16.5 Surface Integrals of Vector Fields
Lukas Geyer
Montana State University
M273, Fall 2011
Lukas Geyer (MSU)
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
1 / 19
Parametrized Surfaces
Definition
An orientation on a surface S is a continuous choice of a unit normal
vector en (P) at each point P os S.
Example
The xy -plane has two orientations, one given by en = k (pointing up), the
other by en = −k (pointing down).
Example
The sphere kxk = R has two orientations, one given by the outward
x
pointing vector en (x) = kxk
, the other by the inward pointing normal
vectors −en (x).
Lukas Geyer (MSU)
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
2 / 19
The Möbius Strip
Caution
Not all surfaces are orientable. The most popular example of a
non-orientable surface is the Möbius strip depicted below.
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16.5 Surface Integrals of Vector Fields
M273, Fall 2011
3 / 19
Vector Surface Integral
Definition
The normal component of a vector field F at a point P on an oriented
surface S is
F(P) · en (P)
Definition
The vector surface integral of a vector field F over a surface S is
ZZ
ZZ
F · dS =
(F · en )dS.
S
S
It is also called the flux of F across or through S.
Applications
Flow rate of a fluid with velocity field F across a surface S.
Magnetic and electric flux across surfaces. (Maxwell’s equations)
Lukas Geyer (MSU)
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
4 / 19
Parametrized Vector Surface Integral
Calculating Parametrized Surface Integrals
A regular parametrization G (u, v ) (i.e., a parametrization with n(u, v ) 6= 0
n
for all u, v ) induces an orientation by en = knk
. We get
ZZ
ZZ
F·S=
S
F(G (u, v )) · n(u, v ) du dv .
D
Differential Version
dS = n(u, v ) du dv
Orientation matters
Reversing the orientation of S changes the sign of the integral.
Lukas Geyer (MSU)
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
5 / 19
Vector Surface Integral Example I
Example
ZZ
F · dS, where F = h0, 0, zi and S is the oriented surface
Find
S
parametrized by G (u, v ) = (u cos v , u sin v , v ), where 0 ≤ u ≤ 1,
0 ≤ v ≤ 2π.
Tu = hcos v , sin v , 0i
Tv
= h−u sin v , u cos v , 1i
n = Tu × Tv = hsin v , − cos v , ui
Is the integral positive, negative, or
0? Positive!
Lukas Geyer (MSU)
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
6 / 19
Vector Surface Integral Example II
Example
ZZ
F · dS, where F = h0, 0, zi and S is the oriented surface
Find
S
parametrized by G (u, v ) = (u cos v , u sin v , v ), where 0 ≤ u ≤ 1,
0 ≤ v ≤ 2π.
Evaluating the integral
ZZ
n = hsin v , − cos v , ui
Z 1 Z 2π
F · dS =
h0, 0, v i · hsin v , − cos v , ui dv du
S
0
0
Z 1
Z 2π
Z 1 Z 2π
vu dv du =
u du
v dv
=
0
=
Lukas Geyer (MSU)
0
0
0
4π 2
1
·
= π2.
2 2
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
7 / 19
Flow Rate Example I
Example
A fluid flows with constant velocity v = 3k(m/s). Calculate the flow rate
(in m3 /s) through the part of the elliptic paraboloid z = x 2 + y 2 with
z ≤ 4 and upward pointing normal vector.
Parametrize Surface
At height z the trace is a circle of radius r =
parameters:
G (r , θ) = (r cos θ, r sin θ, r 2 ),
Tr
√
z. Using r and θ as
0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π
= hcos θ, sin θ, 2r i
Tθ = h−r sin θ, r cos θ, 0i
n = Tr × Tθ = h−2r 2 cos θ, −2r 2 sin θ, r i
Lukas Geyer (MSU)
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
8 / 19
Flow Rate Example II
Example
A fluid flows with constant velocity v = 3k(m/s). Calculate the flow rate
(in m3 /s) through the part of the elliptic paraboloid z = x 2 + y 2 with
z ≤ 4 and upward pointing normal vector.
Integrate
G (r , θ) = (r cos θ, r sin θ, r 2 ),
2
0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π
2
n = h−2r cos θ, −2r sin θ, r i
ZZ
Z
2 Z 2π
h0, 0, 3i · h−2r 2 cos θ, −2r 2 sin θ, r i dθ dr
Z 2 Z 2π
Z 2
=
3r dθ dr = 6π
r dr = 12π.
F · dS =
S
Lukas Geyer (MSU)
0
0
0
0
0
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
9 / 19
Another Example I
Example
Let S be the boundary of the solid cone with base x 2 + y 2 ≤ 4 in the
xy -plane and apex (0, 0, 4), with outward-pointing normal vector. Find the
flux of the vector field F = h−y , x, zi through S.
Parametrizing S
We have to parametrize both the base and the boundary cone, calculate
the flux through both and add the results.
Parametrizing the base
Standard parametrization of a disk of radius 2:
G (r , θ) = (r cos θ, r sin θ, 0),
Lukas Geyer (MSU)
0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π.
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
10 / 19
Another Example II
Example
Let S be the boundary of the solid cone with base x 2 + y 2 ≤ 4 in the
xy -plane and apex (0, 0, 4), with outward-pointing normal vector. Find the
flux of the vector field F = h−y , x, zi through S.
Parametrizing the base
G (r , θ) = (r cos θ, r sin θ, 0),
Tr
0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π,
= hcos θ, sin θ, 0i
Tθ = h−r sin θ, r cos θ, 0i
n = Tr × Tθ = h0, 0, r i
Problem: r ≥ 0, so n points up, into the cone, not out. Solution: Change
the sign of n.
