I.31 Now given a circular field, the circumference is 30 bu and the diameter 10 bu. Question: What is the area? Answer:

Transcription

1 Chapter 9 Areas of circular regions 9.1 Problems I I.31 Now given a circular field, the circumference is 30 bu and the diameter 10 bu. I.3 Given another circular field, the circumference is 181 bu and the diameter 60 1 bu. 3 I.33 Now given a bowl-like field, the circumference of the base is 30 bu and the middle path is 16 bu. I.34 Given another bowl-like field, the circumference of the base is 99 bu and the middle path is 51 bu. I.35 Now given a field, a segment of a circle, whose chord is 30 bu and sagitta 15 bu. I.36 Given another field, segment of a circle, whose chord is bu. 9 bu and sagitta I.37 Now given an annular field, whose inner and outer circumferences are 9 bu and 1 bu respectively. The width is 5 bu. 1 I.31: 75 [square] bu. I.3: 11 mu 90 1 [square] bu. I.33: 10 [square] bu. I.34: 5 mu [square] bu. I.35: 1 mu [square] bu. I.36: mu [square] bu. I.37: mu 55 [square] bu. I.38: 4 mu [square] bu.

2 50 Areas of circular regions I.38 Given another annular field, whose inner and outer circumferences are bu and bu respectively. The width is 1 bu. 3 Question: what is the area? Answer: 9. The rules 9..1 The rule for circular fields Multiplying half the circumference by the radius yields the area of the cirle in [square] bu. Another rule: One fourth the product of the circumference and the diameter. Another rule: One fourth the product of three times the diameter squared. Another rule: The circumference squared and divided by The rule for Bowl-like fields A quarter of the product of the middle path by the circumference The rule for segmental fields Take the chord multiplying the sagitta; square the sagitta; add, then halve The rule for annular fields Add the inner and outer circumferences, then halve; multiplying by the width yields the product in [square] bu. A more precise rule: Lay down the integral parts of the inner and outer circumferences above, and the corresponding numerators and denominators [of the fractional parts] below. Multiply the integral part by the common denominator and add the product to the corresponding numerator. Add the two sums and then halve. Alternatively, multiply the integral part of the width by its denominator and then add the product to its numerator. Multiply the sum by the circumference as dividend; multiply the denominator as divisor. Divide giving the product of bu. The remainder is expressed by a fraction. Dividing again by the number of [square] bu in a mu yields the required number of mu. 9.3 Liu Hui s commentary on the rule for circular fields The answers given to Problems I.31 and I.3 are respectively 75 [square] bu and 11 mu bu. Liu Hui corrected these by 71 [square] bu and 10 mu

3 9.3 Liu Hui s commentary on the rule for circular fields 51 [square] bu. These were based on a more accurate estimate of the circumference and diameter ratio. (c) Liu: Taking half the circumference as length and the radius as width; hence multiplying the two gives the area in [square] bu. Given a circle of diameter chi, consider its inscribed regular hexagon. The rate of the diameter of the circle to its perimeter is 1 to 3. As in the figure, 3 times the product of the side of the hexagon by the radius yields the area of the inscribed regular dodecagon. Dividing again; then 6 times the product of the side of the dodecagon by the radius yields the area of a 4- gon. The larger the number of sides, the smaller the difference between the area of the circle [and that of its inscribed polygons]. Dividing again and again until it cannot be divided further yields a regular polygon coinciding with the circle, with no portion whatever left out. Outside a regular polygon there are co-apothems. Multiply the sides by the co-apothem giving the area [of the polygon plus these co-apothem rectangles] is larger than that of the circle. As the number of sides increases [without limit] the polygon coincides with the circle and no co-apothem exists, so that [the polygon plus those co-apothem rectangles] is no longer larger than the circle. Take the perimeter [of the inscribed n-gon] to multiply the radius; compare that with the area of the inscribed n-gon; it is twice as large. Hence the rule Multiplying half the circumference by the radius yields the area of the circle in [square] bu. According to this method, the more precise rate is different from a rate of 3 to 1, which is the perimeter of the [inscribed] hexagon [to the radius]. The difference between a polygon and a circle is just like that between the bow [arc] and its chord, which can never coincide. Yet such a tradition has been passed down from generation to generation and no one cares to check it. So many scholars followed the tradition that their error has persisted. It is hard to accept without a convincing derivation. Planar forms are either curvilinear or rectilinear. A study of the rates for a square and its inscribed circle may be trivial, but it will lead to far-reaching consequences. By elaborating on this point, one gets wide applications. Let us verify the facts by diagrams for a more precise rate. Assertions without facts are not sound, so notes and comments in detail are given as follows LIU Hui s dissection of a circle Problems I.31,3 of Jiuzhang Suanshu compute the area of a circle as the product of half-circumference and half-diameter, and adopted the formula diameter one, circumference 3. The third century commentator LIU Hui pointed out the inadequacy of this, and explained the mensuration of the circle by the method of dissection of the circle.

