NPTEL C OURSE ON
M ATHEMATICS IN I NDIA :
F ROM V EDIC PERIOD TO M ODERN TIMES
L ECTURE 37
Proofs in Indian Mathematics - Part 2
K. Ramasubramanian
IIT Bombay
Outline
Proofs in Indian Mathematics - Part 2
Volume of a sphere
I
I
I
I
A couple of theorems
I
I
I
The principle involved
Area of a circle
Volume of a slice of a given thickness
Total volume
Jy
a-sam
aya (Theorem on product of chords)
. varga-ny
Jy
a-varg
antara-ny
aya (diff. of the squares of chords)
Jy
a-sam
varga-ny
a
ya
Jyotpatti
.
The cyclic quadrilateral
I
I
I
Expressing diagonals in terms of sides
Area in terms of sides
Circumradius in terms of sides
Volume of a sphere
The principle involved
I To find the volume of a sphere, it is divided into large number (say n) of
slices of equal thickness.
I In the figure the sphere NASBN is divided into slices by planes parallel
to the equatorial plane AGB.
I Then the volume of each slice is
obtained.
I
I
Volume = Area thickess.
Area is obtained by finding the average
radius of circles at the top and bottom
of the slice.
If d is diameter, then the thickess of the
slice = dn .
I Thus, first we need to obtain an
expression for finding the the area of a
circle.
I Sum up the elementary volumes of the
slices, that will add up to the sphere.
Volume of a sphere
Obtaining the area of a circular slice
Figure: Circular slice cut into two pieces.
I In the figure above we have indicated a circular slice being turned into a
rectangle by appropriately sectioning it, and inserting one half of the
circular slice into the other.
I The length of this rectangular strip corresponds to half the
circumference C. If the radius is r 0 , then the area of this slice is
Area =
1
C r 0.
2
(1)
Volume of a sphere
Obtaining the elementary volume of the sphere (= volume of the circular slice
I Let r be the radius of the sphere and C the circumference of a great
circle on this sphere..
I The radius of the j-th sliceinto which the sphere has been divided
intocan be conceived of as the half-chord Bj (bhuj
a).
I Now, the circumference of this slice is given by
Cjth slice =
C
r
C
r
Bj
I Hence the area of this circular slice is
Ajth slice
1
=
2
Bj Bj
.
I Therefore, the elementary volume is given by
V =
1
2
C
r
where is the thickness of the slices.
Bj2 ,
(2)
Volume of a sphere
Summing up the elementary volumes of the circular slices)
I The volume of the sphere is obtained by finding the sum of the
elementary volumes constituted by the slices:
V =
V =
n
X
1 C
Bj2
2 r
j=1
I In the above expression, the
thickness can be expressed in
2r
terms of the radius as =
.
n
I If Bj can also be expressed in terms
of r , then we can get V = V (r ).
I For this we invoke the
Jy
a-sara-sam
aya given by
. varga-ny
Aryabhat
a.
.
(3)
Jya-sara-sam
aya of Aryabhat
. varga-ny
.a
Theorem on the square of chords
In his Aryabhat
.ya, Aryabhat
. a has presents the theorem on the
product of chords as follows (in half
ary
a):
vxa.ea Za.=;sMa:va:gRaH A:DRa.$ya.a:va:gRaH .sa Ka:lu ;Da:nua:Sa.eaH
(Aryabhat
.ya, Gan.ita 17)
I
The words varga and qsamvarga refer to square and product
respectively.
Similarly, dhanus and sara refer to arc and arrow respectively.
A
Using modern notations the above ny
aya
may be expressed as:
product of saras = Rsine
DE EB
= AE 2
C
Volume of a sphere
Summing up the elementary volumes of the circular slices)
I It was shown that the volume of the sphere may be expressed as:
V =
n
X
1 C
Bj2
2 r
j=1
In the figure AP = PB = Bj is the jth
half-chord, starting from N.
Applying jy
a-
sara-sam
aya,
. varga-ny
we have
Bj2
=
=
=
AP PB = NP SP
1
[(NP + SP)2 (NP 2 + SP 2 )]
2
1
[(2r )2 (NP 2 + SP 2 )].
(4)
2
I It may be noted in the figure that the j-th Rversine NP = j and its
complement SP = (n j). Hence, while summing the squares of the
Rsines Bj2 , both NP 2 and SP 2 add to the same result.
Volume of a sphere
Summing up the elementary volumes of the circular slices)
I Thus the expression for the volume of the sphere given by
n
X
1 C
Bj2 ,
V
2 r
j=1
reduces to
V
C
2r
"
1
[n. (2r )2
2
2r
n
2
2. [1 + 2 + ...n ]
2r
n
.
I It was known to Kerala mathematicians, that for large n
12 + 22 + ...n2 =
n3
.
3
I Using this, the expression for the volume of the sphere becomes
C
2r
C
6
=
=
4r 3
d 2.
8 3
r
3
(5)
Jyasam
aya and Jy
a-varg
antara-nyaya
. varga-ny
Theorem on product of chords and the difference of the square of chords
I In the figure DM is the perpendicular
from the vertex D onto the diagonal
AC of the cyclic quadrilateral.
