12-1
Chapter 13 Vector Integral Calculus
Discussions：
1. Integrals of vector fields over curves and surfaces
2. Relationships between the two integrals
13.1. Line Integrals
○ A curve in 3-D
Parametric equations：
C : x  x(t ), y  y(t ), z  z (t ),
at b
x( t ) , y (t ) ,z t ( ) : c o o r d i n a t e f u n c t i o n s
Initial point: ( x(a), y (a), z (a))
Terminal point: ( x(b), y(b), z(b))
Closed： ( x(a), y(a), z (a)) = ( x(b), y(b), z(b))
○ Example 13.1：
C : x(t )  2cos t , y(t )  2sin t , z  4, 0  t  2
t=0
12-2
○ Continuous: x(t ), y(t ), z (t ) are continuous
Differentiable: x(t ), y(t ), z(t ) are differentiable
Smooth: x(t ), y (t ), z (t ) have continuous
derivatives, which are not all zero
for the same t  has a continuous
tangent vector x '(t )i  y '(t ) j  z '(t )k
◎ Definition 13.1:
C:
Line integral
smooth curve
f ( x, y, z ), g ( x, y, z ), h( x, y, z) : continuous on C
 C f ( x, y, z )dx  g ( x, y, z )dy  h( x, y , z )dz
dx
dy
dz 
b
 a  f ( x, y, z )  g ( x, y, z )  h( x, y, z ) dt
dt
dt
dt 

○ Riemann integral
Steps: 1. Substitute x(t ), y(t ), z (t ) into
f ( x, y, z ), g ( x, y, z ), h( x, y, z )
Obtain F (t ), G(t ), H (t )
2. Substitute dx, dy, dz with
dx(t )
dy (t )
dz (t )
dt ,
dt ,
dt
dt
dt
dt
C fdx  gdy  hdz becomes Riemann integral
12-3
○ Example 13.2:
f ( x, y, z)  x, g ( x, y, z )   yz, h( x, y, z )  e z
C: x  t 3 , y  t , z  t 2 , 1  t  2
dt 3
d (t )
 dx 
dt  3t 2 dt , dy 
dt  dt
dt
dt
d (t 2 )
dz 
dt  2tdt
dt
z
C fdx  gdy  hdz  C xdx  yzdy  e dz
 1 (t 3 ) (t32 ) t ( t )2 ( )( et1 ) t (dt ) ( 2 )
111
2
 1 (3t 5  t 3 2tet )dt 
 e 4 e
4
2
2
2
◎ Path integral
Piecewise smooth: x '(t ), y '(t ), z '(t ) continuous,
and not all zero for the same t ,
at all but finitely many t
Path:
piecewise smooth curve.
 C fdx  gdy  hdz
 C fdx  gdy  hdz 
1
 C fdx  gdy  hdz
n
12-4
○ Example 13.5:
C1 : x 2  y 2  1
P: 
C2 : line
from (1,0) to (0,1)
from (0,1) to (2,1)
2
2
Compute P x ydx  y dy
(a) Parameterize curves
C1 : x c o ts ,y
s ti n , t
0
C2 : x s, y 1 ,  0s
/2
2
＊ Different curves use different parameters
(b) Replace (dx, dy) with (dt , ds)
dx
d cos t
dt 
dt   sin tdt
dt
dt
dy
d sin t
dy  dt 
dt  cos tdt
dt
dt
C1: dx 
dx
ds
ds 
ds  ds
ds
ds
dy
d1
dy  ds  ds  0
ds
ds
C2: dx 
12-5
(c) Integral
C1 : C x 2 ydx  y 2 dy
1

 0 2 [cos 2 t sin t ( sin t )  sin 2 t cos t ]dt

 0 2 [ cos 2 t sin 2 t  sin 2 t cos t ]dt
  /16  1/ 3
C2 : C x 2 ydx  y 2dy  0 s 2ds  8/ 3
2
2
(d) P x 2 ydx  y 2 dy   /16  3
◎ Think of a line integral in terms of vector operations
Consider C fdx  gdy  hdz
Form F ( x, y, z )  f ( x, y, z )i  g ( x, y, z) j  h( x, y, z)k
Curve
C: x(t ), y(t ), z(t )
Form R(t )  x(t )i  y(t ) j  z (t )k (position vector)
t=a
C
t
R(a)
t=b
R(t)
R(b)
Origin
 dR(t )  dxi  dyj  dzk
F dR  f ( x, y, z )dx  g ( x, y, z )dy  h( x, y, z )dz
C f ( x, y, z )dx  g ( x, y, z )dy  h( x, y, z )dz
 C F dR
12-6
○ Example 13.6:
Force: F  i  yj  xyzk moves a particle from
(0,0,0) to (1,-1,1) along curve:
x  t , y  t 2 , z  t , 0  t  1
Work: c F dR  c dx  ydy  xyzdz
 0 (1  t 2 ( 2t )  t 4 ) dt  0 (1  2t 3  t 4 ) dt
 3/10
1
1
◎ Theorem 13.1:
1. C ( F  G )dR  C F dR  C G dR
2. C  F dR   C F dR
◎ Definition 13.2:
C: x(t ), y(t ), z(t )
a t b
 C : x(t )  x( a  b  t ), y (t )  y ( a  b  t ),
z (t )  z (a  b  t ), a  t  b
Initial point:
( x (a ) ,y (a ) z, a( ) ) x ( b ( y) ,b ( z ) b, ( ) )
Terminal point: ( x(b), y(b), z(b))  ( x(a), y( a), z( a))
e.g., C : x(t )  t , y(t )  t 2 , z(t )  t , a  t  b
C : x(t )  a  b  t , y (t )  (a  b  t ) 2 ,
z (t )  a  b  t , a  t  b
12-7
◎ Theorem 13.2:
C fdx  gdy  hdz   C fdx  gdy  hdz
Proof：
dx

