Mass Problem.: M Lim X F (X, Y, Z) S F (X, Y, Z) Ds
Mass Problem.: M Lim X F (X, Y, Z) S F (X, Y, Z) Ds
Mass Problem.: M Lim X F (X, Y, Z) S F (X, Y, Z) Ds
1 Line Integrals
Mass problem. Find the mass M of a very thin wire whose linear
density function (the mass per unit length) is known.
We model the wire by a smooth curve C between two points P and Q
in 3-space. Given any point (x, y, z) on C, we let f (x, y, z) denote the
corresponding value of the density function.
1
• The mass M of the wire is
ˆ
M= f (x, y, z) ds
C
• If C is a curve
´ in the xy-plane and f (x, y) is nonnegative function
on C, then C f (x, y) ds is equal to the area of the “sheet” that
is swept out by a vertical line segment that extends upward from
(x, y) to a height f (x, y) and moves along C
ˆ
∆Ak ≈ f (x∗k , yk∗)∆sk ⇒ A = f (x, y) ds
C
2
Evaluating line integrals
ˆ n
X
f (x, y) ds = lim f (x∗k , yk∗)∆sk
C max ∆sk →0
k=1
Xn
= lim f (x(t∗k ), y(t∗k ))|~r 0(t∗k )|∆tk
max ∆tk →0
k=1
ˆ b
= f (x(t), y(t))|~r 0(t)| dt
a
ˆ b
s
2 2
dx dy
= f (x(t), y(t)) + dt
a dt dt
1. C : 21 (t + t2)~i + 21 (t + t2) ~j , 0 ≤ t ≤ 1
2. C : (2 − 2t)~i + (1 − t) ~j , 0 ≤ t ≤ 1
3
Similarly, if C is a curve in 3-space smoothly parametrised
~r = x(t)~i + y(t) ~j + z(t) ~k , then
ˆ ˆ b
f (x, y, z) ds = f (x(t), y(t), z(t))|~r 0(t)| dt
C a
ˆ b
s
2 2 2
dx dy dz
= f (x(t), y(t), z(t)) + + dt
a dt dt dt
´
Example. Find C (xy + z 3)ds for
ˆ n
X
f (x, y, z) dy = lim f (x∗k , yk∗, zk∗)∆yk
C max ∆sk →0
k=1
ˆ n
X
f (x, y, z) dz = lim f (x∗k , yk∗, zk∗)∆zk
C max ∆sk →0
k=1
4
The sign of these line integrals depends on the orientation of C.
Reversing the orientation changes the sign.
Thus, one should find parametric equations for C in which
the orientation of C is in the direction of increasing t, and then
ˆ ˆ b
f (x, y, z) dx = f (x(t), y(t), z(t)) x0(t) dt
C a
´
Example. Find C (1 + x2y)dy for
1. C : 12 (t + t2)~i + 21 (t + t2) ~j , 0 ≤ t ≤ 1
2. C : (2 − 2t)~i + (1 − t) ~j , 0 ≤ t ≤ 1
while ˆ ˆ
f (x, y) ds = + f (x, y) ds
C −C
Convention
ˆ ˆ ˆ
f (x, y) dx + g(x, y) dy = f (x, y) dx + g(x, y) dy
C C C
We have
ˆ ˆ b
f (x, y) dx+g(x, y) dy = (f (x(t), y(t)) x0(t) + g(x(t), y(t)) y 0(t)) dt
C a
5
Integrating a vector field along a curve
One can say that a vector field is a vector-valued function with the
number of components equal to the number of independent variables
(coordinates).
ˆ ˆ
F~ · d~r = (f ~i + g ~j + h ~k) · (dx~i + dy ~j + dz ~k)
C ˆC
= f (x, y, z) dx + g(x, y, z) dy + h(x, y, z) dz
C
6
Example.
The curve C is a line segment connecting the points (−π/2 , π) and
(3π/2 , −2π/3). Parameterise C, and evaluate
ˆ
F~ · d~r
C
where
3y
F~ (x, y) = (−6xy + 3π 3 sin 3x)~i − (3x2 + 2π 3 cos ) ~j
2
´
Answer : F~ · d~r = 47 π 3 ≈ 121.441
C 12
= F~ · T~ ds
C
where T~ = ~r 0(s) is a unit tangent vector along C.
7
Line integrals along piecewise smooth curves
where
3y
F~ (x, y) = (−6xy + 3π 3 sin 3x)~i − (3x2 + 2π 3 cos ) ~j
2
8
2 Independence of path; Conservative vector fields
´
The curve C in C F~ · d~r is called the path of integration.
If F~ is conservative, i.e.
~ = ∂φ ~i + ∂φ ~j ,
F~ = ∇φ ~ = ∂φ ~i + ∂φ ~j + ∂φ ~k
or F~ = ∇φ
∂x ∂y ∂x ∂y ∂z
or, equivalently
ˆ
~ · d~r = φ(x1, y1) − φ(x0, y0)
∇φ
C
10
A test for conservative vector fields
A connected set D is
simply connected
if no simple closed curve in D
encloses points that are not in D
(has no holes).
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Theorem. (Conservative Field Test)
∂f ∂g
=
∂y ∂x
at each point in D.
Conversely, if D is simply connected and fy = gx holds at each point
in D, then F~ (x, y) = f (x, y)~i + g(x, y) ~j is conservative.
If
~ = ∂φ ~i + ∂φ ~j
F~ = ∇φ
∂x ∂y
then
∂f ∂ ∂φ ∂ ∂φ ∂g
= = =
∂y ∂y ∂x ∂x ∂y ∂x
Test.
