Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 6 Vector analysis (벡터 해석)
Lecture 18 Basic vector analysis
1. Introduction
Vector function, Vector calculus, ex. Gauss’s law
2. Application of vector multiplication (벡터곱의 응용)
1) Dot product
 
A  B  AB cos   A x B x  A y B y  A z B z
 
AB
2) Cross product
i
 
A  B  Ax
Ay
Az ,
Bx
By
Bz
- Example
a) Work
b) Torque
j
k
 
A B
 
A  B  AB sin 
 
W  Fd cos   F  d
 
dW  F  d r



v
 
 rF

c) Angular velocity
 

v r
r sin 


r
3. Triple products (삼중곱)
1) Triple scalar product (삼중 스칼라곱)
Ax
  
A  (B  C )  Bx
Ay
Az
By
Bz
Cx
Cy
Cz
height  A cos 
“Volume of the parallelepipe”
cf. volume of unit cell for reciprocal vectors









a2  a3
a 3  a1
a2  a3
b1  2   
 , b 2  2  
 , b3  2  

a1  a 2  a 3
a1  a 2  a 3
a1  a 2  a 3
Ax
  
A  (B  C )  Bx
Ay
Az
By
Bz
Cx
Cy
Cz
- An interchange of rows changes just the sign of a determinant.
  
  
A  (B  C )  ( A  B) C
  
 C (A  B)
  
 ( A  C )  B
So, it does not matter where the dot and cross are.
  
 
A  (B  C )  ( ABC )
2) Triple vector product (삼중 벡터곱)

 
A  (B  C )


 

some vector in
A  ( B  C )  aB  bC
  
  
 ( A  C ) B  ( A  B )C
the plane of B and C
Prove this!
(Vector equation is true independently of the coordinate system.)

B  Bxi

C  C xi  C y j

A  Ax i  A y j  Azk
3) Application of the triple scalar product
“Torque”
 
 rF

This question is in one special case, namely when r and F
are in a plane perpendicular to the axis.
   
  n rF n

 

4) Application of the triple vector product
Angular momentum
Centripetal acceleration
  
L r p
 
 mr  v
 

 m r  (  r )

 

a    (  r )



v
r sin 


r
4. Differentiation of vectors (벡터의 미분)
1) Differentiation of a vector

A  ( Ax  A y  Az )

dA y
d A dA x
dA z

i
j
k
dt
dt
dt
dt
Example 1.

r  ix  jy  k z ,

 dr
dx
dy
v 
i
j
k
dt
dt
dt

2
2
 dv
d r
d x
a 

i

2
2
dt
dt
dt
dz
,
dt
2
j
d y
dt
2
2
k
d z
dt
2
2) Differentiation of product



d
da
dA
(aA) 
Aa
,
dt
dt
dt





d
dB dA 
(A  B)  A 

 B,
dt
dt
dt





d
dB dA 
(A  B)  A 

B
dt
dt
dt
(careful of order! )
Example 2. Motion of a particle in a circle at constant speed
 
2
r  r  r  const .,
 
2
v  v  v  const .
Differentiating the above equations,

 dr
2r 
0
dt

 dv
2v 
0
dt
or
or
 
r  v  0,
 
v a  0
“two vectors are perpendicular”
 
r  v  0,
   
Differenti ating this, r  a  v  v  0
 
2
r  a  v
 
 
 
2
r a v
 r  v  0 & v  a  0,
a 
v
2
r
3) Other coordinates (e.g., polar)
( i , j)
rectangula
 (e r , e θ )
r coord.
polar
coord.
( i , j) :
constant in magnitude and direction
(e r , e θ ) :
constant in magnitude, but
directions changes
e r  i cos   j sin 
e θ   i sin   j cos 
der
  i sin 
dt
  i cos 
dt
Example 3.