dt
dy
dz
g ( x(t ), y (t ), z (t ))
 h( x(t ), y (t ), z (t )) ]dt
dt
dt
C fdx  gdy  hdz  a [ f ( x(t ), y (t ), z (t ))
b
Let s  a  b  t  b  s  a, ds  dt
x(t )  x(a  b  t )  x(s), y(t )  y(s), z (t )  z (s)
d x(t ) d
d
dx( s ) ds
dx( s )
 x(a  b  t )  x( s ) 

dt
dt
dt
ds dt
ds
d y (t )
dy ( s )

,
dt
ds
d z (t )
dz ( s )

dt
ds
 C fdx  gdy  hdz
dx
dy
 g ( x( s ), y ( s ), z ( s ) )
ds
ds
dz
 h( x( s), y ( s), z ( s) )]( ds)
ds
 b [  f ( x( s ), y ( s ), z ( s ))
a
  a [ f ( x( s ), y ( s ), z ( s ))
b
dx
dy
 g ( x ( s ), y ( s ), z ( s ))
ds
ds
dz
]ds
ds
dx
dy
b
  a [ f ( x(t ), y (t ), z (t ))
 g ( x(t ), y (t ), z (t ))
dt
dt
dz
 h( x(t ), y (t ), z (t )) ]dt   C fdx  gdy  hdz
dt
 h( x( s ), y ( s ), z ( s ))
12-8
13.1.1. Line Integral w.r.t. Arc Length
◎ Definition 13.3:
C: x(t ), y(t ), z (t ), a  t  b
Length: s(t )  a x( )2  y( )2  z( )2 d
t
 ds  x(t ) 2  y(t ) 2  z(t ) 2 dt
 : real-valued function continuous on C
 C  ( x, y, z )ds
b
 a  ( x(t ), y(t ), z(t )) x(t )2  y(t ) 2  z(t ) 2 dt
○ Example 13.8：
C: x  4cos t , y  4sin t, z  3, 0  t 

2
 x  4sin t , y  4cos t , z  0
2
ds  x(t ) 2  y(t ) 2 z(t ) dt
 16sin 2 t  16cos 2 tdt = 4dt

C xyds  0 2 4cos t (4sin t )4dt

 64 0 2 cos t sin tdt  32
○ Example 13.9：
C : x  2cos t , y  2sin t , z  3, 0  t   / 2
Density function:  ( x, y, z)  xy 2 g/cm
12-9
Mass：
m  C  ( x, y, z )ds  C xy 2 ds

 0 2 2cos t (2sin t ) 2 4sin 2 t  4cos 2 tdt

= 0 2 16cos t sin 2 tdt  16 / 3 g
Center of mass：
1
 x ( x, y, z )ds
m C
3 
= 0 2 (2cos t ) 2 (2sin t ) 2 4sin 2 t  4cos 2 tdt
16
x

=6 0 2 cos 2 t sin 2 tdt  3 / 8
1
C y ( x, y, z )ds
m
3 2
= 0 (2cos t )(2sin t )3 4sin 2 t  4cos 2 tdt
16
y

=6 0 2 cos t sin 3 tdt  3/ 2
1
C z ( x, y, z )ds
m
3 2
 0 3(2cos t )(2sin t ) 2 4sin 2 t  4cos 2 tdt
16
z

 9 0 2 sin 2 t cos tdt  3
12.2
Green’s Theorem
Piecewise smooth closed curve
C : x(t ), y(t ), z(t ), a  t  b
12-10
。 Positive oriented:
Move around C counterclockwise as t varies
e.g., C : x  cos t , y  sin t , 0  t  2
Negative oriented:
Move around C clockwise as t varies
e.g., C : x   cos t , y  sin t , 0  t  2
。 Simple curve:
 x(t1 )  x(t2 )
only if t1  t2 except closed curve

 y (t1 )  y (t2 )
e.g.,
curve
not simple
12-11
。 Jordan curve theorem：
A simple closed curve C separates the plane into
two regions having C as common boundary
◎ Theorem 13.3: Green Theorem
C: simple, closed, positively oriented
D = C + intention
f, g ,
g f
,
x y
:
c o n t i nDu o u s o n
 C f (x y, dx
)  g x (y ,dy ) D
g f

