Download 1 Surface Integrals Mass problem. Find the mass M of a curved

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Surface Integrals
Mass problem. Find the mass
M of a curved lamina σ whose
density function f (x, y, z)
(the mass per unit area)
is known.
1. Divide σ into n small patches σk
with areas ∆Sk , k = 1, ..., n.
2. Let (x∗k , yk∗, zk∗) be a sample point
in the k-th patch with
∆Mk ≈ f (x∗k , yk∗, zk∗)∆Sk the mass
of the corresponding section.
3.
M=
n
X
k=1
∆Mk = lim
n→∞
n
X
¨
f (x∗k , yk∗, zk∗)∆Sk =
f (x, y, z) dS
σ
k=1
The last term is the notation for the limit of the Riemann sum,
and it is called the surface integral of f (x, y, z) over σ.
4. If f = 1 the we get the area of σ
¨
n
X
S=
dS =
∆Sk
σ
k=1
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Evaluating surface integrals
Theorem. Let σ be a smooth parametric surface whose vector equation is ~r = x(u, v)~i + y(u, v) ~j + z(u, v) ~k where (u, v) varies over a
region R in the uv-plane. If f (x, y, z) is continuous on σ, then
¨
¨
∂~r ∂~r f (x(u, v), y(u, v), z(u, v)) × dA
∂u ∂v
R
f (x, y, z) dS =
σ
This formula follows by approximating ∆Sk as
∂~r ∂~r ∆Sk ≈ × ∆Ak
∂u ∂v
2
2
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Example. σ: x + y + z = 1.
˜
σ
y 2 dS =? Answer : 4π/3.
Surface integrals over z = g(x, y), y = g(x, z), x = g(y, z)
If σ is a surface of the form z = g(x, y), we take
x = u , y = v ⇒ ~r = u~i + v ~j + g(u, v) ~k
s
2 2
∂~r ∂~r × = 1 + ∂z + ∂z
⇒ (u → x , v → y)
∂u ∂v ∂u
∂v
s
2 2
¨
¨
∂z
∂z
f (x, y, z) dS =
f (x, y, g(x, y)) 1 +
+
dA
∂x
∂y
σ
R
The region R lies in the xy-plane and is the projection of σ on the
xy-plane. Analogously for y = g(x, z), x = g(y, z).
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Example. Find the mass of the lamina that is the portion of x2 +
y 2 − 3z − 1 = 0 inside x2 + y 2 = 9/4. The density is
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δ(x, y, z) = δ0( + x2 + y 2)
4
√
81π
Answer : M = 40 δ0 (4 2 − 1) ≈ 29.6256 δ0
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Flux and so on
F~ (~r) represents the velocity of
a fluid particle at (x, y, z).
Velocity vectors are tangent to
streamlines.
F~ (~r) represents the electric field
of two particles with opposite
charges.
Both vector field involve some
type of “flow” – of a fluid or
of charged particles in
an electrostatic field.
They are flow fields.
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Oriented surfaces
We study flows of vector fields
through permeable surfaces
placed in the field.
A normal surface has two sides.
This is a Möbius strip which has
only one side; a bug can traverse
the entire surface without
crossing an edge.
A two-sided surface is said to be
orientable.
One-sided is nonorientable.
To distingush between the two sides
of σ consider a unit normal vector ~n
at each point.
~n and −~n point to opposite sides of σ
and can be used to distinguish
between two sides.
If σ is a smooth orientable surface then it is always possible to choose
the direction of ~n at each point so that ~n = ~n(x, y, z) varies continuously over the surface.
This unit vectors are said to form an orientation of the surface.
A smooth orientable surface has only two possible orientations.
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Orientation of a smooth parametric surface
σ : ~r = x(u, v)~i + y(u, v) ~j + z(u, v) ~k
The unit normal
~n = ~n(u, v) =
∂~r
∂u
∂~r
∂u
∂~r
× ∂v
∂~r × ∂v
is a continuous vector-valued function.
