Chapter 2 (part 1)
Review of Vector Calculus
Outline
1. Basic vector algebra
2. Orthogonal coordinate systems
3. Transformation of coordinate systems
4. Line, surface, volume integrals
5. Gradient of a scalar field
6. Divergence of a vector field
7. Curl of a vector field
8. Laplacian operator
Basic Vector Algebra
❑ A scalar is a quantity that has only magnitude.
❑ A vector is a quantity that has both magnitude and direction.
❑ For vector
  

A = a A = Aa
where
 


A = a x Ax + a y Ay + a z Az


 A A
a=  =
A A
A = Ax2 + Ay2 + Az2
 


 A a x Ax + a y Ay + a z Az
a= =
A
Ax2 + Ay2 + Az2
 
 
A
=
B
a
❑ If A = B, it means that
and also = b .
Vector Addition & Subtraction
Parallelogram rule
  
C = A+ B
  
D = A− B
Head-to-tail rule
Position & Distance Vectors
❑ The position vector (or radius vector) of point P
is defined as the directed distance from the
origin O to P.
❑ The position vector of point P is useful in
defining its position in space.



R12 = R2 − R1
❑ The distance vector is the displacement
from one point to another.
Vector Scalar Product
❑ Scalar (or dot) product
 
A  B = AB cos AB
 
A  B = Ax Bx + Ay B y + Az Bz
  2
A  A = A = A2
   
A B = B  A
(
)
Commutative
      
A B + C = A B + AC
Distributive
Vector Cross Product (i)
❑ Vector (or cross) product
  
A  B = an ( AB sin  AB )
 
A B

B
AB

A
(a)
(a) Right-hand rule
(b) Right-handed-screw rule
(b)
Vector Cross Product (ii)


ax a y
 

A  B = AB sin  AB an = Ax Ay
Bx B y

= (Ay Bz − Az B y )a x + ( Az Bx

az
 Ay Az  Az Ax  Ax
Az = a x
+ ay
+ az
B y Bz
Bx
Bz Bx
Bz


− Ax Bz )a y + (Ax B y − Ay Bx )a z
 
A A = 0
 
A B
 
 
A  B = −B  A
(
)
Anti-commutative
      
A B + C = A B + A C

B
Distributive

A
Ay
By
Vector Triple Product
(
)
(
)
(
        
A B  C = B  C  A = C  A B
(
) (
(
) (
) (
        
A B  C = B AC − C A B
)
  
  
A B  C  A B  C
)
)
Bac-cab rule
NOT associative
Cartesian Coordinates
❑ A point P can be represented as (x, y, z).
❑ The ranges of the coordinate variables are
− x
− y 
− z 
𝑎Ԧ 𝑦
𝑎Ԧ 𝑧
𝑎Ԧ𝑥
Cylindrical Coordinates
❑ A point P can be represented as (, , z).
❑  is called the azimuthal angle and is measured from the x-axis in
the xy plane.
❑ The ranges of the coordinate variables are
0  
0    2
− z 

❑ Note that a is not in degrees.

❑ The magnitude of vector A is

A = A2 + A2 + Az2
Transformation between Cartesian and
Cylindrical Coordinates (I)
 = x + y ,  = tan
2
2
x =  cos  ,
−1
 tan −1 ( y / x )
 −1
tan ( y / x ) + 
 tan −1 ( y / x ) − 
 =
 /2


− /2

 undefined
y
, z=z
x
y =  sin  , z = z
x0
x0& y0
x0& y0
x=0& y 0
x=0& y0
x=0& y =0
z
y
x
Transformation between Cartesian and
Cylindrical Coordinates (II)
 A   cos 
 A  = − sin 
  
 Az   0
 Ax  cos 
 A  =  sin 
 y 
 Az   0
sin 
cos 
0
0  Ax 
0  Ay 
1  Az 
− sin 
cos 
0
0  A 
0  A 
1  Az 
Cylindrical Coordinates
  
❑ The base vectors (a x , a y , a z ) are independent of the location of a point.
 
