16
VECTOR CALCULUS
VECTOR CALCULUS
16.7
Surface Integrals
In this section, we will learn about:
Integration of different types of surfaces.
SURFACE INTEGRALS
The relationship between surface integrals
and surface area is much the same as
the relationship between line integrals and
arc length.
SURFACE INTEGRALS
Suppose f is a function of three variables
whose domain includes a surface S.
 We will define the surface integral of f over S
such that, in the case where f(x, y, z) = 1,
the value of the surface integral is equal to
the surface area of S.
SURFACE INTEGRALS
We start with parametric surfaces.
Then, we deal with the special case
where S is the graph of a function of two
variables.
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.
PARAMETRIC SURFACES
Then, the surface S
is divided into
corresponding
patches Sij.
PARAMETRIC SURFACES
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:
 f ( x, y, z) dS 
S
m
lim
m, n
n
 f (P ) S
i 1 j 1
*
ij
ij
SURFACE INTEGRALS
Notice the analogy with:
 The definition of a line integral
(Definition 2 in Section 16.2)
 The definition of a double integral
(Definition 5 in Section 15.1)
SURFACE INTEGRALS
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
u
S
D
 rv | dA
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
u
S
D
 rv | dA  A( S )
SURFACE INTEGRALS
Formula 2 allows us to compute a surface
integral by converting it into a double integral
over the parameter domain D.
 When using this formula, remember that
f(r(u, v) is evaluated by writing
x = x(u, v), y = y(u, v), z = z(u, v)
in the formula for f(x, y, z)
SURFACE INTEGRALS
Example 1
Compute the surface integral  x dS ,
S
where S is the unit sphere
2
x2 + y2 + z2 = 1.
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
SURFACE INTEGRALS
Example 1
As in Example 10 in Section 16.6,
we can compute that:
|rΦ x rθ| = sin Φ
Example 1
SURFACE INTEGRALS
Therefore, by Formula 2,
 x dS
2
S
  (sin  cos  ) | r  r | dA
2
D

2
0


0
(sin  cos  sin  d d
2
2
Example 1
SURFACE INTEGRALS
2

  cos  d  sin  d
2
0

2
0

1
2
0
1
2

(1  cos 2 ) d  (sin   sin  cos  ) d
 
4

3
3
2
0
1
2

sin 2 0   cos   cos  
0
2
1
3
3
APPLICATIONS
Surface integrals have applications
similar to those for the integrals we have
previously considered.
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).
MASS
Then, the total mass of the sheet
is:
m    ( x, y , z ) dS
S
CENTER OF MASS
The center of mass is:
 x, y , z 
where
1
x   x  ( x, y , z ) dS
m S
1
y   y  ( x, y , z ) dS
m S
1
z   z  ( x, y , z ) dS
m S
MOMENTS OF INERTIA
Moments of inertia can also be
defined as before.
 See Exercise 39.
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,
g
g
rx  ry   i 
jk
x
x
and
2
 z   z 
| rx  ry |       1
 x   y 
2
Formula 4
GRAPHS
Therefore, in this case, Formula 2
becomes:

f ( x, y, z ) dS
S
 
D
2
 z   z 
f ( x, y, g ( x, y ))       1 dA
 x   y 
2
GRAPHS
Similar formulas apply when it is
more convenient to project S onto
the yz-plane or xy-plane.
GRAPHS
For instance, if S is a surface with
equation y = h(x, z) and D is its projection
on the xz-plane, then
 f ( x, y, z) dS
S
 
D
 y   y 
f ( x, h( x, z ), z )       1 dA
 x   z 
2
2
Example 2
GRAPHS
Evaluate  y dS where S is the surface
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 2 dy
 2
1
4

2
3
(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
The surface S is shown.
 We have changed
the usual position
of the axes to get
a better look at S.
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
1
0
0
 2
2
0

