Now that we know how to parameterize surfaces, we`re in a position

Now that we know how to parameterize surfaces, we’re in a position to define and
compute surface integrals. We begin with the following scenario. Suppose f(x, y, z)
is a continuous function giving density at (x, y, z). Let S be a smooth (bounded)
surface in R3 . What’s the mass of S? We could approximate the mass of S via the
following: Partition S into sub-surfaces, say Si,j . Let pi,j ∈ Si,j be a sample point.
Then the mass of Si,j is approximately
f(pi,j )A(Si,j ),
where A(Si,j ) is the surface area of Si,j . So the mass of S is approximately
X
f(pi,j )A(Si,j ).
i,j
In fact, the mass of S is precisely
n,m
X
lim
n,m→∞
f(pi,j )A(Si,j ).
i,j
We now make the following definition.
Definition 1. Let S be a smooth (bounded) surface in R3 , and let f be a continuous
function. Then we call the limit
lim
n,m
X
n,m→∞
f(pi,j )A(Si,j ),
i,j
with notation being as above, the surface integral (scalar) of f over S. We
denote this limit by
ZZ
f dS.
S
Now, how do we compute these integrals? We compute them with the help of
parameterizations, of course. Suppose S is parameterized by r(u, v) for (u, v) ∈ D.
As usual, partitioning D into Di,j induces a partition of S into Si,j . Let (ui , vj ) ∈
Di,j so that we can take
pi,j = r(ui , vj ).
We know from the previous lecture that
A(Si,j ) ≈ kru (ui , vj ) × rv (ui , vj )k∆u∆v.
In fact, we have
ZZ
f dS
=
S
=
lim
n,m→∞
lim
n,m→∞
ZZ
n,m
X
f(pi,j )A(Si,j )
i,j
n,m
X
f(r(ui , vj ))kru (ui , vj ) × rv (ui , vj )k∆u∆v
i,j
f(r(t))kru × rv k du dv.
=
D
1
2
In short, if S is parameterized by r(u, v) for (u, v) ∈ D, then
ZZ
ZZ
f dS =
f(r(t))kru × rv k du dv .
S
D
RR
Example 1. Compute
S
f dS where S is the surface of the cone z =
p
x2 + y2
below the plane z = 1, and f(x, y, z) = 1 − z.
Solution. Notice that S is the graph of a function above the unit circle in the
(x, y) plane. So we parameterize S via
x(u, v) = v cos u
y(u, v) = v sin u
where (u, v) ∈ [0, 2π] × [0, 1]. Notice that
i
j
ru × rv = −v sin u v cos u
cos u
sin u
So
z(u, v) = v,
k
0 = hv cos u, v sin u, −vi.
1
√
kru × rv k = v 2.
Hence
ZZ
Z 2π Z 1
f dS
=
0
S
√
(1 − v)(v 2) dv du
0
2
1
√
v
v3
= 2 2π
−
2
3 0
√
2π
=
.
3
Now, suppose S is a smooth bounded surface in R3 .
Definition 2. S is oriented by the vector field n means that n has the following
properties:
(i) kn(x, y, z)k = 1 for all (x, y, z).
(ii) For (x, y, z) ∈ S, n(x, y, z) is perpendicular to S at (x, y, z).
(iii) n(x, y, z) is continuous on S.
If no such vector field n exists for a given S, then we say that S cannot be oriented.
In layman’s terms, if S is oriented by n, then S has two distinct sides and n
points in the direction of what we consider the “positive” side. A famous example
of a non-orientable surface is the Möbius band:
3
But, for surfaces that can be oriented, we can talk about the so-called flux of a
vector field F through S.
Let F be a continuous vector field in R3 , and let S be a smooth, bounded, oriented
surface S. Suppose F represents the velocity of a fluid. What volume of fluid passes
through S per unit of time? We call this rate the flux of F through S. We can
approximate flux by the usual process that we’ve become accustomed to. Namely,
partition S into sub-surfaces, say Si,j , and let pi,j ∈ Si,j be a sample point. Let n
be the vector field orienting S. Then the flux of F through Si,j is approximately
F(pi,j ) · n(pi,j )A(Si,j )1.
So the flux of F through S is approximately
X
F(pi,j ) · n(pi,j )A(Si,j ).
i,j
In fact, the flux of F through S is precisely
n,m
X
lim
n,m→∞
F(pi,j ) · n(pi,j )A(Si,j ).
i,j
Definition 3. Let S be a smooth, bounded, oriented surface in R3 , and let F be a
continuous vector field in R3 . Then we call the limit
n,m
X
F(pi,j ) · n(pi,j )A(Si,j ),
lim
n,m→∞
i,j
with notation being as above, the surface integral (vector) of F over S. We
denote this limit by
ZZ
F · dS.
S
Remark. Note that
ZZ
ZZ
F · dS =
S
F · n dS
S
where n orients S.
Now, how do we compute these integrals? As before, we compute them with the
help of parameterizations. Suppose r(u, v) for (u, v) ∈ D parameterizes S. Then,
notations as they were for surface integrals of scalar functions we take
1The dot-product here is approximately the rate of flow perpendicular to S .
i,j
4
• pi,j = r(ui , vj ).
• A(Si,j ) ≈ kr(ui , vj ) × rv (ui , vj )k∆u∆v.
r(ui , vj ) × rv (ui , vj ) 2
• n(pi,j ) = ±
.
kr(ui , vj ) × rv (ui , vj )k
Putting everything together, we arrive at the following formula:
ZZ
ZZ
F · dS =
F(r(u, v)) · (±ru × rv ) du dv .
S
D
Example 2. Find the flux of F(x, y, z) = i across the upper unit hemisphere with
outward orientation.
Solution. We parameterize the upper unit hemisphere with


 x = sin φ cos θ
r(θ, φ) =
(θ, φ) ∈ [0, 2π] × [0, π/2] = D.
y = sin φ sin θ


z = cos φ
Note that
rθ
= h− sin φ sin θ, sin φ cos θ, 0i
rφ
= hcos φ cos θ, cos φ sin θ, − sin φi.
So
rθ × rφ = h− sin2 φ cos θ, ?, ?i.3
Hence
ZZ
ZZ
F · dS
h1, 0, 0i · h− sin2 φ cos θ, ?, ?i dA
=
D
S
Z π/2 Z 2π
− sin2 φ cos θ dθ dφ
=
0
0
= 0.
2The ± is determined by which direction n points.
3Since F has no j or k component, the only component here that matters is i.

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