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Vector Fields.
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2
1
0
-1
-2
-3
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-4
-3
-2
-1
0
1
2
3
4
5
What is a vector field ? e.g. Gravitational field.
(Note that the above diagram is not an illustration of the gravitational vector field, why not?).
Definition
A vector field is a function that associates
a unique vector F (P ) with each point P in a
region of 2-space or 3-space.
1
Thus, a vector field F (·) is either
F : IR2 7→ IR2 ,
or
F : IR3 7→ IR3 .
It maps a position (“fixed”) vector, to a (“nonfixed”) vector in space. In other words, it assigns every point in space with a vector.
Example :






F (x, y) = yi = yi + 0j =
y 

0
6.0
4.8
3.6
2.4
1.2
0.0
-1.2
-2.4
-3.6
-4.8
-6.0
-5
-4
-3
-2
-1
0
2
1
2
3
4
5



.
Example : F (x, y) = xi + 0j
5
4
3
2
1
0
-1
-2
-3
-4
-5
-3
-2
-1
0
1
2
3
2.67
4.00
Example : F (x, y) = −yi + xj
4
-4
-4.00
-2.67
-1.33
0.00
3
1.33
Definition
If r is a radius vector in 2-space or 3-space, and
if c is a constant, then a vector field of the form
c
F (r) =
3r
||r||
is called an inverse-square field.
The radius vector r is just


 x 













r = y
z









= xi + yj + zk.
Thus, in terms of x, y, and z, we have :
c
F (x, y, z) =
3 (xi + yj + zk).
(x2 + y 2 + z 2) 2
Examples : gravitational field, electrostatic field.
4
Gradient Fields.
We recall that if φ is a scalar function, then
∂φ
∂φ
i +
j, or
∇φ =
∂x
∂y
∂φ
∂φ
∂φ
i +
j +
k.
∇φ =
∂x
∂y
∂z
∇φ is called the gradient of φ. Note that the
gradient itself defines a vector field, and it is
called the gradient field of φ.
Example : φ(x, y) = x + y.
The gradient of φ is
∇φ = i + j.
5.00
3.89
2.78
1.67
0.56
-0.56
-1.67
-2.78
-3.89
-5.00
-5.00
-3.57
-2.14
-0.71
0.71
5
2.14
3.57
5.00
Definition
A vector field F is said to be conservative in a
region if it is the gradient field for some function
φ in that region. The function φ is called a
potential function for F in the region.
Example : Inverse-square fields are conservative
in any region that does not contain the origin. E.g., the function
φ(x, y, z) = −
c
(x2 + y 2 + z 2)
1
2
where
(x, y, z) 6= (0, 0, 0).
is a potential function for
c
F (x, y, z) =
3 (xi + yj + zk).
(x2 + y 2 + z 2) 2
6
cx
∇φ(x, y, z) =
3
2
i +
(x2 + y 2 + z 2)
cz
cy
3 j +
3 k
(x2 + y 2 + z 2) 2
(x2 + y 2 + z 2) 2
=
c
(x2 + y 2 + z 2)
= F (x, y, z).
Definition
3
2
(xi + yj + zk)
Suppose
F (x, y, z) = f (x, y, z)i + g(x, y, z)j
+h(x, y, z)k,
where f , g, and h are scalar functions of x, y,
and z, then we define the divergence of F ,
(written as div F or ∇ · F ), by
∂f ∂g ∂h
+
+
.
∇ · F = div F =
∂x ∂y ∂z
7
Definition
Suppose
F (x, y, z) = f (x, y, z)i + g(x, y, z)j
+h(x, y, z)k,
where f , g, and h are scalar functions of x, y,
and z, then we define the curl of F , (written
as curl F or ∇ × F ), by
∂ h ∂ g 
 i
∇ × F = curl F =
−

∂y ∂ z 
∂ h 
∂ f

 j
−
+ 

∂ z ∂ x

∂ f 
∂ g

 k.

−
+

∂x ∂y






Note that by taking the gradient, we bring a scalar-valued function up
to becomes a vector-valued function. However, the divergence brings
a vector-valued function down to a scalar-valued function, whereas
the curl maps a vector-valued function down to another vector-valued
function.
8
grad f , −→ vector-valued function
div F , −→ scalar-valued function
curl F , −→ vector-valued function
Note also that the notation is suggestive :




∂
 f 




 ∂ x 




∇·F =









∂
∂y
∂
∂z


















· g ,
h



∂

i
 f 

 ∂ x 














 ∂ 






∇ × F =  ∂ y  ×  g  = ∂∂x








 ∂ 


f
h
∂z










j
∂
∂y
g
k
∂ .
∂z
h
The del operator :

∂


 ∂ x 




∂
∂
∂


 ∂ 


∇ =  ∂ y  =
i+
j+
k.


