EE 340 MATH HELP SESSION
This document contains a math help session for the benefit of EE 340 students. It consists
two parts. Part 1 (Review and Theoretical Background) and Part 2 (Solved Problems). Part 1
is to be read by students in order to review some basic concepts in vector calculus and related
theory. It is also needed in preparation for Part 2, which will be delivered by the course
instructor in a problem session format outside the classroom. Thus, EE 340 students are
strongly advised to review Part 1 before the problem session. The last two pages contain a
summary of vector operators and differentials used in this document and throughout the
course.
Part 1 (Review and Theoretical Background)
Cartesian systems
(Right Hand Rule Sequence)
, , • Point in the System
,
example: (3, 4, 5)
• Unit Vectors
In place of , , use , , • Vectors in the Cartesian System
e.g.
3 4 7
• Vector Connecting Two Points ! , " , # , $ !$ , "$ , #$ • Length of a Vector (Magnitude of a Vector):
Let where , , are numbers or functions
∴ || e.g. 1:
let 3 7 2
∴ || √3 7 2
e.g. 2:
let ' ( ln ∴ || + ' ( ln • Addition and Subtraction of Vectors
3 7 9
8 2 3 3 8 7 2 9 3 3 8 7 2 9 3 • Multiplication of Vectors
⋅ |||| cos 234
a) Dot product
note that ⋅ 1 1
⋅ 0
e.g.:
⋅ 0
3 4 9
2 7 6
⋅ 93 4 9 : ⋅ 92 7 6 :
6 0 0 0 28 0 0 0 54 76
Result is a scaler. The math operation may be written as:
⋅ ⟨3,4, 9⟩ ⋅ ⟨2, 7,6⟩ 6 28 54 76
• Angle between two vectors
cos 2 ⋅
76
||||
10.296 ? 9..434
0.7824 141.48°
Notice that the angle 2 is taken such that the two vectors are directed either out or into the
common point.
• Projection of a Vector P on Another Vector Q
e.g.:
Define a unit vector in the direction of a vector , say if 3 4 9 . The unit vector in
the direction is given by
3 1
3 4 9 || √3 4 9
Let 2 7 6
Express as ∥B CB , then;
∥B 93 ⋅ :3
this is the component of in the direction of ∴ D∥B EFG B HFIJKLFMG N
∥B 1
√106
√3
1
4 9
6 28 543 To get CB ∥B
CB 92 7 6 : O
Notice: E E E∥B E ECB E .
93 4 9 : ⋅ 92 7 6 :R 3
76
S
1
√106 √106
3,4, 9T 76
3 4
9 106
76
U 93 4 9 : 4.15 4.13 0.45
106
Let the point in space be 9V , V , V :,
• Common Problems: the normal vector from point P onto a line AB
and the line in space be 2 4 , 2
find |W| & Y (see Fig.)
1. take two points & on the line
2. get 3. Z34 9 ⋅ 34 ::34
4. Then Z W |W|Y Z34
Special Case: when the line is in parallel to one of the axis.
e.g.: Let the equation of the line be ] , ] , and the point R is V , V , 0..
Notice that Z is parallel to [\.
∴ W 9V ] : 9V ] :
: ∴ ^ 9V ] : 9V ] :
: 9] V : 9] V :
? _|| || ⋅ sin 234 a^
b) Cross Product
? b
b 9 : 9 :
Cross products of unit vectors
? , ? , ? Application: normal distance between a point and a line using cross product
1- Solution Using Cross Product
Let the vector in the line direction be 34 and the point be
c c , c , c . The normal from c on the line in the direction away
from the line is given by
W 34 c 34
where c is the vector connecting any point on the line to point c.
Exercise:
Let the line equation be 2 4, 2. Find the normal on the line from the
point 1, 2, 3.
Solution:
Consider the two points on the line. One has 1, 2, 3; ∴ 1, 2, 3,
and the other has 2, 0, 4; ∴ 2,0, 4.
∴ 2 1 0 2 4 3 ⟨1,2, 1⟩
1
⟨1,2, 1⟩
34 √6
c 1 1 2 2 3 3 ⟨0,4,6⟩
1
1
⟨1,2, 1⟩ ? ⟨0,4,6⟩ ?
⟨1,2, 1⟩
W
√6
√6
1 1
⟨1,2, 1⟩
e1
2 1e ?
√6
√6
0
4
6
1
1
⟨16, 6,4⟩ ?