Lukas Geyer (MSU)
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
11 / 19
Another Example III
Example
Let S be the boundary of the solid cone with base x 2 + y 2 ≤ 4 in the
xy -plane and apex (0, 0, 4), with outward-pointing normal vector. Find the
flux of the vector field F = h−y , x, zi through S.
Integrating over the base
G (r , θ) = (r cos θ, r sin θ, 0),
ZZ
0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π,
n = −h0, 0, r i
Z 2 Z 2π
F · dS =
h−r sin θ, r cos θ, 0i · h0, 0, −r i dθ dr = 0
S1
Lukas Geyer (MSU)
0
0
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
12 / 19
Another Example IV
Example
Let S be the boundary of the solid cone with base x 2 + y 2 ≤ 4 in the
xy -plane and apex (0, 0, 4), with outward-pointing normal vector. Find the
flux of the vector field F = h−y , x, zi through S.
Parametrizing the cone
Trace in z = k is a circle of radius (4 − k)/2 for 0 ≤ k ≤ 4, so
4−z
4−z
cos θ,
sin θ, z , 0 ≤ z ≤ 4, 0 ≤ θ ≤ 2π.
G (z, θ) =
2
2
1
1
Tz =
− cos θ, − sin θ, 1
2
2
4−z
4−z
sin θ,
cos θ, 0
Tθ =
−
2
2
z −4
z −4
z −4
n = Tz × Tθ =
cos θ,
sin θ,
.
2
2
2
Lukas Geyer (MSU)
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
13 / 19
Another Example V
Example
Let S be the boundary of the solid cone with base x 2 + y 2 ≤ 4 in the
xy -plane and apex (0, 0, 4), with outward-pointing normal vector. Find the
flux of the vector field F = h−y , x, zi through S.
Integrating over the cone
4−z
4−z
G (z, θ) =
cos θ,
sin θ, z , 0 ≤ z ≤ 4, 0 ≤ θ ≤ 2π.
2
2
z −4
z −4
z −4
cos θ,
sin θ,
n =
.
2
2
2
(z − 4)/2 ≤ 0, so n points down, i.e., into the cone. Again we have to
change the sign to fix the orientation.
Lukas Geyer (MSU)
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
14 / 19
Another Example VI
Example
Let S be the boundary of the solid cone with base x 2 + y 2 ≤ 4 in the
xy -plane and apex (0, 0, 4), with outward-pointing normal vector. Find the
flux of the vector field F = h−y , x, zi through S.
Integrating over the cone, correct orientation
4−z
4−z
cos θ,
sin θ, z , 0 ≤ z ≤ 4, 0 ≤ θ ≤ 2π.
2
2
4−z
4−z
4−z
cos θ,
sin θ,
.
n =
2
2
2
G (z, θ) =
F·n=
z −4
4−z
sin θ,
cos θ, z
2
2
Lukas Geyer (MSU)
4−z
4−z
4−z
·
cos θ,
sin θ,
2
2
2
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
15 / 19
Another Example VII
Example
Let S be the boundary of the solid cone with base x 2 + y 2 ≤ 4 in the
xy -plane and apex (0, 0, 4), with outward-pointing normal vector. Find the
flux of the vector field F = h−y , x, zi through S.
Integrating over the cone, correct orientation
G (z, θ) =
4−z
4−z
cos θ,
sin θ, z
2
2
z −4
4−z
F·n=
sin θ,
cos θ, z
2
2
z(4 − z)
4z − z 2
=
=
2
2
Lukas Geyer (MSU)
,
0 ≤ z ≤ 4, 0 ≤ θ ≤ 2π.
4−z
4−z
4−z
·
cos θ,
sin θ,
2
2
2
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
16 / 19
Another Example VIII
Example
Let S be the boundary of the solid cone with base x 2 + y 2 ≤ 4 in the
xy -plane and apex (0, 0, 4), with outward-pointing normal vector. Find the
flux of the vector field F = h−y , x, zi through S.
Integrating over the cone, correct orientation
G (z, θ) =
F·n =
4−z
4−z
cos θ,
sin θ, z
2
2
4z − z 2
2
ZZ
Z
Lukas Geyer (MSU)
0 ≤ z ≤ 4, 0 ≤ θ ≤ 2π.
,
4 Z 2π
F · dS =
S2
Z
4 Z 2π
F · n dθ dz =
0
0
0
16.5 Surface Integrals of Vector Fields
0
4z − z 2
dθ dz
2
M273, Fall 2011
17 / 19
Another Example IX
Example
Let S be the boundary of the solid cone with base x 2 + y 2 ≤ 4 in the
xy -plane and apex (0, 0, 4), with outward-pointing normal vector. Find the
flux of the vector field F = h−y , x, zi through S.
Integrating over the cone, correct orientation
ZZ
Z 4
4z − z 2
4z − z 2 dz
dθ dz = π
2
0
0
0
z=4
3
z
64
32π
2
= π 2z −
= π 32 −
=
.
3 z=0
3
3
Z
4 Z 2π
F · dS =
S2
Lukas Geyer (MSU)
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
18 / 19
Another Example X
Example
Let S be the boundary of the solid cone with base x 2 + y 2 ≤ 4 in the
xy -plane and apex (0, 0, 4), with outward-pointing normal vector. Find the
flux of the vector field F = h−y , x, zi through S.
Adding up the results
ZZ
ZZ
ZZ
F · dS =
S
F · dS +
S1
= 0+
Lukas Geyer (MSU)
F · dS
S2
32π
32π
=
.
3
3
16.5 Surface Integrals of Vector Fields
M273, Fall 2011
19 / 19