31 surface integrals and flux

Surface Integrals and Flux
A flux is a fluid flow.

A flow is a steady flow if the directions and the
magnitudes (the volume) of the flow are constants,
unchanged in time.

assumed to be steady flow.
unchanged in time. All flows in this section are

We want to calculate the amount of fluid flows
"across" a surface in one unit of time.

In order to flow "across" a surface, the surface
must have two distinct sides.

In order to flow "across" a surface, the surface
must have two distinct sides.
The Mobius strip is an
example of a surface only
has one side.
Mobius strip

A two-sided surface is one
where we can select the normal
vectors of the surface in such a
way they are continuous,
without abrupt change of
directions.

directions. Such surface are
called orientable surfaces.

normal vectors on
the Mobius strip
The normal vectors on the
Mobius strip can't be selected
without abrupt change in the
directions at some location.

normal vectors on
the Mobius strip

Such surfaces are said to be
non-orientable.
normal vectors on
the Mobius strip

A orientable surface with
selected unit normal vectors
such that they form a continuous
vector field (pointing to one side
of the surface) is called an
oriented surface.

oriented surface.
An oriented surface

oriented surface.
An oriented surface
The selected unit normal vectors is said to be an
orientation for the surface, it is denoted as N.

oriented surface.
An oriented surface
As noted before, an orientable surface has exactly
two orientations.

oriented surface.
All the surfaces in the following discussion are
assumed to be orientable.
An oriented surface
As noted before, an orientable surface has exactly
two orientations.

A 3D flow may be represented by a vector field
F(x, y, z) = f(x, y, z)i + g(x, y, z)j + h(x, y, z)k
or simply as
F = fi + gj + hk.

or simply as
F = fi + gj + hk.
A 3D flow
F = fi + gj + hk

or simply as
F = fi + gj + hk.
Each vector F(x, y, z) gives the
direction of the flow and the volume
of the flow (per unit of time). A 3D flow
F = fi + gj + hk

or simply as
F = fi + gj + hk.
unchanged in time.
F = fi + gj + hk

or simply as
F = fi + gj + hk.
assumed to be steady flows.
F = fi + gj + hk

or simply as
F = fi + gj + hk.
assumed to be steady flows. We want to calculate
the flux or the rate of the amount of fluid that flow F
across a surface σ.
F = fi + gj + hk

Let σ = z(x, y) be an oriented
surface over R with N as a unit
normal field.

normal field.
σ
Pi
R
N

normal field.
Partition σ into small patches P's
and let Pi be a typical patch.
σ
Pi
Pi
R
N

normal field.
Let Ni be a unit normal at some
point of Pi. σ
Ni
Pi
R
N
Pi

normal field.
point of Pi. σ
Pi
Ni
Pi
Let Ti be a tile normal to Ni that is
tangent to and approximates Pi.
Ti
R
N

normal field.
point of Pi. σ
Pi
Ni
Pi
Let Ti be a tile normal to Ni that is
tangent to and approximates Pi.
Ti
R
Fi
Let Fi be the flow vector at the
tangent point.
N
F

Hence the volume Vi flows
through Pi is approximately the
volume created by Ti displaced
through space by Fi.
Pi
Ti
Fi

The height of the box is the
projection of Fi onto the unit
normal Ni which is Fi•Ni.
Pi
Ni
Ti
Fi
Vi
Ti
Fi
Fi•Ni
Ni

Therefore Vi = (Fi•Ni)*(area of Ti)
Pi
Ni
Ti
Fi
Vi
Ti
Fi
Fi•Ni
Ni

Therefore Vi = (Fi•Ni)*(area of Ti)
= (Fi•Ni)*zx(xi, yi)+zy(xi, yi)+1 ΔxΔy
2 2
where (xi, yi, zi) is the tangential
point.
Pi
Ni
Ti
Fi
Vi
Ti
Fi
Fi•Ni
Ni

The total flow F across the surface σ is:
V = Σ Fi• Nizx(xi, yi)+zy(xi, yi)+1 ΔxΔy
2 2
lim Δx0
lim Δy0

2 2
lim Δx0
lim Δy0
= ∫R
∫ F•Nzx+zy+1 dA where dA = dxdy or dydx
2 2

2 2
lim Δx0
lim Δy0
= ∫R
2 2
Or in simpler notation of surface integral,
integrating over the surface σ against dS:
∫ ∫ F•N dS
σ
where dS = zx(xi, yi)+zy(xi, yi)+1 ΔxΔy

2 2
lim Δx0
lim Δy0
= ∫R
2 2
Or in simpler notation of surface integral,
integrating over the surface σ against dS:
∫ ∫ F•N dS
σ
where dS = zx(xi, yi)+zy(xi, yi)+1 ΔxΔy
These are the flux integrals of flow F across σ.
They measure the amount of the flow F across σ
in a fixed time.

