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 x2  f y2 dxdy or S   1  f x2  f y2 dydx
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 
F
  nds    Qdx  Pdy    x  y dA   divFdA
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
 
F
  ndS    FdV , that is
 
F
  ndS  divFdV 
S
S
T

where n 
T
 f xi  f y j  k
1  f x2  f y2
-1-
  P Q R  
   x  y  z   dV

T 
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
 x  y dS    x  y 
S
0 0
1 y
1

x2
 3  
 xy 
0 2


0
1 1 y
1  (1)  (1) dxdy  3 
2
2
 ( x  y )dxdy
0 0
1
1 1  y 2

1 2y  y2  2y  2y2 
dy  3  
 (1  y ) y  dy  3  
 dy
2
2
0

 0

1

31
3 
y 3 
3  1
3 2
3
2

1  y dy 
y


1








2 0
2 
3 
2  3 2  3 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
111
 P Q R 
   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
S3
S6
S2
S1
S5
S4
y
S1: left face
S2: right face
S3: top face
S4: bottom face
S5: front face
S6: back face
x
 
F
  ndS 
S
 
F
  ndS 
 
 
F

n
dS

F

  ndS 
 
 
F

n
dS

F

  ndS 
 
F
  ndS
S1
S2
S4
S6
S3
S5
y 0
 


F

n
dS

(
xi

2
yj

3
zk
)


j
dxdz

1 1


S1
S1
 
F
  n2 dS2 
 ( xi  2 yj  3zk )   j dxdz
S2
S2
 
F
  n3 dS3 
 ( xi  2 yj  3zk )  k dxdy
S3
S3
11
11
00
y 1 1 1
00
11
  (2ydxdz    0dxdz  0
  (2ydxdz  2  dxdz  2

00
z 1 1 1

00
11
  (3zdxdy  3  dxdy  3
00
 
F
  n3 dS3 
 ( xi  2 yj  3zk )   k dxdy
S4
S4
 
F
  n5 dS5 
 ( xi  2 yj  3zk )  i dydz 
S5
S4
00
z 0 1 1
11
00
00
  (3zdxdy  0  dxdy  0

x 1 1 1
 
F
  n5 dS5 
 ( xi  2 yj  3zk )   i dydz
S6
S6
11
  ( xdydz    dydz  1
00
x 0 1 1

00
  ( xdydz  0  dydz  0
00
 
Therefore,  F  ndS  0  2  3  0  1  0  6   divFdV
S
T
Problems 2 and 3 verify the Divergence Theorem.
-3-
11
00
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
S2: back triangle, n2=-i
y+z=1
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

n
dS

F

  ndS
S
S2
S1
S3
y 0
 
 F  n1dS1   ((x  y)i  (y  z)j  (x  z)k )   j dxdz 
11
S1
00
S1
  zdxdz    z (1  z )dz   
1 1 z
1
1
0 0
0
0

1

 z3 z2 
1
2
 z  z dz      
2 0
6
 3
 
F
  n2 dS2 
 ((x  y)i  (y  z)j  (x  z)k )   i dxdz
S2
S2
x 0 1 1

  (x  y )dydz
00
  ydydz    z (1  z )dz   
1 1 z
1
1
0 0
0
0

  ( y  z )dxdz
1

 z3 z2 
1
 z  z dz      
2 0
6
 3
2
z 0 1 1
 
 F  n3 dS3 
 ((x  y)i  (y  z)j  (x  z)k )   k dxdy 
  (x  z )dxdy
S3
S2
00
  xdydz    y (1  y )dy   
1 1 y
1
1
0 0
0
0

-4-

1
 y3 y2 
  1
 y  y dy  

2  0
6
 3
2
Problem 4 (continuation)
z 1 x  y
 
i  j k 
F

n
dS
  ((x  y)i  (y  z)j  (x  z)k )  
 1  1  1 dxdy
3
3

3


S4
S4
1 1 y
1

y 2 

  2 x  2 y  2 z  dxdy    2 dxdy  2  (1  y )dy  2 y 
 1
2
S4
0 0
0

0
 
1 1 1
1 1
Therefore,  F  ndS      1  1  
6 6 6
2 2
S
1
1 1 x 1 x  y
 divFdV   1  1  1dV  3 
T
T
0 0
1 x

1 1 x
dzdydx  3
0
 (1  x  y )dydx
0 0
1

y 2 
1  x 2 


 3  y  xy 
 dx  3 1  x  x(1  x)  2 dx
2
0
0
0

1


2  2x  2x  2x 2  1  2x  x 2
31
 3
dx   1  2 x  x 2 dx
2
20
0
1
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
  ndS for the solid
S
x  y  4 , z  0 and z  2 if F  3xi  y j  3z k
2
2
divF 
2
P Q R


 3  2 y  6z
x y z
2
 
 F  ndS   divFdV 4
S
T
4 x 2 2
0
2 4 x 2

 (3  2 y  6 z )dzdydx  4

0
0
0
2
 6  4 y  12 dydx  4 18 y  2 y 
4 x
2
 4
0
2
2
0
2
 72 
0
2
 4  x dx  84  x 
2
0
2 2
0
4 x
0
2

2 2


dx  8 9 4  x 2  4  x 2 dx
 72( )  8(0)  72
-5-
2
0
3z  2 yz  3z 
0
0
dydx
Problem 5 (continuation)
 
 
 
 
F

n
dS

F

n
dS

F

n
dS

F
1
2



  ndS 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
The vector xi  yj  0k is normal to the cylinder x 2  y 2  4, n3 

xi  yj  ok
x2  y2

2
1 4 x 2
z 0 1 4 x
 
2
2
2
 F  ndS1   3xi  y j  3z k   k dS1  4  3z dydx  4 12  0dydx  0
S1
S
0

0

0
0
1 4 x 2
z  2 1 4 x
 
2
2
2
 F  ndS 2   3xi  y j  3z k  k dS1  4  3z dydx  4 12  dydx  48
S2
2
S
0
0
0
0
y
 j i

f
j

f
k

i

y
z
For S 3 , Let x  f ( y, z )  4  y 2  0 z , then n 4 
 x

2
2
2
1 f y  f k
y
1
x2


 
3x 2  y 3
 xi  yj  ok 
2
2
F

n
dS

3
xi

y
j

3
z
k

dS

2 

 2 2

2
2 
x

y
S2
S
S x y




2 2
2
2 2
3x 2  y 3
 x dzdy  2 
2 0
 /2
4


12  3 y 2  y 3
4 y
2 0
12  12 sin 2   8 sin 3 
4  4 sin 2 
 / 2
 /2

2
2
dzdy  4 
12  3 y 2  y 3
2
2 cos d  4
1
 /2
4 y
2
y2
x2
yk  xi
x2  y2
dzdy
dy


 1  cos 2 
2
(12  12
  8 1  cos  sin d 
2


 / 2

 /2


3
8 cos 3  
 4  [6  6 cos 2  8 1  cos  sin  ]d  46  sin 2  8 cos  

2
3 


 / 2
 / 2
 24
Therefore
 
2
 
 
 
 F  ndS   F  ndS1   F  ndS 2   F  ndS 3  0  48  24  72
S
S1
S2
S3
Therefore, the divergencetheorem is verified
-6-