∂u × ∂r
∂v
z
Recall that normal vectors to a plane can point in two opposite
directions. By an outward unit normal vector to a surface Σ,
we will mean the unit vector that is normal to Σ and points away
from the “top” (or “outer” part) of the surface. This is a hazy
y
definition, but the picture in Figure 4.4.4 gives a better idea of
0
what outward normal vectors look like, in the case of a sphere.
With this idea in mind, we make the following definition of a
x
surface integral of a 3-dimensional vector field over a surface:
Figure 4.4.4
Definition 4.4. Let Σ be a surface in R3 and let f( x, y, z) = f ( x, y, z)i
( x, y, z)j
( x, y, z)k
1
+ f 2
+ f 3
be a vector field defined on some subset of R3 that contains Σ. The surface integral of f
over Σ is
f · dσ =
f · n dσ ,
(4.30)
Σ
Σ
where, at any point on Σ, n is the outward unit normal vector to Σ.
Note in the above definition that the dot product inside the integral on the right is a
real-valued function, and hence we can use Definition 4.3 to evaluate the integral.
Example 4.10. Evaluate the surface integral
f· dσ, where f( x, y, z) = yzi+ xzj+ xyk and Σ
Σ
is the part of the plane x + y+ z = 1 with x ≥ 0, y ≥ 0, and z ≥ 0, with the outward unit normal
n pointing in the positive z direction (see Figure 4.4.5).
z
Solution: Since the vector v = (1,1,1) is normal to the plane x + y + z = 1
1
(why?), then dividing v by its length yields the outward unit normal
n
vector n = 1 , 1 , 1 . We now need to parametrize Σ. As we can see
3
3
3
Σ
y
from Figure 4.4.5, projecting Σ onto the xy-plane yields a triangular
0
region R
1
= {( x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x }. Thus, using ( u, v) instead of
1
x + y + z = 1
( x, y), we see that
x
Figure 4.4.5
x = u, y = v, z = 1 − ( u + v), for 0 ≤ u ≤ 1,0 ≤ v ≤ 1 − u
4.4 Surface Integrals and the Divergence Theorem
161
is a parametrization of Σ over R (since z = 1 − ( x + y) on Σ). So on Σ,
1
1
1
1
f · n = ( yz, xz, xy) ·
,
,
=
( yz + xz + xy)
3
3
3
3
1
1
=
(( x + y) z + xy) =
(( u + v)(1 − ( u + v)) + uv)
3
3
1
=
(( u + v) − ( u + v)2 + uv)
3
for ( u, v) in R, and for r( u, v) = x( u, v)i + y( u, v)j + z( u, v)k = ui + vj + (1 − ( u + v))k we have
∂r
∂r
∂r
∂r
×
= (1,0,−1) × (0,1,−1) = (1,1,1)
⇒
×
= 3 .
∂u
∂v
∂u
∂v
Thus, integrating over R using vertical slices (e.g. as indicated by the dashed line in Figure
4.4.5) gives
f · dσ =
f · n dσ
Σ
Σ
∂r
∂r
=
(f( x( u, v), y( u, v), z( u, v)) · n)
×
dv du
∂u
∂v
R
1
1− u 1
=
(( u + v) − ( u + v)2 + uv) 3 dv du
0
0
3
1
( u + v)2
( u + v)3
uv 2 v=1− u
=
−
+
du
0
2
3
2
v=0
1 1
u
3 u 2
5 u 3
=
+
−
+
du
0
6
2
2
6
u
u 2
u 3
5 u 4 1
1
=
+
−
+
=
.
6
4
2
24
8
0
Computing surface integrals can often be tedious, especially when the formula for the
outward unit normal vector at each point of Σ changes. The following theorem provides an
easier way in the case when Σ is a closed surface, that is, when Σ encloses a bounded
solid in R3. For example, spheres, cubes, and ellipsoids are closed surfaces, but planes and
paraboloids are not.
162
CHAPTER 4. LINE AND SURFACE INTEGRALS
Theorem 4.8. (Divergence Theorem) Let Σ be a closed surface in R3 which bounds a
solid S, and let f( x, y, z) = f ( x, y, z)i
( x, y, z)j
( x, y, z)k be a vector field defined on some
1
+ f 2
+ f 3
subset of R3 that contains Σ. Then
f · dσ =
div f dV ,
(4.31)
Σ
S
where
∂ f
∂ f
∂ f
div f =
1 + 2 + 3
(4.32)
∂x
∂y
∂z
is called the divergence of f.
The proof of the Divergence Theorem is very similar to the proof of Green’s Theorem, i.e. it
is first proved for the simple case when the solid S is bounded above by one surface, bounded
below by another surface, and bounded laterally by one or more surfaces. The proof can then
be extended to more general solids.3
Example 4.11. Evaluate
f · dσ, where f( x, y, z) = xi + yj + zk and Σ is the unit sphere Σ
x 2 + y 2 + z 2 = 1.
Solution: We see that div f = 1 + 1 + 1 = 3, so
f · dσ =
div f dV =
3 dV
Σ
S
S
4 π(1)3
= 3
1 dV = 3vol( S) = 3 ·
= 4 π .
3
S
In physical applications, the surface integral
f · dσ is often referred to as the flux of f
Σ
through the surface Σ. For example, if f represents the velocity field of a fluid, then the flux
is the net quantity of fluid to flow through the surface Σ per unit time. A positive flux means
there is a net flow out of the surface (i.e. in the direction of the outward unit normal vector
n), while a negative flux indicates a net flow inward (in the direction of −n).
