Transcript surface integral

SECTION 13.7
SURFACE INTEGRALS
SURFACE INTEGRALS
The relationship between surface integrals and
surface area is much the same as the relationship
between line integrals and arc length.
13.7 P2
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.
We start with parametric surfaces.
Then, we deal with the special case where S is
the graph of a function of two variables.
13.7 P3
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
We first assume that the parameter domain D is
a rectangle and we divide it into subrectangles
Rij with dimensions ∆u and ∆v.
13.7 P4
PARAMETRIC SURFACES
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
 ij ) Sij
i 1 j 1
13.7 P5
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
max ui , v j 0
n
 f ( P ) S
i 1 j 1
*
ij
ij
13.7 P6
SURFACE INTEGRALS
Notice the analogy with:


The definition of a line integral (Definition 2 in
Section 13.2)
The definition of a double integral (Definition 5 in
Section 12.1)
13.7 P7
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.
13.7 P8
SURFACE INTEGRALS
In our discussion of surface area in Section 13.6,
we made the approximation
∆Sij ≈ |ru × rv| ∆u ∆v
where:
x
y
z
x y
z
ru  i  j  k rv  i  j  k
u u
u
v v v
are the tangent vectors at a corner of Sij.
13.7 P9
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
v
D
13.7 P10
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:
1dS   | r  r
u
S
v
| dA  A( S )
D
13.7 P11
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)
13.7 P12
Example 1
Compute the surface integral
 x
2
dS ,
S
where S is the unit sphere x2 + y2 + z2 = 1.
13.7 P13
Example 1 SOLUTION
As in Example 4 in Section 13.6, we use the
parametric representation
x = sin f cos , y = sin f sin , z = cos f
0 ≤ f ≤ p, 0 ≤  ≤ 2 p

That is,
r(f, ) = sin f cos  i + sin f sin  j + cos f k
13.7 P14
Example 1 SOLUTION
As in Example 9 in Section 13.6, we can
compute that:
|rf × r| = sin f
13.7 P15
Example 1 SOLUTION
Therefore, by Formula 2,
2
2
x
dS

(sin
f
cos

)
| rf  r | dA


S
D

2p
0

p
0
2p
sin 2 f cos 2  sin f df d
p
  cos  d  sin f df
2
0

2p
0

1
2
3
0
1
2
p
(1  cos 2 ) d  (sin f  sin f cos f ) df
 
2
0
1
2
sin 2 0
2p
4p
  cos f  cos f  
0
3
1
3
3
p
13.7 P16
APPLICATIONS
Surface integrals have applications similar to
those for the integrals we have previously
considered.
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 r(x, y, z).
13.7 P17
MASS
Then, the total mass of the sheet is:
m   r ( x, y, z ) dS
S
13.7 P18
CENTER OF MASS
The center of mass is  x , y , z , where
1
x   x r ( x, y, z ) dS
m S
1
y   y r ( x, y, z ) dS
m S
1
z   z r ( x, y, z ) dS
m S
Moments of inertia can also be defined as
before.

See Exercise 35.
13.7 P19
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 
13.7 P20
GRAPHS
Thus,
g g
rx  ry   i 
jk
x
x
and
2
 z   z 
| rx  ry |       1
 x   y 
2
13.7 P21
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
13.7 P22
GRAPHS
Similar formulas apply when it is more
convenient to project S onto the yz-plane or xyplane.
13.7 P23
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
13.7 P24
Example 2
Evaluate
 y dS where S is the surface
S
z = x + y2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2
See Figure 2.
SOLUTION
z
z
 1 and
 2y
x
y
13.7 P25
Example 2 SOLUTION
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
13.7 P26
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 
S1
  f ( x, y, z ) dS
Sn
13.7 P27
Example 3
Evaluate
 z dS , where S is the surface whose:
S



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.
13.7 P28
Example 3 SOLUTION
The surface S is shown in Figure 3.