Let σ be given by z = z(x, y).

There are two unit normal fields, the upward-unit-
normal and the downward-unit normal.

upward-unit-normal:
<-zx, -zy,1>
zx+zy+1
2 2 (z is positive)

upward-unit-normal:
downward-unit normal:
<-zx, -zy,1>
zx+zy+1
2 2
zx+zy+1
2 2
<zx, zy, -1>
(z is positive)
(z is negative)

F•N
upward-unit-normal:
<-zx, -zy,1>
For upward-unit normal N, the integrand
zx+zy+1
2 2
zx+zy+1
2 2
zx+zy+1 =
2 2
<zx, zy, -1>
(z is positive)
(z is negative)

F•N
upward-unit-normal:
<-zx, -zy,1>
F•
zx+zy+1
2 2
zx+zy+1
2 2
zx+zy+1 =
2 2
zx+zy+1
2 2 zx+zy+1
2 2
<zx, zy, -1>
<-zx, -zy,1>
(z is positive)
(z is negative)

F•N
upward-unit-normal:
<-zx, -zy,1>
F• = F•
zx+zy+1
2 2
zx+zy+1
2 2
zx+zy+1 =
2 2
zx+zy+1
2 2 zx+zy+1
2 2
<zx, zy, -1>
<-zx, -zy,1>
<-zx, -zy,1>
(z is positive)
(z is negative)

F•N
upward-unit-normal:
<-zx, -zy,1>
F• = F•
So ∫R
∫F•N dA = ∫R
∫F•
zx+zy+1
2 2
zx+zy+1
2 2
zx+zy+1 =
2 2
zx+zy+1
2 2 zx+zy+1
2 2
zx+zy+1
2 2
<zx, zy, -1>
<-zx, -zy,1>
<-zx, -zy,1>
<-zx, -zy,1> dA
(z is positive)
(z is negative)

Summary: Let σ be surface defined by z = z(x, y),
a differentiable function over the domain R.
Let F = fi + gj + hk be a continuous 3D flow.

Then the volume of the flow F across σ or the flux
I. in the upward direction is:
∫R
∫F• <-zx, -zy,1> dA

∫R
∫F• <-zx, -zy,1> dA
II. in the downward direction is:
∫R
∫F• <zx, zy, -1> dA

∫R
∫F• <-zx, -zy,1> dA
II. in the downward direction is:
∫R
∫F• <zx, zy, -1> dA
There are four more versions of these integrals
depending the defining functions of the surfaces as
x = x(y, z) or y = y(x, z).

III. in the positive-y-direction is:
∫R
∫F• <-yx, 1, -yz> dA
Let σ be defined by y = y(x, z), then the flux integral
of F across σ

∫R
∫F• <-yx, 1, -yz> dA
IV. in the negative-y-direction is:
∫R
∫F•
of F across σ
<yx, -1, yz> dA

∫R
∫F• <-yx, 1, -yz> dA
∫R
∫F•
of F across σ
<yx, -1, yz> dA
V. in the positive-x-direction is:
∫ ∫F• <1,-xy , -xz> dA
Let σ be defined by x = x(y, z), then the flux integral
of F across σ

∫R
∫F• <-yx, 1, -yz> dA
∫R
∫F•
of F across σ
<yx, -1, yz> dA
V. in the positive-x-direction is:
∫R
∫F• <1,-xy , -xz> dA
VI. in the negative-x-direction is:
∫R
∫F•
Let σ be defined by x = x(y, z), then the flux integral
of F across σ
<-1, xy , xz> dA

Example: Let σ be the surface defined by z = x2 + y2
below z = 1 oriented by the downward unit normal.
Let F = yj + k. Find the volume of the flow F across σ.

z = 1
R
σ
x2 + y2 =1
x
y

The integrand for the downward direction is
F• <zx, zy, -1>
z = 1
R
σ
x2 + y2 =1
x
y

F• <zx, zy, -1> = <0, y, 1><2x, 2y, -1>
z = 1
R
σ
x2 + y2 =1
x
y

F• <zx, zy, -1> = <0, y, 1><2x, 2y, -1> = 2y2 – 1
z = 1
R
σ
x2 + y2 =1
x
y

∫R
∫F• <zx, zy, -1> dA
F• <zx, zy, -1> = <0, y, 1><2x, 2y, -1> = 2y2 – 1
z = 1
R
σ
x2 + y2 =1
x
Hence the flux integral is:
y

∫R
∫F• <zx, zy, -1> dA
F• <zx, zy, -1> = <0, y, 1><2x, 2y, -1> = 2y2 – 1
z = 1
R
σ
x2 + y2 =1
x
= ∫R
∫2y2 – 1 dA
y

