Liu Hui and Zu Gengzhi on the volume of a sphere

1. Introduction

[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 ≈ 3³/₈ (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.

2. The use of Cavalieris theorem by Liu Hui and Zu Gengzhi.

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
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.

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, qi 棋, "block", to mi 幂, "area", so that the first line would be translated, "If areas are piled up to form volumes . . .". The modern editor Qian Baocong 錢寶琮 rejects this emendation, but it fits very well in a statement of Cavalieri's Theorem.[5]

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.

3. Summary of the proof

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   =  r²

SPWT   =  b²

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

NSTV + RVTX + TWQX = r²–b²

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

r² – b² = a²

NSTV + RVTX + TWQX = a²

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 r³/3. The intersection of this pyramid with a plane parallel to its base at a distance a from the apex has area a². 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 2r³/3, and the volume of the whole box-lid is

8 × 2r³/3 = 8 × 2(d/2)³/3 =2d³/3

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 × 2d³/3

    = πd³/6

4. The attribution of Liu Huis commentary

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.

5. Note on the translation

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.

6. Translation

[6.1. Problems 23 and 24 of the Jiu zhang suanshu]

[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.

[6.2.] Method for the extraction of the spherical root:

The term kai liyuan 開立圓 is obviously analogous to such terms as kai fang 開方, "extraction of the square root", and kai 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 wan for "sphere" here; elsewhere liyuan 立圓 is used in the same meaning.

[6.3. Liu Huis comment]

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

[6.3.1. The incorrect derivation of the Jiu zhang suanshu]

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)² ≈ (3/4)² = 9/16

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.

[6.3.2. This derivation is incorrect]

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;

10 cun 寸 = 1 chi 尺.

assemble them to make a cube of 2 cun [as in Figure 1]. Round [gui 規] it to make a cylinder with diameter 2 cun and height 2 cun.

That is, inscribe a cylinder in the 2-cun cube; see Figure 2.

Further round it perpendicularly [you fu heng gui zhi 又復橫規之].

That is, inscribe a second cylinder with axis perpendicular to that of the first.

Then its form resembles the square lid of a mouhe 牟合[=盒].

A mouhe seems to have been some sort of box. In the following the expression mouhe fanggai 牟合方蓋 is abbreviated to hegai 合蓋 and I translate this as "box-lid".[10]

Each of the eight blocks now resembles a yangma 陽馬, rounded.

See Figure 5; a yangma was a pyramid with rectangular base and with one lateral edge perpendicular to the base (cf. Wagner 1979: 166).

As 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.

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.

[6.3.3. The formula is nevertheless not too bad]

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 < 9d³/16

Liu Hui has proved that V  ≠  9d³/16, and below he states that he is unable to find the correct formula. How then does he know that V < 9d³/16 ? It may be that he knew the correct formula,V  =  πd³/6, 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.

[6.3.4. The geometer’s frustration]

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.

[6.4. Zhang Heng on the sphere]

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.

[6.5. Li Chunfeng’s comment]

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:

[6.5.1. Zu Gengzhi’s derivation]

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

Since Zu Gengzhi proves below that V  =  πd³/6, 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 the Jiu zhang suanshu text.

What is the reasoning behind this?

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 [r²] 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 [r²] 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.

    r² – a² = b² = SPWT

    r² = NPQR

    NPQR – SPWT = TSNRQW

Therefore

    TSNRQW = a²

Regardless of the height [a], the situation is always like this. This being the case, there is something which "comes to the same thing but by a different path".[13] Thus one can
Go far afield to expound the situation,
Borrow a comparison to analyze the subtleties.

In the following see Figure 9.

Take a yangma 陽馬 [with square base and] with [the sides of] the square and the height equal [and equal to r].

A yangma is a pyramid with rectangular base and with one edge perpendicular to the base (Wagner 1979: 166).

Invert it, cut horizontally [along the shaded plane], and remove the upper part. Then the product of the height [of the cut, a] by itself, and the area of the upper cut, are similarly equal.

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

Now,
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 a yangma is sth/3, where s and t are the sides of the rectangular base and h is the height.

If 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.

[6.5.2. The geometer’s triumph]

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.

[6.5.3. Using a better value of π]

Using the Exact Proportions [milü 密率, i.e. π  ≈  3¹/₇, the volume of this sphere is originally the product of the diameter by itself twice, multiplied by 11 and divided by 21.

, with π ≈ 3¹/₇.

Given [yue 約] this volume, the original volume [i.e. the volume of the circumscribed cube] is required. Therefore one multiplies by 21 and divides by 11. Now when something is multiplied by itself twice, then extracting the cube root restores the original number; therefore extracting the cube root gives the diameter of the sphere.

7. References