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VECTOR CALCULUS
16.7
Surface Integrals
In this section, we will learn about:
Integration of different types of surfaces.
PARAMETRIC SURFACES
Suppose a surface S has a vector equation
r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k
(u, v) D
PARAMETRIC SURFACES
•We first assume that the parameter
domain D is a rectangle and we divide
it into subrectangles Rij with dimensions
∆u and ∆v.
•Then, the surface S is divided into
corresponding patches Sij.
•We evaluate f at a point Pij* in each
patch, multiply by the area ∆Sij of the
patch, and form the Riemann sum
m
n
 f ( P ) S
i 1 j 1
*
ij
ij
Equation 1
SURFACE INTEGRAL
Then, we take the limit as the number
of patches increases and define the surface
integral of f over the surface S as:

S
f ( x, y, z ) dS  lim
m , n 
m
n
 f ( P ) S
i 1 j 1
*
ij
ij
 Analogues to: The definition of a line integral
(Definition 2 in Section 16.2);The definition of a double
integral (Definition 5 in Section 15.1)
 To evaluate the surface integral in Equation 1, we
approximate the patch area ∆Sij by the area of an
approximating parallelogram in the tangent plane.
SURFACE INTEGRALS
In our discussion of surface area in
Section 16.6, we made the approximation
∆Sij ≈ |ru x rv| ∆u ∆v
where:
x
y
z
ru  i 
j k
u u
u
x y
z
rv  i  j  k
v v
v
are the tangent vectors at a corner of Sij.
Formula 2
SURFACE INTEGRALS
If the components are continuous and ru and
rv are nonzero and nonparallel in the interior
of D, it can be shown from Definition 1—even
when D is not a rectangle—that:
 f ( x, y, z) dS   f (r(u, v)) | r  r | dA
u
S
D
v
SURFACE INTEGRALS
This should be compared with the formula
for a line integral:

C
b
f ( x, y, z ) ds   f (r(t )) | r '(t ) | dt
a
Observe also that:
1
dS

|
r

r
|
dA

A
(
S
)
u
v


S
D
SURFACE INTEGRALS
Example 1
Compute the surface integral
where S is the unit sphere
x2 + y2 + z2 = 1.
,
x
dS

2
S
SURFACE INTEGRALS
Example 1
As in Example 4 in Section 16.6,
we use the parametric representation
x = sin Φ cos θ, y = sin Φ sin θ, z = cos Φ
0 ≤ Φ ≤ π, 0 ≤ θ ≤ 2π
 That is,
r(Φ, θ) = sin Φ cos θ i + sin Φ sin θ j + cos Φ k
 we can compute: |rΦ
x rθ| = sin Φ
Example 1
SURFACE INTEGRALS
Therefore, by Formula 2,
2
x
 dS
S
  (sin  cos  ) 2 | r  r | dA
D

2
0


0
2
(sin 2  cos 2  sin  d d

  cos  d  sin 3  d
2
0

2
0
0
1
2
(1  cos 2 ) d  (sin   sin  cos 2  ) d
 

1
2

4
3

0
1
2

sin 2 0   cos   cos  
0
2
1
3
3
APPLICATIONS
For example, suppose a thin sheet
(say, of aluminum foil) has:
 The shape of a surface S.
 The density (mass per unit area)
at the point (x, y, z) as ρ(x, y, z).
CENTER OF MASS
Then, the total mass of the sheet
is: m    ( x, y, z ) dS
S
The center of mass is:
 x, y , z 
1
x   x  ( x, y, z ) dS
m S
where y  1 y  ( x, y, z ) dS
m 
S
1
z   z  ( x, y, z ) dS
m S
GRAPHS
Any surface S with equation z = g(x, y)
can be regarded as a parametric surface
with parametric equations
x=x
y=y
z = g(x, y)
 So, we have:
 g 
rx  i    k
 x 
 g 
ry  j    k
 y 
Equation 3
GRAPHS
•Thus, r  r   g i  g j  k
x
y
x
x
2
 z   z 
and | rx  ry |       1
 x   y 
2
•Formula 2 becomes:
 f ( x, y, z ) dS
S
 
D
2
 z   z 
f ( x, y, g ( x, y ))       1 dA
 x   y 
2
Example 2
GRAPHS
Evaluate
where
S
is
the
surface
y
dS

S

z
1
x
and
z
 2y
y
z = x + y2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2
Example 2
GRAPHS
So, Formula 4 gives:
2
 z   z 
S y dS  D y 1   x    y  dA
2

