Download - Homework 1 Solutions - University of Massachusetts Amherst · 26 CHAPTER 1 1-36. d z x y c2 = x 2 + y 2 The form of the integral suggests t h e use of the d ivergence theorem. (1)

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1

Homework 1 Solutions

1.

2 CHAPTER 1

1-2.

a)

x1

A

B

C

D

α

βγ

O

E

x2

x3

From this diagram, we have

cosOE OAα =

cosOE OBβ = (1)

cosOE ODγ =

Taking the square of each equation in (1) and adding, we find

2 2 2 2cos cos cos OA OB ODα β γ+ + = + +

2 2 2OE (2)

But

2 2

OA OB OC+ =2 (3)

and

2 2

OC OD OE+ =2

(4)

Therefore,

2 2 2

OA OB OD OE+ + =2 (5)

Thus,

2 2 2cos cos cos 1α β γ+ + = (6)

b)

x3

A A′x1

x2O

ED

C

C′

B′E′D′

First, we have the following trigonometric relation:

2 2

2 cosOE OE OE OE EEθ2′ ′+ − = ′ (7)

2MATRICES, VECTORS, AND VECTOR CALCULUS 3

But,

2 2 22

2 2

2

cos cos cos cos

cos cos

EE OB OB OA OA OD OD

OE OE OE OE

OE OE

β β α

γ γ

′ ′ ′ ′= − + − + −

′ ′= − + −′ ′

′+ −′

α

(8)

or,

2 2 22 2 2 2 2 2

2 2

cos cos cos cos cos cos

2 cos cos cos cos cos cos

2 cos cos cos cos cos cos

EE OE OE

OE OE

OE OE OE OE

α β γ α β

α α β β γ γ

γ

α α β β γ

′ ′= + + + + +′ ′ ′

′− + +′ ′ ′

′= + − + +′ ′ ′ γ ′ (9)

Comparing (9) with (7), we find

cos cos cos cos cos cos cosθ α α β β γ γ= + +′ ′ ′ (10)

1-3.

x1e3′

x2

x3

O

e1e2

e3

A e2′

e1′e2

e1

e3

Denote the original axes by , , , and the corresponding unit vectors by e , , . Denote the new axes by , , and the corresponding unit vectors by

1x 2x 3x 1 2e 3e

1x′ 2x′ 3x′ 1′e , 2′e , e . The effect of the rotation is e e , , e . Therefore, the transformation matrix is written as:

3′

1 3′→ 2 1′e e→ 3 → 2′e

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

1 1 1 2 1 3

2 1 2 2 2 3

3 1 3 2 3 3

cos , cos , cos , 0 1 0cos , cos , cos , 0 0 1

1 0 0cos , cos , cos ,

′ ′ ′

λ ′ ′ ′= =

′ ′ ′

e e e e e e

e e e e e e

e e e e e e

1-4.

a) Let C = AB where A, B, and C are matrices. Then,

ij ik kjk

C A= B∑ (1)

( )tji jk ki ki jkij

k k

C C A B B A= = =∑ ∑

..

3

2.

4

3.26 CHAPTER 1

1-36.

d

z

x

y

c2 = x2 + y2

The form of the integral suggests the use of the divergence theorem.

(1) S V

d⋅ = ∇ ⋅∫ ∫A a A dv

Since ∇⋅ , we only need to evaluate the total volume. Our cylinder has radius c and height d, and so the answer is

1=A

(2) 2

Vdv c dπ=∫

1-37.

z

y

x

R

To do the integral directly, note that A , on the surface, and that . 3rR= e

5

rd da=a e

3 3 24 4S S

d R da R R Rπ π⋅ = = × =∫ ∫A a (1)

To use the divergence theorem, we need to calculate ∇⋅A . This is best done in spherical coordinates, where A . Using Appendix F, we see that 3

rr= e

( )22

15rr

r r2r

∂∇ ⋅ = =

∂A A (2)

Therefore,

( )2 2 2

0 0 0sin 5 4

R

Vdv d d r r dr R

π π 5θ θ φ π∇ ⋅ = =∫ ∫A∫ ∫ (3)

Alternatively, one may simply set dv in this case. 24 r drπ=

4. ..

24 CHAPTER 1

1-32. Note that the integrand is a perfect differential:

( ) (2 2d d

a b a bdt dt

⋅ + ⋅ = ⋅ + ⋅r r r r r r r r� � �� � )� (1)

Clearly,

( ) 2 22 2 cona b dt ar br⋅ + ⋅ = + +∫ r r r r �� � �� st. (2)

1-33. Since

2

d r rdt r r r 2

rr

−= = −

r r r r r� �� � (1)

we have

2

r ddt dt

r r dt r− =∫ ∫

r r r�� (2)

from which

2

rdt

r r r− = +∫

r r rC

�� (3)

where C is the integration constant (a vector).

1-34. First, we note that

( )ddt

× = × + ×A A A A A A� � � �� (1)

But the first term on the right-hand side vanishes. Thus,

( ) ( )ddt dt

dt× = ×∫ ∫A A A A�� � (2)

so that

( )dt× = × +∫ A A A A C�� � (3)

where C is a constant vector.

1

5.

24 CHAPTER 1

1-32. Note that the integrand is a perfect differential:

( ) (2 2d d

a b a bdt dt

⋅ + ⋅ = ⋅ + ⋅r r r r r r r r� � �� � )� (1)

Clearly,

( ) 2 22 2 cona b dt ar br⋅ + ⋅ = + +∫ r r r r �� � �� st. (2)

1-33. Since

2

d r rdt r r r 2

rr

−= = −

r r r r r� �� � (1)

we have

2

r ddt dt

r r dt r− =∫ ∫

r r r�� (2)

from which

2

rdt

r r r− = +∫

r r rC

�� (3)

where C is the integration constant (a vector).

1-34. First, we note that

( )ddt

× = × + ×A A A A A A� � � �� (1)

But the first term on the right-hand side vanishes. Thus,

( ) ( )ddt dt

dt× = ×∫ ∫A A A A�� � (2)

so that

( )dt× = × +∫ A A A A C�� � (3)

where C is a constant vector.