❑ But, a  , a are not because



a = a x cos  + a y sin 



a = −a x sin  + a y cos 
❑ Caution for vector addition and integral in cylindrical coordinates



f ( ) a  d  a  f ( )d

𝑎Ԧ𝜙
𝑎Ԧ 𝑧
𝑎Ԧ𝜌
Spherical Coordinates
❑ A point P can be represented as (r, , ).
❑  is called the colatitude and is the angle between the z-axis and the
position vector of P.
𝑎Ԧ𝜃
❑ The ranges of the coordinate variables are
0r 
0  
0    2
❑ The magnitude of vector A is

A = Ar2 + A2 + A2
𝑎Ԧ𝜙
𝑎Ԧ𝑟
Transformation between Cartesian and
Spherical Coordinates (I)
r = x 2 + y 2 + z 2 ,  = tan −1
x2 + y2
y
,  = tan −1
z
x
x = r sin  cos 

y = r sin  sin 
z = r cos 

Transformation between Cartesian and
Spherical Coordinates (II)
 Ar   sin  cos 
  
 A  = cos  cos 
 A   − sin 
 
sin  sin 
cos  sin 
cos 
cos    Ax 
− sin    Ay 
0   Az 
 Ax  sin  cos 
 A  =  sin  sin 
 y 
 Az   cos 
cos  cos 
cos  sin 
− sin 
− sin    Ar 
 
cos    A 
0   A 
Summary of Coordinate Transformations
Cartesian to
cylindrical
Cylindrical to
Cartesian
Cartesian to
spherical
Spherical to
Cartesian
Cylindrical to
spherical
Spherical to
cylindrical
Coordinate
variables
Basis
vectors
 = x2 + y2
 = x cos  + y sin 



 = − x sin  + y cos 

 = tan −1 ( y / x )
z=z

 
z=z
y =  sin 
 
z=z
z=z
(
 = tan −1 x 2 + y 2 / z
 = tan −1 ( y / x )
x = r sin  cos 
y = r sin  sin 
z = r cos
 = r2 + z2
 = tan −1 (r / z )
 =
)
A = Ax cos  + Ay sin 
A = − Ax sin  + Ay cos 
Az = Az

 
x =  cos  −  sin 

 
y =  sin  +  cos 
x =  cos 
r = x2 + y2 + z 2

Vector
components
Ax = A cos  − A sin 
Ay = A sin  + A cos 
Az = Az
 


r = x sin  cos  + y sin  sin  + z cos Ar = Ax sin  cos  + Ay sin  sin  + Az cos
 


 = x cos cos  + y cos sin  − z sin  A = Ax cos cos  + Ay cos sin  − Az sin 



A = − Ax sin  + Ay cos 
 = − x sin  + y cos 


 
x = r sin  cos  +  cos cos  −  sin  Ax = Ar sin  cos  + A cos cos  − A sin 


 
y = r sin  sin  +  cos sin  +  cos  Ay = Ar sin  sin  + A cos sin  + A cos 

 
Az = Ar cos − A sin 
z = r cos −  sin 
 

Ar = A sin  + Az cos
r =  sin  + z cos
 

A = A sin  − Az sin 
 =  sin  − z sin 


 =
A = A
 = r sin 
 = r sin  +  cos
A = Ar sin  + A cos
 =
 =
z = r cos
 
z = r cos −  sin 






A = A
Az = Ar cos − A sin 
Constant-Coordinate Surfaces
❑ Surfaces in Cartesian, cylindrical, or
spherical coordinate systems are
easily generated by keeping one of
the coordinate variables constant and
allowing the other two to vary.
Vector Area
❑ The vector area (or area vector) is defined as


S = S an

an
Direction of this unit normal vector
Surface area = S
❑ For a closed surface, the plus sign
typically indicates the outward vector.
Differential Length, Area, and Volume
(Cartesian coordinate system)
❑ Differential normal surface area

dy dz a x

 

dS = dS an = dx dz a y
dx dy a
z

❑ Differential displacement




dl = dx a x + dy a y + dz a z
❑ Differential volume
dv = dx dy dz
Differential Length, Area, and Volume
(Cylindrical coordinate system)
❑ Differential normal surface area

  d dz a 
 

dS =  d dz a
  d d a
z

❑ Differential displacement




dl = d a  +  d a + dz a z
❑ Differential volume
dv =  d d dz
Differential Length, Area, and Volume
(Spherical coordinate system)
❑ Differential normal surface area

r sin  d d ar
 

dS =  r sin  dr d a
 r dr d a


2
❑ Differential displacement




dl = dr ar + r d a + r sin  d a
❑ Differential volume
dv = r 2 sin  dr d d

Line Integral
❑ The line integral

A

 
A  dl
is the integral of the
L
tangential component of along curve L.

