# Summary 7 The divergence theorem

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```					Calculus III                         Summary 7-The divergence theorem            Sanchez

Surface area in rectangular coordinates:
The surface area S determined by a region R in the xy-plane is given by
S   1  f x  f y dxdy or S   1  f x  f y dydx
2     2                      2     2

R                                   R
Let z=f(x, y).

Line integrals and Surface integral:
The integral of a function z=f(x, y) along a curve in the xy plane or the integral of a function
w=f(x, y, z) on a curve in space are called line integrals and are denoted by
 f ( x, y)ds or  f ( x, y, z )ds
C                   C

______________________________________

Similarly, a surface integral is defined by
2           2
 h   h 
 f ( x, y, z )dS   f ( x, y, z ) 1   x    y  dxdy, where the surface S
   
 
S                    D
is described by h( x, y ) and D is the domain of the function w  f ( x, y, z )

The divergence Theorems
Suppose that C is a closed piecewise smooth curve that bounds the closed region R. Let F=Pi +
Qj be a vector field with component functions that have continuous first-order partial derivatives
on R. Let n be the outer normal vector to C. Then
                             P Q 
                                x  y dA   divFdA
F  n ds    Qdx  Pdy   



C              C                     R                  R
________________________________________________

Similarly,
Suppose that S is a closed piecewise smooth surface that bounds the space region T. Let F=Pi +
Qj + Rk be a vector field with component functions that have continuous first-order partial
derivatives on T. Let n be the outer unit normal vector to S. Then
                                                                        P Q R  
 F  ndS     FdV , that is            F  ndS   divFdV       x  y  z   dV
               
S                  T                         S              T             T                 

      f xi  f y j  k
where n 
1  f x2  f y2

-1-
Problem 1:
Evaluate the surface integral    f ( x, y, z )dS   if f(x, y, z)=x+y and S is the first octant part of

the plane x+y+z=1
2         2
 h   h 
 f ( x, y, z )dS   f ( x, y, z ) 1       dxdy, where the surface S
 
 x   y 
S                    D
is described by h( x, y ) and D is the domain of the function w  f ( x, y, z )
z  1  x  y  x  y  1in the xy  plane
1 1 y                                             1 1 y
 x  y dS    x  y      1  (1)  (1) dxdy  3                ( x  y)dxdy
2           2

S                 0 0                                                0 0
1 y
1 2
x              1 1  y 2                      1
1 2y  y2  2y  2y2 
 3 
 2  xy  dy  3  
                      (1  y ) y  dy  3                         dy
0         0       0
      2                  
        0           2           
1


31
2 0

 1  y dy  2  y  3   2 1  3   2  3   3
2

3

y3 

3  1
     
3 2
 
3
        0

Problem 2
Calculate    FdV where F=xi + 2yj + 3zk and T is the space region with boundaries S
T
given by the first octant cube with opposite vertices (0, 0, 0) and (1, 1, 1)

Z

Y

x

 P Q R          111
   FdV   divFdV    x  y  z dV     1  2  3dxdydz
              
T             T            T                     000
111
 6    dxdydz  6(1  1  1)  6
000

-2-
Problem 3.
Calculate the outward flux of the vector field F=xi + 2yj +3zk across the surface S given by the
boundary of the first octant cube with opposite vertices (0, 0 ,0) and (1, 1, 1).
z
S1: left face
S2: right face
S3              S3: top face
S6              S4: bottom face
S2       S5: front face
S1       S5              y   S6: back face
S4

x

 
 F  n dS 
S
                                                                                        
 F  ndS           F  ndS         F  ndS     F  ndS                F  ndS      F  n dS
S1                   S2                 S3             S4                        S5              S6

                                               y 0                   11                11
 F  n1dS1   ( xi  2 yj  3 zk )   j dxdz                         (2ydxdz    0dxdz  0
S1                     S1                                                00                  00
                                                             y 1 1 1                  11
 F  n2 dS 2            ( xi  2 yj  3zk )   j dxdz               (2ydxdz  2  dxdz  2
S2                        S2                                           00                   00
                                                             z 1 1 1                  11
 F  n3 dS 3            ( xi  2 yj  3zk )  k dxdy             (3zdxdy  3  dxdy  3
S3                        S3                                           00                   00
                                                                 z 0 1 1                   11
 F  n3 dS 3            ( xi  2 yj  3zk )   k dxdy                (3zdxdy  0  dxdy  0
S4                        S4                                                00                   00
                                                          x 1 1 1                11
 F  n5 dS 5            ( xi  2 yj  3zk )  i dydz             ( xdydz    dydz  1
S5                        S4                                          00               00
                                                              x 0 1 1                 11
 F  n5 dS 5            ( xi  2 yj  3zk )   i dydz               ( xdydz  0  dydz  0
S6                        S6                                               00               00
 
Therefore,  F  ndS  0  2  3  0  1  0  6   divFdV
S                                                           T
Problems 2 and 3 verify the Divergence Theorem.

-3-
Problem 4. Verify the divergence theorem if F=(x + y)i + (y + z)j +(x + z)k and S is the surface
of the tetrahedron bounded by the three coordinate planes and the plane x + y + z =1

z
S1: left triangle, n1 =-j

y+z=1                S2: back triangle, n2=-i
x+z=1
y   S3: bottom triangle, n3=-k
x+y=1

x
S4:triangle determined by the plane x+y+z=1, n4=i+j+k which is
the normal to the plane.

