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SECTION 13.7
SURFACE INTEGRALS
P213.7
SURFACE INTEGRALS
The relationship between surface integrals and surface area is much the same as the relationship between line integrals and arc length.
P313.7
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.
P413.7
PARAMETRIC SURFACES
Suppose a surface S has a vector equationr(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.
P513.7
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
*
1 1
( )m n
ij iji j
f P S
P613.7
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:
*
max , 01 1
( , , ) lim ( )i j
m n
ij iju vi jS
f x y z dS f P S
P713.7
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)
P813.7
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.
P913.7
SURFACE INTEGRALS
In our discussion of surface area in Section 13.6, we made the approximation
∆Sij ≈ |ru × rv| ∆u ∆v
where:
are the tangent vectors at a corner of Sij.
u v
x y z x y z
u u u v v v
r i j k r i j k
P1013.7
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:
( , , ) ( ( , )) | |u v
S D
f x y z dS f u v dA r r r
P1113.7
SURFACE INTEGRALS
This should be compared with the formula for a line integral:
Observe also that:
( , , ) ( ( )) | '( ) |b
C af x y z ds f t t dt r r
1 | | ( )u v
S D
dS dA A S r r
P1213.7
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)
P1313.7
Example 1
Compute the surface integral ,
where S is the unit sphere x2 + y2 + z2 = 1.
2
S
x dS
P1413.7
Example 1 SOLUTION
As in Example 4 in Section 13.6, we use the parametric representation
x = sin cos , y = sin sin , z = cos 0 ≤ ≤ , 0 ≤ ≤ 2
That is, r(, ) = sin cos i + sin sin j + cos k
P1513.7
Example 1 SOLUTION
As in Example 9 in Section 13.6, we can compute that:
|r × r| = sin
P1613.7
Example 1 SOLUTION
Therefore, by Formula 2,2 2
2 2 2
0 0
(sin cos ) | |
sin cos sin
S D
x dS dA
d d
r r
2 2 3
0 0
2 2120 0
2 31 1 12 2 3 0
cos sin
(1 cos 2 ) (sin sin cos )
4sin 2 cos cos
3
d d
d d
0
P1713.7
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 (x, y, z).
P1813.7
MASS
Then, the total mass of the sheet is:
( , , )S
m x y z dS
P1913.7
CENTER OF MASS
The center of mass is , where
Moments of inertia can also be defined as before. See Exercise 35.
, , x y z
1 1( , , ) ( , , )
1( , , )
S S
S
x x x y z dS y y x y z dSm m
z z x y z dSm
P2013.7
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:
x y
g g
x y
r i k r j k
P2113.7
GRAPHS
Thus,
and
x y
g g
x x
r r i j k
22
| | 1x y
z z
x y
r r
P2213.7
GRAPHS
Therefore, in this case, Formula 2 becomes:
22
( , , )
( , , ( , )) 1
S
D
f x y z dS
z zf x y g x y dA
x y
P2313.7
GRAPHS
Similar formulas apply when it is more convenient to project S onto the yz-plane or xy-plane.
P2413.7
GRAPHS
For instance, if S is a surface with equation y = h(x, z) and D is its projection on the xz-plane, then
2 2
( , , )
( , ( , ), ) 1
S
D
f x y z dS
y yf x h x z z dA
x z
P2513.7
Example 2
Evaluate where S is the surface
z = x + y2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2See Figure 2.
SOLUTION
S
y dS
1 and 2z z
yx y
P2613.7
Example 2 SOLUTION
So, Formula 4 gives:
22
1 2 2
0 0
1 2 2
0 0
22 3/ 21 2
4 30
1
1 1 4
2 1 2
13 22 (1 2 )
3
S D
z zy dS y dA
x y
y y dy dx
dx y y dy
y
P2713.7
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:
1
( , , )
( , , ) ( , , )n
S
S S
f x y z dS
f x y z dS f x y z dS
P2813.7
Example 3
Evaluate , where S is 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.
S
z dS
P2913.7
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.
