Mathematical Methods for Physics and Engineering : A Comprehensive Guide

NUMERICAL METHODS

h(ξ 1 )>ξ 2. The fraction of times that this inequality is satisfied estimates the value

of the integral (without the scaling factors (B−A)(b−a)) since the expectation

value of this fraction is the ratio of the area below the curvey=h(z)tothearea

of a unit square.

To illustrate the evaluation of multiple integrals using Monte Carlo techniques,

consider the relatively elementary problem of finding the volume of an irregular

solid bounded by planes, say an octahedron. In order to keep the description

brief, but at the same time illustrate the general principles involved, let us suppose

that the octahedron has two vertices on each of the three Cartesian axes, one on

either side of the origin for each axis. Denote those on thex-axis byx 1 (<0) and

x 2 (>0), and similarly for they-andz-axes. Then the whole of the octahedron

can be enclosed by the rectangular parallelepiped

x 1 ≤x≤x 2 ,y 1 ≤y≤y 2 ,z 1 ≤z≤z 2.

Any point in the octahedron lies inside or on the parallelepiped, but any point

in the parallelepiped may or may not lie inside the octahedron.

The equation of the plane containing the three vertex points (xi, 0 ,0),(0,yj,0)

and (0, 0 ,zk)is

x

xi

+

y

yj

+

z

zk

=1 fori, j, k=1, 2 , (27.53)

and the condition that any general point (x, y, z) lies on the same side of the

plane as the origin is that

x

xi

+

y

yj

+

z

zk

− 1 ≤ 0. (27.54)

For the point to be inside or on the octahedron, equation (27.54) must therefore

be satisfied forall eightof the sets ofi, jandkgiven in (27.53).

Thus an estimate of the volume of the octahedron can be made by generating

random numbersξfrom the usual uniform distribution and then using them in

sets of three, according to the following scheme.

With integermlabelling themth set of three random numbers, calculate

x=x 1 +ξ 3 m− 2 (x 2 −x 1 ),

y=y 1 +ξ 3 m− 1 (y 2 −y 1 ),

z=z 1 +ξ 3 m(z 2 −z 1 ).

Define a variablenmas 1 if (27.54) is satisfied for all eight combinations ofi, j, k

values and as 0 otherwise. The volumeVcan then be estimated using 3Mrandom

numbers from the formula

V

(x 2 −x 1 )(y 2 −y 1 )(z 2 −z 1 )

=

1

M

∑M

m=1

nm.