The Chemistry Maths Book, Second Edition

5.4 The integral calculus 145

The integral is evaluated in Example 6.10; , and the area of the circle is four

times this.

Method 2.Divide the area of the circle into concentric

circular strips as in Figure 5.13. Given that the circumference

of a circle of radius ris 2 πr(see Example 5.12), the area

of the strip between rand r 1 + 1 ∆rlies between 2 πr∆rand

2 π(r 1 + 1 ∆r)∆r:

2 πr∆r 1 < 1 ∆A 1 < 12 π(r 1 + 1 ∆r)∆r (5.35)

and, when∆ris small enough,∆A 1 ≈ 12 πr∆r. The total area

is then

0 Exercise 48

We note that the indefinite integral can be viewed as a special case of the definite

integral, equation (5.34), in which the upper limit of integration, b, has been replaced

by the variable x, whilst the lower limit of integration, a, is arbitrary. The quantity

F(a)is therefore an arbitrary constant, and can be replaced by the symbol C:

(5.36)

The use of differentials

A convenient alternative approach to the definite integral as an area makes use of the

concept of the differential introduced in Section 4.12. This approach is widely used in

the application of the calculus to the formulation of physical problems.

Consider the expression (5.35) in Example 5.11. Division by ∆rgives

and, in the limit ∆r 1 → 10 ,

The quantitydA 2 dris the rate of change of the areaA(r)of the circle with respect

to the radius r. The corresponding differentialdA 1 = 12 πr 1 dris, in the language of

differentials, the area of a circular strip of radius rand ‘infinitesimal’ width dr; it is

an element of areaor differential area. The ‘sum’ of these elements is the integral

AdA rdr

A a

==ZZ

00

2 π

∆

A

r

dA

dr

→= 2 πr

22 ππr

A

r

<< +rr

∆

()∆

ZZfxdx fxdx Fx Fa Fx C

a

x

() ==−=+() () () ()

Ardra

a

==Z

0

2

2 ππ

Aa=π

2

4

a

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Figure 5.13