The geometer's triumph" is clearly an answer to "The geometer's frustration". It seems most likely, therefore, that at least part of this passage is not by Liu Hui; only the first sentences, up to the statement translated "but this reasoning is incorrect", can be his.

The passage beginning with this statement and ending with "The geometer's
frustration" is closely related, both in style and methodology, to the passage
by Zu Gengzhi. Therefore it seems very possible that it is also by him, or by
his father, Zu Chongzhi. Zu Chongzhi wrote a commentary on the *Jiu zhang
suanshu*, now lost;[9] it seems quite
possible that Li Chunfeng incorporated it into his edition, and that then or
later Liu Hui's commentary and this one were jumbled.

In the next section I translate the original problems from the *Jiu
zhang suanshu* and the relevant parts of the commentaries by Liu Hui and Li
Chunfeng. I have had considerable help in understanding this material from an
article by Li Yan (1963: 59-61), and the Figures are redrawn directly from this
article. The text used is that of Qian Baocong's punctuated edition (1963: 154,
line 18 – p. 158, line 10). In the translation mathematical comments are
indented in the text, while philological comments are given in the footnotes.

[Problem 23.] Consider a volume of [*V* =] 4500 [cubic] *chi* 尺.
If it is a sphere, what is the diameter [*d*]?

Answer: [*d* =] 20 *chi*.

[Problem 24.] Consider a volume of [*V* =] 1,644,866,437,500 [cubic] *chi*. If it is a sphere, what is the diameter [*d*]?

Answer: [*d* =] 14,300 *chi*.

The termkai liyuan開立圓 is obviously analogous to such terms askai fang開方, "extraction of the square root", andkai lifang開立方, "extraction of the cube root"; therefore I have translated it as "extraction of the spherical root". Note also problems 17 and 18 of chapter 4:kai yuan開圓, "extraction of the circular root" (Qian Baocong 1963: 152).

Lay out the number of [cubic] *chi* in the volume; multiply by 16; divide
by 9; extract the cube root of the result; this is the diameter of the sphere
[*wan丸*].

Note the use of the word

wanfor "sphere" here; elsewhereliyuan立圓 is used in the same meaning.

*Liyuan* [lit., "standing circle"] means *wan* [lit., "pellet"].

The creator of the method used the proportions circumference 3, diameter 1. If it is supposed that the area of a circle occupies [π/4 ≈] 3/4 of the area of a [circumscribed] square, then a cylinder also occupies 3/4 of a [circumscribed] cube. Further, supposing [incorrectly] that the cylinder has the proportion of a square, 12, and that the proportion of the sphere is 9, then the sphere also occupies 3/4 of the [circumscribed] cylinder. Laying out 4 parts and multiplying by itself gives 16 parts; multiplying 3 by itself gives 9; thus the sphere would occupy 9/16 of the cube.

(π/4)Therefore multiplying the volume by 16 and dividing by 9 would give the volume of the [circumscribed] cube. The diameter of the sphere and [the edge of] the cube are equal; thus extracting the cube root would give the diameter.^{2}≈ (3/4)^{2}= 9/16

But this reasoning is incorrect [*ran ci yi fei ye *然此意非也].
How is this to be verified? Take eight cubical blocks [*qi* 棋],
letting each have [the edge of] the cube equal to 1 *cun* 寸;

10assemble them to make a cube of 2cun寸 = 1chi尺.

That is, inscribe a cylinder in the 2-Further round it perpendicularly [cuncube; see Figure 2.

That is, inscribe a second cylinder with axis perpendicular to that of the first.Then its form resembles the square lid of a

AEach of the eight blocks now resembles amouheseems to have been some sort of box. In the following the expressionmouhe fanggai 牟合方蓋is abbreviated tohegai合蓋 and I translate this as "box-lid".[10]

See Figure 5; aAs to the box-lid, it has the proportion of a square. The sphere is inside it [i.e., is inscribed in it], so it has the proportion of a circle.yangmawas a pyramid with rectangular base and with one lateral edge perpendicular to the base (cf. Wagner 1979: 166).