(
dA )
x y
where dA  dxdy
Proof：
 C : y h( x) , x f r oam
C: 1
C2 : y k ( x) , x f r obm
i,
D  ( x, y ) | a  x  b, h( x)  y  k ( x)
C f ( x, y )dx  C f ( x, y )dx  C f ( x, y )dy
1
2
 a f ( x, h( x))dx  b f ( x, k ( x))dx
b
a
= a [ f ( x, h( x))  f ( x, k ( x)]dx
b
----- (A)
bto
ato
12-12
D
f
b k ( x ) f
dA  a h ( x ) dydx
y
y
 b f ( x, y ) |kh (( xx )) dx
a
= a [ f ( x, k ( x))  f ( x, h( x)]dx
b
----- (B)
From (A) and (B),
C f ( x, y )dx   D
f
dA
y
----- (C)
ii,
C3
C4
C : x  f ( y ), y from d to c
C: 3
C4 : x  g ( y ), y from c to d
C g ( x, y )dx  D
g
dA
x
----- (D)
From (C) and (D)
C f ( x, y )dx  g ( x, y )dx  D (
g f
 )dA
x y
○ Example 13.10：
Force: F ( x, y)  ( y  x 2e x )i  (cos2 y 2  x) j
12-13
Work: C F dR


(cos 2 y 2  x)  ( y  x 2e x )]dA
x
y
 D 2dA  2(area of D)  4
 D [
12.2.1. An Extension of Green’s Theorem
Suppose there are finite points enclosed by C, at which
f , g,
f
g
are not continuous or undefined
, or
y
x
*
(a)
(b)
D*
From Green’s theorem,
C* f ( x, y)dx  g ( x, y )dy  D* (
g f
 )dA -- (12.2)
x y
(d)
(c)
(12.2)  C* f ( x, y )dx  g ( x, y )dy
n
 C f ( x, y) dx  g ( x, y) dy    Kj f ( x, y) dx
j 1
 g ( x, y) dy  D (
g f
 )dA
x y
Reverse the orientations on circles
12-14
C f ( x, y )dx  g ( x, y )dy
g f
   Kj f ( x, y )dx  g ( x, y )dy + D (  )dA
j 1
x y
n
○ Example 13.12:
 f ( x, y ) 
C
y
x
dx

dy
x2  y 2
x2  y 2
y
x
,
g
(
x
,
y
)

x2  y 2
x2  y 2
g
y 2  x2
f

 2

x ( x  y 2 ) 2 y
f , g,
----- (A)
f g
,
: not continuous at (0,0)
y x
Case 1： C does not enclose (0,0)
C
y
x
g f
dx

dy

(

)dA  0

D
x2  y 2
x 2 y 2
x y
Case 2： C encloses (0,0)
)  g x( y , dy)
C f ( x ,y dx
  K f x (y ,dx ) g x y( dy, )D
g f
+  ( dA
x y
  K f x (y ,dx ) g x y( dy, )
k : x r c o s ,y r s i n , 0 
f ( x ,y )
y
r s i n  s i n


2
2
2
x y
r
r
2
)
12-15
g ( x ,y )
x
r c o s
c o s


x2  y 2
r 2
r
d x  rs i n d
 , d y
r co sd
  K f ( x, y )dx  g ( x, y )dy
cos
2   sin 
= 0 [
(r sin  ) 
(r cos )]d
r
r
2
2
= 0 [sin 2   cos 2  ]d  0 d  2
Conclusion:
C f ( x, y)dx  g ( x, y)dy
0 C does not enclose (0,0)

2 C encloses (0,0)
12.2. Independence of Path and Potential
Theorem in the Plane
○ Definition 13.4:
F ( x, y ) : conservation vector field on D
if  a real-value potential function  (x,y ) in D
s.t. F  
◎ Theorem 13.4: Let F  
 (a) C F dR is independent of path in D
(b) C F dR  0
Proof:
F   


i
j
x
y
12-16
 C F dR  C


 dy
b  dx
dx 
dy  a (

)dt
x
y
x dt y dt
d ( x(t ), y (t ))
dt   ( x(t ), y (t )) |ba
dt
  ( x(b), y (b))   ( x(a ), y (a ))
 a
b
  (terminal point of C )   (initial point of C )
○ If F ( x, y)  f ( x, y)i  g ( x, y) j is conservative
  s.t. F   
Then,


i
j
x
y


 f ( x, y ),
 g ( x, y )
x
y
To find  ,


 f ( x, y ) or
 g ( x, y )
x
y
2) Integrate w.r.t. the variable
3) Use the second equation to find 
1) Begin with
○ Example 13.14:
F ( x, y)  2 x cos 2 yi  (2 x 2 sin 2 y  4 y 2 ) j
 f ( x, y)  2 x cos 2 y, g ( x, y)  (2 x 2 sin 2 y  4 y 2 )
Find  s.t.
and

 f ( x, y)  2 x cos 2 y,
x

 g ( x, y )  (2 x 2 sin 2 y  4 y 2 )
y
Choose

 2 x cos 2 y
x
12-17
Integrate w.r.t. x,
 ( x, y )   2 x cos 2 ydx  x 2 cos 2 y  c( y )
  2
2
y

y
( x cos 2 y  c( y ))  2 x sin 2 y  4 y 2
 2 x 2 sin 2 y  c( y )  2 x 2 sin 2 y  4 y 2
4
 c( y )  4 y 2 , c( y )   y 3
3
4
   x 2 cos 2 y  y 3
3
◎ Theorem 13.5:
F ( x, y)  f ( x, y)i  g ( x, y) j : conservative
iff
g f
(see Theorem 13.7 for proof)