P3013.7
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 ≤ ≤ 2 0 ≤ z ≤ 1 + x = 1 + cos
P3113.7
Example 3 SOLUTION
Therefore,
and
sin cos 0 cos sin
0 0 1z
i j k
r r i j
2 2| | cos sin 1z r r
P3213.7
Example 3 SOLUTION
Thus, the surface integral over S1 is:
1
2 1 cos
0 0
2 2120
21 12 20
231 102 2 4
| |
(1 cos )
1 2cos (1 cos 2 )
32sin sin 2
2
z
S D
z dS z dA
z dz d
d
d
r r
P3313.7
Example 3 SOLUTION
Since S2 lies in the plane z = 0, we have:
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.
2 2
0 0S S
z dS dS
P3413.7
Example 3 SOLUTION
3
22
2 1
0 0
2 1 2
0 0
21 12 30
2
0
(1 ) 1
(1 cos ) 1 1 0
2 ( cos )
2 cos
sin2 2
2 3
S D
z zz dS x dA
x y
r r dr d
r r dr d
d
P3513.7
Example 3 SOLUTION
Therefore,
1 2 3
32
30 2
2
2
S S S S
z dS z dS z dS z dS
P3613.7
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).
P3713.7
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.
P3813.7
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.
P3913.7
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.
P4013.7
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.
P4113.7
ORIENTED SURFACES
From now on, we consider only orientable (two-sided) surfaces.
P4213.7
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.
P4313.7
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.
P4413.7
POSSIBLE ORIENTATIONS
There are two possible orientations for any orientable surface.
P4513.7
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
As the k-component is positive, this gives the upward orientation of the surface.
22
1
g gx y
g gx y
i j kn
P4613.7
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
The opposite orientation is given by –n.
| |u v
u v
r r
nr r
P4713.7
ORIENTATION
For instance, in Example 4 in Section 13.6, we found the parametric representation
r(, ) = a sin cos i + a sin sin j + a cos k
for the sphere x2 + y2 + z2 = a2
P4813.7
ORIENTATION
Then, in Example 9 in Section 13.6, we found that:
r × r = a2 sin2 cos i + a2 sin2 sin j
+ a2 sin cos kand
|r × r | = a2 sin
P4913.7
ORIENTATION
So, the orientation induced by r(, ) is defined by the unit normal vector
| |
sin cos sin sin cos
1( , )
a
r rn
r r
i j k
r
P5013.7
POSITIVE ORIENTATION
Observe that n points in the same direction as the position vector—that is, outward from the sphere.
See Figure 8.
Fig. 17.7.8, p. 1122
P5113.7
NEGATIVE ORIENTATION
The opposite (inward) orientation would have been obtained (see Figure 9) if we had reversed the order of the parameters because r × r = –r × r
P5213.7
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.
P5313.7
SURFACE INTEGRALS OF VECTOR FIELDS
Suppose that S is an oriented surface with unit normal vector n.
Then, imagine a fluid with density (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.
P5413.7
SURFACE INTEGRALS OF VECTOR FIELDS
Then, the rate of flow (mass per unit time) per unit area is v.
If we divide S into small patches Sij , as in Figure 10 (compare with Figure 1), then Sij is nearly planar.
P5513.7
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
( v · n)A(Sij)
where , v, and n are evaluated at some point on Sij. Recall that the component of the vector v in the
direction of the unit vector n is v · n.
P5613.7
VECTOR FIELDS
Summing these quantities and taking the limit, we get, according to Definition 1, the surface integral of the function v · n over S:
This is interpreted physically as the rate of flow through S.
( , , ) ( , , ) ( , , )
S
S
dS
x y z x y z x y z dS
v n
v n
P5713.7
VECTOR FIELDS
If we write F = v, then F is also a vector field on .
Then, the integral in Equation 7 becomes:
S
dSF n
3
P5813.7
FLUX INTEGRAL
A surface integral of this form occurs frequently in physics—even when F is not v.
It is called the surface integral (or flux integral) of F over S.
P5913.7
Definition 8
S S
d dS F S F n
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
This integral is also called the flux of F across S.
P6013.7
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).
P6113.7
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:
where D is the parameter domain.