                                                  
 F  n dS   F  ndS               F  ndS      F  n dS
S                S1                    S2              S3
                                                            y 0                      11
 F  n1dS1   ((x  y)i  (y  z)j  (x  z)k )   j dxdz                            ( y  z )dxdz
S1                 S1                                                                      00
1

  zdxdz    z (1  z )dz                      
1 1 z                    1                     1                 z3 z2  1
                                                       z  z dz      
2
 3
0 0                       0                     0                    2 0
 6
                                                                                  x 0 1 1
 F  n2 dS 2        ((x  y)i  (y  z)j  (x  z)k )   i dxdz                     (x  y )dydz
S2                    S2                                                                   00
1

  ydydz    z (1  z )dz                      
1 1 z                    1                     1                   z3 z2  1
                                                       z  z 2 dz      
 3
0 0                       0                     0                      2 0
 6
                                                                                  z 0 1 1
 F  n3 dS 3        ((x  y)i  (y  z)j  (x  z)k )   k dxdy                      (x  z )dxdy
S3                    S2                                                                    00
1 1 y                                                                                1

  xdydz    y (1  y )dy                          
1                     1                 y3 y2     1
                                                       y  y dy  
2           
 3  2   6
0 0                       0                     0                       0

-4-
Problem 4 (continuation)
z 1 x  y
                                                   i  j k 
 F  n3 dS 3   ((x  y)i  (y  z)j  (x  z)k )            1  1  1 dxdy
S4                        S4                                 3 
1
1 1 y                 2   1
 y  y  1
  2 x  2 y  2 z  dxdy    2 dxdy  2  (1  y )dy  2
S4                           0 0            0                     2 0
            1 1 1             1 1
Therefore,  F  ndS      1  1  
S                6 6 6              2 2
1 1 x 1 x  y                         1 1 x
 divFdV   1  1  1dV                      3                  dzdydx  3                (1  x  y )dydx
T                              T                        0 0      0                             0 0
1 x
1           2      1
 2 
 y  xy  y  dx  3 1  x  x(1  x)  1  x dx
 3 
0            2 0    0
                      2 
2  2x  2x  2x 2  1  2x  x 2
                 
1
31
 3                                   dx   1  2 x  x 2 dx
0                2                      20
1
31               3 (1  x) 3    1           1
  1  x 2 dx                0  1 
20               2     3 01    2           2
 
Therefore,  F  n dS   divFdV which verifies the convergence theorem.
S             T

 
Problem 5. Verify the Divergence theorem in the evaluation of                                        F  n dS for the solid
S
x  y  4 , z  0 and z  2 if F  3xi  y j  3z k
2            2                                                                2         2

P Q R
divF                    3  2 y  6z
x y z

3z  2 yz  3z 
                     2                     4 x 2 2                                             2 4 x 2
2 2
 F  ndS   divFdV 4                             (3  2 y  6 z )dzdydx  4                                              0
dydx
S                           T             0        0    0                                            0   0

 6  4 y  12dydx  4 18 y  2 y 
4 x
                        
2
2                                          2                       4 x   2         2
 4                                                  2
dx  8 9 4  x 2  4  x 2 dx
0
0        0                                 0                                        0

 72
2

0
 4  x dx  84  x 
2            2 2
0
 72( )  8(0)  72

-5-
Problem 5 (continuation)
                                        
 F  n dS   F  n dS1   F  n dS 2   F  n dS 3
S                    S1                       S2                     S3

S1 : is the circular base of the cylinder, n1  k , z  0

S 2 : is the circular top face of the cylinder, n 2  k , z  2
S 3 : is the lateral surface of the cylinder
xi  yj  ok
The vector xi  yj  0k is normal to the cylinder x 2  y 2  4, n3 
x2  y2

              z 0 1 4  x

 
2
1 4 x 2
 F  ndS1   3xi  y j  3z k   k dS1  4  3z dydx  4 12  0dydx  0
2      2                            2

S1                      S                                                          0       0                               0        0

             z  2 1 4 x

                                                               1 4 x 2                   2

 F  ndS 2   3xi  y j  3z k  k dS1  4  3z dydx  4 12  dydx  48
2      2                        2

S2                        S                                                    0       0                               0       0
y
      f y j  fzk i  x j i                                                        yk  xi
For S 3 , Let x  f ( y, z )  4  y 2  0 z , then n 4                           
2    2         2
1 f y  f k        y                                                          x2  y2
1
x2
              

 
          xi  yj  ok 
F  n dS 2   3 xi  y j  3 z k  
2       2
 x2  y2     

dS  
3x 2  y 3
1
y2
dzdy
S x y
2     2                                                    x2
S2              S                                     
2 2                                     2 2                                             2
3x 2  y 3                               12  3 y 2  y 3                             12  3 y 2  y 3
2         x dzdy  2                                                dzdy  4                                   dy
2                                        2
2 0                                     2 0         4 y                           2           4 y
 /2                                                                  /2
4     
12  12 sin 2   8 sin 3 
2 cosd  4           
 1  cos 2 
(12  12
      2     
2

  8 1  cos  sin d           
 / 2                  4  4 sin 
2
 / 2
 /2                                                                                                                              /2
            2  3

 4  [6  6 cos 2  8 1  cos  sin  ]d  46  sin 2  8 cos 
8 cos3  

 / 2                                      
    2                      3     / 2
 24
                                                                
Therefore            F  ndS   F  ndS1   F  ndS 2   F  ndS 3  0  48  24  72
S                    S1                     S2                     S3
Therefore, the divergencetheorem is verified

-6-

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