Notes
Surface Integrals of Vector Fields
§16.5
15 November 2013
Clicker Question:
Notes
A river is flowing at a constant rate of 5 m/s. We take a
rectangular net that is 6 m wide and 3 m tall and place it in
the river so that a vector perpendicular to the net (a normal
vector) is parallel to the velocity of the water. What is the
rate at which water flows through the net?
A. 0 m3 /s
D. 90 m3 /s
B. 15 m3 /s
E. None of the above
C. 30 m3 /s
receiver channel: 41
session ID: bsumath275
Rate of Flow Across Surface.
Notes
If F represents the velocity field of a fluid flow, then at what
rate is fluid flowing across a surface M?
normal component
represents flow
across the surface
F
n
tangential component
represents flow along the surface
Length of normal comp. = kFk cos θ = F · en
(en = n/knk)
Clicker Question:
Notes
A river is flowing at a constant rate of 5 m/s. We take a
rectangular net that is 6 m wide and 3 m tall and place it in
the river so that there is a 60 degree angle between a vector
perpendicular to the net (a normal vector) and the velocity of
the water. What is the rate at which water flows through the
net?
A. 0 m3 /s
D. 90 m3 /s
B. 45 m3 /s
E. None of the above
C. 78 m3 /s
receiver channel: 41
session ID: bsumath275
Rate of Flow Across Surface, continued.
Notes
At a given point,
rate of flow across surface
= length of normal component = F · en
So
ZZ
F · en dS
rate of fluid flow across M =
M
This is a scalar surface integral like in the previous section.
We can work it out using a parametrization of M.
Flux Integral.
Notes
When M parametrized by (x, y , z) = G (u, v ),
(u, v ) in some domain D, then remember:
n = Tu × Tv ,
en =
n
,
knk
dS = knk du dv
We define
dS = en dS = n du dv ,
Flux Integral
ZZ
ZZ
F · dS =
M
F · n du dv
D
If F represents the velocity field of a fluid flow, then this integral
gives the rate of fluid flowing across M.
In general this is called the flux of F across M.
Orientation.
Notes
If we reverse the orientation of M
I use −n instead of n
then the flux is negated.
Example.
Notes
Example: What is the flux of F(x, y , z) = h0, 0, xi across the
portion M of the plane x + y + z = 1 in the first octant
x, y , z ≥ 0, oriented with upward pointing normal?
Answer: Parametrize with G (u, v ) = (u, v , 1 − u − v ),
u, v ≥ 0, u + v ≤ 1. Then
Tu = h1, 0, −1i,
Tv = h0, 1, −1i
so
n = Tu × Tv = h1, 1, 1i
Example, continued.
Notes
Therefore the flux is
dS
z
}|
{
F
n
z }| { z }| {
h0, 0, ui · h1, 1, 1i dv du
flux =
0
0
Z 1 Z 1−u
Z 1
1 1
1
=
u dv du =
u(1 − u) du = u 2 − u 3 2
3 0
0
0
0
1
= .
6
Z
1
Z
1−u
Worksheet.
(For #2b:
Notes
R 2π
0
Worksheet #1–2
R 2π
sin θ dθ = 0 cos3 θ dθ = 0.)
3
Clicker Question:
Notes
Which vector field has a positive flux through the surface
below?
z
A. F = x ̂
B. F = y ̂
C. F = −zı̂
y
x
receiver channel: 41
D. F = (z + x)ı̂
session ID: bsumath275
Notes