 
A  dl =

b

A cos dl
a
L
❑ Closed contour integral:

 
A  dl
L
❑ When we represent the path L by a parametric
Path L in space
representation, we have
(  t   )
L ( t ) = x ( t ) ax + y ( t ) a y + z ( t ) az
❑ Then, a line integral of a vector

A
over a path L
z
L (t )
y
x
is given by

b
a
A  dL = 


dL
A  dt
dt


t
Exercise (i)
❑ Calculate

P2
P1
y
 
 
3

F  dl along L, where F = a x xy − a y 2 x
P2
L
(1) In Cartesian coordinates
 

dl = a x dx + a y dy

P2
 
F  dl =
P1
P1




(
)
(
a
xy
−
a
2
x

a
dx
+
a
dy ) = 

P2
x
P1
3
O
y
x
0
3
3
0
y
P2
xydx − 2 xdy
P1
x2 + y 2 = 9
=  x 9 − x 2 dx −  2 9 − y 2 dy
0
3
1

2 2
= − ( 9 − x ) −  y 9 − y 2 + 9sin −1
3

3
y
 
=
−
9
1 + 
3  0
 2
3
(2) In cylindrical coordinates
 F   cos 
 F  = − sin 
  
 Fz   0

P2
P1
 
F  dl =
sin 
cos 
0

P2
P1
0  xy 
0 2 x 
1  0 
 

F = a  (xy cos  − 2 x sin  ) − a (xy sin  + 2 x cos  )
 
x = 3 cos  , y = 3 sin 
dl = a 3d
− 3(9 sin 2  cos  + 6 cos 2  )d = − 9(sin 3  +  + sin  cos  ) 0
 
= −91 + 
2

 /2
x
Exercise (ii)
❑ Calculate

P2
P1
 
 
3

F  dl along L, where F = a x xy − a y 2 x
y
P2
L
(3) In Cartesian coordinates using the parametric representation
l = ax 3cos ( t ) + a y 3sin ( t ) ,
(0  t   / 2)
l
O
P1
3
Since x ( t ) = 3cos ( t ) and y ( t ) = 3sin ( t ),
F = ax 9 cos ( t ) sin ( t ) − a y 6 cos ( t )

P2
P1
F  dl = 
 /2
=
 /2
=
 /2
0
0
0
F
dl
dt
dt
 ax 9 cos ( t ) sin ( t ) − a y 6 cos ( t )    −ax 3sin ( t ) + a y 3cos ( t )  dt
 −27 cos ( t ) sin 2 ( t ) + 18cos 2 ( t )  dt
 /2
 27 sin 3 ( t )
9sin ( 2t ) 
= −
+ 9t +

3
2

0
 
= −9 1 + 
 2
x
Exercise

❑ Calculate
P2
P1
y
 
 
3

F  dl along L, where F = a x xy − a y 2 x
P2
L
(1) In Cartesian coordinates
 

dl = a x dx + a y dy

P2
 
F  dl =
P1

P1
(ax xy − a y 2 x )(ax dx + a y dy ) =
P2
P1
O

P2
xydx − 2 xdy
P1
y = −x + 3
=  x ( − x + 3) dx −  2 ( 3 − y ) dy
0
3
3
0
3
3
3 
1 
 1

=  − x3 + x 2  − 2 3 y − y 2  = −13.5
2 0
2 0
 3

(2) Alternative method
x (t ) = 3 − t

P2
P1
y (t ) = t
3
F  dl =  F 
0
l = ax ( 3 − t ) + a y t ,
dl
dt
dt
=   ax ( 3 − t ) t − a y 2 ( 3 − t )    −ax + a y  dt
0
3
=  ( t 2 − 3t ) + 2 ( t − 3) dt = −13.5
0
3
( 0  t  3)
3
x
Surface Integral (i)
❑ The surface integral or flux is defined as

S
 
A  dS =

b
Surface S

A cos dS
a
❑ For a close surface, we have

 
A  dS ,
S
referred to as the net outward flux of
which is

A
from S.
Surface S in space
❑ When we represent the surface S by a parametric
representation, we have
S ( t ) = x ( u , v ) ax + y ( u , v ) a y + z ( u , v ) a z
z
(u, v in R)
y
S ( u, v )
x
❑ The normal vector of the surface S at point P is a
v
vector perpendicular to the tangent plane of S at P.
R
u

an
Surface Integral (ii)
dS
dv
❑ Then, the normal vector of the surface can
be expressed as
dS dS
n=

du dv
S
❑ The unit normal vector is the normalized n vector.
Thus,
dS dS
an =

du dv
dS dS

du dv
❑ The surface integral can be obtained by
 A  dS =  A  n dudv
S
R
dS
du
Exercise