1

2
0 0
y 1  1  4 y 2 dy dx
1
2
0
0
  dx 2  y 1  2 y dy
 2
1
4

2
3
2
(1  2 y )
2 3/ 2
13
2
 
0
3
2
GRAPHS
If S is a piecewise-smooth surface—a finite
union of smooth surfaces S1, S2, . . . , Sn that
intersect only along their boundaries—then
the surface integral of f over S is defined by:
 f ( x, y, z) dS
S
  f ( x, y, z ) dS     f ( x, y, z ) dS
S1
Sn
GRAPHS
Evaluate
Example 3
z
dS
 , where S is
S
the surface whose:
 Sides S1 are given by the cylinder x2 + y2 = 1.
 Bottom S2 is the disk x2 + y2 ≤ 1 in the plane z = 0.
 Top S3 is the part of the plane z = 1 + x that
lies above S2.
GRAPHS
Example 3
For S1, we use θ and z as parameters
(Example 5 in Section 16.6) and write its
parametric equations as:
x = cos θ
y = sin θ
z=z
where:
 0 ≤ θ ≤ 2π
 0 ≤ z ≤ 1 + x = 1 + cos θ
Example 3
GRAPHS
Therefore,
i
r  rz   sin 
0
j
cos 
0
k
0  cos  i  sin  j
1
and
| r  rz | cos   sin   1
2
2
Example 3
GRAPHS
Thus, the surface integral over S1 is:
 z dS   z | r  r
z
S1
| dA
D

2

2

0
0
1
2
0

1
2
1 cos

(1  cos  ) d
2
0
z dz d
2
1
1

2
cos



2 (1  cos 2 )  d
3
    2sin   sin 2  
2
1
2
3
2
1
4
2
0
GRAPHS
Example 3
Since S2 lies in the plane z = 0,
we have:
z
dS

S2
  0 dS
S2
0
GRAPHS
Example 3
S3 lies above the unit disk D and is
part of the plane z = 1 + x.
 So, taking
g(x, y) = 1 + x
in Formula 4
and converting to
polar coordinates,
we have the following
result.
Example 3
GRAPHS
2
 z   z 
S z dS D (1  x) 1   x    y  dA
3
2
2
1
0
0



 (1  r cos ) 1  1  0 r dr d

2   (r  r cos  ) dr d

2    cos   d
2
1
0
0
2
0
1
2
2
1
3
2
 sin  
 2 
 2

3 0
2
Example 3
GRAPHS
Therefore,
 z dS   z dS   z dS   z dS
S
S1
S2
3

0 2
2


3
2

 2 
S3
SURFACE INTEGRALS OF VECTOR FIELDS
Suppose that S is an oriented surface with
unit normal vector n.
Then, imagine a fluid with density ρ(x, y, z)
and velocity field v(x, y, z) flowing through S.
 Think of S as an imaginary surface that doesn’t
impede the fluid flow—like a fishing net across
a stream.
Then, the rate of flow (mass per unit time) per
unit area is ρv.
SURFACE INTEGRALS OF VECTOR FIELDS
If we divide S into small patches Sij ,
then Sij is nearly planar.
SURFACE INTEGRALS OF VECTOR FIELDS
So, we can approximate the mass of fluid
crossing Sij in the direction of the normal n
per unit time by the quantity
(ρv · n)A(Sij)
where ρ, v, and n are
evaluated at some point on Sij.
 Recall that the component of the vector ρv
in the direction of the unit vector n is ρv · n.
VECTOR FIELDS
Equation 7
Summing these quantities and taking the limit,
we get, according to Definition 1, the surface
integral of the function ρv · n over S:
  v  n dS
S
   ( x, y, z ) v( x, y, z )  n( x, y, z ) dS
S
 This is interpreted physically as the rate of flow
through S.
VECTOR FIELDS
If we write F = ρv, then F is also a vector
field on R3. Then, the integral in Equation 7
becomes:
F

n
dS

S
A surface integral of this form occurs
frequently in physics—even when F is not ρv.
It is called the surface integral (or flux integral)
of F over S.
Definition 8
FLUX INTEGRAL
If F is a continuous vector field defined
on an oriented surface S with unit normal
vector n, then the surface integral of F over S
is:
F

d
S

F

n
dS


S
S
 This integral is also called
the flux of F across S.
FLUX INTEGRAL
If S is given by a vector function r(u, v),
then n is given by Equation 6.
 Then, from Definition 8 and Equation 2,
we have (D is the parameters’ domain):
ru  rv
S F  dS  S F  ru  rv dS

ru  rv 
  F(r (u, v)) 
 ru  rv dA
ru  rv 
D 
 So,
F

d
S

F

(
r

r
)
dA
u
v


S
D
FLUX INTEGRALS
Example 4
•Find the flux of the vector field
F(x, y, z) = z i + y j + x k
across the unit sphere :x2 + y2 + z2 = 1
•Using the parametric representation:
r(Φ, θ) = sin Φ cos θ i + sin Φ sin θ j + cos Φ k
0≤Φ≤π
0 ≤ θ ≤ 2π
F(r(Φ, θ)) = cos Φ i + sin Φ sin θ j + sin Φ cos θ k
FLUX INTEGRALS
Example 4
From Example 10 in Section 16.6,
rΦ x rθ = sin2 Φ cos θ i + sin2 Φ sin θ j + sin Φ cos Φ k
Therefore, F(r(Φ, θ)) · (rΦ x rθ) = cos Φ sin2 Φ
cos θ + sin3 Φ sin2 θ + sin2 Φ cos Φ cos θ
Then, by Formula 9, the flux is:
 F  dS
S
  F  (r  r ) dA
D

