The Chemistry Maths Book, Second Edition

9.11 Change of variables 287

The integral (9.59) is then

EXAMPLE 9.31Evaluate the integral off(r, 1 θ) 1 = 1 e

−r

2

sin

2

1 θover (i) the area of a

circle of radius aand (ii) the whole plane.

(i)

Because the integrand is the product of a function of rand a function of θ,

e

−r

2

sin

2

θr 1 = 1 (e

−r

2

r)(sin

2

θ)

and because the limits of integration are constants, the double integral can be

factorized:

Because

and, using the substitutionu 1 = 1 r

2

, du 1 = 12 rdr,

Therefore,

(ii) Lettinga 1 → 1 ∞,

0 Exercises 69–71

ZZ

0

2

0

2

2

2

π

π

∞

erdrd

−r

sin θθ=

Ie

a

=−

(

)

−

π

2

1

2

ZZ

00

0

2

2

2

1

2

1

2

1

2

1

a

r

a

uu

a

erdr edu e

−−−

==−













=−ee

−a

( )

2

ZZ

0

2

2

0

2

1

2

12

1

2

1

2

2

ππ

sin θθdd=− =−( cos )θ θ θ sin θ

















=

0

2 π

π

sin ( cos ),

2 1

2

θθ=− 12

Ierdr d

a

r

=×

−

ZZ

00

2

2

2

π

sin θθ

ZZZ

R

e dA e rdrd

r

a

−−r

=

22

2

0

2

0

2

sin θθsin θ

π

I e rdrd d e rdr e

a

r

a

ra

==×=−

−−−

ZZ Z Z

0

2

00

2

0

21

ππ

θθ π (aa+













1.