⟨1,2, 1⟩
√6
√6
1 1
10
19
e16 6 4 e 6
3
3
3 1
2 1
2- Solution Using Cross Product
We have solved for 34 and c
∴ Z c ⋅ 34 34 f
√g
⟨0,4,6⟩ ?
√g
⟨1,2, 1⟩h 34 √g
8 634 ⟨1,2, 1⟩
1
1 2 1
10
19
∴ W Zc c Z ⟨0,4,6⟩ ⟨ , , ⟩ 3 3 3
3
3
3
i
Cylindrical Coordinate System:
Point is j, k, 0ljl∞
0 l k l 2n
∞ l l ∞
Notice that j is never negative.
1- o
2- p
3- • Unit vectors
o ⋅ o 1
p ⋅ p 1
• Dot and Cross Product
⋅ 1
,
,
o ⋅ p 0
,
o ⋅ 0
• How to write a vector connecting two points: q , r , # ,
$ q$ , r$ , #$ ?
Is it: j j o k k p Answer is NO
The way to write the vector is as follows:
1. Transform the points and into Cartesian.
2. Write the vector in Cartesian
3. Transform the vector from Cartesian into Cylindrical
Conversion between Cartesian/Cylindrical
Coordinates
Define two matrices:
cos k
s t sin k
0
sin k
cos k
0
0
0u
1
cos k
s v t sin k
0
Let a vector be written in the two systems
o o p p sin k
cos k
0
0
0u
1
∴
o
w x _sa wp x
and
e.g.:
o
wp x _s v a w x
Write the vector connecting the points 3, 30°, 5, 5, 50°, 12
Transformation laws:
j cos 2
,
j + ,
j sin k
k tanv
,
,
Next, obtain the , , coordinates of , j 3
j 5
Now,
k 30°
k 50°
is same
is same
5
12
j cos k 3 cos 10 2.598
j sin k 3 sin 10 1.5
5
∴ 2.598,1.5,5
Similarly:
j cos k 5 cos 50 3.21
j sin k 5 sin 50 3.83
12
∴ 3.21,3.83,12
∴ 0.612 2.33 7
Next
o
cos k
wp x t sin k
0
sin k
cos k
0
0 0.612
0u t 2.33 u
7
1
∴ o 0.612 cos k 2.33 sin k
7
p 0.612 sin k 2.33 cos k
∴ 0.612 cos k 2.33 sin k
k o 0.612 sin k 2.33 cos kp 7
Notice: || must be the same in both systems.
Exercise: Change back from cylindrical to Cartesian.
Spherical Coordinate System
Point is {, 2, k
{ cos 2
{ sin 2 cos k
{ sin 2 cos k
0l{l∞
0l2ln
0 l k l 2n
Surface of Coordinate Systems
• Cartesian Systems:
Consider the point (3,4,5)
To locate it, either: move 3 along , 4 along , 5 along Or:
3 is a plane with unit normal vector . Similarly 4 is a plane with unit normal vector .
And 5 is a plane with unit normal vector .
The intersection of the above three planes defines the point (3,4,5).
Surfaces j constant, constant
constant and k constant are described as
• Cylindrical Coordinates
j constant → cylinder of radius j
Unit normal vector to it is o
constant is a plane parallel to - plane with unit normal vector .
constant & j constant
constant intersect in a circle.
k constant is a semi-infinite
infinite plane with:
0 | j | ∞, ∞ | | ∞
follows:
• Spherical System
Unit vectors are:
Along { → ~
Along 2 → 
Along k → p
{ constant is the surface of a sphere.
2 constant is the surface of a cone.
k constant is a plane
Exercise: imagine the intersection of
1- { ] , 2 ]
circle
2- { ] , k ]
semi-circle
3- 2 ] , k ]
a line along the cone surface
p € _k constanta (€ to the plane)
plane
• Unit Vectors
~ € _{ constanta (€ to the sphere)
And  is a unit vector € to the cone surface in the direction of increasing 2.
• Vectors in Spherical Coordinate System
~ ~   p p
{ cos k ~ ln { ( 'p  √{ sin k sin 2p
e.g.
• Transformations Relations
{ + k tanv \
{ sin 2
{ sin 2 sin k
sin 2 cos k
w x t sin 2 sin k
cos 2
cos 2 cos k
cos 2 sin k
sin 2
Also,
~
sin 2 cos k
w  x t sin 2 sin k
p
cos 2
cos 2 cos k
cos 2 sin k
sin 2
2 tanv
+ ‚ ƒ‚
{ cos 2
sin k ~
cos k u w x
p
0
sin k v cos k u w x
0
~ ~   p p
|| ~  p
• Transformation of Unit Vectors
The same matrices given earlier applies to unit vectors transformation
cos k
w x t sin k
0
o
cos k
wp x tsin k
0
sin k
cos k
0
sin k
cos k
0
0 o
0u wp x
1 0 0u w x
1 The above will be needed extensively when performing integration involving unit vectors under the
integration sign.
