The Chemistry Maths Book, Second Edition

9.12 Exercises 289

because the value of a definite integral does not depend on the symbol used for the

variable of integration. Then

and the double integral is over the whole xy-plane. A transformation to polar

coordinates then gives

Therefore,

(9.64)

9.12 Exercises

Section 9.1

1.Find the value of the functionf(x, y) 1 = 12 x

2

1 + 13 xy 1 − 1 y

2

1 + 12 x 1 − 13 y 1 + 14 for

(i)(x, y) 1 = 1 (0, 1) (ii)(x,y) 1 = 1 (2, 0) (iii)(x,y) 1 = 1 (3, 2)

2.Find the value off(r, θ, φ) 1 = 1 r

2

sin

2

1 θ 1 cos

2

φ 1 + 121 cos

2

1 θ 1 − 1 r

3

sin 12 θ 1 sin 1 φfor

(i)(r, θ, φ) 1 = 1 (1, π 2 2, 0) (ii)(r, θ, φ) 1 = 1 (2, π 2 4, π 2 6) (iii)(r, θ, φ) 1 = 1 (0, π, π 2 3).

Section 9.3

Find and for

3.z 1 = 12 x

2

1 − 1 y

2

4.z 1 = 1 x

2

1 + 12 y

2

1 − 13 x 1 + 12 y 1 + 13 5.z 1 = 1 e

2 x+ 3 y

6.z 1 = 1 sin(x

2

1 − 1 y

2

) 7.z 1 = 1 e

x

2

cos(xy)

Find all the nonzero partial derivatives of

8.z 1 = 1 x

2

1 − 13 x

2

y 1 + 14 xy

2

9.u 1 = 13 x

2

1 + 1 y

2

1 + 12 xy

3

Find all the first and second partial derivatives of

10.z 1 = 12 x

2

y 1 + 1 cos(x 1 + 1 y) 11.z 1 = 1 sin(x 1 + 1 y)e

x−y

Show thatf

xy

1 = 1 f

yx

for

12.f 1 = 1 x

3

1 − 13 x

2

y 1 + 1 y

3

13.f 1 = 1 x

2

cos(y 1 − 1 x) 14.

Show thatf

xyz

1 = 1 f

yzx

1 = 1 f

zxy

for

15.f 1 = 1 cos(x 1 + 12 y 1 + 13 z) 16.f 1 = 1 xye

yz

17.Ifr 1 = 1 (x

2

1 + 1 y

2

1 + 1 z

2

)

122

find

18.Ifφ 1 = 1 f(x 1 − 1 ct) 1 + 1 g(x 1 + 1 ct), where cis a constant, show that.

19.Ifxyz 1 + 1 x

2

1 + 1 y

2

1 + 1 z

2

1 = 10 , find.

z

y

x

∂

∂













∂

∂

=

∂

∂

2

22

2

2

φφ 1

xct

∂

∂

,

∂

∂

,

∂

∂

r

x

r

y

r

z

.

f

xy

xy

=

• 
  

22

∂

∂

z

y

∂

∂

z

x

Z

0

2

2

∞

edx

−x

=

π

Ierdrdderdr

2 rr

0

2

00

2

0

1

4

1

44

22

== =

−−

ZZ Z Z

ππ

π

∞∞

θθ

Iedxedy e

2 xy xy

1

4

1

4

22 2

==

−

−

−

−

−−

−+

ZZ ZZ

∞

∞

∞

∞

∞

∞

∞

∞

(

22

)

dxdy