The Chemistry Maths Book, Second Edition

10.4 Volume integrals 301

EXAMPLE 10.6Find the volume∆vin Figure 10.6 and show that it reduces to the

approximate expression (10.10) for small ∆-values.

When∆ris small enough, the terms in(∆r)

2

and(∆r)

3

can be neglected. When∆θis

small,

cos(θ 1 + 1 ∆θ) 1 = 1 cos 1 θ 1 cos 1 ∆θ 1 − 1 sin 1 θ 1 sin 1 ∆θ 1 ≈ 1 cos 1 θ 1 − 1 sin 1 θ∆θ

sincecos 1 ∆θ 1 → 11 andsin 1 ∆θ 1 → 1 ∆θas∆θ 1 → 10. Therefore, for small ∆-values,

∆v 1 ≈ 1 r

2

sin 1 θ∆r 1 ∆θ 1 ∆φ

We note that this does not provide a proof of (10.9), since the volume element is used

for the construction of the volume integral.

EXAMPLE 10.7Evaluate the integral of the functionf(r, 1 θ, 1 φ) 1 = 111 + 1 r

2

1 cos

2

1 θ 1 sin

2

1 φ

over a sphere of radius aand centre at the origin.

The integral can be evaluated in two parts, as in Example 10.5:

The ranges of integration arer 1 = 101 → 1 a, θ 1 = 101 → 1 π, andφ 1 = 101 → 12 π. Then

(i)

and this is the volume of the sphere.

(ii)

0 Exercises 18, 19

=××

a

5

5

2

3

π

=ZZ Z

0

4

0

2

0

2

2

a

rdr d d

ππ

cos sinθθθ sin φφ

ZZZZ

V

rd r

a

22 2

0

2

00

42 2

cos sinθθφφv= cos sin sinθ

ππ

ddr d dθ φ

==ZZ Z ××=

0

2

00

2

3

3

3

22

4

3

a

rdr d d

a

a

ππ

sinθθ φ ππ

ZZZZ

V

drdrdd

a

v=

0

2

00

2

ππ

sinθθφ

ZZZ

VVV

fdvv=+d r dv

22 2

cos sinθ φ

=+ +

















−+



rrrr r

223

1

3

∆∆ ∆() () cos cos( )θθθ∆











∆φ

==







+∆ +∆ +∆

+

ZZ Z

r

rr

r

rr

rdr d d

r

2

3

3

θ

θθ

θθ

φ

φφ

sin φ

∆











−

























++

θ

θθ

θ

φ

φφ

φ

∆∆

cos

∆v=

+∆ +∆ +∆

ZZZ

φ

φφ

θ

θθ

θθφ

r

rr

rdrdd

2

sin