∂x
∂y
∂z


 ∂ 
∂z

9
Example : F (x, y, z) = x2yi + 2y 3zj + 3zk.
∂f ∂g ∂h
+
+
∇ · F = div F =
∂x ∂y ∂z
= 2xy + 6y 2z + 3.
∂ 3z ∂ 2y 3z 
 i
∇ × F = curl F =
−

∂y
∂z 


2
∂ x y
∂ 3z 


 j
−
+ 

∂z
∂x


2
3
 ∂ 2y z
∂ x y 


 k
−
+ 

∂x
∂y 








= −2y 3i − x2k.
Laplacian operator
2
2
2
∂
∂
∂
+
+
.
∇2 = ∇ · ∇ =
2
2
2
∂x
∂y
∂z
∇2φ is the divergence of the grad of φ, not the
grad of the div.
10
Line Integrals.
Let C be the graph in the xy-plane of a smooth
vector-valued function r(t) = x(t)i+y(t)j, and
let f (x(t), y(t)) be continuous and non-negative
for a ≤ t ≤ b.
f(x,y)
Q(x(t),y(t),f(x(t),y(t)))
r(b)
P(x(t),y(t),0)
r(a)
The area of the surface swept out by the vertical
line segment from the point P (x(t), y(t), 0) to
the point Q(x(t), y(t), f (x(t), y(t))) as t varies
from a to b is denoted by
A =
Z
C
f (x, y) ds =
Z b
a
f (x(t), y(t))
11
v
u
u
u
u
t
d x 2  d y 2
+
dt
dt
dt



Recall that if s is an arc length parameter for
C, then
ds
=
dt
v
u
u
u
u
t
d x 2  d y 2
+
dt
dt



which is commonly expressed as
ds =
v
u
u
u
u
t
d x 2  d y 2
+
dt
dt
dt



Because ds is so closely related to arc length, the
line integral is sometimes called the integral
of f over C with respect to arc length.
Example : Find the area of the surface extending upward from the circle x2 + y 2 = 1 to the
parabolic cylinder z = 1 − x2.
12
Denote the circle by C and represent it as
r(t) = cos(t)i + sin(t)j
The area is thus given by
2
(1
−
x
) ds
C
Z
A =
Z 2π
=
0
Z 2π
=
0
2
(0 ≤ t ≤ 2π)
r
(1 − cos (t)) (− sin(t))2 + (cos(t))2 dt
1 Z 2π
2
(1 − cos(2t)) dt = π.
sin (t) dt =
2 0
Line integrals with respect to x and y
Z
f (x, y) dx =
Z
g(x, y) dy =
C
C
Z b
a
Z b
a
f (x(t), y(t)) x′ (t) dt,
g(x(t), y(t)) y ′ (t) dt.
Note that line integrals with respect to x along
a segment parallel to the y axis, whereas line
integrals with respect to y along a segment parallel to the x axis.
Short hand
Z
C
f (x, y) dx + g(x, y) dy =
Z
C
13
f (x, y) dx +
Z
C
g(x, y) dy.
Example : Evaluate
Z
C
2xy dx + (x2 + y 2) dy
over the circular arc C given by
x = cos(t), y = sin(t), (0 ≤ t ≤ π2 ).
d

(2
cos(t)
sin(t))
(cos(t)) dt
2xy
dx
=
0
C
dt
Z π
= −2 02 sin2(t) cos(t) dt
π
2
2 3 2
= − sin (t) = − .
0
3
3


Z π
Z
d
2
2
2
2
2
(cos
(t)
+
sin
(t))  (sin(t)) dt
(x
+
y
)
dy
=
0
C
dt
Z π
2
Z
=
Z π
2
0


π
2
cos(t) dt = sin(t)]0 = 1.
Thus,
Z
C
1
2
2xy dx + (x2 + y 2) dy = − + 1 = .
3
3
14
Line integrals in 3-space
If C is a curve in 3-space represented by a smooth
vector-valued function
(a ≤ t ≤ b)
r(t) = x(t)i + y(t)j + z(t)k
then, integrate f (x, y, z) w.r.t. arc-length is
given by
Z
C
f (x, y, z) ds =
Z b
a
f (x(t), y(t), z(t))
v
u
u
u
u
t
d x 2  d y 2  d z 2
+
+
dt.
dt
dt
dt





If f (x, y, z), g(x, y, z), and h(x, y, z) are continuous functions of t on C, and when
Z
C
Z
C
f (x, y, z) dx =
Z
C
g(x, y, z) dy =
Z
C
h(x, y, z) dy =
Z b
f (x(t), y(t), z(t)) x′ (t) dt,
a
Z b
g(x(t), y(t), z(t)) y ′ (t) dt,
a
Z b
h(x(t), y(t), z(t)) z ′ (t) dt
a
f (x, y, z) dx + g(x, y, z) dy + h(x, y, z) dz =
Z
C
f (x, y, z) dx +
Z
C
g(x, y, z) dy +
15
Z
C
h(x, y, z) dz.
Evaluate
Z
C
(xy + z 3) ds
where C is the portion of the helix
x = cos(t), y = sin(t), and z = t
(0 ≤ t ≤ π).
v
u
u
u
u
u
u
u
t
d x 2
2
d y 


d z 2

 dt
+   +
ds =


dt
dt
dt
v
√
u
u
t
2
2
= (− sin(t)) + (cos(t)) + 1 dt = 2 dt.