That is, the ratio of the volumes of the box-lid and the sphere is equal to the ratio of the areas of a square and an inscribed circle, 4:π. See Section 2 above.From this consideration it is clearly incorrect to let the proportion of the cylinder be that of a square.

That is, it is incorrect to assume that the ratio of the volumes of a cylinder and an inscribed sphere is 4:π. Here Liu Hui has shown that this ratio holds for the box-lid and the inscribed square, and the box-lid is clearly smaller than the cylinder.

If the proportions of a circle are taken to be circumference 3, diameter 1 [i.e., if πis taken to be 3], then the [calculated] area of the circle is too small. If it is supposed that the cylinder has the proportion of a square, then the [calculated] volume of the sphere is too large. [These errors] make up for each other; therefore the proportions 9 and 16 are accidentally close to the true [proportions], but [the calculated volume of] the sphere is still too large.

V< 9d^{3}/16Liu Hui has proved that

V≠ 9d^{3}/16, and below he states that he is unable to find the correct formula. How then does he know thatV< 9d^{3}/16 ? It may be that he knew the correct formula,V= πd^{3}/6, but was unable to prove it. I am more inclined to think that he arrived at this inequality by an impressionistic comparison of the magnitudes of the two errors mentioned above.

Liu Hui has now solved part of the problem. The difficulty that remains is to find the volume of the box-lid. He concludes with the following bit of doggerel.Look inside the cube

And outside the box-lid;

Though the diminution increases,

It doesn't quite fit.

The marriage preparations are complete;

But square and circle
wrangle,

Thick and thin make treacherous plots,

They are incompatible.

I wish to give my humble reflections,

But fear that I will miss the correct
principle;

I dare to let the doubtful points stand,

Waiting for one who
can expound them.

I have here ignored a long discussion of Zhang Heng's 張衡(A.D. 78-139) calculations concerning the volumes of a sphere and the inscribed and circumscribed cubes (Qian Baocong 1963: 156, line 7 - p. 157, line 4). This discussion is interesting, as it gives us a glimpse of Zhang Heng's mathematics; but it is not relevant to our present concern.

Your subjects[11] [Li] Chunfeng and others respectfully comment. Zu Gengzhi says that both Liu Hui and Zhang Heng took the cylinder to have the proportion of a square and the sphere to have the proportion of a circle, and that he therefore established a new method. Zu Gengzhi's "Extraction of the spherical root" says:

Multiplying the volume by 2 and extracting the cube root gives the diameter of the sphere.

Since Zu Gengzhi proves below thatWhat is the reasoning behind this?V= πd^{3}/6, this statement implies the use of the value π ≈ 3. Zu Gengzhi knew that π ≠ 3; his father Zu Chongzhi calculated the value of π to eight places.[12] Possibly he uses the value π ≈ 3 in order to facilitate comparison with theJiu zhang suanshutext.

Take a cubical block. Place the pivot [of a drawing compass] at the left back
lower corner [point D in Figure 3]. Draw an arc [HC] vertically. Remove the
right upper side-piece [*lian*廉,
i.e. the curved object HGCBFE]. Place the pieces together again, draw an arc
[HA] horizontally, and remove the front upper side-piece [i.e. the curved
object HEABFG]. Now the cubical block is divided into four. There is one block
inside the arcs [the curved object HABCD; see Figure 5], called the inner
block. There are three blocks outside the arcs [HEAB, BEFGH, and HGCB; see
Figures 6, 7, and 8 respectively], and these are called the outer blocks.

In the following see Figure 4.Again put the blocks together, and cut horizontally [on any plane NPQR parallel to ABCD].

Explaining in terms of right triangles [i.e., using the Pythagorean Theorem]:
Let the remaining height [*a* = AN] be the base [of a right triangle]. Let
[the edge *b* = ST of] the upper cut square [SPWT] of the inner block be
the height. The magnitude [i.e., the length of the side, *r*]* *of
the original cube is the hypotenuse. [The right triangle is ANV.]

