# Chapter 12 Vector Integral Calculus by 2Jc849

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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
t
x( t ) , y ( ) ,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:         Line integral

C:    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
b               dx              dy             dz 
 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
C fdx  gdy  hdz  C xdx  yzdy  e dz
z

 1 (t 3 ) (t32 ) t ( t )  )( et1 ) t (dt ) ( 2 )
2                                    2
2
                       (              
111
 1 (3t 5  t 3 2tet )dt           
e 4 e
2                2

4
◎ 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    from (1,0) to (0,1)
P: 
C2 : line             from (0,1) to (2,1)

Compute P x ydx  y dy
2       2

(a) Parameterize curves
C1 : x c o t ,y
s                   

s ti n , t 0       /2

C2 : x s, y 1 ,  0s           2
＊ Different curves use different parameters

(b) Replace (dx, dy) with (dt , ds)

dx      d cos t
C1: dx    dt          dt   sin tdt
dt        dt
dy      d sin t
dy  dt            dt  cos tdt
dt        dt

dx      ds
C2: dx     ds     ds  ds
ds      ds
dy      d1
dy  ds  ds  0
ds      ds
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
t=b
R(a)
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
1                              1

 3/10

◎ 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:                 a , (              )          ,
( x (a ) ,y ( ) 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
C fdx  gdy  hdz  a [ f ( x(t ), y (t ), z (t ))        
b

dt
dy                            dz
g ( x(t ), y (t ), z (t ))     h( x(t ), y (t ), z (t )) ]dt
dt                            dt
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 )           d z (t )    dz ( s )
          ,                  
dt         ds                  dt         ds
 C fdx  gdy  hdz
dx                              dy
 b [  f ( x( s ), y ( s ), z ( s))     g ( x( s), y ( s), z ( s) )
a

ds                              ds
dz
 h( x( s), y ( s), z ( s) )]( ds)
ds
dx                              dy
  a [ f ( x( s ), y ( s ), z ( s ))       g ( x( s), y ( s), z ( s))
b

ds                              ds
dz
 h( x( s ), y ( s ), z ( s ))     ]ds
ds
dx                              dy
  a [ f ( x(t ), y (t ), z (t ))     g ( x(t ), y (t ), z (t ))
b

dt                              dt
dz
 h( x(t ), y (t ), z (t )) ]dt   C fdx  gdy  hdz
dt
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
 a  ( x(t ), y(t ), z (t )) x(t )2  y(t )2  z(t )2 dt
b

○ Example 13.8：

C: x  4cos t, y  4sin t, z  3, 0  t 
2
 x  4sin t , y  4cos t , z  0
ds  x(t ) 2  y(t ) 2 z(t ) dt
2

 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 ( x, y, z ) ds
m C
3 
= 0 2 (2cos t ) 2 (2sin t ) 2 4sin 2 t  4cos 2 tdt
16

=6 0 2 cos 2 t sin 2 tdt  3 / 8
1
y     C y ( x, y, z )ds
m
3 2
= 0 (2cos t )(2sin t ) 3 4sin 2 t  4cos 2 tdt
16

=6 0 2 cos t sin 3 tdt  3/ 2

1
z   C z ( x, y, z )ds
m
3 2
 0 3(2cos t )(2sin t ) 2 4sin 2 t  4cos 2 tdt
16

 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
g f
f, g ,        ,          :                  u
c o n t i nD o u s o n
x y
g f
dy 
 C f (x y dx  g x (y , ) D
, )                                   
(   dA )
x y
where dA  dxdy

Proof：
 C : y h( x) , x f r oam                 bto
i,                    C : 1
C2 : y k ( x) , x f r ob
m                 ato

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)
12-12

f       b k ( x ) f
D      dA  a h ( x ) dydx
y                 y
 b f ( x, y ) |k (( x )) dx
a                 x
h

= a [ f ( x, k ( x))  f ( x, h( x)]dx
b
----- (B)