which yields
Z
C
3
(xy + z ) ds =
Z π
0
√
√
(cos(t) sin(t) + t ) 2 dt
3
π
2
t4 
 sin (t)
+ 
= 2
2
4 0
√ 4
2π
.
=
4

16
Line integrals over piecewise smooth
curves
Z
C
=
Z
C1
+
Z
C2
+··· +
Z
Cn
.
Example : Evaluate
Z
C
x2y dx + xdy
in a counter-clockwise direction around the triangular path as shown
B(1,2)
C2
C3
C1
C1 = r(t) = (1 − t)
A(1,0)






0 

0
17



+t






1 

0



=






t 

0



C2 = r(t) = (1 − t)
C3 = r(t) = (1 − t)



0

 1 






2



+t
+t





1 

=



2

 0 






0
=









1






2t
 1 − t





2 − 2t
d
[0] dt = 0.
dt
d
x y dx + xdy = 0 (1 )(2t) [1] dt +
dt
Z 1
d
[2t] dt = 2.
dt
Z 1
2
Z
1 

Z 1
C1
C2






d
x y dx + xdy = 0 (t )(0) [t] dt +
dt
Z 1
2
Z
Z

C3
2
2
(t)
0
(1)
0
x2y dx + xdy =
d
((1
−
t)
)(2
−
2t)
[1 − t] dt +
0
dt
d
[2 − 2t] dt
dt
Z 1
Z 1
1
3
3
= 2 0 ((t − 1) ) dt + 2 0 (t − 1) dt = − − 1 = − .
2
2
Z 1
2
Z 1
(1 − t)
0
Thus,
Z
C
x2y dx + xdy = 0 + 2 −
18
1
3
= .
2
2






Change of parameter in line integrals
Theorem
Independence of Parametrization
If C is a smooth parametric curve, then the
value of any line integral over C is unchanged
by a smooth change of parameter that preserves
the orientation of C.
Theorem Reversal of Orientation
If C is a smooth parametric curve, then a smooth
change of parameter that reverses the orientation of C changes the sign of a line integral over
C with respect to x, y, or z, but leaves the
value of a line integral over C with respect to
arc length unchanged.
Z
f (x, y) ds = −
Z
f (x, y) ds
f (x, y, z) ds = −
Z
f (x, y, z) ds
−C
Z
−C
19
C
C
Arc-length as a line integral
Let C be a smooth parametric curve
(a ≤ t ≤ b)
r(t) = x(t)i + y(t)j + z(t)k
Then,
Z
C
ds =
v
u
Z b u
u
u
t
a
d x 2  d y 2  d z 2
+
+
dt.
dt
dt
dt





Thus, the arc-length L of a smooth parametric
curve C can be expressed as
L =
Z
C
ds
Mass of a wire
Consider an ideal non-homogeneous bent wire
(sufficiently thin) represented by a curve C, with
density function δ(x, y, z). The total mass
M is then given by
M =
Z
C
δ(x, y, z) ds.
Example : A thin wire shaped in the form of a
helix with parametric equations
x = cos(t),
y = sin(t),
z = 2t,
(0 ≤ t ≤ π)
with density function δ(x, y, z) = kz (k > 0).
20
The total mass is thus given by
M =
Z
C
kz ds
Z π
r
2t (− sin(t))2 + (cos(t))2 + 22 dt
= k
√0 Z π
√
= 2 5k
t dt = 5 kπ 2.
0
Work done
If a particle (or point charge, etc etc) is under
a force field F (x, y, z), then the work done W
required to move it along a path C is given by
W =
Z
C
F (x, y, z) · T (x, y, z) ds.
where T (x, y, z) gives the tangent vector of the
path at (x, y, z). Suppose the path C has parametric form
(a ≤ t ≤ b),
r(t) = x(t)i + y(t)j + z(t)k
then, the tangent vector can be expressed as
dr
T =
=
ds
 dx
 ds

 dy

 ds


dz
ds
21








=
′
x
(t) 








′
y (t)
z ′(t)







dt
ds
Hence, work done W can be expressed as
W =
=
Z
F (x, y, z) · T (x, y, z) ds =
C
x′(t) 


Z
C






′
F (x, y, z) · y (t)
z ′(t)







′
x
(t) 








Z
′
F (x, y, z) · y (t)
C
z ′(t)







dt
ds
ds
dt
If
F (x, y, z) =
F (x, y, z) 
 1







F2(x, y, z)
F3(x, y, z)







= F1(x, y, z)i + F2(x, y, z)j + F3(x, y, z)k,
then
W =
Z
C
′
Z
F1(x, y, z) x (t) dt +
C
+
Z
Z
F2(x, y, z) y ′(t) dt
C
F3(x, y, z) z ′(t) dt
Z
=
Z
F1(x, y, z) dx +
=
Z
F1(x, y, z) dx + F2(x, y, z) dy + F3(x, y, z) dz.
C
C
C
F2(x, y, z) dy +
Short hand :
W =
Z
C
F (x, y, z) · d r
22
C
F3(x, y, z) dz
Example : Suppose a force field in 2-D given by
F (x, y) =



x3 y
(x − y)

= x3yi + (x − y)j,


acts on a particle that moves along the parabola
y = x2 from (−2, 4) to (1, 1).
5
4
3
2
1
0
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
If we use x = t as the parameter, the path is
represented by
r(t) = ti + t2j =



t
2
t

(−2 ≤ t ≤ 1).