Example 1: „ p‚
sin k o
p†
Wk
ALL UNIT VECTORS ARE VARIABLES EXCEPT , , .
Now o cos k sin k p‚
p‚
p‚
∴ „ ‡ sin k 9cos k sin k :Wk ˆ‡ sin k cos k‰ ˆ‡ sin k Wk‰ p†
p†
p†
Example 2:
Let Š { sin k ~ sin 2  { p
~
Determine:
i.
ii.
iii.
D at 10, 150°, 330°
Component of Š tangent to the spherical surface { 10 at Unit vector at € to Š and tangent to the cone 2 150°
Solution:
i.
ii.
iii.
DŠE
‹,Œ‹,ii‹
10 sin 330 ~ ‹ sin 150 cos 330  100 p ⟨5, 0.043,100⟩
Tangent to the sphere are Š and Šp
at : Š 0.043 , Šp 100p
The vector is € to Š and € to 2 , i.e. € to 
So the vector is € to Ž and  but Ž ?  is € to both Ž ,  . Then, this vector
may be obtained from DŽ ?  | .
Thus Ž ?  gives the vector ^
~

p
∴ ^ e5 . 043 100e 100~ 5p
0
1
0
1
9100~ 5p :.
∴ ^ √100 5
Another solution:
Also, the vector is € to 2, so it must be in the form  ±9’~ “p :
Now Š and  are perpendicular to each other, then Š ⋅  0
∴ 5~ 0.043 100p ⋅ 9’~ “p : 0
∴ 5’ 100“ 0
∴
1
+20“
“
∴ ’ 20“
920“~ “p : ±
1
√401
20~ p +'‚ ƒ” ‚
Differential Calculus
• Cartesian System
a) Differential lengths:
Differential length is a vector
W› W or W or W Generally, W› W W W Surfaces are: ] or ] or ]i
b) Differential Surface:
For ] , surface: the unit normal to it is (or possibly )
its differential area is WW
On ] surface: W[ WW .
On ]i surface: W[ WW .
Notice again: surface is a vector.
W[ WW± W[ WW± W[i WW± W WWW NOT A VECTOR
c) Differential Volume:
• Cylindrical System
W› Wjo jWkp W
Specifically, W› in o direction is DW› Eo Wjo ,
W› in p direction is DW› Ep jWk p ,
and W› in direction is DW› E W .
and
Differential Surfaces
W[: Here again the surfaces coincide with the system surfaces:
i.e.
[ : j ] cylindrical surface W[ DW›|p ⋅ DW›| jWk ⋅ Wo
[ : k ] plane
[i : ]i plane
For [ : W[ _jWkWao
W[ DW›|o ⋅ DW›| Wj ⋅ Wp
W[i DW›|o ⋅ DW›|p Wj ⋅ jWk
Differential volume: Multiplication of the three W›’s
• Differential Volume
∴ W Wj ⋅ jWk ⋅ W NOT A VECTOR
• Spherical Differential Elements
a) Differential Lengths
W› W{~ {W2 { sin 2 Wk p
{ ] is a sphere: W[ {W2{ sin 2 Wk ~
b) Differential Surfaces
2 ] is a conical surface. The normal to the cone surface is in the  .
.
∴ DW[E^ž W{ { sin 2 Wk
Finally:
k ]i is a plane in the p direction.
W[ W{ ⋅ {W2 p
• Application:
e.g. 1: Considerr the quarter cylinder shown,
lengths AD and AB are given by
the
‹
Ÿ3Ž ‡ W
‹
¡/
Ÿ34 ‡
p ‹
Dj|o
ΠWk
5 ⋅ n/2
2
Integrate unit area in direction
The Area FDC:
∴ W[ Wj ⋅ jWk
25n
j
¡/
∴ [ ‡ ‡ jWjWk £ ¤ _ka‹ 4
2 ‹
o ‹ p ‹
Œ
^/
Œ
e.g. 2:
Surface area of a section
on of a cone 2 30°, { 5,
k 20° to 50°
Œ
[ ‡
Œ‹°
1 n
{
‡ W{ ⋅ { sin 2 Wk £ ¤ sin 30° _kaŒ‹°
‹° 25 ? ?