From (A) and (B),
f
C f ( x, y )dx   D             dA                 ----- (C)
y
ii,
C : x  f ( y ), y from d to c
C3               C4      C: 3
C4 : x  g ( y ), y from c to d

g
C g ( x, y)dx  D              dA                ----- (D)
x
From (C) and (D)
g f
C f ( x, y )dx  g ( x, y )dx  D (         )dA
x y
○ Example 13.10：

Force: F ( x, y)  ( y  x 2e x )i  (cos 2 y 2  x) j
12-13

Work: C F dR

                   
 D [   (cos 2 y 2  x)  ( y  x 2e x )]dA
x                  y
 D 2dA  2(area of D)  4

12.2.1. An Extension of Green’s Theorem
Suppose there are finite points enclosed by C, at which
f      g
f , g,      , or    are not continuous or undefined
y      x
*

(a)               D*                (b)

From Green’s theorem,
g f
C* f ( x, y )dx  g ( x, y )dy  D* (      )dA -- (12.2)
x y

(c)                                 (d)

(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 f
 g ( x, y )dy  D (      )dA
x y
Reverse the orientations on circles
12-14

C f ( x, y )dx  g ( x, y )dy
n                                          g f
   Kj f ( x, y)dx  g ( x, y )dy + D (      )dA
j 1                                       x y
y            x
○ Example 13.12:           C            dx  2     dy
x2  y 2     x  y2
y                      x
 f ( x, y )              , g ( x, y )  2
x2  y 2               x  y2
g    y2  x2     f
     2                                         ----- (A)
x ( x  y 2 ) 2 y
f g
f , g,        ,   : not continuous at (0,0)
y x
Case 1： C does not enclose (0,0)
y           x            g f
C              dx  2   dy  D (     )  0
dA
x2  y 2     x y 2         x y
Case 2： C encloses (0,0)

C f ( x ,y dx  g x( y , dy)
)
g f
dx 
  K f x (y , ) g x y( dy D
, )          +  ( dA           )
x y
dx 
  K f x (y , ) g x y( dy )
,

k : x r c o s ,y r  i n ,  
   s         0                    2
y              
r s i n  s  n
i

f ( x ,y )                      
x2  y 2      r 2     r
12-15

x     rc os       c s
o

g ( x ,y )                  
x y
2   2
r 2
r
s  d
d x  r i n  , d 
y            r c sd
o
  K f ( x, y )dx  g ( x, y )dy
2   sin                cos
= 0 [          (r sin  )       (r cos )]d
r                    r
= 0 [sin 2   cos 2  ]d  0 d  2
2                          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

             b  dx     dy
 C F dR  C           dx     dy  a (             )dt
x      y           x dt y dt
d ( x(t ), y(t ))
 a                  dt   ( x(t ), y(t )) |b
b
a
dt
  ( x(b), y (b))   ( x(a), y (a))
  (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                    i    j
x    y
               
Then,        f ( x, y ),     g ( x, y )
x               y
To find  ,
                   
1) Begin with        f ( x, y ) or     g ( x, y )
x                   y
2) Integrate w.r.t. the variable
3) Use the second equation to find 
○ 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.              f ( x, y)  2 x cos 2 y,
x

and             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
        ( x cos 2 y  c( y ))  2 x sin 2 y  4 y 2
2

y       y
 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
g f
iff           (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,               (b) Green’s theorem
g f
(c)        ,            (d) Potential function 
x y
12-18

○ Let D：all points in the plane except (0,0)
y        x
F ( x, y )            i 2     j  f ( x, y ) i  g ( x, y ) j
x y
2    2
x y 2

f g    y2  x2         y2  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 
 f ( x, y )                              sin 
x y      cos   sin 
2    2      2         2

x           cos
g ( x, y )  2                        cos
x y   2
cos   sin 
2         2

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
i nh
of patD

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

A：a domain
S：not a domain          M：not 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          x0
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 dR2

 C f ( x, y )dx  g ( x, y )dy
2

 0 f (a  x, b)xdt
1

 (a  x, b)   (a, b)
                                  0 f (a  t x, b)dt
1

x
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)
lim
x0 
x
 lim f (a  x, b)  f (a, b)
x0
12-22