dA
dt

dA
dt
 er
dA r
 er
dA r
dt
dt
 Ar
der
dt
 e θ Ar
 eθ
d
dt

dA
dt
 eθ
dt
 A
dA 
dt
d
dt
?
dA 
 j cos 
dt
deθ

A  Ar e r  A e θ ,
d
deθ
dt
 e r A
d
dt
.
d
dt
 j sin 
d
dt
 eθ
d
,
dt
 er
d
dt
.
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 6 Vector analysis
Lecture 19 Directional derivative; Gradient
5. Fields (장)
Field: region + the value of physical quantity in the region
ex) electric field, gravitational field, magnetic field
6. Directional derivative: gradient (방향 도함수 ; 기울기벡터)
T ( x, y, z )
The change of temperature with distance depends on
the direction.  directional derivative dT
 T for  s
ds
1) definition of directional derivative
 ( x , y , z ) : scalar
function
   grad   i
d
ds
  u

x
j

y
k

z
,
(directional derivative for u: directional unit vector)
Example 1. Find the directional derivative
  x y  xz at (1,2,-1)
2
direction

A  ( 2 ,  2 ,1)

A
1
u    ( 2 ,  2 ,1)
3
A
  i

x
j

y
k
  (1, 2 ,  1)  ( 3 ,1,1)
 u 
5
3

z
 ( 2 xy  z ) i  x j
2

y
 xk ,
2) Meaning of gradient : along it the change (slope)
is fastest (steepest).
3) Relation between scalar function and gradient
   0,
lim
s 0

s

s
0
 0   u
“The vector grad. is perpendicular to the surface =const.”
Example 3. surface x^3y^2z=12. find the tangent plane and normal line at (1,-2,3)
w x y z
3
2
 w  i 3 x y z  j 2 x yz  k x y  36 i  12 j  4 k
2
2
3
3
2
9 ( x  1)  3 ( y  2 )  ( x  3 )  0 ,
x 1
9

y2
3

z3
1
4) other coordinates (e.g., polar)
  i

x
 j

y
 er

r
 eθ
1 
r 
cf. Cylindrical & Spherical coord.
T 
T 
T
s
T
r
rˆ 
rˆ 
1 T
r 
φˆ 
T
z
zˆ
1 T ˆ
1
T
θ
φˆ
r 
r sin   
cylindrical
spherical
7. Some other expressions involving grad. (  을 포함하는 다른 표현들)
1) vector operator
 i

x
   (i
2) divergence of V
j

x
y
j
k


z
k
y

z
)  i

x
j

y
k


  
  V  div V  (
,
, )  (V x , V y , V z )
x y z

3) curl of V

V x
x

V y
y

V z
z


  
  V  curl V  (
,
, )  (V x , V y , V z )
x y z

i

j

k

x
Vx
y
Vy
z
Vz

z
4) Laplacian
        div grad 
2
(

,

  0
2

x y z
 
2

,
x
2
 
)(
2

y
2
  
,
,
)
x y z
 
2

z
2
is Laplace' equation.
1  
2
  
2
a
  
2
2
t
1 
a
2
t
2
is the wave equation.
is the diffusion
equation
or equation
of heat conduction
5) and etc.



  (   V )   (   V )  (    )V


2
  (  V )   V



  ( V )  V      (   V )
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 6 Vector analysis
Lecture 20 Line integral & Green’s theorem
8. Line integrals (선적분)
1) definition
integrating along a given curve.
only one independent variable!

F  dr
Example 1.
F=(xy)i-(y2)j, find the work from (0,0) to (2,1)
 
dW  F  d r

d r  i dx  j dy
 
2
F  d r  xydx  y dy
W 
 ( xydx  y dy )
2
path 1 (straight line)
y
1
2
x , dy 
1
2
W1 
dx
2
path 2 (parabola)
2
 xydx
 y dy 
2
0
y
1
0
1
x , dy 
2
4
xdx
2
2
W2 
 xydx
0

2
1  1
x  xdx   x 
dx  1
2
2  2
1
2
 y dy 
2

0
2
2
1 2 1
2
x  x dx   x 
xdx 
4
3
4
 2
1
path 3 (broken line)
1
1) y  0
( 0  y  0  y dy )  
2
1
3
2
2)  ( x  1  dx
y0
 1  0)  2,
W3  
1
2
3
5
3
2)
1)
path 4 (parameter) x=2t^2, y=t^2
x: (0,2)  t: (0,1)
1
W4 

xydx  y dy 
2
  2 tdt
 2 t  t  4 tdt  2 t
2
0
2
2 2

7
6
dx  0
dy  0
Example 2. Find the value of
I 

xdy  ydx
x  y
2
2
path 1 (polar coordinate ) r=1 (constant)
so, only d may be considered.
x  cos  , dx   sin  d 
y  sin  , dy  cos  d  ,
xdy  ydx
x  y
2
2