2 ‹
2 6
~ ‹ p ‹°
Œ
25n ¥ .
12
Observation:
To obtain the differential area of a surface, determine the direction of the surface, then take the
other two differential elements of the lengths.
e.g: Area of a spherical surface: [
[ ~ {W2 ⋅ { sin 2 Wk~
Area of a conical surface: [
 W2 ⋅ { sin 2 Wk
• Line Integral
Definition:
Let there be a force F and one wishes to evaluate the
following
ollowing integral over certain path from ¬ to ¬ .
§‚
„ ‡ ¦ ⋅ W›
§†
e.g.:
Let ¦ and we wish to integrate
„ „34 „4¨ „¨Ž „Ž3
¦ ⋅ W› over the path ABCDA
Where,
D„34 |
Y© YBª
‹
‡ 9 : ⋅ W
‹
‡ W £
D„4¨ |
Y© YB«
i
1
¤ 3 3
‹
‡ D|
‹
‹, ‹ W
0
Next, motion along CD:
„¨Ž : Here the plane of motion is constant and .
∴ W› W W
∴ „¨Ž ‡ W ‡ W
‹
‹
‡ W ‡ D |
‹
‹
W £
i
1
2
¤ 1_a‹ 1 3 ‹
3
3
Finally, D„Ž3 |Y©
YB« ƒYB­
‹
‹
‹ D
| W
‹
W
Here, we need to relate y to z. This comes from the equation of the line DA. The equation of this line
is obtained from the equation of the line connecting points D(1,1,1) and A(1,0,0). It is ∴ „Ž3
i
1 ‡ W ‡ W £ ¤ £ ¤
2 3 Exercise:
Find
4
¦
3
‹
⋅ W› with ¦ 3 7
‹
A(0,0,0) and 5,3, 2 by moving along: 1. to to ,
2. to to • Cylindrical System
Example of line integral in cylindrical system:
Let ¦ j cos k o j sin kp ( ®p Find „ §‚
¦
§†
⋅ W›
where 3,30°, 2, 5,60°, 6
Part 1: move from to along j axis o then along p direction, then along direction:
∴ „ ‡ ¦ ⋅ Wj o ‡ ¦ ⋅ jWk p ‡ ¦ ⋅ W Œ
‡ Dj cos k|p
o i
i‹°, Wj
g‹°
‡ Dj sin k ⋅ j|o
p i‹°
Next, Part 2: move from to along → o → p
g
„ ‡ D( ®p ⋅ Eo
i,p i‹°
Œ
W ‡ Dj cos k|
o i
g,p i‹°
Wj Œ, Wk
g
‡ D( ®p ⋅ Ep
g‹°
‡ Dj sin k ⋅ j|o
p i‹°
Exercise p → → o : move along p followed by then o .
• Del Operator¯:
Œ, g Wj
The Del operator is given the symbol ∇. In the Cartesian system ∇ ± ± ± The operation of ∇ can be classified into:
±
1- Gradient of a scalar
Let ² be a scalar function
e.g. ² , , ( '
³²
³²
³²
∴ ∇² ³
³
³ And THE RESULT IS A VECTOR.
Usage
1- ´ ∇ basic law we shall need later
2- Obtaining a unit vector ^ normal to a surface [.
Example: [ is normally described by an equation in the form:
 Š 0
∴ Let ²  Š
∴ ^ ± |∇µ|
∇µ
which is the unit vector normal to the surface f.
±
±
g‹°,o Œ
W
e.g. 1:
Let [ ¶ 3 7
∇[ ¶ 3 ∴ ^ ±
vBª ƒiB« vB­
√‚ ƒi‚ ƒ‚
±
√
3 e.g. 2:
Given 7 4 9
Find C and ∥ to the above plane [′
Solution:
C 9 ⋅ ^ :^
Parallel to ^ and € to the plane
∥ C
Parallel to the plane and € to ^
Notice that we obtain the € component then we multiply by ^
• Surface Integral
Definition: „· ¦ ⋅ W[
The integral is either open surface or closed surface
Example:
The vector ¸ is given by ¸ 10(( v jo ,
Integrate over the closed surface of a cylinder
W[ Wj ⋅ jWk
W[ Wj ⋅ jWk W[i jWkW o
Notice the direction is negative here
∴ „º»~µBž ‡ ¸ ⋅ W[ ‡ ¸ ⋅ W[ ‡ ¸ ⋅ W[i
¡
B
‡ ‡ D¸ ⋅ WjjWk E
p ‹o ‹
¡ ©
‡ ‡ D¸ ⋅ jWkWo E
p ‹ ‹
©
¡
B
‡ ‡ D¸ ⋅ jWjWk E
o B
p ‹o ‹
the result is zero.