Similarly,
 ( a   x, b )  ( a , b )

lim                         f a ( , )
b
x0
x

      ( ,b  f ( ,b )
a      )     a
x

L i k e w i s ea, b  ( ,a )
f      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 g
 F : conservative on D iff           
y x

Proof :
i. (If) Given F: conservative with potential ,
    
 F = 
               i    j  f x( y,  ) g x( jy,
i                  )
x    y
                
 f ( x, y )       , g ( x, y ) 
x                y
f    2    2   g
                
y xy yx y
12-23

f g
ii. (Only if)    Given      ,
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
J F dR  D (  )dA  0

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
cov , 
s                  v
sin
au      bu
x 2      y
(  )  ( ) 2  cos v  sin v2  1
2

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

x              y             z
Tv       (u0 , v0 )i  (u0 , v0 ) j  (u0 , v0 ) k
0
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

x               y             z
Tu       (u0 , v0 ) i  (u0 , v0 ) j  (u0 , v0 ) k
0
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            k
x            y               z
       (u0 , v0 )     u
( 0 ,v0     )   ( ,
u0 v0 )
u            u              v
x            z              z
(u0 ,v 0 )    u( 0v , 0        v
) u (0 , 0 )
v            v              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 and g

f          f
 ( f , g ) u           v f g g f
                    
 (u, v) g              g u v u v
u          v
12-26

 The Normal vector
 ( y, z )  ( z , x)     ( x, y )
N ( p0 )            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 
v   6
4 4 2

The Jacobians :
 ( y, z )          y z z y 
      
 (u , v) ( 1 , )  u v u v  ( 1 , )
 26
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        1
 ( y, z )  ( y, z )                    S
                       S        S  
 (u , v)  ( x, y )                     x
x        y
S     S
 ( z , x)                S  ( x, y ) 1 0
 x     y   ,               1
 ( x, y )                y  ( x, y ) 0 1
1      0
The normal at p0 : ( x0 , y0 , S ( x0 , y0 )) :
S                S
N ( p0 )    ( x0 , y0 ) i  ( x0 , y0 ) j  k
x                y
z                z
  ( x0 , y0 )i  ( x0 , y0 ) j  k
x                y
○ Example 13.25: Cone: z  S ( x, y)  x 2  y 2
S            x         S         y
                       ,      
x       x2  y 2       y      x2  y 2
The Normal vector at P0  (3,1, 10)
3    1
N ( P0 )         i    j  k (inner normal)
10    10

3     1
The outer normal : -N ( P0 )             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)       0   0
 (u, v)     0    0

 ( x, y )
[              ]( u ,v ) ( z  z0 )  0
 (u, v)      0    0

This is the equation of the tangent plane
12-29

○ Example 13.26:

3 b 1
The normal vector at P0 : ( a     , , )
4 4 2
on the elliptical cone: x  au cos v, y  bu sin v, z  u
3b a       ab
is N     i j k
4     4     2
 The tangent plane :
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 :
S                  S
N ( P0 )  (      )( 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
S                        S
or z  z0       ( 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):
S 2 S 2                 z      z
D 1  (      )  ( ) 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:

 : z  S ( x, y )  9  x  y with x  y  9
2   2        2    2

D : all points on or inside the circle of radius 3
about the origin in (x,y ) plane
12-31

z        x           x
Compute                        
x    9  x2  y 2    z
z        y           y
              
y    9  x2  y 2    z
x2  y 2  z 2
 Area of   D            2
dxdy
z
3
 D                  dxdy
9 x  y
2    2

Convert to polar coordinates :
x  r cos , y  r sin  , 0  r  3, 0    2
3                                3
rdrd
2
 D                           dxdy  0 0
3

9 x  y
2         2
9r   2

1            1
3
 6 0                   dr  6 [(9  r ) ]  6 [9 ]  18
3                                    2   2 3          2

9r       2                            0

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 :
c f ( x, y, z )ds  a f ( x(t ), y (t ), z (t )) x '(t )  y '(t )  z '(t ) dt
b                                  2         2         2