( ( , ))
u v
u vS S
u vu v
u vD
d dS
u v dA
r rF S F
r r
r rF r r r
r r
P6213.7
FLUX INTEGRAL
Thus, we have:
( )u v
S D
d dA F S F r r
P6313.7
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
P6413.7
Example 4 SOLUTION
Using the parametric representation
r(, ) = sin cos i + sin sin j + cos k
0 ≤ ≤ 0 ≤ ≤ 2
we have:F(r(, )) = cos i + sin sin j + sin cos k
P6513.7
Example 4 SOLUTION
From Example 9 in Section 13.6,r × r
= sin2 cos i + sin2 sin j + sin cos kTherefore,
F(r(, )) · (r × r )= cos sin2 cos + sin3 sin2 + sin2 cos cos
P6613.7
Example 4 SOLUTION
Then, by Formula 9, the flux is:
2 2 3 2
0 0
( )
(2sin cos cos sin sin )
S
D
d
dA
d d
F S
F r r
P6713.7
Example 4 SOLUTION
This is by the same calculation as in Example 1.
22
0 0
23 2
0 0
23 2
0 0
2 sin cos cos
sin sin
0 sin sin
4
3
d d
d d
d d
P6813.7
FLUX INTEGRALS
Figure 11 shows the vector field F in Example 4 at points on the unit sphere.
P6913.7
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, 4/3, represents: The rate of flow through the unit sphere in units of
mass per unit time.
P7013.7
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:
( ) ( )x y
g gP Q R
x y
F r r i j k i j k
P7113.7
VECTOR FIELDS
Thus, Formula 9 becomes:
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.
S D
g gd P Q R dA
x y
F S
P7213.7
Example 5
Evaluate 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.
S
dF S
P7313.7
Example 5 SOLUTION
S consists of: A parabolic top surface S1.
A circular bottom surface S2. See Figure 12.
P7413.7
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.
P7513.7
Example 5 SOLUTION
On S1,P(x, y, z) = y Q(x, y, z) = xR(x, y, z) = z = 1 – x2 – y2
Also,
2 2g g
x yx y
P7613.7
Example 5 SOLUTION
So, we have:
1
2 2
2 2
[ ( 2 ) ( 2 ) 1 ]
(1 4 )
S
D
D
D
d
g gP Q R dA
x y
y x x y x y dA
xy x y dA
F S
P7713.7
Example 5 SOLUTION
2 1 2 2
0 0
2 1 3 3
0 0
2140
14
(1 4 cos sin )
( 4 cos sin )
( cos sin )
(2 ) 0
2
r r r dr d
r r r dr d
d
P7813.7
Example 5 SOLUTION
The disk S2 is oriented downward.
So, its unit normal vector is n = –k and we have:
since z = 0 on S2.
2 2
( ) ( ) 0 0S S D D
d dS z dA dA F S F k
P7913.7
Example 5 SOLUTION
Finally, we compute, by definition,as the sum of the surface integrals of F over the pieces S1 and S2:
S
dF S
1 2
02 2S S S
d d d
F S F S F S
P8013.7
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.
P8113.7
ELECTRIC FLUX
For instance, if E is an electric field (Example 5 in Section 13.1), the surface integral
is called the electric flux of E through the surface S.
S
dE S
P8213.7
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:
where 0 is a constant (called the permittivity of free space) that depends on the units used. In the SI system, 0 ≈ 8.8542 × 10–12 C2/N · m2
0
S
Q d E S
P8313.7
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 = 40/3
P8413.7
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 ∇uwhere K is an experimentally determined constant called the conductivity of the substance.
P8513.7
HEAT FLOW
Then, the rate of heat flow across the surface S in the body is given by the surface integral
S S
d K u d F S S
P8613.7
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.
P8713.7
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.
P8813.7
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 2 22( )
KCx y z
a F n
P8913.7
Example 6 SOLUTION
However, on S, we have:x2 + y2 + z2 = a2
Thus, F · n = –2aKC
P9013.7
Example 6 SOLUTION
Thus, the rate of heat flow across S is:
2
3
= = 2
= 2 ( )= 2 (4 )
= 8
S S S
d dS aKC dS
aKCA S aKC a
KC a
F S F n