Thus, the work done is given by
W =
=
=
=
Z
C
Z 1
−2
Z 1
−2
Z 1
−2
′
F1(x, y) x (t) dt +
x3y · (1) dt +
3
2
(t) (t ) dt +
Z 1
Z 1
−2
−2
Z
C
F2(x, y) y ′(t) dt
(x − y) · (2t) dt
((t) − (t2)) 2t dt
(t5 + 2t2 − 2t3) dt = 3.
23
Path Independence.
Line integrals in general, are path dependent.
However, for some special kind of vector field,
the value of the line integrals may be independent of the path of integration.
Example : Consider the vector field
y


F (x, y) = yi + xj =

x


,
5
4
3
2
1
0
-1
-2
-3
-4
-5
-5
-4
-3
-2
-1
0
24
1
2
3
4
5
The value of C F · dr along the straight line
from (0, 0) to (1, 1) is given by
Z
Z 1
0
t · 1 dt +
Z 1
0
t · 1 dt = 1.
The value of C F · dr along the parabola y = x2
from (0, 0) to (1, 1) is given by
Z
Z 1
t2 · 1 dt +
0
Z 1
0
t · 2t dt = 1.
The value of C F · dr along the parabola y = x3
from (0, 0) to (1, 1) is given by
Z
Z 1
0
3
t · 1 dt +
Z 1
0
t · 3t2 dt = 1.
These results are not coincidental. There is
something special about the vector field F in
order to have these results.
Recall that a vector field F is called conservative if there exists a scalar function of vector
variable φ (called potential) such that F = ∇φ.
Now, consider the following theorem.
25
Theorem
(The Fundamental Theorem of Line Integral)
Suppose that F (x, y) = F1(x, y)i + F2(x, y)j,
where F1 and F2 are continuous in some open
region containing the points (x0, y0) and (x1, y1).
If
F (x, y) = ∇φ(x, y)
at each point of this region, then for any piecewise smooth curve C that starts at (x0, y0), and
ends at (x1, y1), and lies entirely in the region,
we have
Z
C
F (x, y) · dr = φ(x1, y1) − φ(x0, y0).
The theorem simply says if the field is conservative, the line integrals depends only on the
start and end of the path. In the above example, the field F (x, y) = yi + xj is conservative
since F (x, y) = ∇φ(x, y) where φ(x, y) = xy.
Moreover, φ(1, 1) − φ(0, 0) = 1 · 1 − 0 · 0 = 1.
26
From the above argument, for a given conservative vector field, if we have a closed path C, i.e.
the starting point and the end point of the path
are the same, then the value of the line integral
Z
C
F (x, y) · dr = φ(x0, y0) − φ(x0, y0) = 0.
How about the converse ?
We need the concept of a connected region. A
region D is called connected if any two points
in D can be connected by a piecewise smooth
curve that lies entirely in D.
Example : All inverse-square fields are defined
in connected regions. Only at the origin that
the field is not defined, thus, for every 2 points
in space (other than the origin) we can find a
piecewise smooth curve joining them.
27
Theorem
If F is continuous on an open connected region,
then the following are equivalent :
• F is a conservative vector field in the region.
•
•
Z
C
F · dr = 0
for every piecewise smooth closed
curve C in the region.
Z
C
F · dr
is independent of path for every piece-
wise smooth curve C in the region.
A curve is said to be simple if it does not intersect itself anywhere between its end points.
A closed curve is called a simple closed curve
if there are no “knots”.
Simple and closed
Closed but not simple
28
A region D is called simply connected if no
simple closed curve in D encloses points not in
D. Informally stated, a simple connected set is
connected and has no holes.
Example : All inverse-square fields are not defined in simply connected regions since the field
is not defined at the origin. Thus, inverse-square
fields are defined in connected regions but not
simply connected regions.
Theorem (The conservative field test)
Let F (x, y) = F1(x, y)i + F2(x, y)j, where F1
and F2 have continuous first partial derivatives
in an open simply connected region. Then F is
a conservative vector field on that region if and
only if
∂ F1
∂ F2
=
∂y
∂x
at each point of the region.
29
Example : Let F (x, y) = 2xy 3i + (1 + 3x2y 2)j.
F is conservative, because 2xy 3 and (1+3x2y 2)
have continuous first partial derivatives in entire
2-space, and
∂ 2xy 3
∂ (1 + 3x2y 2)
=
= 6xy 2.
∂y
∂x
Since F is conservative, there exists φ, such that
F = ∇φ, or
2xy 3 =
∂φ
∂x
(1 + 3x2y 2) =
and
∂φ
.
∂y
From the first expression, we have
φ =
Z
2xy 3 dx = x2y 3 + k(y)
where k(y) is extra term involving only of y.
Since ∂∂ φy = (1 + 3x2y 2), thus
∂
(x2y 3 + k(y)) = (1 + 3x2y 2)
∂y
3x2y 2 + k ′(y) = (1 + 3x2y 2).
This suggests that k ′(y) = 1, or k(y) = y + K
where K is a constant. Hence,
φ = x2y 3 + y + K.
30
Of course, we can start off with
φ =
Z
(1 + 3x2y 2) dy = y + x2y 3 + k(x)
where k(x) is the extra term involving only of
x. Since ∂∂ φx = 2xy 3, thus
∂
(y + x2y 3 + k(x)) = 2xy 3
∂x
2xy 3 + k ′(x) = 2xy 3.
This suggests that k ′(x) = 0, or k(x) = K
where K is a constant. Hence,
φ = y + x2y 3 + K.
Since F is conservative, thus, if we have a path
C (whatever it is) starts at (1, 4) and ends at
(3, 1)
Z
C
2xy 3 dx + (1 + 3x2y 2) dy = φ(3, 1) − φ(1, 4)
= (10 + K) − (68 + K) = −58.
Note that all the theories above can be extended
to 3-space.
Note also that physicists always call V = −φ
the potential energy.
31
Note that line integrals are usually harder to
evaluate. For one, we have to express the path
in parametric form. If we have a closed curve
path in a simply connected region, do we have
some short-cut to evaluate the line integral ?
We know that if the field is conservative, this
line integral should be zero. How about if the
field is not conservative ?
Theorem (Green’s Theorem)
Let R be a simply connected plane region whose
boundary is a simple, closed, piecewise smooth
curve C oriented counterclockwise. If f (x, y)
and g(x, y) have continuous first partial derivatives on some open set containing R, then
I
C
f (x, y) dx + g(x, y) dy =
32
Z Z
R
∂g ∂f

