MATH 100
Lecture 22 Introduction to surface integrals
Mass of a bent lamina:
let thedensitybe  x, y, z ,
then
Def : If a curvedlamina with density x, y, z  has theequation z  f x, y ;
and if theprojectionof thislaminaon thexy  planeis theregion R, then
themass M of thelaminais defined by
M    x, y, f x, y  f x2  f y2  1dA
R
2006 Fall MATH 100 Lecture 22
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Definition of density function:
M
 x, y, z   lim
S
where S is thesamll sectionof thearea containingx, y, z 
Remark: when   1, themass equals to thesurface area,
and thearea is
S  
R
f x2  f y2  1dA
2006 Fall MATH 100 Lecture 22
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MATH 100
Lecture 22
Introduction to surface integrals


  x , y , f x , y 
M k   xk , yk , zk S k

k

k

k

k




f x2 xk , yk  f y2 xk , yk  1Ak
2006 Fall MATH 100 Lecture 22
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MATH 100
Lecture 22
Introduction to surface integrals
T hus
n
M   M k
k 1
n


   xk , yk , f xk , yk
k 1





f x2 xk , yk  f y2 xk , yk  1Ak
   x, y, f x, y  f x2  f y2  1dA
R
2006 Fall MATH 100 Lecture 22
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MATH 100
Lecture 22 Introduction to surface integrals
Ex1
z  x2  y 2 , 0  z  1,  x, y, z    0 , findits mass
Sol :
R : x 2  y 2  1,
f x  2 x, f y  2 y
2 x 2  2 y 2  1dA
M    0
R
 0 
2
0
1

0
4r 2  1rdrd
1
4u  1du
0 2
1
 2 0 
3
2 1
  0   4u  12
3 4

1
0

 5  1

6 

 0 
3
2
2006 Fall MATH 100 Lecture 22
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MATH 100
Lecture 22 Introduction to surface integrals
Surface integral: Let  be a surface with finitesurface and g x, y, z  a continuous
functiondefined on  . Subdivide  into i i 1 with surface area
n
S in1 and sum theproduct up
 g x , y , z S
n
k 1

k

k

k
k
Itslimit gives thesurfaceintegralof g over :
n







g
x
,
y
,
z
dS

lim
g
x
,
y
,
z
S k

k
k
k


n 
k 1
2006 Fall MATH 100 Lecture 22
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MATH 100
Lecture 22 Introduction to surface integrals
(a) If g   and z  f x, y , then
2
2




gdS


x
,
y
,
f
x
,
y
f

f
x
y  1dA



R
T heexpression has wider implication
(b) If  is defined by y  f  x, z  and R is theprojectionontoxz  plane,then
2
2




gdS

g
x
,
f
x
,
z
,
z
y

y
 1dA
x
z



R
(c) If x  f  y, z , then
2
2




gdS

g
f
y
,
z
,
y
,
z
y

y
x
z  1dA



R
2006 Fall MATH 100 Lecture 22
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MATH 100
Ex 2
Lecture 22 Introduction to surface integrals
Evaluate
 xzdS over the first octact of x  y  z  1
z  1  x  y, R  x, y  : 0  y  1  x,0  x  1
Sol :
1 1 x
 xzdS  


0 0
x1  x  y   1   1  1dxdy
2
2
1 x
1 2

2
 3   xy  x y  xy  dx
0
2

0
1
2
x

x
3
2


 3    x  dx 
0 2
2
24

1
2006 Fall MATH 100 Lecture 22
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MATH 100
Lecture 22 Introduction to surface integrals
Alternative solution ot Ex 2 :
y  1  x  z , R   x,   : 0    1  x,0  x  1
1 1 x
 xzdS  


0 0
xz  1   1  1dxdz
2
2
3 1
2



x
1

x
dx

0
2
3

24
2006 Fall MATH 100 Lecture 22
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MATH 100
Ex :
Lecture 22 Introduction to surface integrals
Evaluate yz 2 dS over σ : z  x 2  y 2 ,0  z  2

