Chapter 12 Vector Integral Calculus

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					                                                              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
               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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