We have changed the usual position of the axes to
get a better look at S.
13.7 P29
Example 3 SOLUTION
For S1, we use  and z as parameters (Example 5
in Section 13.6) and write its parametric
equations as:
x = cos 
y = sin 
z=z
where:


0 ≤  ≤ 2p
0 ≤ z ≤ 1 + x = 1 + cos 
13.7 P30
Example 3 SOLUTION
Therefore,
i
r  rz   sin 
0
j
cos 
0
k
0  cos  i  sin  j
1
and
| r  rz | cos 2   sin 2   1
13.7 P31
Example 3 SOLUTION
Thus, the surface integral over S1 is:
 z dS   z | r  r
z
S1
| dA
D

2p

2p
0
0

1
2

1 cos
0
1
2
z dz d
(1  cos  ) 2 d
2p
1
1

2
cos



2 (1  cos 2 )  d
0
3p
    2sin   sin 2  
2
1
2
3
2
1
4
2p
0
13.7 P32
Example 3 SOLUTION
Since S2 lies in the plane z = 0, we have:
 z dS   0 dS  0
S2
S2
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.
13.7 P33
Example 3 SOLUTION
2
 z   z 
S z dS D (1  x) 1   x    y  dA
3
2

2p
0
1
 (1  r cos  )
0
 2
2p
 2
2p
0
0
1
 (r  r
0
2
1  1  0 r dr d
cos  ) dr d
 12  13 cos   d
2p
 sin  
 2 
 2p

3 0
2
13.7 P34
Example 3 SOLUTION
Therefore,
 z dS   z dS   z dS   z dS
S
S1
S2
S3
3p

0 2p
2


3
2

 2 p
13.7 P35
ORIENTED SURFACES
To define surface integrals of vector fields, we
need to rule out nonorientable surfaces such as
the Möbius strip shown in Figure 4.

It is named after the German geometer August
Möbius (1790–1868).
13.7 P36
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.
13.7 P37
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.
13.7 P38
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.
13.7 P39
MOBIUS STRIP
Therefore, a Möbius strip really has only one
side.

You can graph the Möbius strip using the parametric
equations in Exercise 28 in Section 13.6.
13.7 P40
ORIENTED SURFACES
From now on, we consider only orientable (twosided) surfaces.
13.7 P41
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).
See Figure 6.
13.7 P42
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.
13.7 P43
POSSIBLE ORIENTATIONS
There are two possible orientations for any
orientable surface.
13.7 P44
UPWARD ORIENTATION
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 the unit normal
vector
g
g

n
x
i
y
 g   g 
1     
 x   y 
2

jk
2
As the k-component is positive, this gives the
upward orientation of the surface.
13.7 P45
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.
13.7 P46
ORIENTATION
For instance, in Example 4 in Section 13.6, we
found the parametric representation
r(f, )
= a sin f cos  i + a sin f sin  j + a cos f k
for the sphere x2 + y2 + z2 = a2
13.7 P47
ORIENTATION
Then, in Example 9 in Section 13.6, we found
that:
rf × r = a2 sin2 f cos  i
+ a2 sin2 f sin  j
+ a2 sin f cos f k
and
|rf × r | = a2 sin f
13.7 P48
ORIENTATION
So, the orientation induced by r(f,  ) is defined
by the unit normal vector
n
rf  r
| rf  r |
 sin f cos  i  sin f sin  j  cos f k
1
 r (f ,  )
a
13.7 P49
POSITIVE ORIENTATION
Observe that n points in the same direction as
the position vector—that is, outward from the
sphere.
See Figure 8.
13.7 P50
Fig. 17.7.8, p. 1122
NEGATIVE ORIENTATION
The opposite (inward) orientation would have
been obtained (see Figure 9) if we had reversed
the order of the parameters because
r × rf = –rf × r
13.7 P51
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.
13.7 P52
SURFACE INTEGRALS OF VECTOR
FIELDS
Suppose that S is an oriented surface with unit
normal vector n.
Then, imagine a fluid with density r(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.
13.7 P53
SURFACE INTEGRALS OF VECTOR
FIELDS
Then, the rate of flow (mass per unit time) per
unit area is r v.
If we divide S into small patches Sij , as in
Figure 10 (compare with Figure 1), then Sij is
nearly planar.
13.7 P54
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
(r v · n)A(Sij)
where r, v, and n are evaluated at some point on
Sij.

Recall that the component of the vector r v in the
direction of the unit vector n is r v · n.
13.7 P55
VECTOR FIELDS
Summing these quantities and taking the limit,
we get, according to Definition 1, the surface
integral of the function r v · n over S:
 r v  n dS
S
  r ( x, y, z ) v( x, y, z )  n( x, y, z) dS
S