dA.
−
∂x ∂y


Example : Use Green’s Theorem to evaluate
I
C
x2y dx + xdy
in a counter-clockwise direction around the triangular path as shown
B(1,2)
C2
C3
C1
I
C
A(1,0)
x2y dx + x dy =
=
=
Z Z
R



Z 1 Z 2x
0 0
Z 1
0
2

∂ x ∂ x y
 dA
−
∂x
∂y
(1 − x2) dydx
(2x − 2x3) dx =
1
.
2
Thus, we obtain exactly the same answer as before while we evaluate the line integrals directly.
33
Example : The Work Done by the force field
F (x, y) = (ex − y 3)i + (cos(y) + x3)j
on a particle that travels once around the unit
circle
x2 + y 2 = 1
in the counterclockwise direction is given by
I
C
F (x, y) · dr =
I
(ex − y 3) dx + (cos(y) + x3) dy
C
Z Z
3
∂ (ex − y 3) 
 ∂ (cos(y) + x )
 dA

−
=
R
∂x
∂y
Z Z
Z Z
2
2
=
(3x + 3y ) dA = 3
(x2 + y 2) dA
R
R
Z 2π Z 1
Z 2π
3
3π
= 3
(r2) rdrdθ =
dθ =
.
0
0
4 0
2


Finding area using Green’s Theorem
I
C
Z Z
f (x, y) dx + g(x, y) dy =
R
∂g ∂f

dA.
−
∂x ∂y


If we take f = 0 and g = x, we have
I
C
x dy =
Z Z
dA = A.
R
Or if we take f = −y and g = 0, we have
I
C
(−y) dx =
Z Z
R
34
dA = A.
By adding these two, we can have
1I
(−y) dx + x dy =
2 C
I
C
(−y) dx =
I
C
x dy = A.
Example : The area of the ellipse
x2 y 2
+
= 1
a2 b2
can be computed by one of these formula. The
ellipse (taking counterclockwise) can be represented parametrically by
x = a cos(t),
y = b sin(t)
(0 ≤ t ≤ 2π)
Thus,
A =
=
=
=
1I
(−y) dx + x dy
2 C
1 Z 2π
(−b sin(t))(−a sin(t)) + (a cos(t))(b cos(t)) dt
2 0
1 Z 2π
ab
(sin2(t) + cos2(t)) dt
0
2
Z
2π
1
dt = π ab.
ab
0
2
35
Multiply Connected Regions
Regions of 2-space that are connected but have
finitely many holes are called multiply connected.
The Green’s Theorem can be extended to multiply connected regions.
C1
R
R’
C2
R ’’
Z Z
R
∂g ∂f