• Divergence of a vector
Let a vector Š Š Š Š ∴∇⋅Š ³Š ³Š ³Š
³
³
³
Result is a SCALAR
∇ ⋅ Š in the cylindrical system is given by
∇⋅ 1 ³
1 ³p ³
jo j ³j
j ³j
³
Example: let Š 2 7( ' ∴ ∇ ⋅ Š 4 7( ' 2
• The Divergence Theorem: the theorem states that
¼ ¦ ⋅ W½ ¾
·
¿©»Àž
∇ ⋅ ¦ W
∯· indicates closed surface, W W› W› W›i
Exercise:
Let ¸ 10( v jo 10( v , then
∇⋅¸ •
1
⋅ 10( v 2j 20( v 0
j
∇ ? is a vector, defined in the Cartesian system as follows
The Curl of a Vector
‹
∇?Â
±
±
±
Â
±
±
±
Result in a vector
There are similar expression for cylindrical & spherical coordinates.
• Stokes Theorem
∮© ¦ ⋅ W› ∬·9∇ ? ¦ : ⋅ W[
where [ is the open surface defined by the closed path ›.
Example:
Let ¦ 2
Å
34¨Ž
¦ ⋅ W› ‡D W|
‹
³
∇ ? ¦ ÂÂ
³
‹
‹
i
i
‡D2W|
‡D2W|
Next,
‹
‹
‹
‡D W|
i
i
D
D
D
Æ 4 Æ Æ 0
3 ‹
2 ‹ 3 8
8
2 ? 9 18
3
3
³
³
2
i
i
‹
³Â
0 0 2 0 2
³ Â
0
RHR gives W[ WW
i
∴ ‡ ∇ ? ¦ ⋅ W[ ‡ ‡ 2WW _ a,i
‹,‹ 2 ⋅ 3 18
‹ ‹
• Laplacian Operator ¯ $
∇ (A vector → no need, a scalar →∇ ² ±‚ µ
± ‚
±‚ µ
± ‚
±‚ µ
).
± ‚
Part 2 (Solved Problems)
Problem 1
y
Evaluate the line integral
2
→
F
∫ .dl over the closed path l
C
l
shown in figure, where:
F = x 2 ( y − 1)ax + 2 xy ( z 2 + 1)ay − 10z ( x 2 + y 2 )az
l
B
1
A
Solution:
y
2
Divide the integral into three parts (see figure):
C
3
→
→
→
→
∫ F .dl = ∫ F .dl + ∫ F .dl + ∫ F .dl
l1
l
x
l2
2
l
l3
B
1
1
−x 2
I 1 = ∫ Fx dx = ∫ x 2 ( y − 1) y =0 dx = ∫ x 2 ( −1)dx =
3
l1
x =0
x =0
2
I 2 = ∫ Fy dy =
l2
0
−
=
3
A
2
∫
2
2 xy ( z + 1)
x =1, z =0
∫
dy =
y =0
I3 =
l3
∫
∫
2(1) y (0 + 1)dy =
x
2 ydy = y 2
2
0
=4
y =0
0
⇒
1
1
2
2
y =0
I 3 = ∫ Fx dx + ∫ Fy dy
l3
1
0
x 2 ( y − 1)
dx +
y =2 x
x =1
∫
2 xy ( z 2 + 1)
x = y / 2, z =0
dy
y =2
→
(Notice that the limits follow the path. They are high to low in this case). Also dl for l 3 has
x and y components, which causes I 3 to break into two parts. Simplifying and integrating,
we get:
0
I3 =
∫
0
x 2 (2 x − 1)dx +
x =1
I3 =
x4 x3
−
2
3
∫
y =2
0
1
I = I1 + I 2 + I 3
+
y3
3
0
2
⇒
2
y
y (0 + 1)dy
2
⇒
0
⇒
I3 =
0
∫
(2 x 3 − x 2 )dx +
x =1
∫
y 2dy
y =2
1 1 8
I 3 = − + − = −2.833
2 3 3
I = −0.3333 + 4 − 2.833
⇒
∴ I = 0.8337 (answer)
z
Problem 2
Evaluate the line integral
z =4
→
F
∫ .dl over the path l
D
l
l
shown in figure, where:
F = ρ cos φaρ + 0.5ρ 2 sin φ (z 2 + 4)aφ − 3ρ z 2 sin 2 φaz
C
A
4
B
x
Solution:
z
z =4
Divide the integral into three parts (see figure):
→
→
→
→
F
.