○ 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 ),
s 2 s 2
  f ( x, y, z )da  D f ( x, y, s ( x, y )) 1  (      )  ( ) 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
(4  x  y )dy  (4  x) y  y 2 1
1
0                                    0
2
1 7
 4 x   x
2 2
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
c f ( x, y )dx  g ( x, y )dy  D (    )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
○ 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

dy             dx
○ F  N  g ( x, y)         f ( x, y)
ds             ds
dx        dy
c f ( x ,y dx  g x( y , dy) c f x y( , )g x y ( , )
)                        [                ds           ]
ds        ds
  F  Nds
○ Conservation of energy
c
D i v e r g e n c e o f v et o r f i e l d      F at point P
l     e tt
= f l o w o f t h e v e c t o r f i e f d v F c a o rP f l u x o
f i e l d F
N
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        j     K
                g f
Cure :   F                          (     )
K
x     y     z    x y
f     g      0
g f
 F  K 
(      )            
x y
○ Unit tangent T (s)  x '(s)i  y '(s) j 3-D
dx dy
F  Tds  [ f ( x, y )i  g ( x, y ) j ]  (     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：

1: z  x  y , x  y  1
2   2   2   2


 2 : x  y  1, z  1
2   2

Unit outer normal to 1:
1 x      y
N1     ( i  j  k)
2 z     z
12-36

1 x    y
 F  N1  ( xi  yj  zk )      ( i  j  k)
2 z   z
1 x2 y 2
=    (   z)  0
2 z   z
  1
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 1 ]dydx
1 1
0

= 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



(x,y,z)
M

Heat energy leaving M across  in t :
(  (k u )  Nda)t, i,e., the flux of k u across  in t
u
The change in temperature at (x,y,z) in t :           t
t
u
Heat loss ( M dv)t
t
 The change in heat energy in M over t
12-38

= The heat change across 
u
(  (k u )  Nda )t)  ( M       dv)t
t
By Gauss's theorem
u
M   (k u )dv  M       dv
t
u
 M (          (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
u
=k  u  k  2u            k  u  k 2u -(A)
t
i, If k : constant
u k 2
( A )        u
t 
12-39

u     2u           k
ii, 1-D,     k 2 , where k=
t    x            
iii, Steady-state 2u  0         Laplaces' equation

12.7.4 Green’s Identities
i, Green's first identities :

 f g Nda  M ( f  g  f  g )dv -(A)
2

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  2u
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

 ( f g  gf ) Nda  M ( f  g  g f )dv  ( D )
2      2
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)
 (y,z) (z,x)     (x,y)
 Normal vector :           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     j   K
           
 F 
x      y   z
f      g    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
N
 (  F )  n  (  F ) 
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

1 h  g  ( y, z )
 (  F ) n d  D
a         [(     )       
N y  z  ( u , v)
f  h  ( z x  g   ( , )
, )          f   x y
(         )       +(        )       N
] dudv
z x  (u ,v ) x y  u( v , )
h  g  ( y z)  f   ( ,z )
,          h       x
 D [ (      )       (      )        +
y  z  ( u v  z   ( u )
, )        x    , 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
: z  x  y
2    2

D : x2  y 2  9
z z
N o r m a l v e c t  : i 
N or          jk
x y
x     y
= i jk
z     z
12-43

x   y         x2 y 2      1

N   i  j  k  ( 2  2  1) 2
z   z         z   z
x2  y 2  ( x2  y2 )
=                         2
x2  y2
N   1
Unit normal vector : n                  (  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            j     h
                  
 F                            xzi  yzj  2k
x           y    z
y            x    xyz
1                          1 2
(  F )  n       ( x2 z  y 2 z  2z)     ( x  y 2  2)
2z                         2
 (  F )  nda  D [(  F )  n] N dxdy
1 2
 D      ( 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
2   3                     2   3

1             1
 [ sin 2 ]0 [ r 4 ]3  2 [ r 2 ]3  18
2
0             0
2             4
12.8.1 Curl :
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 Cr
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
  F ( p0 )  n     (circulation of F about C r )
r 2

= circulation of F per unit area in the plane
normal to 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
 . p  ,  s
1       D               D, s: a sphere contain P
 .1 p2 p
2       ,        
D , p        D , p : a 1p a t h2 f r o m p   to
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

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