dA =
−
∂x ∂y


Z Z
R′
∂g ∂f

dA
−
∂x ∂y


∂g ∂f

+
dA
−
R′′
∂x ∂y
f (x, y) dx + g(x, y) dy
Z Z
=
I


B’dary ofI R’
f (x, y) dx + g(x, y) dy
+
B’dary of R”
=
I
C1
f (x, y) dx + g(x, y) dy +
36
I
C2
f (x, y) dx + g(x, y) dy
Example : Evaluate the integral
I
C
−y dx + x dy
x2 + y 2
if C is a piecewise smooth simple closed curve
oriented counterwise such that
• (a) C does not enclose the origin,
• (b) C encloses the origin.
4.90
3.91
2.92
1.93
0.94
-0.05
-1.04
-2.03
-3.02
-4.01
-5.00
-5.00
-4.01
-3.02
-2.03
-1.04
-0.05
37
0.94
1.93
2.92
3.91
4.90
(a) Let
x
y
and g(x, y) = 2
f (x, y) = − 2
2
x +y
x + y2
Note that
∂g
∂f
−
= 0.
∂x
∂y
Thus, as long as C does not enclose the origin,
the given integral is zero by the Green’s Theorem.
(b) Since the f and g are not defined at the origin, we are having the multiply connected case.
C
C
I
C
−y dx + x dy
+
x2 + y 2
I
−Ca
a
−y dx + x dy
=
x2 + y 2
38
Z Z
R
0 dA = 0.
I
C
−y dx + x dy
=
x2 + y 2
I
Ca
−y dx + x dy
x2 + y 2
=
Z 2π
(−a sin(t))(−a sin(t)) dt + (a cos(t))(a cos(t)) dt
(a cos(t))2 + (a sin(t))2
=
Z 2π
1 dt = 2π.
0
0
Surface integrals
Theorem Surface area formula.
If g has continuous partial derivatives on a closed
region R of the xy-plane, then the area S of that
portion of the surface z = g(x, y) that projects
onto R is given by
S =
v
u
u
u
u
t
Z Z
R
2
∂
z
∂ z 2
+   + 1 d A.
∂x
∂y



To integrate a function f (x, y, z) over the surface, we have
Z Z
σ
f (x, y, z) dS =
Z Z
R
f (x, y, g(x, y))
39
v
u
u
u
u
t
∂ z 2  ∂ z 2
+
+ 1 d A.
∂x
∂y



Theorem
• (a) Let σ be a surface with equation z =
g(x, y) and let R be its projection on the
xy-plane. If g has continuous first partial
derivatives on R and f (x, y, z) is continuous
on σ, then
Z Z
f (x, y, z) dS =
σ
Z Z
R
f (x, y, g(x, y))
v
u
u
u
u
t
∂ z 2  ∂ z 2
+
+ 1 d A.
∂x
∂y



• (b) Let σ be a surface with equation y =
g(x, z) and let R be its projection on the
xz-plane. If g has continuous first partial
derivatives on R and f (x, y, z) is continuous
on σ, then
Z Z
f (x, y, z) dS =
σ
Z Z
R
f (x, g(x, z), z)
v
u
u
u
u
t
∂ y 2  ∂ y 2
+
+ 1 d A.
∂x
∂z



• (c) Let σ be a surface with equation x =
g(y, z) and let R be its projection on the yzplane. If g has continuous first partial derivatives on R and f (x, y, z) is continuous on σ,
then
Z Z
σ
f (x, y, z) dS =
Z Z
R
f (g(y, z), y, z)
40
v
u
u
u
u
t
∂ x 2  ∂ x 2
+
+ 1 d A.
∂y
∂z



Example : Evaluate the surface integral
Z Z
xz dS
σ
where σ is the part of the plane x + y + z = 1
that lies in the first octant.
z
y
y
R
x
x
Now, σ be a surface with equation z = g(x, y) =
1 − x − y and let R be its projection on the xyplane.
Z Z
xz dS =
σ
Z Z
R
=
√
=
√
=
√
r
x (1 − x − y) (−1)2 + (−1)2 + 1 dA
3
Z 1 Z 1−x
3
Z 1
3
0
0
Z 1
0
0
(x − x2 − xy) dydx
xy − x2y −
xy
2
2
1−x


0
dx
√
1

2
3
3
4
√
x
x
x
x
x
3


 dx =
3  −
− x2 + 
+  =
.
2
2
4
3
8 0
24


41
Example : Evaluate the surface integral
Z Z
y 2z 2 dS
σ
s
where σ is the part of the cone z = x2 + y 2
that lies between the planes z = 1 and z = 2.
z
y
x
y
x
Z Z
y 2z 2 dS =
σ
=
Z Z
R
Z Z
=
√
=
√
√
R
2
2
v
u
u
2 u
u
t
y 2 (x2 + y )
y 2 (x2 + y 2)
Z 2π Z 2
1
0
Z 2π Z 2
0
Z 2π
1
√
2
2
y
x