dl
=
F
.
dl
+
F
.
dl
+
F
∫
∫
∫ .dl
∫
l1
l
I 1 = ∫ Fρd ρ =
l1
2
=
l2
ρ cos φ
=0
ρ
=0
φ=
π
dρ
π
3
C
A
2
∫
ρ
D
l
l3
2
∫
ρ
y
π
4
4
x
1
2
B
2
ρ
1
2
dρ =
=
= 1.414
2
2 20
2
Length of Straight Line
from A to B = 2
π
2
I 2 = ∫ Fφ ρd φ =
l2
∫π 0.5ρ
φ=
2
sin φ ( z 2 + 4) ρd φ
π
2
∫π
φ=
∫π
φ=
4
l3
sin φd φ = −32 cos φ π2 =
4
32
= 22.63
2
4
4
∫
− 3ρ z 2 sin 2 φ
π
φ = , ρ =2
dz
2
z =0
4
∫
π
2
0.5(2) 2 sin φ (0 + 4)(2)d φ = 32
I 3 = ∫ Fz dz =
=
ρ =2, z =0
4
π
=
4
− 3(2)z 2 sin 2
z =0
I = I1 + I 2 + I 3
y
π
dz = −6 ∫ z 2dz = −2z 3
2
z =0
⇒
4
0
= −128
I = 1.414 + 22.63 − 128
⇒
∴ I = −103.96 (answer)
Problem 3
z
The cube shown in the figure is 2m on the side.
→
Evaluate the closed surface integral F
∫ .ds over
2m
S
y
the surface of the cube, where:
F = x 2 ( x − 2) yzax + 2z (4 − y 2 )ay − 10xyz 2az
x
Solution
z
→
F
∫ .ds =
→
F .ds +
∫
back
S
→
F .ds +
+∫
left
→
F .ds
∫
front
→
F .ds +
∫
right
→
F .ds +
∫
∫
bottom
2m
→
F .ds
y
top
x
For closed surface integrals, use outer normal.
∫
2
2
2
x =0 z = 0
left
2
2
∫ ∫
=−
2
2z (4 − 0)dxdz = −8
x = 0 z =0
2
x =0 z = 0
→
F .ds =
2
Fy dxdz = −
2
∫ ∫
2
2
F .( −dxdyaz ) = −
→
2
2
2
Fy dxdz = −
→
left
top
)dxdz
y =0
= −32
∫ ∫ 2z (4 − y
2
2
)dxdz
y =2
=0
2
−10xyz 2dxdy
z =0
=0
x =0 y =0
2
Fz dxdy =
2
∫ ∫
Fz dxdy = −
∫ F .ds = ∫ F .ds + ∫ F .ds = −32 − 160 = −192
S
2
2
2
x =0 y =0
→
=0
x =0 z = 0
∫ ∫
∫ ∫
2
0
2
x =0 y =0
F .( +dxdyaz ) =
∫ ∫
∫ ∫ 2z (4 − y
z2
2
zdxdz = −8(2)
∫ ∫
x =2
2
2
x =0 z =0
x =0 y =0
top
2
2
=0
z =0 y =0
x = 0 z =0
x =0 y =0
bottom
∫
2
x 2 ( x − 2) yzdydz
x =0 z = 0
F .( +dxdzay ) =
∫ ∫
→
F .ds =
∫
2
x =0
2
∫ ∫
x = 0 z =0
→
F .ds =
right
2
Fx dydz =
∫ ∫
∫ ∫
x 2 ( x − 2) yzdydz
z =0 y =0
z =0 y =0
2
2
∫ ∫
Fx dydz = −
2
∫ ∫
F .( −dxdzay ) = −
∫ ∫
2
∫ ∫
2
F .( +dydzax ) =
∫ ∫
2
→
F .ds =
2
z =0 y =0
z =0 y =0
front
∫
2
F .( −dydzax ) = −
∫ ∫
→
F . ds =
∫
2
z =0 y =0
back
∫
2
→
F . ds =
2
∫ ∫
x =0 y =0
−10xyz 2dxdy
z =2
= −40(2)(2) = −160
Problem 4
z
a
Consider the open conical surface S shown in the figure
→
(thick solid lines). Evaluate the surface integral ∫ F .ds
S
θ
S
over S, where:
y
φ F = r cos θ ar + r 2 sin θ cos( )aθ − sin φaφ
x
θ=
4
π
6
Solution
z
a
→
∫ F .ds = ∫ F .( r sin θdrd φaθ ) = ∫ Fθ (r sin θdrd φ )
S
S
S
S
φ
= ∫ [r 2 sin θ cos( )]( r sin θ drd φ )
4
S
& a =1
aθ
θ h
y
2π
=
h
=0
r 3 sin 2 θ cos( )drd φ
4
r =0
Where h = a / sin θ =
1
sin
→
F
∫ .ds =
2π
2π
2π
π
φ
π
θ=
2
1
=2
0.5
6
2
4
φ
r4
4
4
r 3 cos( )drd φ =
r =0
=
θ=
π
φ
r 3 sin 2 ( ) cos( )drd φ
r =0
2
∫ ∫
φ
=0
2
∫ ∫
φ
=0
S
=
x
φ
∫ ∫
φ
φ
= 4 ∫ cos( )d φ = 16sin( )
4
4
φ =0
2π
2
×
0
∫
φ
=0
2π
0
= 16
φ
cos( )d φ
4
π
6
& a =1
Problem 5
z
Consider the closed spherical surface of radius a = 4 .