1
2
2
 13 cos   d
2
 sin  
 2 

2


2
3

0
Example 3
GRAPHS
Therefore,
 z dS   z dS   z dS   z dS
S2
S1
S
3
0 2

2


3
2

 2 
S3
ORIENTED SURFACES
To define surface integrals of vector fields,
we need to rule out nonorientable surfaces
such as the Möbius strip shown.
 It is named
after the
German
geometer
August
Möbius
(1790–
1868).
MOBIUS STRIP
You can construct one for yourself by:
1. Taking a long rectangular strip of paper.
2. Giving it a half-twist.
3. Taping the short edges together.
MOBIUS STRIP
If an ant were to crawl along the Möbius strip
starting at a point P, it would end up on
the “other side” of the strip—that is, with its
upper side pointing in the opposite direction.
MOBIUS STRIP
Then, if it continued to crawl in the same
direction, it would end up back at the same
point P without ever having crossed an edge.
 If you have
constructed
a Möbius
strip, try
drawing
a pencil line
down the
middle.
MOBIUS STRIP
Therefore, a Möbius strip really has only
one side.
 You can graph the Möbius strip using the parametric
equations in Exercise 32 in Section 16.6.
ORIENTED SURFACES
From now on, we consider
only orientable (two-sided)
surfaces.
ORIENTED SURFACES
We start with a surface S that has a tangent
plane at every point (x, y, z) on S (except
at any boundary point).
 There are two unit
normal vectors
n1 and n2 = –n1
at (x, y, z).
ORIENTED SURFACE & ORIENTATION
If it is possible to choose a unit normal
vector n at every such point (x, y, z) so that n
varies continuously over S, then
 S is called an oriented surface.
 The given choice of n provides S
with an orientation.
POSSIBLE ORIENTATIONS
There are
two possible
orientations for
any orientable
surface.
UPWARD ORIENTATION
Equation 5
For a surface z = g(x, y) given as the graph
of g, we use Equation 3 to associate with
the surface a natural orientation given by
g
g
the unit normal vector
 i
jk
x
y
n
2
2
 g   g 
1     
 x   y 
 As the k-component is positive,
this gives the upward orientation
of the surface.
Equation 6
ORIENTATION
If S is a smooth orientable surface given
in parametric form by a vector function r(u, v),
then it is automatically supplied with
the orientation of the unit normal vector
ru  rv
n
| ru  rv |
 The opposite orientation is given by –n.
ORIENTATION
For instance, in Example 4 in Section 16.6,
we found the parametric representation
r(Φ, θ)
= a sin Φ cos θ i + a sin Φ sin θ j
+ a cos Φ k
for the sphere x2 + y2 + z2 = a2
ORIENTATION
Then, in Example 10 in Section 16.6,
we found that:
rΦ x rθ = a2 sin2 Φ cos θ i
+ a2 sin2 Φ sin θ j
+ a2 sin Φ cos Φ k
and
|rΦ x rθ| = a2 sin Φ
ORIENTATION
So, the orientation induced by r(Φ, θ) is
defined by the unit normal vector
n
r  r
| r  r |
 sin  cos  i  sin  sin  j  cos  k
1
 r ( , )
a
POSITIVE ORIENTATION
Observe that n points in the same
direction as the position vector—that is,
outward from the sphere.
NEGATIVE ORIENTATION
The opposite (inward) orientation would
have been obtained if we had reversed
the order of the parameters
because
rθ x rΦ = –rΦ x rθ
CLOSED SURFACES
For a closed surface—a surface that is
the boundary of a solid region E—the
convention is that:
 The positive orientation is the one for which
the normal vectors point outward from E.
 Inward-pointing normals give the negative
orientation.
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.
SURFACE INTEGRALS OF VECTOR FIELDS
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 ° 3.
Then, the integral in Equation 7
becomes:
F

n
dS

S
FLUX INTEGRAL
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  dS   F  n dS
S
S
 This integral is also called
the flux of F across S.
FLUX INTEGRAL
In words, Definition 8 says that:
 The surface integral of a vector field over S
is equal to the surface integral of its normal
component over S (as previously defined).
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:
ru  rv
S F  dS  S F  ru  rv dS

ru  rv 
  F(r (u, v)) 
 ru  rv dA
ru  rv 
D 
where D is the parameter domain.
Formula 9
FLUX INTEGRAL
Thus, we have:
F

d
S

F

(
r

r
)
dA
u
v


S
D
Example 4
FLUX INTEGRALS
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
FLUX INTEGRALS
Example 4
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
we have:
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,
r Φ x rθ
= sin2 Φ cos θ i + sin2 Φ sin θ j
+ sin Φ cos Φ k
FLUX INTEGRALS
Example 4
Therefore,
F(r(Φ, θ)) · (rΦ x rθ)
= cos Φ sin2 Φ cos θ
+ sin3 Φ sin2 θ
+ sin2 Φ cos Φ cos θ
FLUX INTEGRALS
Example 4
Then, by Formula 9, the flux is:
 F  dS
S
  F  (r  r ) dA
D

2
0


0
(2sin  cos  cos   sin  sin  ) d d
2
3
2
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
F

d
S


P

Q

R
dA


S
D  x y 
 This formula assumes the upward orientation of S.
 For a downward orientation, we multiply by –1.
VECTOR FIELDS
Similar formulas can be worked out if S
is given by y = h(x, z) or x = k(y, z).
 See Exercises 35 and 36.
VECTOR FIELDS
Evaluate
Example 5
F

d
S

S
where:
 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
Example 5
VECTOR FIELDS
So, we have:
 F  dS
S1


g
g
    P  Q
 R  dA
x
y

D 
  [ y (2 x)  x(2 y )  1  x  y ] dA
2
D
  (1  4 xy  x  y ) dA
2
D
2
2
Example 5
VECTOR FIELDS
2
1
0
0
2
1
0
0

 (1  4r cos  sin   r ) r dr d

   (r  r  4r cos  sin  ) dr d

  ( cos  sin  ) d
2
0
1
4
 14 (2 )  0


2
2
3
2
3
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.
HEAT FLOW
Then, the rate of heat flow across
the surface S in the body is given by
the surface integral
F

d
S


K

u

d
S


S
S
HEAT FLOW
Example 6
The temperature u in a metal ball is
proportional to the square of the distance
from the center of the ball.
 Find the rate of heat flow across a sphere S
of radius a with center at the center of the ball.
Example 6
HEAT FLOW
Taking the center of the ball to be at
the origin, we have:
u(x, y, z) = C(x2 + y2 + z2)
where C is the proportionality constant.
Example 6
HEAT FLOW
Then, the heat flow is:
F(x, y, z) = –K ∇u
= –KC(2x i + 2y j + 2z k)
where K is the conductivity of the metal.
HEAT FLOW
Example 6
Instead of using the usual parametrization
of the sphere as in Example 4, we observe
that the outward unit normal to the sphere
x2 + y2 + z2 = a2 at the point (x, y, z) is:
n = 1/a (x i + y j + z k)
Thus,
2 KC 2
2
2
F n  
(x  y  z )
a
Example 6
HEAT FLOW
However, on S, we have:
x2 + y2 + z2 = a2
Thus,
F · n = –2aKC
Example 6
HEAT FLOW
Thus, the rate of heat flow across S
is:
 F  dS   F  n dS  2aKC  dS
S
S
S
 2aKCA( S )
 2aKC (4 a )
2
 8KC a
3