x y
○ Example 13.18:
F ( x, y )  (2 xy 2  y )i  (2 x 2 y  e x y ) j
f
g

 4 xy  1,
 4 xy  e x y
y
x
4 xy  1  4 xy  e x y, F is not coservative
13.2.1. More Critical Look at Theorem 13.5
-- Explore the relationships between
(a) Independence of path,
(c)
g f
 ,
x y
(b) Green’s theorem
(d) Potential function 
12-18
○ Let D：all points in the plane except (0,0)
F ( x, y ) 
y
x
i

j  f ( x, y ) i  g ( x, y ) j
2
2
2
2
x y
x y
f g
y 2  x2
y 2  x2



0
y x ( x 2  y 2 ) 2 ( x 2  y 2 ) 2
However,
i) Let C : x  cos , y  sin  , 0    
y
 sin 

  sin 
2
2
2
2
x y
cos   sin 
x
cos
g ( x, y )  2

 cos
2
2
2
x y
cos   sin 
 f ( x, y ) 
C f ( x, y )dx  g ( x, y )dy
 0 [( sin  )( sin  )  cos  cos ]d

 0 d  

ii) Let C : x  cos , y   sin  , 0    
12-19
 C f (x y, dx
)  g x (y ,dy )
 0 [sin  ( sin  )  cos  (  cos  )]d

  0 d  

This means that
a)
C f ( x, y )dx  g ( x, y )dy : not independent
o f p ai tnD
h
b)
F: not conservative over D
c)
No  for F
○ D is domain if
1. p0  D,  a circle O about p0 ,
s.t. p  O, p  D
2. p1 , p2  D,  a path connecting p1 and p2  D
S：not a domain
M：not a domain
A：a domain
1<x
2
 y2 < 9
◎ Theorem 13.6:
D：a domain,
F：continuous on D
 F : conservative iff
C F dR : independent of path on D
12-20
Proof：
i, (If) Given F : conservative,
  s.t. F   (Definition 13.4)
C F dR   (terminal point)   (initial point)
(Theorem 13.4)
 C F dR : independent of path on D
ii, (Only if)
Given C F dR : independent of path ,
Show ( F : conservative )  Show (   )
D: a domain
  p, p0  D,  C connecting p and p0 on D
C
p( x ,y )
p0 ( x0 ,y0 )
Let  ( x, y)  C F dR
Show  ( x, y) : a potential function
F : continuous,  : continuous
Let F ( x, y )  f ( x, y ) i  g ( x, y ) j
(a, b) : any point in D


S h o w a (b , )f a (b , ) , a b( , g ) a b( , )
x
y
12-21

 (a  x, b)   (a, b)
(a, b)  lim
x0
x
x
p0 ( x0 ,y0 )
C2 : x  a  t x, y  b, o  t  1
 dx  xdt , dy  0
 (a  x, b)   (a, b)  C F dR
2
 C f ( x, y ) dx  g ( x, y ) dy
2
 0 f ( a  x, b)xdt
1

 (a  x, b)   (a, b)
x
 0 f (a  t x, b)dt
1
By the mean value theorem,  , 0    1,
0 f (a  t x, b)dt  f (a  x, b) 1
1
(height) × (width)
 (a  x, b)   (a, b)
x0
x
 lim f (a  x, b)  f (a, b)
lim

x0
12-22
Similarly,
 (a   x, b)  (a , b )
 f a (b , )
lim
x0
x

(a ,b ) f (a ,b )

x

( , )
L i k e w i s ea, b  (f ,a )b
y

○ D: simply connected domain --  simple, closed
path in D encloses only points of D
＊ D contains no hole
◎ Theorem 13.7: D: simple connected domain
F ( x, y)  f ( x, y)i  g ( x, y) j: vector field
f , g,
f g
,
: continuous on D
y x
 F : conservative on D iff
f g

y x
Proof :
i. (If) Given F : conservative with potential  ,
 F = 
 f ( x, y ) 


i
j  f x( y,i  ) g x( jy,
x
y


, g ( x, y ) 
x
y
f
 2
 2
g




y xy yx y
)
12-23
ii. (Only if)
Given
f g
 ,
y x
Show ( F : conservative on D ) 
Show ( c F dR: independent of path in D )
Let
and
D
J
By Greens' theorem,
g f
F
d
R

(
 )dA  0
J
D
x y
= C F dR   K F dR  C F dR  K F dR

1
1
 C F dR  K F dR ,
1
i.e., C F dR: independent of path
1
○ Summary
F : conservative on a simply connected domain
 C F dR : indepedence of path on the domain
g f


x y
13.4. Surfaces and Surface Integrals
Surface:
 : x  x(u, v), y  y(u, v), z  z(u, v)
○ Example 13.19:
 : x  au cos v, y  bu sin v, z  u
12-24
x
y
 c o vs , 
au
bu
sv i n
x 2
y
)  ( ) 2  cos v2  sin v2  1
au
bu
2
2
x
y
 2  2  u2  z2
a
b
(
○ Position vector:
R(u, v)  x(u, v)i  y(u, v) j  z (u, v)k
12.4.1 Normal Vector
 : x  x(u, v), y  y (u, v), z  z (u, v)
Let p0 : ( x(u0 , v0 ), y (u0 , v0 ), z (u0 , v0 )) on 
u0