❑ Calculate
S
F  dS on a parabolic cylinder S,
where F = ax 3z 2 + a y 6 + az 6 xz and
S : y = x 2 , 0  x  2, 0  z  3
By setting x = u, and z = v, we have y = x2 = u2.
S ( t ) = ax u + a y u 2 + az v
n=
( 0  u  2, 0  v  3)
dS dS

= ( ax + a y 2u )  ( az ) = ax 2u − a y
du dv
2
F

dS
=
a
3
v
+ a y 6 + az 6uv )  ( ax 2u − a y ) dudv
(
x


S
R
 ( 6uv − 6 ) dudv
=  (12v − 12 ) dv
3
2
0
0
=
3
0
= 72
2
2
Volume Integral
❑ The volume integral is defined as
  dv
v
v
❑ The volume integral is a triple integral within a volume V. Thus,
  dxdydz
v
v
z
y
v
x
Del Operator
❑ The del operator , also known as the gradient operator, is a vector
differential operator. It is useful in defining


V ,   A,   A,  2V
❑ In Cartesian coordinates,
 
 
 
 = ax + a y + az
x
y
z
❑ In cylindrical coordinates,
=
 
1  
 
a +
a + a z

 
z
❑ In spherical coordinates,
  1  
1
 
 = ar +
a +
a
r
r 
r sin  
Gradient
❑ The gradient of a scalar field V is a vector that represents both magnitude
and the direction of the maximum space rate of increase of V.
V  V  V 
V = grad V =
ax +
ay +
az
x
y
z
❑ The gradient of a scalar field V satisfies the following formulas.
(V + U ) = V + U
(VU ) = VU + UV
V

U
 UV − VU
=
U2

V n = nV n −1V
f ( x, y ) = x 3 − 3 x − 2 y 2
Gradient:
Examples
Divergence (i)

❑ The divergence of A at a given point P is the outward flux per unit volume
as the volume shrinks about P.


  A = div A = lim
v → 0

 
A  dS
S
v

❑ The divergence of A in Cartesian, cylindrical, and spherical coordinates
 Ax Ay Az
 A=
+
+
x
y
z
 1 
1 A Az
(A ) +
 A=
+
 
 
z
 1  2
1

1 A
( A sin  ) +
(r Ar ) +
 A= 2
r r
r sin  
r sin  
Divergence (ii)
❑ The divergence of vector fields satisfies the following formulas.
( )
( )
 


 A+ B = A+B

 
  VA = V  A + A  V
Divergence:
Examples
Red: source
Blue: sink
Divergence Theorem


 
  Adv = A  dS

v
S
❑ The total outward flux of a vector field through the closed surface S
is equal to the volume integral of its gradient over the volume v
enclosed by S
❑ Since the outward flux to one cell is
inward to the neighboring cells, the
cancellation on every interior
Volume v
surface.

S
 
A  dS =

k
Sk
 
A  dS =

k

Volume vk
 
A  dS
Sk
 k
 k
Closed
surface S
Exercise
❑ Verify the divergence theorem for the following closed surface and vector.

dS1
  2  2

D = ar r + a r sin  + a r 2 sin  sin 
  2
dS1 = ar r sin  d d r = 2
 
dS 2 = a r sin  dr d  = / 2
r =2

S
 ( )  ( ) 


 


D  dS = D  dS1 + D  dS 2 = 2 4 sin  d d + r 3 sin 2  dr d
S1
S2

 /2

2
= 16 sin  d d +
0
0

S1

dS 2
S2
2
 r dr  d = 40
2
0
3
0
1  4
1
 2 2
1 r 2 sin  sin  
dv

v   Ddv =   r 2 r ( r ) + r sin   ( r sin  ) + r sin 



2
2
4
r
+
2
r
cos

+
r
cos

r
sin  d d dr = 40
(
)
0  0
=
2
2
0
Curl (i)

❑ The curl of A is an axial (rotational) vector whose magnitude is the

maximum circulation of A per unit area as the area tends to zero and
whose direction is the normal direction of the area when the area is
oriented to make the circulation maximum.
 