√ 2
+ √ 2
+ 1 dA
x + y2
x + y2


2 dA
(r sin(θ))2 (r2) rdrdθ
r5 sin2(θ) drdθ
2
r6
sin2(θ) dθ
6
r=1
2
0
√
21 2 Z 2π
sin2(θ) dθ
=
2√ 0
√

2π
1
21
21 2  1
2π
θ − sin(2θ)
=
.
=
2
2
4
2
θ=0
=
42

Mass of a curved lamina
An ideal curved lamina has negligible thinkness.
The unit of the density function δ(x, y, z) of a
non-homogeneous curved lamina is [kg m−2].
Example : Find the mass of the lamina that is
portion of the surface y 2 = 4 − z between the
planes x = 0, x = 3, y = 0, and y = 3 if the
density is δ(x, y, z) = y.
z
3
y
3
x
Z Z
y dS =
σ
=
Z Z
R
y 0 + (2y)2 + 1 dA
Z 3Z 3
0
0
r
r
y 4y 2 + 1 dydx =
43
√
1
(37 37 − 1).
4
In general, if the surface is in parametric form,
say,
r(u, v) = r1(u, v)i + r2(u, v)j + r3(u, v)k
(IR2 7→ IR3) is a given parametric form of a
surface, then, the surface area is given by
R Z Z
S =
∂r
∂ r ×
d A.
∂u
∂ v In a similar way, we can integrate f (x, y, z) over
a surface expressed in parametric form
Z Z
f (x, y, z) dS =
σ
Z Z
R
∂ r ∂ r f (r1(u, v), r2(u, v), r3(u, v))
×
dA.
∂ u ∂ v Example : Suppose σ is the hemisphere
r(u, v) = 2 sin(u) cos(v)i + 2 sin(u) sin(v)j + 2 cos(u)k
for which 0 ≤ u ≤ π2 , 0 ≤ v ≤ 2π. Integrating
the function e−z over the hemisphere would be
Z Z
e−z dS
σ
=
=
Z Z
R
Z Z
R



−(2 cos(u)) 




e
2 cos(u) cos(v) 







2 cos(u) sin(v) ×
−2 sin(u)
e−(2 cos(u)) 4 sin(u) dA = 4
−2
= 4π(1 − e ).
44
−2 sin(u) sin(v) 








2 sin(u) cos(v)
π
Z 2π Z 2
0
0
0







dA
e−(2 cos(u)) sin(u) dudv
Surface integrals of vector fields : Flux
Consider a surface σ in a region where a vector
field F is defined. For each point on the surface,
we can find “an arrow” associated with the
field F . Moreover, for each point on the surface,
we can find a unit normal vector n orthogonal
to the tangent plane of that point of the surface.
We can of course find the dot product of these
two vector for every point on the surface. We
call the quantity
Φ =
Z Z
σ
F (x, y, z) · n(x, y, z) dS
the flux of F across σ.
45
We recall that for a given surface of the form
z = g(x, y), we can always construct a scalar
function
G(x, y, z) = z − g(x, y)
such that G(x, y, z) = 0 on every point of the
surface. Moreover, the gradient ±∇ G(x0, y0, z0)
gives the directions normal to the tangent plane
of the surface at (x0, y0, z0). Hence,
n(x, y, z) =
±∇ G(x, y, z)
||∇ G(x, y, z)||
The way we choose the sign depends on the orientation of the surface. If the surface has an
upward orientation, then we choose the positive
one. However, if the surface has a downward
orientation, we would choose the negative one
instead. A smooth surface is said to be orientable if it is possible to construct a unit normal vector at each point of the surface in such
a way that the vectors vary continuously as we
traverse any smooth curve on the surface.
46
Note that not all smooth surfaces are orientable.
For example, the famous Mobius strip.
For different surface representations,
z = g(x, y)
y = g(x, z)
− ∂∂ xz i− ∂∂ yz j +k
− ∂∂ xy i+j − ∂∂ yz k
r
r
n = ∂z 2 ∂z 2
n = ∂y 2 ∂y 2
+
+1
(∂ x) (∂ y)
( ∂ x ) +( ∂ z ) +1
x = g(y, z)
i− ∂∂ xy j − ∂∂ xz k
r
n = ∂x 2 ∂x 2
( ∂ y ) +( ∂ z ) +1
positive (up)
positive (right)
∂ z i+ ∂ z j −k
∂y
i−j + ∂ y k
∂
n = r ∂xz 2 ∂ y ∂ z 2
n = r ∂∂xy 2 ∂∂yz 2
( ∂ x ) +( ∂ y ) +1
( ∂ x ) +( ∂ z ) +1
positive (out of paper)
−i+ ∂∂ xy j + ∂∂ xz k
n = r ∂x 2 ∂x 2
( ∂ y ) +( ∂ z ) +1
negative (down)
negative (left)
negative (into the paper)
Example : Suppose F (x, y, z) = xi + yj + zk.
Let σ be the portion of the surface z = 1 − x2 −
y 2 that lies above the xy-plane, and suppose
that σ is oriented upward. Then, the flux of F
across σ is given by
47
Φ =
Z Z
=
Z Z
=
Z
=
Z
=
Z Z
=
σ
F · n dS =
R
F ·n
v
u
u
u
u
t
∂ z 2  ∂ z 2
+
+ 1 dA
∂x
∂y