 v0
Fix v  v0
 Curve  v : x  x(u, v0 ), y  y (u, v0 ), z  z (u, v0 )
0
Tangent vector to  v at p0 :
0
Tv 
0
x
y
z
(u0 , v0 ) i  (u0 , v0 ) j  (u0 , v0 ) k
u
u
u
12-25
Fix u  u0
 Curve u : x  x(u0 , v), y  y (u0 , v), z  z (u0 , v)
0
Tangent vector to u at p0 :
0
Tu 
0
x
y
z
(u0 , v0 )i  (u0 , v0 ) j  (u0 , v0 ) k
v
v
v
The normal to  at p0
N ( p0 )  Tv  Tu
0
0
x
y
z
(u0 , v0 ) i  (u0 , v0 ) j  (u0 , v0 ) k ]
u
u
u
x
y
z
 [ (u0 , v0 ) i  (u0 , v0 ) j  (u0 , v0 ) k ]
v
v
v
[
i

j
x
y
(u0 , v0 )
(u0 ,v0
u
u
x
z
(u0 ,v 0 )
u( 0v , 0
v
v
(
k
 z
)
u(0 v,0 )
v
z
) u (v0 ,0 )
v
y z z y
z x x z

)i  (

)j
u v u v
u v u v
x y y x
(

)k
u v u v
○ Jacobian determinant of
f
 ( f , g ) u

 (u, v) g
u
f and g
f
v f g g f


g u v u v
v
12-26
 The Normal vector
N ( p0 ) 
 ( y, z )  ( z , x)
 ( x, y )
i
j
k
 (u , v)
 (u , v)
 (u , v)
○ Example 13.24:  : x  au cos v, y  bu sin v, z  u
u  1

2
p0 : ( a 3 , b , 1 ) obtained when 
4 4 2
v 


6
The Jacobians :
 ( y, z )
 y z z y 


 (u , v) ( 1 , )  u v u v  ( 1 , )
2 6
2 6
 [b sin v  0  bu cos v] 1    3b
( , )
2 6
4
 ( z , x)
z x x z
[

] 1
 (u , v) ( 1 , )
u v u v ( 2, 6 )
2 6
 [  au sin v  a cos v  0] 1    a
( , )
2 6
4
 ( x, y )
x y y x
[

] 1
 (u , v) ( 1 , ) u v u v ( 2, 6 )
2 6
 [ a cos v(bu cos v)  b sin v(  au sin v)] 1   ab
( , )
2 6
2
The Normal vector: N ( p0 )   3b i a j  ab k
4
4
2
○ A surface is given as z  S ( x, y)
Its coordinate functions can be written as
x  x, y  y, z  S ( x, y ), i.e., think of u  x, v  y
12-27
x
y
x y
 1,
 1,

0
x
y
y x
0
 ( y, z )  ( y, z )


 S
 (u , v)  ( x, y )
x
S
 ( z , x)
 x
 ( x, y )
1
1
S
S  
x
y
S
S  ( x, y ) 1 0
y   ,

1
y  ( x, y ) 0 1
0
The normal at p0 : ( x0 , y0 , S ( x0 , y0 )) :
S
S
( x0 , y0 )i  ( x0 , y0 ) j  k
x
y
z
z
  ( x0 , y0 ) i  ( x0 , y0 ) j  k
x
y
N ( p0 )  
○ Example 13.25: Cone: z  S ( x, y )  x 2  y 2

S

x
x
x2  y 2
,
S

y
y
x2  y 2
The Normal vector at P0  (3,1, 10)
N ( P0 ) 
3
1
i
j  k (inner normal)
10
10
The outer normal : -N ( P0 ) 
3
1
i
jk
10
10
12-28
○ Normal vectors derived from gradient vectors
 : z  S ( x, y )
 A level surface of  ( x, y, z )  z  S ( x, y )
The gradient of  is a normal vector



S
S
 = i 
j k  i
j  k  N (P )
x
y
z
x
y
12.4.2 Tangent Planes
Let N ( P0 ): a normal vector at P0 (x0 , y0 , z0 )
Let (x, y, z ): any point on the tangent plane
 The vector ( x  x0 )i  ( y  y0 ) j  ( z  z0 )k
lies in the tangent plane and is orthogonal to N
N
P(x,y,z)
P0(x0,y0,z0)
 N  [( x  x0 )i  ( y  y0 ) j  ( z  z0 ) k ]  0
 ( y, z )
 ( z , x)
]( u ,v ) ( x  x0 )  [
]( u ,v ) ( y  y0 )
 (u , v)
 (u , v)
 ( x, y )
[
]( u ,v ) ( z  z0 )  0
 (u , v)
This is the equation of the tangent plane
[
0
0
0
0
0
0
12-29
○ Example 13.26:
3 b 1
, , )
4 4 2
on the elliptical cone: x  au cos v, y  bu sin v, z  u
The normal vector at P0 : ( a
3b a
ab
i j k
4
4
2
 The tangent plane :
is N 