2006 Fall MATH 100 Lecture 22
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MATH 100
Lecture 22 Introduction to surface integrals
zx 
Sol :
x y
2
2
y
, zy 
x2  y 2
R  r ,   : 1  r  2,0    2 
z x2  z y2  1  2
so
 y
x
z dS   y
2 2
2

x y
2
2

2
2dA
R
 2
2
0
1 r sin   r
2
2
2
 rdrd  2 
2
0

2
1
r 5 sin 2 drd
2
 r sin  
21 2 2
 2 
sin d
 d 

0
0
6
2

1
21 2 1  cos
21

d


2
2 0
2
2
6
2
2006 Fall MATH 100 Lecture 22
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MATH 100
Lecture 22 Introduction to surface integrals
Surface integral of vector functions, we have studied line (curve)
integral with orientation, now we go to surface with orientation.
In general, a surface is given by
G(x,y,z) = 0
The particular cones are
G  z  z  x, y   0, G   z x , z y ,1
G  y  y  x, z   0, G   y x ,1, y z
G  x  x y, z   0, G  1, x y , x z
2006 Fall MATH 100 Lecture 22
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MATH 100
Lecture 22 Introduction to surface integrals
There are 2 unit normal vectors
A surface has 2 orientation, corresponding to the 2 normal direction.
The orientation should be chosen in the way that there is no sudden
change in the normal direction when we transverse along the surface.
2006 Fall MATH 100 Lecture 22
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MATH 100
Lecture 22 Introduction to surface integrals
The 2 possible orientation: inward normal and outward normal
2006 Fall MATH 100 Lecture 22
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Surface integral




Def : If F x, y, z   f x, y, z i  g x, y, z  j  hx, y, z k has
continuouscomponentson theorientalsurface , and if
 
n  n  x, y, z  is theunit vector of theorientation, then
 
 F  ndS


is called theflux integralof F over ,

or thesurface integralof F over ,
or thesurface interacting of thenormalcomponent

of F over .
2006 Fall MATH 100 Lecture 22
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MATH 100
Ex :
Lecture 22 Introduction to surface integrals
Suppose  is theportionz  1  x 2  y 2 above xy - plane,
let  be orientedby upward normals,


 
 
and let F x, y, z   xi  yj  zk , evaluate F  n dS

Sol :
upward unit normal:
z x  2 x,

 
 zxi  z y j  k
n
z x2  z y2  1
z y  2 y
(continuous next page)
2006 Fall MATH 100 Lecture 22
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MATH 100
Sol :
Lecture 22 Introduction to surface integrals

 

 
  zxi  z y j  k

  z x2  z y2  1dA
F

n
dS

F


R  z 2  z 2  1 
x
y




 
  F   z x i  z y j  k dA

R



  2 x 2  2 y 2  z dA
R


  2 x 2  2 y 2  1  x 2  y 2 dA
R


  x 2  y 2  1 dA
R

2
0

 r
1
0
2

 1 rdrd
3
2
2006 Fall MATH 100 Lecture 22
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MATH 100
Lecture 22 Introduction to surface integrals
T heorem:
If  : z  z x, y , R is theprojectionof  on xy - plane,then
a 


 


 
 F  ndS   F   z x i  z y j  k dA

R
if  is orientedupward
b 


 


 
 F  ndS   F  z x i  z y j  k dA

R
if  is orienteddownward
2006 Fall MATH 100 Lecture 22
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MATH 100
Ex :
Lecture 22 Introduction to surface integrals


 : x  y  z  a oriented outward normals, F  zk , evaluate
2
2
2
2
 
 F  ndS

Sol :
 
 
 
 F  ndS   F  ndS   F  ndS

1
2

on  1 , z  a 2  x 2  y 2 , n upward

 

 
x
y

 F  ndS  R zk   a 2  x 2  y 2 i  a 2  x 2  y 2 j  k dA
1


  zdA
R

2
0

a
0
a 2  r 2 rdrd
2πa3

3
2006 Fall MATH 100 Lecture 22
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MATH 100
Sol :
Lecture 22 Introduction to surface integrals

on  2 , z  a 2  x 2  y 2 , n downward

 
 

y
x

j  k dA
i
F  n dS   zk

2
2
2

 a2  x2  y2
y

x

a
R
2


   zdA
R
 
2
0

a
0
a 2  r 2 rdrd
2πa3

3
2006 Fall MATH 100 Lecture 22
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