 d

2
2
cos  (cos  d  )  sin  (  sin  d  )
1
0
I1 
x  y 1
 
 d
path 2


xdy  ydx
x  y
2
2
2


xdx  ( x  1) dx
x  ( x  1)
2
1
xdy  ydx
x  y
0
2
1
 
0
2
xdx  (1  x ) dx
x  (1  x )
2
2
0
  arctan( 2 x  1) 1  
1

  arctan( 2 x  1) 0  
2

(0,1)
2
y  x 1
 I 2  
y  1 x
1)
(-1,0)
## Question: Would you compare between example 1 and 2?
2)
(1,0)
2) Conservative fields (F or V) (보존장)
- Example 1 : depends on the path. nonconservative field
- Example 2 : does not depend on the path. conservative field

 curl F  0 , necessary
and sufficient
condition
for conservati ve field

W
W
W
F  W  i
 j
k
x
y
z
W
Fx 
, Fy 
x
 W
2
Using
 Fx
y
xy
 W
yx
, Fz 
y
 W
W
z
2

2

W

yx
Fy
,
x

From this,   F  0
,
and similarly
Fy
z

 Fz
y
,
z

 Fz
x
,

If F   W ,
Conversely

F 0

, if   F  0 , we can find W for which
 Fx

F  W .
3) Potential () (퍼텐셜)


 F : conservati ve field
F    , 
  (  W ) : scalar potential

 
    F  ds
B
for A: a proper reference point
A
cf. Electric field, gravitational field  conservative
Example 3. Show that F is conservative, and find a scalar potential.

3
2
2
F  ( 2 xy  z ) i  x j  ( 3 xz  1) k
1) F is conservative.

F 
i

j

k

x
y
z
2 xy  z
3
x
2
 3 xz  1
2
0
2) Scalar potential of F
(x,y,z)
iii) dz
3
2
2
(
2
xy

z
)
dx

x
dy

(
3
xz
 1) dz
(x,y,0)

i) dx
A
A
ii) dy
(0,0,0)
(x,0,0)
a. find the point where the field (or potential) is zero.


W   F  dr 
B
B
b. do line integral to an arbitrary point along the path with which the
integration is easiest.
i) only dx
y  z  0 , dy  dz  0
x
W 
 ( )  0
x0
ii) only dy
x  const ., z  0 , dx  dz  0
y

x dy  x y
2
2
0
iii) only dz
x , y  const ., dx  dy  0
   W   x y  xz  z
2
z

 ( 3 xz  1) dz   xz  z
2
0
3
3
Example 4. scalar potential for the electric field of a point charge q at the origin
E 
q
r
2
er 

q r
r
2

r
 
    E  dr   q
 to r
For
q 
r
3
r

 to r
 
r  dr
r
3
 
 
d ( r  r )  2 r  d r  2 rdr

  q
r
rdr
r
3

q
r


r
q
r
 
cf . A  d A  AdA
9. Green’s theorem in the plane (평면에서의 Green 정리)
1) Definition of Green theorem
- The integral of the derivative of a function is the function.
b
d
 dx
a
f ( x ) dx  f ( b )  f ( a )
 P ( x , y ),
Q ( x , y ) : a function
Area integral:
Q ( x, y )

x
A
with
continuous
d
b
 
dxdy 
Q ( x, y )
x
yc xa
d
Line integral:
first
partial
dxdy   [ Q ( b , y )  Q ( a , y )] dy
c
c
d
d
  [ Q ( b , y )  Q ( a , y )] dy
c


A
Q
x
dxdy 
 Qdy
C
cf.
s
d
 Q ( x , y ) dy   Q ( b , y ) dy   Q ( a , y ) dy
C
derivative
c
Similarly,

P ( x, y )
y
A
d
dxdy 
b
 
P ( x, y )
yc xa
a
y
b
dxdy   [ P ( x , d )  P ( x , c )] dx
a
 P ( x , y ) dx   P ( x , c ) dx   P ( x , d ) dx
C
a
b
d
  [ P ( x , c )  P ( x , d )] dx
a
  