3b
a 3 a
b ab
1
(x 
)  ( y  )  (z  )  0
4
4
4
4
2
2
○ Given  : z  S ( x, y )
The normal vector at P0 :
N ( P0 )  (
S
S
)( x0 , y0 )i  ( )( x0 , y0 ) j  k
x
y
The tangent plane:

S
S
( x0 , y0 )( x  x0 )  ( x0 , y0 )( y  y0 )  ( z  z0 )  0
x
y
or z  z0 
S
S
( x0 , y0 )( x  x0 )  ( x0 , y0 )( y  y0 )
x
y
12.4.3 Smooth Surfaces
Smooth surface: has a continuous normal vector
e.g., sphere
12-30
Piecewise smooth surface: consists of a finite
number of smooth surface
e.g., cube
○ The area of a smooth surface z = S(x,y):
D 1  (
S 2 S 2
z
z
)  ( ) dA or D 1  ( ) 2  ( ) 2 dxdy
x
y
x
y
where D: the set of points in (x,y ) plane, for which
S is defined.
○ The area is the integral of the length of the normal
vector
D N ( x, y ) dxdy
○ Surface: x  x(u, v), y  y (u, v), z  z (u, v)
(u, v) varies over D in the (u, v) plane
D N (u, v) dudv
○ Example 13.27:
2
2
2
2
 : z  S ( x, y )  9  x  y with x  y  9
D : all points on or inside the circle of radius 3
about the origin in (x,y ) plane
12-31
Compute
z
x
x



x
z
9  x2  y 2
z
y
y



y
z
9  x2  y 2
x2  y 2  z 2
 Area of   D
dxdy
z2
3
 D
dxdy
2
2
9 x  y
Convert to polar coordinates :
x  r cos , y  r sin  , 0  r  3, 0    2
3
 D
2
9 x  y
2
 6 0
3
3
2
dxdy  0 0
9r
9r
1
2 3
0
2
rdrd
1
2
dr  6 [(9  r ) ]  6 [9 ]  18
2
2
3
3
12.4.4 Surfaces Integral
Recall that a smooth curve C :
x  x(t ), y  y (t ), z  z (t ), a  t  b
 the arc length : s (t )  a x '( ) 2  y '( ) 2  z '( ) 2 d 
t
and ds  x '(t ) 2  y '(t ) 2  z '(t ) 2 dt
the line integral of a function f along c :
2
2
2
c f ( x, y, z )ds  a f ( x(t ), y (t ), z (t )) x '(t )  y '(t )  z '(t ) dt
b
○ Definition 13.6 : Surface Integral
12-32
Smooth surface  : x  x(u, v), y  y (u, v), z  z (u, v)
for (u, v) in D of the (u, v) plane
 f ( x, y, z )da  D f ( x(u , v), y (u, v), z (u, v) N (u, v) dudv)
Piecewise smooth surface :
 f ( x, y, z )da   f ( x, y, z )da 
1
  f ( x, y, z )da
n
where 1 , ,  n : smooth surface components
If  : z  s ( x, y ),
  f ( x, y, z )da  D f ( x, y, s ( x, y )) 1  (
s 2 s 2
)  ( ) dxdy
x
y
。Example 13.27：
Plane  : x  y  z  4 lies on
rectangle : 0  x  2,0  y  1
Evaluate  zda
Let D : z  s ( x, y )  4  x  y,0  x  2,0  y  1
  zda  D z 1  (1) 2  (1) 2 dxdy
 3 0 0 (4  x  y ) dydx
2 1
First, compute
1 21
(4

x

y
)
dy

(4

x
)
y

y 0
0
2
1 7
 4 x   x
2 2
1
2 7
Next,  zda  3 0 (  x) dx  5 3
2
12.6 Integral theorems of Gauss and Stokes
12-33
◎ Green’s Theorem
g f
 )dA
x y
C : simple, closed, smooth curve
D : the plane region enclosed by C
Vector field : F ( x, y )  g ( x, y )i  f ( x, y ) j
g f
 F 

x y
c f ( x, y )dx  g ( x, y )dy  D (
○ Parametrization of C by arc length
x  x( s), y  y ( s),0  s  L
 unit tangent vector : T ( s)  x '( s )i  y '( s ) j
unit normal vector : N ( s)  y '( s)i  x '( s) j
○ F  N  g ( x, y )
dy
dx
 f ( x, y )
ds
ds
dx
dy
)  g x( y , dy) c f x[ y( , )g x y ( ,ds )
c f ( x ,y dx
ds
ds
  F  Nds
]
○ Conservation of energy
D i v e r g e n c e o f v e
c t o r f i e l d
F at point P
= f l o w o f t h e v e c t o r f i e fl dv eF c at ot r P f l u x o
f i e l dN F
12-34
○ 3-D
 F  Nda   M   Fdv  Green's divergence theorem
○ Vector field : F ( x, y, z )  f ( x, y)i  g ( x, y) j  0k
i

Cure :   F 
x
f
(  F  K
) 
j

y
g
K

g f
 ( 
K
)
z
x y
0
g f

x y
○ Unit tangent T ( s)  x '( s)i  y '( s) j 3-D
F  Tds  [ f ( x, y )i  g ( x, y ) j ]  (
dx dy
i
j )ds
ds ds
=f ( x, y)dx  g ( x, y)dy
○ Green’s Theorem
2 D :  c F  Tds  D (  F )  KdA
3D : c F  Tds   (  F )  Nda
12.7 Divergence Theorem of Gauss
12-35
Green's Theorem :  c F  Nds  D   F dA
Generalization 2D  3D
2D
3D
C