I Let us consider the two triangles
AMD and CMD, that is formed by DM
in the triangle ADC. It can be easily
seen that
AD 2 DC 2 = AM 2 MC 2 .
(6)
I In other words, the difference in the
squares of the jy
as AD, DC is equal
to the difference in the square of the
base segments (
ab
adh
as) AM, MC.
This result may be noted down for later use as
jy
avarg
antara =
ab
adh
avarg
antara
Jyasam
aya and Jy
a-varg
antara-nyaya
. varga-ny
Theorem on product of chords and the difference of the square of chords
I In order to derive a certain property, we
rewrite (6) as
AD 2 DC 2 = (AM + MC)(AM MC)
(7)
I It may be noted that
AM + MC jy
a of sum of the arcs
AM MC jy
a of diff. of the arcs
I Denoting the chords AD and DC as J1 and J2 , and the arcs associated
with them as c1 and c2 , we may rewrite (7) as
J12 J22 = Jy
a(c1 + c2 ) Jy
a(c1 c2 )
I In otherwords,
.$ya.a:va:ga.Ra:nta.=;m,a = . ca.a:pa:d:ya:ya.ea:ga:a.va:ya.ea:ga.$ya.a:sMa:va:gRaH
Jyasam
aya and Jy
a-varg
antara-nyaya
. varga-ny
Theorem on product of chords and the difference of the square of chords
I Recalling the equation,
J12 J22 = Jy
a(c1 + c2 ) Jy
a(c1 c2 )
(8)
I It can also be shown that
c c i2
h
c + c i2 h
1
2
1
2
Jy
a
J1 J2 = Jy
a
2
2
I In otherwords,
.$ya.a:sMa:va:gRa = . ca.a:pa:d:ya:ya.ea:ga:a.va:ya.ea:ga.a:DRa.$ya.a-va:ga.Ra:nta.=;m,a
I The two equations (8) and (9) are equivalent to the trigonometric
relations:
sin2 (1 ) sin2 (2 )
sin(1 )sin(2 )
sin(1 + 2 ) sin(1 2 ),
(1 + 2 )
(1 2 )
sin2
sin2
.
2
2
with our convention that c1 > c2
(9)
The cyclic quadrilateral & the third diagonal
I
Consider the equation
sin 1 sin2 = sin
(1 + 2 )
2 (1 2 )
sin
.
2
2
If we put 1 = (n + 1), and 2 = (n 1) in the above equation,
then we immediately obtain the following equation
sin(n + 1) =
sin2 n sin2
sin(n 1)
This is precisely the equation that is presented in the following
nkara
verse given by Sa
in his Kriy
akramakar:
ta.a.$ya.a:va:gRa:m,a A.a:d;a.$ya.a:va:gRa:h.a:nMa h:=e;t,a :pua:naH
A.a:sa:a.a:Da:~Ta: a.Za:
a*+:
nya.a l+b.Da.a .~ya.a:du.a.=:ea.a.=:a
The cyclic quadrilateral & the third diagonal
I Notation:
I the arcs (AB, BC, CD, DA) c1 , c2 , c3 , c4
I the chords (AB, BC, CD, DA) J1 , J2 , J3 , J4
I the diagonals (BD, AC, DG) K1 , K2 , K3
G
I Having drawn a quadrilateral,
obviously we can have only two
diagonals.
I G is a point chosen such that
BC = AG.
I By swapping any two sides
could be adjacent, or opposite
we would be affecting only one
of the two diagonal of the
quadrilateral, while the other
diagonal remains fixed.
I This new diagonal that gets
generated due to this swapping
of sides is referred to as the
third diagonal or bh
avikarn
.a
The elegant results we would like to prove
I The three diagonals of the cyclic quadrilateral would be referred to as
1. I+:k+.NRa (is..takarn
. a) chosen/first diagonal
2. I+ta.=;k+.NRa (itarakarn
. a) other/second diagonal
3. Ba.a:a.va:k+.NRa (bh
avikarn
. a) future/third diagonal
I The results that we would prove:
1. I+:k+.Na.Ra: a.(ra:ta:Bua.ja:Ga.a:tEa:k+.a:m,a = I+:k+.NRa Ba.a:a.va:k+.NRa
2. I+ta.=;k+.Na.Ra: a.(ra:ta:Bua.ja:Ga.a:tEa:k+.a:m,a = I+ta.=;k+.NRa Ba.a:a.va:k+.NRa
3. Bua.ja:pra: a.ta:Bua.ja:Ga.a:tEa:k+.a:m,a
= I+:k+.NRa I+ta.=;k+.NRa
I The above results essentially express the product of the diagonals in
terms of the sum of the product of the sides jy
as.
I Making use of them we express the diagonals in terms of sides.
I Then by making use of yet another result
product of two sides of a triangle
= the altitude,
circum-diameter
(10)
we show that the area can be expressed in terms of the diagonals and
in turn, in terms of the sides.
Thanks!
T HANK YOU