A
P
y
dxdy 
 Pdx
C
a

C
Pdx  Qdy 

A
(
Q
x

P
y
) dxdy , Green' s theorem
This relation is valid even for an irregular shape!!
“Using Green’s theorem we can evaluate either a line integral around a closed
path or a double integral over the area inclosed, whichever is easier to do.”
Example 1. F=xyi-y2j, find the work from (0,0) to (2,1) and back
 
dW  F  d r

d r  i dx  j dy
 
2
F  d r  xydx  y dy
W 
 ( xydx  y dy )
2
For a closed path, W 2  W 3   1
(previous section)
1
1
path 2 (parabola) y  x 2 , dy  xdx
4
2
2
W2 

2
xydx  y dy 
2
0

0
2
2
1 2 1
2
x  x dx   x 
xdx 
4
3
4
 2
1
path 3 (broken line)
1
1) y  0
2
( 0  y  0  y dy )  
2)  ( x  1  dx
y0
2
1
2)
3
 1  0)  2,
W3  
1
3
2
5
3
1)
dx  0
dy  0
Example 1. F=xyi-y2j, find the work from (0,0) to (2,1) and back
 
dW  F  d r

d r  i dx  j dy
 
2
F  d r  xydx  y dy
W 
 ( xydx  y dy )
2
For a closed path, W 2  W 3   1
Using Green’s theorem,
W 

xydx  y dy 
2
A
A
1

  xdxdy
A

 
2
 


2
(

y
)

(
xy
)

 dxdy
y
 x

y
  xdxdy
 1
y0 x0
cf .
 Pdx  Qdy 
C

A
(
Q
x

P
y
) dxdy
W 
Example 2.
 ( F x dx  F y dy ) 
A
If
Fy
x

Fx
y
0

A
(
Fy
x

Fx
y
) dxdy
( z-component of curl F = 0),
then, W from one point to another point is independent of the path.
(F : conservative field)
cf .
 Pdx  Qdy 
C

A
(
Q
x

P
y
) dxdy
- Two useful way to apply Green’s theorem to the integration of vector functions
a) Divergence theorem
Q  V x , P  V y ,
Q
x

P
y

V x
x

V y
y
where V  i V x  j V y
 div V , with V z  0
d r  i dx  j dy (tangent)
n ds  i dy  j dx (outward
normal),
where
ds 
dx  dy
2
2
d r  n ds  0
Pdx  Qdy  V y dx  V x dy  ( i V x  j V y )  ( i dy  j dx )  V  n ds
 div
V dxdy 
 V  n ds
A
A
 div

Divergence theorem
V dxdydz 
 V  n d 

cf .
 Pdx  Qdy 
C

A
(
Q
x

P
y
) dxdy
b) Stoke’s theorem
Q  V y , P  V x , where V  i V x  j V y
Q
x

P
y

V y
x

V x
y
 (curl V )  k ,
with V z  0
Pdx  Qdy  V x dx  V y dy  ( i V x  j V y )  ( i dx  j dy )  V  d r
 (curl
V )  k dxdy 
A
A
 (curl

 V  dr
V )  nd 
Stoke’s theorem
 V  dr

cf .
 Pdx  Qdy 
C

A
(
Q
x

P
y
) dxdy
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 6 Vector analysis
Lecture 21 Divergence and Divergence theorem
10. Divergence and divergence theorem (발산과 발산정리)


  
  V  div V  (
,
, )  (V x , V y , V z )
x y z

V x
x

V y
y

V z
z
1) Physical meaning of divergence
flow of a gas, heat, electricity, or particles
V  v
: flow of water
amount of water crossing A’ for t
( vt )( A )(  )
vt A    vt  A cos 
v  cos   V cos   V  n

V  (V x , V y , V z )
- Rate at which water flows across surface 1 V x(1) dydz
- Rate at which water flows across surface 2 V x( 2 ) dydz
- Net outflow along x-axis
In this way,
 V x