D
M
 c F  Nds
 D F  N da
D   F dA
 M   Fdv
◎ Theorem 13.8：Gauss’s Divergence Theorem
  F  N d a  M    F d v
○ Example 13.33：

2
2
2
2
1: z  x  y , x  y  1
2
2
 2 : x  y  1, z  1
Unit outer normal to 1:
1 x
y
N1 
( i  j  k)
z
2 z
12-36
 F  N1  ( xi  yj  zk ) 
 
1
1 x
y
( i  j  k)
z
2 z
1 x2 y 2
=
(   z)  0
z
2 z
F  N1da  0
Unit outer normal to  2 :N 2  k
 F  N2  z
  F  N 2 da   zda   1da  
2
2
  F  Nda   F  N1da   F  N 2 da
1
2
=0    



F  x  y  z  3
x
y
z
1
 M   Fdv  M 3dv  3    
3
○ Example 13.34：
F ( x, y, z)  x 2i  y 2 j  z 2k
the flux of F across the cubic faces :
 F  Nda   F  N1da    F  N 6 da
1
6
By Gauss's theorem,   F  2 x  2 y  2 z
  F  Nda  M   Fdv  M (2 x  2 y  2 z )dv
12-37
= 0 0 0 (2 x  2 y  2 z )dzdydx
1 1 1
= 0 0 [(2 x  2 y ) z  z 2 10 ]dydx
1 1
= 0 0 [(2 x  2 y  1)dydx
1 1
1
= 0 (2x+2)dx =3
12.7.2 Heat Equation - Eqaution models heat
coeduciton
Let p ( x, y, z ) : medium density
 ( x, y, z ) : heat
k( x, y, z ) : thermal conductivity
u( x, y, z , t ) : temperature
medium

M
(x,y,z)
Heat energy leaving M across  in t :
(  (k u )  Nda) t, i,e., the flux of k u across  in t
The change in temperature at (x,y,z) in t :
u
dv)t
t
 The change in heat energy in M over t
Heat loss ( M 
u
 t
t
12-38
= The heat change across 
(  (k u )  Nda )t)  ( M 
u
dv)t
t
By Gauss's theorem
M   (k u )dv  M 
 M ( 
u
dv
t
u
   (k u ))dv  0
t
u
   (ku )  0 (x,y,z), t
t
u
 
   (ku )  heat equation
t
u
u
u
  ( k u )    ( k i  k
j  k k)
x
y
z
 u
 u
 u
= (k )  (k )  (k )
x x y y z z
k u k u k u
=


x x y y z z
 
 2u  2u  2u
+k( 2 + 2 + 2 )
x y z
=k  u  k  2u  
i, If k : constant
( A )
u k 2

u
t 
u
 k  u  k  2u -(A)
t
12-39
u
 2u
k
 k 2 , where k=
ii, 1-D,
t
x

iii, Steady-state 2u  0
 Laplaces' equation
12.7.4 Green’s Identities
i, Green's first identities :
2
 f g Nda  M ( f  g  f  g )dv -(A)
proof :
By Green's theorem,
 f g Nda  M   ( f g )dv -(B)
From the heat equation, i,e. ,
u
=   ( k u )  k   u  k  2 u
t
Let k = f,u = g

   ( f g )  f   g  f  2 g
( B)  f g Nda  M ( f  2 g  f  g )dv
ii, Exchange f and g in (A)
  gf Nda  M ( g 2 f  g  f )dv -(C)
(A)-(C)
 f g Nda   gf Nda
 M ( f  2 g  f  g )dv  M ( g 2 f  g  f )dv
 M ( f  2 g  g 2 f )dv
The Green's second identity
2
2
 ( f g  gf ) Nda  M ( f  g  g f )dv  ( D)
12-40
i, If f(x,y,z)=1,
(D)   g Nda  M  2 gdv  ( E )
ii, If 2g=0(g : harmonic function)
(E)   g Nda  0
i, e., the flux of a harmonic function across a surface = 0
12.8 Integral Theorems of Stokes
○ Sneen's theorem : c F  Tds  D (  F )  KdA
T : unit tangent vector to C
Unit normal vector
In 3-D, two issumes :
1. there are 2 normal vectors
2. the direction of c has no meaning
○ Let  :x  x(u, v), y  y (u, v), z  z (u , v)
 Normal vector :
(y,z) (z,x)
(x,y)
i
j
k
(u,v) (u,v)
(u,v)
Define the orientation of C as 
is the left side when along C


12-41
◎ Theorem 13.9: Stokes’s theorem
c F  dR   (  F ) Nda
○ Let F  fi  gj  hk
 c F  dR  c f ( x ,y z, dx
)  g x (y ,z dy
, ) h x y (z dz
, , )
i

 F 
x
f
j

y
g
K

z
h
h g
f h
g f

)i  ( 
j) ( 
k )
y z
z x
x y
The normal vector
 ( y ,z )  z( x , )  x y( , )
N
i
j
k
 (u , v)  (u , v)
 (u, v)
N
The unit normal vector : n =
N
(
 (  F )  n  (  F ) 