Verify the divergence theorem for this surface and the
enclosed volume using the vector field:
F = r 2ar +
a
1 aθ − r sin θ cos φaφ
r sin θ
x
Solution
z
ar
The divergence theorem states that:
→
F
.
ds
=
(
∇
F
∫ )dv
∫
S
S
a
y
V
V
x
→
IS = F
.
ds
=
F
∫
∫ .(ds r ar ) = ∫ Frds r
S
2π
=
S
S
π
2π
∫ ∫
( r 2 )(r 2 sin θ d θ d φ )
φ =0 θ =0
r =4
π
= 256 ∫
∫
π
sin θd θ d φ = 256(2π )
φ =0 θ =0
π
= 256(2π )( − cos θ )
0
∫ sin θd θ
θ =0
= 256(2π )(1 + 1) = 1024π
1 ∂ 2
1
∂
1 ∂Fφ
∇F = 2
( r Fr ) +
( Fθ sin θ ) +
r ∂r
r sin θ ∂θ
r sin θ ∂φ
=
1 ∂ 2 2
1
∂
1
1 ∂( − r sin θ cos φ )
(r r ) +
(
sin θ ) +
2
r ∂r
r sin θ ∂θ r sin θ
r sin θ
∂φ
=
1 ∂ 4
1
∂ 1 − r sin θ ∂ (cos φ ) 4 r 3
(
r
)
+
( )+
= 2 + 0 + sin φ = 4r + sin φ
r 2 ∂r
r sin θ ∂θ r
r sin θ
∂φ
r
I V = ∫ ∇Fdv =
=
π
2π
∫ θ∫ φ∫
r =0
V
4
4
=0
4
(4 r + sin φ )( r 2 sin θ drd θ d φ ) =
=0
π
∫ ∫
r =0 θ = 0
(8π r − 1 + 1)( r 2 sin θ drd θ ) =
∫ θ∫
r =0
4
π
(4 r φ − cos φ )
=0
π
∫ ∫
8π r 3 sin θ drd θ = 8π
r = 0 θ =0
∴ I S = IV = 1024π (which verifies the divergence theorem)
2π
0
44
(2) = 1024π
4
( r 2 sin θ drd θ )
Problem 6
z
Consider the closed circular line l of radius a shown in
the figure. Verify Stockes' theorem this line and the
bounded flat circular area S using the vector field:
S
a
l
F = z 2 ρ 2aρ + z ρ 2aφ + (z − 1)sin φaz
h
x
Solution
z
Stockes' theorem states that:
S
→
→
F
.