[V x( 2 ) V x(1)] dydz  
dx  dydz
 x

 V y


 dzdx ,
dy
 y



along y - axis
 V z

dz  dxdy ,

 z

along z - axis
 V x V y V z

 x  y  z


 dxdydz  div V dxdydz    V dxdydz


“Divergence is the net rate of outflow per unit volume at a point.”
cf. (from ‘Griffiths’)
(a) positive divergence for positive charge (or negative divergence for a negative
charge)
(b) zero divergence
(c) positive divergence along the z-axis
cf. (from ‘Griffiths’)
Example 1.4 in Figure 1.18
v a  r  x xˆ  y yˆ  z zˆ ,
v b  zˆ ,
v c  z zˆ
.......... .... then the ir divergence
s?
 va  3
  vb  0
  v c  1.
2) Example of the divergence 1
 = (source density) minus (sink density)
= net mass of fluid being created (or added via something like a minute
sprinkler system) per unit time per unit volume
 = density of fluid = mass per unit volume
/t = time rate of increase of mass per unit volume
Rate of increase of mass in dxdydz
= (rate of creation) minus (rate of outward flow)

t
dxdydz   dxdydz    V dxdydz

t
   V
1) If there is no source or sinks,
 V 
2)
If

t
 0,

t
 V 
 0,
 0
Equation
of continuity
cf. div
D  ,
div B  0
3) Example of the divergence 2
Consider any closed surface.
d   r sin  d  d 
2
Mass of fluid flowing out through d is Vn d.
Total outflow:
 V  n d 
For volume element d=dxdydz, the outflow from d is
  V d
 V  n d 
  V d 
surface of
d
  V  lim
d  o
1
d
 V  n d 
surface of
d
another definition of divergence
4) Divergence theorem
   V d 
 V  n d 

volume 
surface
inclosing 
5) Example of the divergence theorem
 V  nd
V  ix  jy  k z ,
 V  nd
 V 

x
x
   V d 

y
y
 V  nd 
surface of
cylinder
?

z
z

 3,
  V d 
 3 d   3 a h
3
volume of
cylinder
If we directly evaluate  V  n d 
top : k
curved
surface : n 
xi  yj
a
r
6) Gauss’s law
electric field at r due to a point chage q at (0,0)
E
q
4  r
2
(  : dielectric
er
Coulomb'
constant, 1/4   9  10 in vacuum
9
D  εE ,
d   sin  d  d  
D
1
r
2
closed
surface σ
D  n dσ 
q
4π
For multi-sources,
q
4 r
2
in mks unit)
er
dA
D  n d   D cos  d   DdA 

s law
 d 
total
solid angle
q
4 r
q
4π
r d 
1
2
2
4
 4   q ( q inside  )
 D  n dσ    D
closed
surface σ
qd 
i
closed
surface σ
i
 n dσ 
q
i
i
 D  n dσ
 total charge inside the closed surface
closed
surface σ
 D  n dσ
closed
surface σ

 d
Gauss’s Law
volume
bounded by 
Using the divergence theorem,
 D  n dσ
closed
surface σ
  D   ext

   D
volume
bounded by 

 d
volume
bounded by 
7) Example of gauss’s law. E=?
a) For electrostatic problem, E=0 inside
b) For symmetry, E should be vertical.
D n  D ,
 D  nd
 D  ( surface area)
total charge inside is C(surface area) for C surface charge density.
D  ( surface area)  C  (surface
D C
area)
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 6 Vector analysis
Lecture 22 Curl and Stoke’s theorem
11. Curl and Stoke’s theorem (회전이론과 Stoke’s 정리)