According to the method of right triangles, if the area of [the square on] the
base is subtracted from the area of [the square on] the hypotenuse, then the
difference is the area of [the square on] the height. If the remaining height
[*a*] is multiplied by itself, and this result is subtracted from the area
[*r*^{2}] of [one face of] the original cube, then the difference
is the area of the upper cut square [SPWT] of the inner block.

The area [*r*^{2}] of [one face of] the original cube is the area
of the upper cut [NPQR] of all four inner and outer blocks.

Therefore the remaining height [*a*], multiplied by itself, is the area of
the upper cut [TSNRQW] of the three outer blocks.

Regardless of the height [

r^{2}–a^{2}=b^{2}=SPWT

r^{2}= NPQR

NPQR–SPWT=TSNRQWTherefore

TSNRQW=a^{2}

Go far afield to expound the situation,

Borrow a comparison to analyze the subtleties.

In the following see Figure 9.Take a

AInvert it, cut horizontally [along the shaded plane], and remove the upper part. Then the product of the height [of the cut,yangmais a pyramid with rectangular base and with one edge perpendicular to the base (Wagner 1979: 166).

This is because the cross-section, shaded in Figure 9, is a square with sideNow,a.

If blocks are piled up to form volumes,

And corresponding areas are equal,

Then the volumes cannot be unequal.

This is a statement of Cavalieri's Theorem. See Section 2 above.

Seen in this way, if the three blocks outside the arcs are contracted at the
sides to form one, then they form one *yangma*. If [the volume of] the
cube is divided in three parts, then it is obvious that the *yangma* occupies one part and the inner block occupies two parts.

This follows from the fact, proved by Liu Hui, that the volume of aIf the eight small cubes are united to form a large cube, then the eight inner blocks are united to form a box-lid. The inner blocks occupy two-thirds of the small cubes; therefore the box-lid also occupies two-thirds of the [large] cube, as can be verified by comparison. Laying out 2/3, multiplying by the proportion of a circle, [π≈] 3, dividing by the proportion of a square, 4, and simplifying, gives the proportion of a sphere [2/3 ×3/4 =1/2]. Therefore [I] have said that a sphere occupies one-half of a [circumscribed] cube.yangmaissth/3, wheresandtare the sides of the rectangular base andhis the height.

The proportions are extremely precise

And my heart shines.

Zhang
Heng copied the ancient,

Smiling on posterity;

Liu Hui followed the
ancient,

Having no time to revise it.

Now what is so difficult about
it?

One need only think.

The quotation from Zu Gengzhi probably ends here.

Using the Exact Proportions [*milü* 密率,
i.e. π ≈ 3^{1}/_{7},
the volume of this sphere is originally the product of the diameter by itself
twice, multiplied by 11 and divided by 21.

, with π ≈ 3Given [^{1}/_{7}.

Bai Shangshu 1982 白尚恕

«九章算术»与刘辉的几何理论

(The
geometrical principles of the *Jiu zhang suanshu* and Liu Hui), pp.
137-161 in *"Jiu zhang suanshu" yu Liu Hui* «九章算术» (The *Jiu zhang suanshu* and Liu Hui), ed. by Wu Wenjun 吴文俊,
Beijing Shifan Daxue Chubanshe. English abstract pp. ••
.

Bai Shangshu (ed.) 1983 白尚恕

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*Zhongguo
suanxue shi* 中國算學史