dl
=
(
∇
×
F
). ds
∫
∫
l
→
dS
S
→
Il = ∫ F .dl = ∫ F .(dl )φ aφ = ∫ Fφ (dl )φ
l
l
=
∫ (z ρ
φ
2π
2
)( ρd φ )
ρ =a , z = h
=0
=
∫ ha d φ = 2π ha
φ
3
ρ aφ
az
1 ∂
∇×F =
ρ ∂ρ
Fρ
∂
∂φ
ρ Fφ
∂
1 ∂
=
ρ ∂ρ
∂z
Fz
z 2ρ 2
aρ
ρ aφ
az
∂
∂φ
ρz ρ 2
∂
∂z
( z − 1)sin φ
1 {aρ [( z − 1) cos φ − ρ 3 ] + ρ aφ [2z ρ 2 − 0] + az [3z ρ 2 − 0]}
ρ
=[
(z − 1) cos φ
ρ
− ρ 2 ]aρ + 2z ρ 2aφ + 3z ρ az
→
I S = ∫ (∇ × F ). ds = ∫ (∇ × F ) z (ds ) z =
S
2π
=
3
=0
aρ
=
x
l
2π
S
a
∫ ρ∫
φ =0
=0
2π
a
φ =0
=0
∫ ρ∫ 3z ρ ( ρd ρd φ )
a
3h ρ 2d ρd φ = 3h (2π )
∫
ρ =0
ρ 2d ρ = 6π h
a3
= 2π ha 3
3
∴ I l = I S = 2π ha 3 (which verifies Stokes' theorem)
z =h
a
l
h
Vector Relations
Cartesian Coordinates (x,y,z)
A = Ax a x + Ay a y + Az a z
∂V
∂V
∂V
ax +
ay +
az
∂x
∂y
∂z
∂A ∂Ay ∂Az
∇⋅ A = x +
+
∂x
∂y
∂z
∇V =
ax
∂
∇× A =
∂x
Ax
ay
∂
∂y
Ay
dl = a x dx + a y dy + az dz
ds = a x dydz + a y dxdz + a z dxdy
dv = dxdydz
az
 ∂A ∂ Ay 
 ∂Ay ∂Ax 
∂
 ∂Ax ∂Az 
ay + 
= z −
−
−
a x + 
a z

∂z  ∂y
∂z 
∂x 
∂y 
 ∂z
 ∂x
Az
∂ 2V ∂ 2V ∂ 2V
∇V = 2 + 2 + 2
∂x
∂y
∂z
2
Cylindrical Coordinates (ρ,φ,z)
A = Aρ a ρ + Aφ a φ + Az a z
1 ∂V
∂V
∂V
aρ +
aφ +
az
∂ρ
∂z
ρ ∂φ
1 ∂
(ρAρ ) + 1 ∂Aφ + ∂Az
∇⋅ A =
∂z
ρ ∂ρ
ρ ∂φ
∇V =
aρ
1 ∂
∇× A =
ρ ∂ρ
Aρ
∇ 2V =
ρa φ
∂
∂φ
ρAφ
dl = ar dr + aθ rd θ + aφ r sin θ d φ
ds = ar r 2 sin θ d θ d φ + aθ r sin θ drd φ + aφ rdrd θ
dv = r 2 sin θ drd θ d φ
az
∂Aρ 
 1 ∂Az ∂Aφ 
 ∂Aρ ∂Az 
1∂
∂
=
−
−
a ρ + 
a φ +  (ρAφ ) −
a z
ρ  ∂ρ
∂ρ 
∂φ 
∂z  ρ ∂φ
∂z 
 ∂z
Az
1 ∂  ∂V  1 ∂ 2V ∂ 2V
ρ
+
+
ρ ∂ρ  ∂ρ  ρ 2 ∂φ 2 ∂z 2
dl = ar dr + aθ rd θ + aφ r sin θ d φ
Spherical Coordinates (r,θ,φ)
ds = ar r 2 sin θ d θ d φ + aθ r sin θ drd φ + aφ rdrd θ
A = Ar a r + Aθ a θ + Aφ a φ
∇V =
dv = r 2 sin θ drd θ d φ
∂V
1 ∂V
1 ∂V
ar +
aθ +
aφ
r ∂θ
r sin θ ∂φ
∂r
∇⋅ A=
∂Aφ
1 ∂ 2
1
∂
( Aθ sin θ ) + 1
r Ar +
2
r sin θ ∂θ
r sin θ ∂φ
r ∂r
(
ar
1
∂
∇× A= 2
r sin θ ∂r
Ar
)
ra θ
∂
∂θ
rAθ
(r sin θ )a φ
1
∂
=
r sin θ
∂φ
(r sin θ )Aφ
+
∂Aθ
 ∂
 ∂θ (Aφ sin θ ) − ∂φ

∂A
1∂
(rAθ ) − r a φ

r  ∂r
∂θ 
1 ∂  2 ∂V 
1
∂ 
∂V 
1
∂ 2V
∇V= 2
r
+
 sin θ
+
∂θ  r 2 sin 2 θ ∂φ 2
r ∂r  ∂r  r 2 sin θ ∂θ 
2


1  1 ∂Ar
∂
a r + r  sin θ ∂φ − ∂r (rAφ )a θ