  
  V  curl V  (
,
, )  (V x , V y , V z )
x y z

Ex. v
i

j

k

x
Vx
y
Vy
z
Vz



v
 ωr
r sin 
  v    (ω  r )  (  r )ω  (ω   )r


r
 x y z 
  3 ω
ω (   r )  ω 



x

y

z





 
( ω   ) r    x
i  y
j  z
k  ( i x  j y  k z )  i  x  j  y  k  z  ω
x
y
z 

since
y
x

   v    (ω  r )  2 ω
z
x
0
“Curl v gives the angular velocity.”
1) meaning of curl
circulatio
n
 V  dr
curl V  0
curl V = 0
curl V  0
curl V = 0
curl V  0
cf. (from ‘Griffiths’)
vs.
cf. (from ‘Griffiths’)
Example 1.5
v a   y xˆ  x yˆ (Fig. 1.19a)
v b  x yˆ (Fig. 1.19b)
.......... ...... then, their curls?
  v a  2 zˆ ,
  v a  zˆ .
 V  d r   (curl
around d 
V )  k dxdy 
d
(   V )  n  lim
d  0
1
d
 (curl
V )  nd ,
d
 V  dr
around d 
bad
good
Stoke’s theorem
 V  d r   (   V )  n d 
curve
bounding 
surface 
Example 1. V  4 y i  x j  2 z k ,
 (  V )  n d 
over the hemisphere
?
x  y  z  a , z  0.
2
2
2
2
  V   3k
(a) integrate the expression at it stands
(b) use Stoke’s theorem and evaluate the line integral around the circle
(c) use Stoke’s theorem to say that the integral is the same over any
surface bounded by the circle, for example, the planar area inside the
circle.
nk
  V   n   3k   k
 3
Ampere’s law
 H  dr  I
C
H : magnetic field
C : closed curve
I : current
For a specific case,
2
 H  d r   H rd 
C
 H r  2  I
0
H 
I
2 r
 H  d r  I   J  n d  ,
C
J : current densitiy

cf. I 
 J  n d 
 H  d r   (   H )  n d 
C

 (   H )  n d 


 J  n d 
H J

: one of the Maxwell equations
Conservative fields
‘simply connected’ if a simple closed curve
in the region can be shrunk to a point without
encountering any points not in the region.
If the components of F have continuous first partial derivatives in a simple
connected region, any one implies all the others.
a) curl F = 0
b) closed line integral = 0
c) F conservative
d) F = grad W, W single valued
Vector potential
curl V  0 , V   W
for W : scalar potential
 By definition
,   W  0
div V  0 , V    A for A : vector potential
 By definition
E 0
,    A  0
 B  0
Example 2. V  i ( x 2  yz )  j2 yz  k ( z 2  2 zx ),
i)
div V 

x
( z  yz ) 
2

y
( 2 yz ) 

V    A, A  ?
( x  2 zx )
2
z
 2x  2z  2z  2x  0
ii)
V A 
i

j

k

x
Ax
y
Ay
z
Az
There are many A’s to satisfy this equation. For convenience, set one
component A_x =0.
V A 
i

j

x
0
y
Ay
k

Ay
  Az
 

z
z
 y
Az

  Az 
 i   
j 
 x 

 Ay

 x

 k

x  yz 
2
 Az
y

Ay
 2 yz  
,
z
Az
x
z  2 zx 
2
,
Ay
x
A y  z x  zx  f 1 ( y , z )
2
2
A z  2 xyz  f 2 ( y , z )
x  yz 
2
Az
y

Ay
z
 2 xz 
f 2
y
 2 zx  x 
2
There are many ways to select f1 and f2.
If taking
f1  0 , f 2  
1
2
y z,
2
then, A  j( z x  zx )  k ( 2 xyz 
2
2
1
2
2
y z)
f
1
z

f 2
y
x 
2
f
1
z
Generalization for A
div V  0 , V    A for A : vector potential
For A_x=0, V    A , V y  
A y   V z dx  f ( y , z ),
 Az
x
Vz 
,
x
A z    V y dx  g ( y , z ),
 V y V z
V x

   

y
z

y
z

Az
Ay
Ay

 dx  h ( y , z )


For div V  0 ,
 V y V z
 

z
 y
 V x

,

x

Vx 

V x
x
dx  h ( y , z )
cf. When we know one A, all others are of the form,
A  u,
   u  0
V
   A
cf. Cylindrical coordinate
Gradient
T 
T
s
sˆ 
1 T
s 
φˆ 
T
z
zˆ
Divergence
v 
1 
s s
 sv s  
1  v
s 

v z
z
Curl
v 
sˆ
1 
s φˆ

zˆ

r s
vs

rv 
z
vz
Laplacian
1   T 
1  T
 T
 T 

s
 2
2
2
s s  s  s 
z
2
2
2
cf. Spherical coordinate
Gradient
T 
T
r
rˆ 
1 T ˆ
1
T
θ
φˆ
r 
r sin   
Divergence
v 
1 
r
r
2
r
2

vr 
1

r sin   
sin
 v  
1
 v
r sin   
Curl
v 
rˆ

r θˆ

r sin  φˆ

r sin   r
vr

rv 

r sin  v 
1
2
Laplacian
1   2 T 
1
 
T 
1
 T
 T  2
.
r
 2
 sin 
 2
2
2
r r 
 r  r sin    
   r sin   
2
2
Homework
Chapter 6
3-17, 4-5, 6-10, 10-14, 11-16