A  dl 




 an
  A = curl A =  lim L
 S →0 S 



 max


an
Curl (ii)

❑ The curl of A in Cartesian, cylindrical, and spherical coordinates

ax
 
 A =
x
Ax

ay

y
Ay

a
 1 
 A =
 
A

ar

1

 A = 2
r sin  r
Ar

az
  Az Ay    Ax Az    Ay Ax  
=
−
−
ax + 
 a y +  x − y  a z
z  y
z 

z

x




Az

a


A

ra


rA

az
  1 Az A    A Az  
1   (A ) A  
=
−
a
+
−
a
+
−
  
 

az
z   
z 

z











Az

r sin  a

1   (A sin  ) A
=
−


r sin  


r sin  A
  1  1 Ar  (rA )  1   (rA ) Ar  
−
−
a
 ar + 
 a + 
r  sin  
r 
r  r
 

Curl (iii)
❑ Properties of the curl
( )
( ) (
( )
( )
 


 A+ B =  A+ B
  

 
 


  A B = A   B − B   A + B  A − A B



  VA = V  A + V  A

  A = 0
  V = 0
) (
) (
) (
)
Curl: Examples
Red: counterclockwise
Blue: clockwise
Divergence
versus
Curl
Stokes’s Theorem
(
)
 
 
  A  dS = A  dl

S
L

❑ The circulation of a vector field A around a closed path L is equal to the

surface integral of the curl of A over the open surface S bounded by L,


provided A and   A are continuous on S.
❑ There is cancellation on every interior path, so

dl
the sum of the line integral around the Lk’s is the
same as the line integral around the bounding
curve L.

L
 
A  dl =

k
Lk
 
A  dl =

k

 
A  dl
Lk
S k
S k
L
Exercise
❑ Verify the Stokes’s theorem on the quarter-circle loop shown below.
 

F = a x xy − a y 2 x

ax


 F =
x
xy

ay

y
− 2x
y

az


= −(2 + x )a z
z
0


(
)
(
)


F

d
S
=
−
2
−
x
a



S

z  dxdya z =
S

3
= − 2 9 − y2 +
0
P2
O
P1
P2
3
 
dS = a z dxdy
P2
P1

3
0
9− y
2
0
O
(− 2 − x )dxdy
3
1
(9 − y 2 )dy = −91 +  
2
2

P1
 F  dl =  F  dl +  F  dl +  F  dl
L
O
=  F  ( ax dx + a y dy ) +  F  ( a y dy )
P2
O
P1
P2
=
P2
P1
+  F  ( ax dx ) y =0
P1
x =0
O
 
xydx − 2 xdy =  x 9 − x dx −  2 9 − y 2 dy = −9 1 + 
3
0
 2
0
2
3
x
Laplacian
❑ The Laplacian of a scalar field V, written as  2V , is the divergence of the
gradient of V.
❑ The Laplacian of V in Cartesian, cylindrical, and spherical coordinates
 2V  2V  2V
V= 2 + 2 + 2
x
y
z
2
1   V
 
V=
   
2
 1  2V  2V
 + 2
+ 2
2
z
  
1   2 V 
1
 
V 
1
 2V
 V = 2 r
+ 2
 sin 
+ 2 2
r r  r  r sin   
  r sin   2

❑ It is also possible to define the Laplacian of a vector A .
2
(
)
2 A =    A −     A
=  2 Ax ax +  2 Ay a y +  2 Az az
Classification of Vector Fields


❑ A vector field A is said to be solenoidal (or divergenceless) if   A = 0.



 
❑ If   A = 0 , A  dS = 0 and A =   F .

S


❑ A vector field A is said to be irrotational (or potential) if   A = 0.
❑ An irrotational field is also known as a conservative field.
 


❑ If   A = 0,  A  dl = 0 and A = −V.
L




  A = 0,   A = 0   A  0,   A = 0


  A = 0,   A  0


  A  0,   A  0
Helmholtz’s Theorem
❑ Helmholtz’s theorem states that a vector field is determined to within an
additive constant if both its divergence and its curl are specified
everywhere.
 

A = Airrotational + Asolenoidal

= −V +   B
❑ Since the gradient and curl operators are differential operators, the field

must be obtained by integrating V and B in some manners, which will
lead to constants of integration. The determination of these additive
constants requires the knowledge of some boundary conditions.
Summary
1. Scalar vs. vector
2. Vector addition, subtraction, dot product, and cross product
3. Coordinate systems. Why do we use more than one coordinate
system?
 
4. In cylindrical coordinates, a , a are coordinate-dependent.
5. In spherical coordinates, the coordinate variables are coordinatedependent.
6. Transformations of coordinate systems
7. Line, surface, and volume integrals
8. Gradient, divergence, and curl
9. Divergence theorem and Stokes’s theorem