electromagnetic thepry, Exams of Electrical Engineering

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Problem 3.25 Use the appropriate expression for the differential surface area dsto
determine the area of each of the following surfaces:
(a) r=3; 0
φ
π
/3; 2z2,
(b) 2r5;
π
/2
φ
π
;z=0,
(c) 2r5;
φ
=
π
/4; 2z2,
(d) R=2; 0
θ
π
/3; 0
φ
π
,
(e) 0R5;
θ
=
π
/3; 0
φ
2
π
.
Also sketch the outlines of each of the surfaces.
Solution:
32
∆Φ = π/3
25
y
x
25
(a) (b)
(d) (e)
(c)
Figure P3.25: Surfaces described by Problem 3.25.
(a) Using Eq. (3.43a),
A=Z2
z=2Z
π
/3
φ
=0(r)|r=3d
φ
dz =³(3
φ
z)|
π
/3
φ
=0´¯¯¯
2
z=2=4
π
.
(b) Using Eq. (3.43c),
A=Z5
r=2Z
π
φ
=
π
/2(r)|z=0d
φ
dr =³¡1
2r2
φ
¢¯¯
5
r=2´¯¯¯
π
φ
=
π
/2=21
π
4.
pf2

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Problem 3.25 Use the appropriate expression for the differential surface area d s to

determine the area of each of the following surfaces:

(a) r = 3; 0 ≤ φ ≤ π/3; − 2 ≤ z ≤ 2,

(b) 2 ≤ r ≤ 5; π/ 2 ≤ φ ≤ π; z = 0,

(c) 2 ≤ r ≤ 5; φ = π/4; − 2 ≤ z ≤ 2,

(d) R = 2; 0 ≤ θ ≤ π/3; 0 ≤ φ ≤ π,

(e) 0 ≤ R ≤ 5; θ = π/3; 0 ≤ φ ≤ 2 π.

Also sketch the outlines of each of the surfaces.

Solution:

3 2

∆Φ = π/

5 2

y

x

2 5

(a) (b)

(d) (e)

(c)

Figure P3.25: Surfaces described by Problem 3.25.

(a) Using Eq. (3.43a),

A =

∫ 2

z =− 2

∫ π/ 3

φ = 0

( r )| r = 3

d φ dz =

( 3 φ z )|

π/ 3 φ = 0

2

z =− 2

= 4 π.

(b) Using Eq. (3.43c),

A =

∫ 5

r = 2

∫ π

φ = π/ 2

( r )| z = 0 d φ dr =

1 2

r

2 φ

5

r = 2

π

φ = π/ 2

21 π

(c) Using Eq. (3.43b),

A =

∫ 2

z =− 2

∫ 5

r = 2

φ = π/ 4 dr dz =

( rz )|

2 z =− 2

5

r = 2

(d) Using Eq. (3.50b),

A =

∫ π/ 3

θ = 0

∫ π

φ = 0

R

2 sin θ

R = 2

d φ d θ =

(− 4 φ cos θ )|

π/ 3 θ = 0

π

φ = 0

= 2 π.

(e) Using Eq. (3.50c),

A =

∫ 5

R = 0

∫ 2 π

φ = 0

( R sin θ )| θ = π/ 3 d φ dR =

1 2

R

2 φ sin

π

2 π

φ = 0

5

R = 0

3 π