Thus, this formula defines a positive orientation of σ, and
~n points in the positive direction from σ.
The orientation determined by −~n is the negative orientation of σ,
and ~n points in the negative direction from σ.
Example. ~r = cos u~i + v ~j − sin u ~k
Flux
Fluid is either liquids or gases.
Liquids are regarded to be incompressible.
Gases are compressible.
We consider incompressible fluids which are in a steady state.
That means the velocity of the fluid at a fixed point does not vary with
time.
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Problem. An oriented surface σ is immersed in an incompressible,
steady-state fluid flow, and σ is permeable so that the fluid flows
through it freely. Find the net volume of the fluid Φ that passes through
σ per unit of time.
Net volume is the volume that passes through σ in the positive direction minus the volume that passes through σ in the negative direction.
Velocity of the fluid
F~ (x, y, z) = f (x, y, z)~i + g(x, y, z) ~j + h(x, y, z) ~k
~n is the unit normal toward
the positive side of σ.
(F~ · ~n) ~n is the projection of F~ on ~n.
The sign of F~ · ~n determines
the direction of the flow.
1. Divide σ into n patches σk with area ∆Sk , k = 1, ..., n
2. The approximate net volume of fluid crossing σk in the direction
of ~n per unit of time is
F~ (x∗ , y ∗, z ∗) · ~n(x∗ , y ∗, z ∗) ∆Sk
k
k
k
k
k
k
3. The exact net volume is
n
X
Φ = lim
F~ (x∗k , yk∗, zk∗) · ~n(x∗k , yk∗, zk∗) ∆Sk
n→∞
k=1
4. The quantity Φ defined by this limit is called the flux of F~ across
σ, and is expressed as the surface integral
¨
F~ (x, y, z) · ~n(x, y, z) dS
Φ=
σ
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Evaluating flux integrals
The integrals of the form
˜
σ
F~ · ~n dS are called flux integrals.
To calculate them we use
¨
¨
F~ · ~n dS =
σ
σ
¨
∂~r
∂~r
∂~r ∂~r ×
∂u
∂v
~
~
F · ~n × dA =
F · ∂~r ∂~r × ∂u ∂v
σ
∂u
∂v
¨
F~ ·
=
σ
∂~r ∂~r
×
∂u ∂v
∂~r ∂~r × dA
∂u ∂v dA
Example. F~ = z ~k, σ : x2 + y 2 + z 2 = a2 oriented outward.
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Φ =?. Answer : 4π
3a
Let σ : z = g(x, y), ~r = u~i + v ~j + g(u, v) ~k
This choice of u, v imposes positive and negative orientations on σ.
Write σ as G(x, y, z) = z − g(x, y). Then
~
∇G
,
~n = ~
∇G
~ = ∂~r × ∂~r
∇G
∂u ∂v
~ is ⊥ σ, and ∇G
~ = ~k + · · · and
because ∇G
∂~r
∂u
∂~r
× ∂v
= ~k + · · ·
Thus,
¨
¨
F~ · ~n dS =
σ
¨
~ dA =
F~ · ∇G
σ
σ
∂z
∂z
~j + ~k dA
F~ · − ~i −
∂x
∂y
if σ is oriented up (along the positive z-axis).
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Example. z = 1 − x2 − y 2 and z ≥ 0. F~ = ~r. Φ =?.
Answer : 3π/2.
Similarly, let
σ : y = g(z, x) , ~r = v ~i + g(u, v) ~j + u ~k , G(x, y, z) = y − g(z, x)
It is oriented along the positive y-axis.
or let
σ : x = g(y, z) , ~r = g(u, v)~i + u ~j + v ~k , G(x, y, z) = x − g(y, z)
It is oriented along the positive x-axis.
Then
~
∇G
,
~n = ~ ∇G
~ = ∂~r × ∂~r
∇G
∂u ∂v
and
¨
¨
F~ · ~n dS =
~ dA
F~ · ∇G
σ
σ
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