This is interpreted physically as the rate of flow
through S.
13.7 P56
VECTOR FIELDS
If we write F = r v, then F is also a vector field
on 3.
Then, the integral in Equation 7 becomes:
 F  n dS
S
13.7 P57
FLUX INTEGRAL
A surface integral of this form occurs frequently
in physics—even when F is not r v.
It is called the surface integral (or flux integral)
of F over S.
13.7 P58
Definition 8
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.
13.7 P59
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).
13.7 P60
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.
13.7 P61
FLUX INTEGRAL
Thus, we have:
 F  dS   F  (r  r ) dA
u
S
v
D
13.7 P62
Example 4
Find the flux of the vector field
F(x, y, z) = z i + y j + x k
across the unit sphere
x 2 + y 2 + z2 = 1
13.7 P63
Example 4 SOLUTION
Using the parametric representation
r(f, )
= sin f cos  i + sin f sin  j + cos f k
0≤f≤p
0 ≤  ≤ 2p
we have:
F(r(f, ))
= cos f i + sin f sin  j + sin f cos  k
13.7 P64
Example 4 SOLUTION
From Example 9 in Section 13.6,
r f × r
= sin2 f cos  i + sin2 f sin  j + sin f cos f k
Therefore,
F(r(f, )) · (rf × r )
= cos f sin2 f cos 
+ sin3 f sin2 
+ sin2 f cos f cos 
13.7 P65
Example 4 SOLUTION
Then, by Formula 9, the flux is:
 F  dS
S
  F  (rf  r ) dA
D

2p
0

p
0
(2sin 2 f cos f cos   sin 3 f sin 2  ) df d
13.7 P66
Example 4 SOLUTION
p
2p
 2  sin f cos f df  cos  d
2
0
0
p
2p
  sin f df  sin 2  d
3
0
p
0
2p
 0   sin f df  sin 2  d
0
3
0
4p

3

This is by the same calculation as in Example 1.
13.7 P67
FLUX INTEGRALS
Figure 11 shows the vector field F in Example 4
at points on the unit sphere.
13.7 P68
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, 4p/3, represents:

The rate of flow through the unit sphere in units of
mass per unit time.
13.7 P69
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

13.7 P70
VECTOR FIELDS
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.
Similar formulas can be worked out if S is given
by y = h(x, z) or x = k(y, z).

See Exercises 31 and 32.
13.7 P71
Example 5
Evaluate
where:


 F  dS
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.
13.7 P72
Example 5 SOLUTION
S consists of:



A parabolic top surface S1.
A circular bottom surface S2.
See Figure 12.
13.7 P73
Example 5 SOLUTION
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.
13.7 P74
Example 5 SOLUTION
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
13.7 P75
Example 5 SOLUTION
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  y ) dA
2
2
D
13.7 P76
Example 5 SOLUTION
2p
1
0
0
2
1
0
0


p
   (r  r  4r cos  sin  ) dr d
p
  (  cos  sin  ) d
2
0
(1  4r 2 cos  sin   r 2 ) r dr d
3
3
1
4
 14 (2p )  0

p
2
13.7 P77
Example 5 SOLUTION
The disk S2 is oriented downward.
So, its unit normal vector is n = –k and we have:
 F  dS   F  (k ) dS   ( z) dA   0 dA  0
S2
S2
D
D
since z = 0 on S2.
13.7 P78
Example 5 SOLUTION
Finally, we compute, by definition,  F  dS
as the sum of the surface integrals S
of F over the pieces S1 and S2:
p
p
 F  dS   F  dS   F  dS  2  0  2
S
S1
S2
13.7 P79
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.
13.7 P80
ELECTRIC FLUX
For instance, if E is an electric field (Example 5
in Section 13.1), the surface integral
 E  dS
S
is called the electric flux of E through the
surface S.
13.7 P81
GAUSS’S LAW
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  e 0  E  dS
S
where e0 is a constant (called the permittivity of
free space) that depends on the units used.

In the SI system, e0 ≈ 8.8542 × 10–12 C2/N · m2
13.7 P82
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 = 4pe0/3
13.7 P83
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).
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.
13.7 P84
HEAT FLOW
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
13.7 P85
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.
13.7 P86
Example 6 SOLUTION
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.
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.
13.7 P87
Example 6 SOLUTION
Instead of using the usual parametrization of the
sphere as in Example 4, we observe that the
outward unit normal to the spherex2 + 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
13.7 P88
Example 6 SOLUTION
However, on S, we have:
x 2 + y 2 + z2 = a 2
Thus,
F · n = –2aKC
13.7 P89
Example 6 SOLUTION
Thus, the rate of heat flow across S is:
 F  dS=  F  n dS =  2aKC  dS
S
S
S
=  2aKCA( S )=  2aKC (4p a 2 )
=  8KCp a
3
13.7 P90