N
N
1 h g  ( y, z ) f h  ( z , x)
[(  )
(  )
N y z  (u , v) z x  (u , v)
+(
g f  ( x, y )
 )
]
x y  (u , v)
12-42
a D
 (  F ) n d 
1 h  g  ( y, z )
[( 
)

N y  z  ( u, v)
f  h  ( z, x)  g  f ( ,x )y

)
+( 
)
] dudv
N
z x  (u ,v ) x y  u( v , )
h  g  ( y, z)  f  h ( ,z )x
 D [ ( 
)
 ( 
)
+
y  z  ( u, v)  z  x ( u, )v
g  f  ( x, y)
(
) dudv]
x y  (u ,v )
where
D : a set of points (u,v), over which
x(u,v),y(u,v),z(u,v) are defined.
(
○ Example 13.35：
Let F ( x, y, z )   yi  xj  xyzk
2
2
: z  x  y
D : x2  y 2  9
z z
N o r m a l v e cNt o r : i 
jk
x y
x
y
= i jk
z
z
12-43
1
x
y
x2 y 2
N   i  j  k  ( 2  2  1) 2
z
z
z
z
x2  y 2  ( x2  y2 )
=
 2
x2  y2
Unit normal vector : n 
N
1

( xi  yj  zk )
N
2z
(inner normal)
c : x  3cos t , y  3sin t , z  3 0  t  2
c F  dR  c  ydx  xdy  xyzdz
2
= 0 [-3sin t (-3sin t )  3cos t 3cos t ]dt
= 0 9dt  18
2
i

 F 
x
y
j

y
x
h

  xzi  yzj  2k
z
 xyz
1
1 2
( x2 z  y 2 z  2z) 
( x  y 2  2)
2z
2
 (  F )  nda  D [(  F )  n] N dxdy
(  F )  n 
 D
1 2
( x  y 2  2) 2dxdy  D ( x 2  y 2  2)dxdy
2
 0 0 (r 2 cos 2   r 2 sin   2)rdrd
2
3
 0 0 r 3 cos 2 drd  0 0 2rdrd
1
1
 [ sin 2 ]02 [ r 4 ]30  2 [ r 2 ]30  18
2
4
2
3
12.8.1 Curl :
2
3
12-44
let F ( x, y, z ) : the veloity of a fluid
Cr F  dR   (  F )  nda
r
R '(t ) : a tangent component of F about C r
F  R ' : the tangent component of F about C r
Cr F  dR : the circulation of field about C r
as r  0
Cr F  dR   (  F )( p0 )  nda
r
=(  F )( p0 )  n(area of  r )
= r 2 (  F )( p0 )  n
1
(circulation of F about C r )
2
r
= circulation of F per unit area in the plane
normal to n
  F ( p0 )  n 
※ the curl of F is a measure of rotation of the fluid at a
point.
12.8.2 Potential Theory in 3D
F ( x, y, z ) is conservative if  a potential function  ,
s.t. F  
 C F  dR   ( p1 )   ( p0 )
12-45
Where p0 , p1 : the initial area the termal points of c
i.e., the line integral of a conservative vector field is
independent of pth
○ Example 13.26：
Let F ( x, y, z )  ( yze xyz  4 x)i  ( xze xyz  z  cos y ) j
+( xye xyz  y )k
Find 


 yze xyz  4 x,
 xze xyz  z  cos y
x
y

 xye xyz  y
x

Begin with  dz   ( xye xyz  y )dz
z
 ( x, y, z )  e xyz  yz  k ( x, y )


 yze xyz  4 x  [e xyz  yz  k ( x, y )]
x
x
k
=yze xyz 
x


 4 x and k ( x, y )  2 x 2  c( y )
y
 ( x, y, z )  e xyz  yz  2 x 2  c( y )


 xze xyz  z  cos y  [e xyz  yz  2 x 2  c( y )]
y
y
= xze xyz  z  c '( y )
 c '( y )  cos y and c( y ) 
 ( x, y, z )  e xyz  yz  2 x 2  sin y
12-46
○ Test to determine when F has 
D: a d o m a i n i f
1 . p 
D , s D , s : a s p h e r e c o n t a i n P
2 .1 p2 
, p D , p D , p : a 1p a t h2 f r o m p t o
On a domain,

= independence of path
◎ Theorem 13.10:
c F  dR independent of path on D
iff F : conservative
Choose any p0  D
If any p  D, c  D from p0 to p
Define  ( p)  c F  dR
 F  
◎ Theorem 13.11:
D: a simple connected domain(i.e., simple,
closed path in D is the boundary of a piecewise
smooth surface in D)
 F : conservative iff   F  0
i.e., F is an irrotational vector field
Proof :
(1)() F : conservative,
 , s.t. F  
  F    ( )  0
12-47
(2)() To prove F : conservative
= To prove c F  dR : independent of path
Let c,k : curves from p0 to p1
L  c  (-k ) : closed curve
 L F  dR  c F  dR  k F  dR
=  (  F )  nda  0
 c F  dR  k F  dR
i.e., the line integral is independent of path