4 5 Areas of circular regions Consider a circle of radius R. Denote by A its area. Inscribe in the circle a regular polygon of n sides, each of length a n. For the regular n gon, denote by (i) p n the perimeter, (ii) A n the area, (iii) d n the distance from the center to a side, and (iv) c n = R d n. Since a 6 = R, it is easy to see that d 6 = ( a6 ), R c 6 = R d 6, ( a 1 = c 6 + a6 ) ( = R R ) ( a6 ) + ( a6 ). LIU Hui made use of a circle of radius 1. Beginning with a 6 =1, he obtained n a n d n c n p n = n a n More generally, LIU made use of ( a n = R R to iterate the process several times and obtained ) ( an ) + ( an ), n a n d n c n p n = n a n (a) (b) (c) (d) (e) The rounding off of the 6th digit after the decimal point is not correct. To 10 places, these are (a) ; (b) ; (c) ; (d) ; (e)

5 9.3 Liu Hui s commentary on the rule for circular fields 53 LIU Hui actually computed a n by taking the square root of a n these values: and he recorded a 1 = , a 4 = , a 48 = , a 96 = Exercise Show that a n satisfies the recurrence relation and deduce that a 6 k = + + a n = 4 a n, (k fold square root). Considering the area A n of the inscribed regular n gon, LIU Hui made use of the following relations to estimate the area A of the circle: A 1 = 3R, A n = 1 n R a n, A n A n = 1 n a n c n, A n < A < A n +(A n A n ). n a n A n A n A n A n +(A n A n ) From these, LIU Hui concluded that the area of the circle is 3.14 correct to two places of decimal.

6 54 Areas of circular regions In the fifth century, ZU Congzi gave the value of π correct to 6 decimal places, as between and His manuscript Sushu was lost. But it is generally believed that he carried out LIU Hui s program further to regular polygons of sides 6 11 = 188 and 6 1 = 5576: n a n A n A n A n A n +(A n A n ) (d) Liu: The circumference is still the perimeter of a polygon. The area of a circle should be the product of half the circumference by its radius, therefore the product of the circumference by its diameter is to be divided by 4. According to Liu s rate, multiplying the given circumference by 50, then dividing it by 157, get the diameter [and conversely] multiplying the given diameter by 157, and dividing it by 50, get the circumference. The new rate should still be slightly smaller - the diameter obtained by means of the circumference is too long, while the circumference obtained by means of the diameter is too short. The areas obtained by means of the diameter are somewhat smaller, while areas obtained by means of the circumference will be slightly larger. 9.4 The bowl-like and the segmental fields The bowl-like field LIU Hui s commentary: The rule is false. So [we] give a counter-example by considering a square-based pyramid. Given a right pyramid with a base 6 chi square and an altitude of 4 chi. Let 4 chi be the gu and half a side of the base 3 chi be the gou, then the slant height 5 chi is the hypotenuse of a right-angled triangle. Four times the product of the gou by the hypotenuse yields 60 [square] chi, i.e. the area of the four lateral faces of the pyramid. Suppose again a [right circular] cone is inscribed in the [square-based] pyramid, then the rate for the area of the lateral faces of the pyramid to that of the cone is the same as the rate for the area of the square to that of the [inscribed] circle. As to the square-based pyramid, the side of its base is 6 chi, and its base perimeter 4 chi. 5chi times half the perimeter also equals the area of the lateral faces of the pyramid. Similarly, to find the area of the lateral surface

7 9.4 The bowl-like and the segmental fields 55 of the cone, multiply half the middle path by half the circumference of the circular base. Since the middle path of the bowl-like field is curvilinear, using the same rule as with a cone the area is too small. The exact rule is complicated to use, so here an approximate rule is used for estimating large areas of farmland only. The rule for finding the area of the lateral surface of a cone is similar to the one for a circle. So the Rule calls for a quarter of the product of the middle path by the circumference, just like the Rule for Circular Fields. In the comments on the Diameter of a Sphere Rule, the various rates between circles and squares are discussed in detail; this can be referred to The segmental field LIU Hui s commentary: In a square inscribe a circle, the area of its inscribed dodecagon is three quarters of the area of the circumscribed square, and that of the inscribed square is half [that ofj the outer square. Therefore the red area is a quarter of the outer square. Considered as a semicircle, [the area of] a segment may be calculated as follows: Take the chord multiplying the sagitta, add, then halve is the yellow area; square on the sagitta, then halve is twice the blue area. Joining the yellow and blue areas gives a polygon, the area of which is half a dodecagon inscribed in a circle. The sides of the polygon do not go outside the segment, so the area is too small. A rate of 3 to 1 for the Rule for Circular Fields yields the area of a dodecagon, which is also too small. Similarly, here one gets approximately the area of half a regular dodecagon inscribed in a semicircle. If the segment is less than a semicircle, the Rule is even rougher. It would then be appropriate to apply the Rule for the calculation of a cylindrical pillar. let the saw length be the chord, and the depth be the sagitta of the segment, and then find the diameter. Now halve the arc when the diameter is known. Let half the chord be the gu, the sagitta be the gou and find the hypotenuse which is the chord of the small [half] arc. Again let half the chord of the half arc be the gou, the radius as the hypotenuse, then find the gu. Subtracting that gu from the radius, we get the sagitta of the small arc. When we divide the arc again and again, we render it smaller and smaller. A series of sums of products of chords by sagittas would approach the exact value. But it is necessary to study the errors carefully in calculating. As for the measurement, it is convenient to apply the Rule to get an approximate value.

8 56 Areas of circular regions