∂f ∂g ∂f ∂h ∂g ∂h
= , = , =
∂y ∂x ∂z ∂x ∂z ∂y
12
3 Green’s Theorem
ˆ ¨ ˆ ¨
∂f ∂g
f (x, y) dx = − dA , g(x, y) dy = dA
C R ∂y C R ∂x
13
We have
ˆ ˆ
f (x, y) dx = f (x, y) dx
C
ˆ C1
+ f (x, y) dx
C2
C1 : x = t , y = g1(t);
−C2 : x = t , y = g2(t);
Thus
ˆ ˆ b ˆ b
f (x, y) dx = f (t, g1(t)) dt − f (t, g2(t)) dt
C
ˆ b a a
¨
∂g ∂f
A= dA =⇒ − =1
R ∂x ∂y
Simplest solutions
‰
i) f = 0 , g = x =⇒ A = x dy
C
‰
ii) f = −y , g = 0 =⇒ A = − y dx
C
‰
y x 1
iii) f = − , g = =⇒ A = −y dx + x dy
2 2 2 C
Positive orientation of R:
the outer boundary is
oriented counterclockwise
the inner boundaries
are oriented clockwise.
Divide R into two simply
connected regions R0 and R00.
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¨ ¨ ¨
∂g ∂f ∂g ∂f ∂g ∂f
− dA = − dA + − dA
R ∂x ∂y R0 ∂x ∂y R00 ∂x ∂y
‰ ‰
= f (x, y) dx + g(x, y) dy + f (x, y) dx + g(x, y) dy
boundary boundary
of R0 of R00
‰
= f (x, y) dx + g(x, y) dy + f (x, y) dx + g(x, y) dy
C1 C2
˛
= f (x, y) dx + g(x, y) dy
boundary
of R
Example. Evaluate ‰
−y dx + x dy
C x2 + y 2
if (a) C does not enclose the origin, and (b) C encloses the origin.
16
4 Surface Integrals
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
17
Evaluating surface integrals
2 2 2
˜
Example. σ: x + y + z = 1. σ y 2 dS =? Answer : 4π/3.
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
σ R ∂x ∂y
5 Flux and so on
19
Oriented surfaces
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.
21
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 direc-
tion minus the volume that passes through σ in the negative direction.
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Evaluating flux integrals
˜
The integrals of the form σ F~ · ~n dS are called flux integrals.
¨
∂~r ∂~r
= F~ · × dA
R ∂u ∂v
~
∇G
~n = , ~ = ∂~r × ∂~r
∇G
~
∇G
∂u ∂v
because ∇G ~ = ~k + · · · and
~ is ⊥ σ, and ∇G ∂~r ∂~r
× ∂v = ~k + · · ·
∂u
Thus,
¨ ¨ ¨
∂z ∂z ~j + ~k dA
F~ · ~n dS = F~ · ∇G
~ dA = F~ · − ~i −
σ R R ∂x ∂y
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 = , ~ = ∂~r × ∂~r
∇G
~
∇G ∂u ∂v
and
¨ ¨
F~ · ~n dS = F~ · ∇G
~ dA
σ R
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6 The Divergence Theorem
∂f ∂g ∂h
divF~ = + +
∂x ∂y ∂z
˚ ¨ "ˆ g2 (x,y)
#
∂h ∂h
dV = dz dA
G ∂z R g1 (x,y) ∂z
¨
= [h(x, y, g2(x, y)) − h(x, y, g1(x, y))] dA
R
¨ ¨ ¨
h ~k · ~n dS = h ~k · ~n dS + h ~k · ~n dS
σ
¨σ1 σ2
∂z ∂z ~ ~
= h(x, y, g1(x, y)) ~k · ( ~i + j − k) dA
¨ R ∂x ∂y
∂z ∂z ~ ~
+ h(x, y, g2(x, y)) ~k · (− ~i − j + k) dA
R ∂x ∂y
where we have taken into account that
¨
~k · ~n = 0 ⇒ h ~k · ~n dS = 0
σ 3
σ3
26
Example. Find the flux of the vector field
F(x, y, z) = (−3x+12xy 2+4z 3) i−(2y 2−y+4x2) j+(4+3z+5yz−4y 3) k
across the surface σ: x2 + y 2 + z 2 = 9 with outward orientation.
Answer : flux = 40685 π
¨ ˚ ˚
Φ(G) = F~ ·~n dS = divF~ dV ≈ divF~ (P0) dV = divF~ (P0)V
σ(G) G G
Thus
Φ(G)
divF~ (P0) ≈
vol(G)
Φ(G)
where vol(G) is called the outward flux density of F~ across G.
Taking the limit vol(G) → 0 ( → 0), we get
¨
Φ(G) 1
divF~ (P0) = lim = lim F~ · ~n dS .
vol(G)→0 vol(G) vol(G)→0 vol(G) σ(G)
27
Sources and sinks
Consider an incompressible fluid which means that its density is con-
stant. Consider a point P0 in fluid, and a small sphere G centred at
P0. If divF~ (P0) > 0 then Φ(G) > 0, and more fluid goes out through
the sphere than comes in. Since the fluid is incompressible this can
only happen if fluid is entering the flow at P0 otherwise the density
would decrees. Similarly, if divF~ (P0) < 0 then Φ(G) < 0, then fluid
is leaving the flow at P0. In an incompressible fluid, points at which
divF~ (P0) > 0 are called sources, and points at which divF~ (P0) < 0
are called sinks. If there are no sources and sinks we have
divF~ (P ) = 0 if P 6= O. Thus
˜ ˜
~ · ~n dS + ~ na dS = 0
σ F σa F · ~
~na = −~r /r ⇒
˜ ˜
~ · ~n dS = ~ ~r
σ F σa F · r dS
˜
= ac2 σa dS = 4πc
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7 Stoke’s Theorem
30