2
0


0
(2sin 2  cos  cos   sin 3  sin 2  ) d d
Example 4
FLUX INTEGRALS

2
 2 sin  cos  d  cos  d
2
0
0

2
  sin  d  sin  d
3
0

2
2
0
 0   sin  d  sin  d
0
3
2
0
4

3
 This is by the same calculation as in Example 1.
FLUX INTEGRALS
The figure shows the vector field F in
Example 4 at points on the unit sphere.
VECTOR FIELDS
If, for instance, the vector field in Example 4
is a velocity field describing the flow of a fluid
with density 1, then the answer, 4π/3,
represents:
 The rate of flow through the unit sphere
in units of mass per unit time.
VECTOR FIELDS
In the case of a surface S given by a graph
z = g(x, y), we can think of x and y as
parameters and use Equation 3 to write:
 g

g
F  (rx  ry )  ( P i  Q j  R k )    i 
j k 
y
 x

VECTOR FIELDS
Formula 10
Thus, Formula 9 becomes:


g
g
S F  dS  D   P x  Q y  R  dA
 This formula assumes the upward orientation of S.
 For a downward orientation, we multiply by –1.
VECTOR FIELDS
Evaluate
where:
Example 5
F

d
S

S
 F(x, y, z) = y i + x j + z k
 S is the boundary of the solid region E
enclosed by the paraboloid z = 1 – x2 – y2
and the plane z = 0.
VECTOR FIELDS
S consists of:
 A parabolic top surface S1.
 A circular bottom surface S2.
Example 5
VECTOR FIELDS
Example 5
Since S is a closed surface, we use the
convention of positive (outward) orientation.
 This means that S1 is oriented upward.
 So, we can use Equation 10 with D being
the projection of S1 on the xy-plane, namely,
the disk x2 + y2 ≤ 1.
Example 5
VECTOR FIELDS
On S1,
P(x, y, z) = y
Q(x, y, z) = x
R(x, y, z) = z = 1 – x2 – y2
Also,
g
  2x
x
g
 2 y
y
VECTOR FIELDS
Example 5
So, we have:
 F  dS
S1


g
g
    P  Q
 R  dA
x
y

D 
  [ y (2 x)  x(2 y )  1  x 2  y 2 ] dA
D
  (1  4 xy  x 2  y 2 ) dA
D
Example 5
VECTOR FIELDS

2

2
0
0
2
1
 (1  4r
0
1
 (r  r
0
3
2
cos  sin   r ) r dr d
2
 4r cos  sin  ) dr d
3
  ( cos  sin  ) d
0
1
4
 14 (2 )  0


2
Example 5
VECTOR FIELDS
The disk S2 is oriented downward.
So, its unit normal vector is n = –k
and we have:
 F  dS   F  (k ) dS   ( z) dA
S2
S2
D
  0 dA  0
D
since z = 0 on S2.
Example 5
VECTOR FIELDS
Finally, we compute, by definition,
 F  dS
as the sum of the surface integrals S
of F over the pieces S1 and S2:
 F  dS   F  dS   F  dS
S
S1
S2


2
0

2
APPLICATIONS
Although we motivated the surface integral
of a vector field using the example of fluid
flow, this concept also arises in other physical
situations.
ELECTRIC FLUX
For instance, if E is an electric field
(Example 5 in Section 16.1), the surface
integral
E

d
S

S
is called the electric flux of E through
the surface S.
GAUSS’S LAW
Equation 11
One of the important laws of electrostatics is
Gauss’s Law, which says that the net charge
enclosed by a closed surface S is:
Q   0  E  dS
S
where ε0 is a constant (called the permittivity
of free space) that depends on the units used.
 In the SI system, ε0 ≈ 8.8542 x 10–12 C2/N · m2
GAUSS’S LAW
Thus, if the vector field F in Example 4
represents an electric field, we can conclude
that the charge enclosed by S is:
Q = 4πε0/3
HEAT FLOW
Another application occurs in
the study of heat flow.
 Suppose the temperature at a point (x, y, z)
in a body is u(x, y, z).
HEAT FLOW
•Then, the heat flow is defined as
the vector field F = –K ∇u
where K is an experimentally determined
constant called the conductivity of the
substance.
•Then, the rate of heat flow across
the surface S in the body is given by
the surface integral
 F  dS   K  u  dS
S
S