(The
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[1]Since 1978 a number of publications have
touched on the problem of the volume of a sphere in Chinese and world
mathematics. In Western languages see especially van der Waerden 1983: 204-207;
Lam Lay-yong & Shen Kangsheng 1985; Martzloff 1988: 269-279; Li Yan &
Du Shiran 1988: 85-87; Seidenberg 1989; in Chinese, Hong Wansheng 1981: 28-30;
Bai Shangshu 1982: 150-155; 1983: 120-135. I have also been informed of several
earlier studies: Mikami 1932: 74-84; Li Yan 1937: 30-32; 1954: 29-31; T. Kiang
1972. The first modern writer to discuss Zu Gengzhi's treatment of the sphere
was Xu Youren 徐有壬 (1800-1860), whose *Jie qiu jieyi* 截球解義 was published in 1872 (see Li Yan 1954: 29-31).
Dr. Alexei Volkov has informed me of the following works in Russian which touch
on the problem: Volkov 1988 (cf. 1991); Beriozkina 1957; 1980; Yushkevich 1982.
He also notes a complete translation of Liu Hui's commentary on the *Jiu
zhang suanshu* by Kawahara Hideki, which I have not seen.

[2]*Jin shu* 1974, **16**: 491; *Sui
shu* 1973, **16**: 409. On some doubts concerning the attribution see
Section 4 below and Wagner 1978.

[3]On the interpretation of the term *he
gai* see footnote 10 below.

[4]Cf. Baron 1969: 126. Cavalieri published the theorem in 1635, but its roots go back at least as far as Archimedes (3rd cent. B.C.).

[5]See Qian Baocong 1963: 158, n. 5, where Li Huang is cited.

[6]Other uses of the assumption can be seen in
Liu Hui's commentary on *Jiu zhang suanshu*, chapter 1, problems 33-34;
chapter 5, problems 9, 11, 13. See Qian Baocong 1963: 108, lines 6-15; 163,
lines 7-12; 164, line 10 - 165, line 4; 166, lines 1-8.

[7]On *Zhui shu* see Mikami 1934: 84-97;
Needham 1959: 35-36; Li Yan 1963: 56-58, 65; Qian Baocong 1964: 86-90;
Martzloff 1988: 39 n. 3. Zu Chongzhi's calculation of π is quoted by Li
Chunfeng in his commentary on *Jiu zhang suanshu*, chapter 1, problem 32;
see Qian Baocong 1963: 104, line 5 - 106, line 9; cf. Wagner 1978: 206-208.

[8]In his commentary on *Jiu zhang suan
shu*, chapter 5, problem 15 (Qian Baocong 1963: 167, line 1 - 168, line 4);
see Wagner 1979.

[9]*Nan Qi shu* 1972, **52**: 902; *Nan shi* 1975, **72**: 1774; Li Yan 1963: 64; Wagner 1978: 211.

[10]Here I arbitrarily continue to follow Li
Yan's interpretation (1963: 59) of the phrase *mouhe fanggai*. An
anonymous referee of this article for *Chinese science* suggested an
alternative interpretation: "a combination of a pair of covers on a common
square base", *mou* meaning "double" and *he* having its usual
meaning, "to combine". This still leaves open the question of what sort of
"covers" these might be.
The late Prof. Kurt Vogel pointed out to me the similarity of the geometric
form under consideration here with the ancient bronze or ceramic vessel type
called a *fang* 鈁. In the phrase *mouhe fanggai* the character *fang* 方 could be a loan character for this *fang* 鈁, and
the phrase could be translated, "a combined pair of *fang* 鈁-covers".

Bai Shangshu (1983: 123) interprets the cover as an "umbrella" (*san* 傘). Therefore Crossley's translation of Li Yan and Du Shiran (1987: 74,
85) translates *mouhe fanggai* as "two square umbrellas".
J.-C. Martzloff (1988: 270) interprets *gai* 蓋 as "vault" (in the
architectural sense) rather than "cover", so that the phrase could be
translated, "a double vault".

[11] Li Chunfeng's commentary was, in a formal sense, directed to the Emperor.

[12]*Sui shu* 1973, **16**: 388. See
also Needham 1959: 101; Juschkewitsch 1964: 58.

[13]A reference to the *Book of Changes*:
"Everything under heaven comes to the same but by a different path" (*Zhou
yi*, *Xi ci*, *xia* 周易繫辭下, *Sibu
congkan* 四部叢刊 edition, **8**: 3b).