∫R
∫F• <zx, zy, -1> dA
F• <zx, zy, -1> = <0, y, 1><2x, 2y, -1> = 2y2 – 1
z = 1
R
σ
x2 + y2 =1
x
= ∫R
∫2y2 – 1 dA in polar form
= ∫
R=0
∫(2r2sin2() – 1) rdrd
1
=0
2π
y

∫R
∫F• <zx, zy, -1> dA
F• <zx, zy, -1> = <0, y, 1><2x, 2y, -1> = 2y2 – 1
z = 1
R
σ
x2 + y2 =1
x
= ∫R
∫2y2 – 1 dA in polar form
= ∫
R=0
∫(2r2sin2() – 1) rdrd = -π/2
1
=0
2π
y

Example: Let σ be the surface of the tetrahedron
with vertices at the origin (1, 0, 0),
(0,1 ,0) and (0, 0,1) oriented by the outward unit
normal. Let F = xi + yj + zk. Find the volume of the
flow F across σ.

x
y
(1,0,0)
(0,1,0)
(0,0,1)
flow F across σ.

flow F across σ.
x
y
(1,0,0)
(0,1,0)
(0,0,1)
There are four surfaces and they are defined by:
x = 0, y = 0, z = 0, x + y + z = 1.

x
y
(1,0,0)
(0,1,0)
(0,0,1)
x = 0, y = 0, z = 0, x + y + z = 1.
The outward normal fields needed are:
I. z = 0 is <zx, zy, -1> = <0, 0, -1> = N1
flow F across σ.

x
y
(1,0,0)
(0,1,0)
(0,0,1)
x = 0, y = 0, z = 0, x + y + z = 1.
I. z = 0 is <zx, zy, -1> = <0, 0, -1> = N1
II. y = 0 is <yx, -1, yz> = <0, -1, 0> = N2
flow F across σ.

x
y
(1,0,0)
(0,1,0)
(0,0,1)
x = 0, y = 0, z = 0, x + y + z = 1.
I. z = 0 is <zx, zy, -1> = <0, 0, -1> = N1
II. y = 0 is <yx, -1, yz> = <0, -1, 0> = N2
III. x = 0 is <-1, 0, 0> = N3
flow F across σ.

x
y
(1,0,0)
(0,1,0)
(0,0,1)
x = 0, y = 0, z = 0, x + y + z = 1.
I. z = 0 is <zx, zy, -1> = <0, 0, -1> = N1
II. y = 0 is <yx, -1, yz> = <0, -1, 0> = N2
III. x = 0 is <-1, 0, 0> = N3
IV. z = -x – y + 1 is <1, 1, 1> = N4
flow F across σ.

The integrand for the surface z = 0 is
F•N1 = <x, y, z>•<0, 0, -1>
x
y
(1,0,0)
(0,1,0)
(0,0,1)
R

F•N1 = <x, y, z>•<0, 0, -1> = -z
x
y
(1,0,0)
(0,1,0)
(0,0,1)
R

F•N1 = <x, y, z>•<0, 0, -1> = -z = 0, so the flow is 0
across z = 0.
x
y
(1,0,0)
(0,1,0)
(0,0,1)
R

across z = 0.
x
y
(1,0,0)
(0,1,0)
(0,0,1)
Similar the flow through y = 0 and x = 0 are 0's also.
R

across z = 0.
x
y
(1,0,0)
(0,1,0)
(0,0,1)
Note the flow is within the planes, so the volume is 0.
R

across z = 0.
x
y
(1,0,0)
(0,1,0)
(0,0,1)
The integrand for x + y + z = 1
F•N1 = <x, y, z>•<1, 1, 1>
R

across z = 0.
x
y
(1,0,0)
(0,1,0)
(0,0,1)
R
F•N1 = <x, y, z>•<1, 1, 1>
= x + y + z

across z = 0.
x
y
(1,0,0)
(0,1,0)
(0,0,1)
R
F•N1 = <x, y, z>•<1, 1, 1>
= x + y + z
= x + y + (1 – x – y) = 1

across z = 0.
x
y
(1,0,0)
(0,1,0)
(0,0,1)
R
F•N1 = <x, y, z>•<1, 1, 1>
= x + y + z
= x + y + (1 – x – y) = 1
Integrate this over the domain
R = {0 < x < 1; 0 < y < 1 – x }.

across z = 0.
x
y
(1,0,0)
(0,1,0)
(0,0,1)
∫R
∫1 dA
R
F•N1 = <x, y, z>•<1, 1, 1>
= x + y + z
= x + y + (1 – x – y) = 1
R = {0 < x < 1; 0 < y < 1 – x }.

across z = 0.
x
y
(1,0,0)
(0,1,0)
(0,0,1)
∫R
∫1 dA = area of R = ½
R
F•N1 = <x, y, z>•<1, 1, 1>
= x + y + z
= x + y + (1 – x – y) = 1
R = {0 < x < 1; 0 < y < 1 – x }.

31 surface integrals and flux

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