Number Theory: An Introduction to Mathematics

1 Elliptic Integrals 495

An example is the determination of the arc length of alemniscate. This curve,
which was studied by Jacob Bernoulli (1694), has the form of a figure of eight and is
the locus of all pointsz∈Csuch that| 2 z^2 − 1 |=1 or, in polar coordinates,

r^2 =cos 2θ(−π/ 4 ≤θ≤π/ 4 ∪ 3 π/ 4 ≤θ≤ 5 π/ 4 ).

If−π/ 4 ≤Θ≤0, the arc lengths(Θ)fromθ=−π/4toθ=Θis given by

s(Θ)=

∫Θ

−π/ 4

[r^2 +(dr/dθ)^2 ]^1 /^2 dθ

=

∫Θ

−π/ 4

[r^2 +( 1 −r^4 )/r^2 ]^1 /^2 dθ

=

∫ R

0

( 1 −r^4 )−^1 /^2 dr.

If we make the change of variablesx =

√

2 r/( 1 +r^2 )^1 /^2 , then on account of
dx/dr=

√

2 /( 1 +r^2 )^3 /^2 we obtain

s(Θ)= 2 −^1 /^2

∫X

0

[( 1 −x^2 / 2 )( 1 −x^2 )]−^1 /^2 dx.

Another example is the determination of the surface area of an ellipsoid. Suppose
the ellipsoid is described in rectangular coordinates by the equation

x^2 /a^2 +y^2 /b^2 +z^2 /c^2 = 1 ,

wherea>b>c>0. The total surface area is 8S,whereSis the surface area of the
part contained in the positive octant. In this octant we have

z=c[1−(x/a)^2 −(y/b)^2 ]^1 /^2

and hence

1 +(∂z/∂x)^2 +(∂z/∂y)^2 =[1−(αx/a)^2 −(βy/b)^2 ]/[1−(x/a)^2 −(y/b)^2 ],

where

α=(a^2 −c^2 )^1 /^2 /a,β=(b^2 −c^2 )^1 /^2 /b.

Consequently

S=

∫a

0

∫b( 1 −(x/a) (^2) ) 1 / 2
0
[1−(αx/a)^2 −(βy/b)^2 ]^1 /^2 [1−(x/a)^2 −(y/b)^2 ]−^1 /^2 dydx.
If we make the change of variables
x=arcosθ, y=brsinθ,
with JacobianJ=abr, we obtain
S=ab
∫π/ 2
0
dθ

∫ 1

0

( 1 −σr^2 )^1 /^2 ( 1 −r^2 )−^1 /^2 rdr,