The term divergence comes from interpreting div f as a measure of how much a vector
field “diverges” from a point. This is best seen by using another definition of div f which is
equivalent4 to the definition given by formula (4.32). Namely, for a point ( x, y, z) in R3, 1
div f( x, y, z) = lim
f · dσ ,
(4.33)
V →0 V Σ
3See TAYLOR and MANN, § 15.6 for the details.
4See SCHEY, p. 36-39, for an intuitive discussion of this.
4.4 Surface Integrals and the Divergence Theorem
163
where V is the volume enclosed by a closed surface Σ around the point ( x, y, z). In the
limit, V → 0 means that we take smaller and smaller closed surfaces around ( x, y, z), which
means that the volumes they enclose are going to zero. It can be shown that this limit is
independent of the shapes of those surfaces. Notice that the limit being taken is of the
ratio of the flux through a surface to the volume enclosed by that surface, which gives a
rough measure of the flow “leaving” a point, as we mentioned. Vector fields which have zero
divergence are often called solenoidal fields.
The following theorem is a simple consequence of formula (4.33).
Theorem 4.9. If the flux of a vector field f is zero through every closed surface containing a
given point, then div f = 0 at that point.
Proof: By formula (4.33), at the given point ( x, y, z) we have
1
div f( x, y, z) = lim
f · dσ for closed surfaces Σ containing ( x, y, z), so
V →0 V Σ
1
= lim
(0) by our assumption that the flux through each Σ is zero, so
V →0 V
= lim 0
V →0
= 0 .
QED
Lastly, we note that sometimes the notation
f ( x, y, z) dσ
and
f · dσ
Σ
Σ
is used to denote surface integrals of scalar and vector fields, respectively, over closed sur-
faces. Especially in physics texts, it is common to see simply
instead of
.
Σ
Σ
Exercises
A
For Exercises 1-4, use the Divergence Theorem to evaluate the surface integral
f · dσ of
Σ
the given vector field f( x, y, z) over the surface Σ.
1. f( x, y, z) = xi + 2 yj + 3 zk, Σ : x 2 + y 2 + z 2 = 9
2. f( x, y, z) = xi + yj + zk, Σ : boundary of the solid cube S = {( x, y, z) : 0 ≤ x, y, z ≤ 1}
3. f( x, y, z) = x 3i + y 3j + z 3k, Σ : x 2 + y 2 + z 2 = 1
4. f( x, y, z) = 2i + 3j + 5k, Σ : x 2 + y 2 + z 2 = 1
164
CHAPTER 4. LINE AND SURFACE INTEGRALS
B
5. Show that the flux of any constant vector field through any closed surface is zero.
6. Evaluate the surface integral from Exercise 2 without using the Divergence Theorem, i.e.
using only Definition 4.3, as in Example 4.10. Note that there will be a different outward
unit normal vector to each of the six faces of the cube.
7. Evaluate the surface integral
f · dσ, where f( x, y, z) = x 2i + xyj + zk and Σ is the part of Σ
the plane 6 x + 3 y + 2 z = 6 with x ≥ 0, y ≥ 0, and z ≥ 0, with the outward unit normal n
pointing in the positive z direction.
8. Use a surface integral to show that the surface area of a sphere of radius r is 4 πr 2. ( Hint:
Use spherical coordinates to parametrize the sphere. )
9. Use a surface integral to show that the surface area of a right circular cone of radius R
and height h is πR h 2 + R 2. ( Hint: Use the parametrization x = r cos θ, y = r sin θ, z = h r, R
for 0 ≤ r ≤ R and 0 ≤ θ ≤ 2 π. )
10. The ellipsoid x 2
a 2 + y 2
b 2 + z 2
c 2 = 1 can be parametrized using ellipsoidal coordinates
x = a sin φ cos θ , y = b sin φ sin θ , z = c cos φ , for 0 ≤ θ ≤ 2 π and 0 ≤ φ ≤ π.
Show that the surface area S of the ellipsoid is
π
2 π
S =
sin φ
a 2 b 2 cos2 φ + c 2( a 2 sin2 θ + b 2 cos2 θ)sin2 φ dθ dφ .
0
0
(Note: The above double integral can not be evaluated by elementary means. For specific
values of a, b and c it can be evaluated using numerical methods. An alternative is to
express the surface area in terms of elliptic integrals.5)
C
11. Use Definition 4.3 to prove that the surface area S over a region R in R2 of a surface
z = f ( x, y) is given by the formula
2
2
S =
1 + ∂f
d A .
∂x
+ ∂f
∂ y
R
( Hint: Think of the parametrization of the surface. )
5 BOWMAN, F., Introduction to Elliptic Functions, with Applications, New York: Dover, 1961, § III.7.
4.5 Stokes’ Theorem
165
4.5 Stokes’ Theorem
So far the only types of line integrals which we have discussed are those along curves in R2.
But the definitions and properties which were covered in Sections 4.1 and 4.2 can easily be
extended to include functions of three variables, so that we can now discuss line integrals
along curves in R3.
Definition 4.5. For a real-valued function f ( x, y, z) and a curve C in R3, parametrized by
x = x( t), y = y( t), z = z( t), a ≤ t ≤ b, the line integral of f ( x, y, z) along C with respect to arc length s is
b
f ( x, y, z) ds =
f ( x( t), y( t), z( t))
x ′( t)2 + y′( t)2 + z ′( t)2 dt .
(4.34)
C
a
The line integral of f ( x, y, z) along C with respect to x is
b
f ( x, y, z)