________________
Shri Mahavir Jain Aradhana Kendra
www.kobatirth.org
CHAPTER VIII-CALCULATIONS REGARDING EXCAVATIONS. 265
The rule for arriving at the value of the cubical contents of a spherically bounded space :
28. The half of the cube of half the diameter, multiplied by nine, gives the approximate value of the cubical contents of a sphere. This (approximate value), multiplied by nine and divided by ter. on neglecting the remainder, gives rise to the accurate value of the cubical measure.
An example in illustration thereof.
29. In the case of a sphere measuring 16 in diameter, calculate and tell me what the approximate value of (its) cubical measure is, and also the accurate measure (thereof).
The rule for arriving at the approximate value as well as the accurate value of the cubical contents of an excavation in the form of a triangular pyramid, (the height whereof is taken to be equal to the length of one of the sides of the equilateral triangle forming the base)
:
Acharya Shri Kailassagarsuri Gyanmandir
301. The cube of half the square of the side (of the basal equilateral triangle) is multiplied by ten; and the square root (of the resulting product is) divided by nine. This gives rise to the approximately calculated value (required). (This approximate) value, when multiplied by three and divided by the square root of
X
28. The volume of a sphere as given here is (1) approximately=
de
and (2) accurately==
9 9 2 x 10' cal contents of a sphere is 4 r3; and this becomes cemparable with the
3
above value, if is taken to be
10. Both the MSS. read Hið
, making it appear that the accurate value is of the approximate
10 9
X
()
The correct formula for the cubi
a
value; but the text adopted is which makes the accurate value 9 of the approximate one. It is easy to see that this gives a more accurate result in regard to the measure of the cubical contents of a sphere than the other reading.
10
30. Algebraically represented the approximate value of the cubical contents of a triangular pyramid according to the rule comes to x 1/5,
as 18
as i.e., 12
as
12
XV2; where 34
20
; and the accurate value becomes equal to
9
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