∂z
∂z
−
i
−
j + k dA
F
·
R
∂x
∂y


Z
∂
z
∂
z
(xi + yj + zk) · − i −
j + k dA
R
∂ xZ Z ∂ y
Z
(2x2 + 2y 2 + (1 − x2 − y 2)) dA
(2x2 + 2y 2 + z) dA =
R
R
R


(x2 + y 2 + 1) dA
Z 2π Z 1
0
Z Z
0
(r2 + 1) rdrdθ =
3π
.
2
Theorem
Let σ be a smooth surface of the form z =
g(x, y), y = g(x, z), or x = g(y, z), and suppose that the equation is rewritten as G(x, y, z) =
0 by taking g to the left side. Let R be the projection of σ on the xy-plane if z = g(x, y), on
the xz-plane if y = g(x, z), and on the yz-plane
if x = g(y, z). If g is continuous, and has continuous first partial derivative on R, then
Z Z
σ
F · n dS = ±
Z Z
R
F · ∇G dA
where the + sign is used if σ has positive orientation and the − sign if it has negative orientation.
48
Example : Suppose F (x, y, z) = zk. Let σ be
the sphere x2 +y 2 +z 2 = a2 oriented by outward
normals. Thus
Z Z
σ
F · n dS =
Z Z
σ1
F · n dS +
Z Z
σ2
F · n dS
where σ1 is the upper hemisphere and σ2 is the
lower hemisphere.
For the upper hemisphere,
s
we have z = a2 − x2 − y 2.
Z Z
σ1
F · n dS
y
x
√
√
i
+
j + k dA
(zk)
·
=
2
2
2
2
2
2
R
a − rx − y
a −x −y
Z Z
Z Z
z dA =
=
a2 − x2 − y 2 dA
R
R
Z 2π Z a √
2πa3
2
2
.
=
a − r rdrdθ =
0
0
3
Z Z


s
For the lower hemisphere, z = − a2 − x2 − y 2
Z Z
σ2
F · n dS
y
x
√
√
i
+
j − k dA
(zk)
·
2
2
2
2
2
2
R
a −x −y
a −x −y
3
r
Z Z
Z
2πa
−z dA =
a2 − x2 − y 2 dA =
.
R
R
3
=
Z Z
=
Z


Thus,
Z Z
σ
4πa3
.
F · n dS =
3
49
The Divergence (Gauss’) Theorem
Theorem
Let G be a solid with surface σ oriented outward. If
F (x, y, z) = f (x, y, z)i + g(x, y, z)j + h(x, y, z)k
where f , g, and h have continuous first partial
derivative on some open set containing G, then
Z Z
σ
F · n dS =
Z Z Z
G
∇ · F dV.
The theorem simply relates a triple integral over
a volume to its boundary as some surface integral. Just like the Green’s Theorem that relates
a double integral over a region to its boundary
as some line integral, or the Fundamental Theorem of Calculus that relates a single integral
over an interval to its end points.
50
Example : Recall from the last example from
Unit 9 Part 2 that : Suppose F (x, y, z) = zk
and σ is the sphere x2 + y 2 + z 2 = a2 oriented
by outward normals. The value of the surface
integral (flux) is given by
Z Z
σ
4πa3
F · n dS =
.
3
If we make use of the Divergence Theorem, we
have
Z Z
σ
F · n dS =
Z Z Z
=
Z Z Z
=
G
Z Z G
Z
∇ · F dV
∂f ∂g ∂h
+
+
dV
∂x ∂y ∂z
0 + 0 + 1 dV
G
4πa3
.
= volume of the sphere =
3
Thus, we obtain the same result as before with
simplier calculations.
Example s: Suppose we only have half a sphere
now z = a2 − x2 − y 2, with a vector field given
by
F (x, y, z) = x3i + y 3j + z 3k,
51
Using the divergence theorem, the flux is thus
given by
Z Z
σ
F · n dS =
∂f ∂g ∂h
+
+
dV
∂x ∂y ∂z
3x2 + 3y 2 + 3z 2 dV
Z Z Z
=
Z Z G
Z
=
Z 2π Z π Z a
2
G
0
= 3
= 3
0
3ρ2 ρ2 sin(φ)dρdφdθ
0
Z 2π Z π Z a
2
0
0
0
ρ4 sin(φ)dρdφdθ
a

5
Z 2π Z π
2 

0
3a5
=
5
3a5
=
5
3a5
=
5
ρ
dφdθ
sin(φ)
5
ρ=0
0
Z 2π Z π
2
0
0
sin(φ)dφdθ
π
2
Z 2π
[− cos(φ)]φ=0 dθ
Z 2π
6πa5
.
dθ =
5
0
0
Example : Let σ be the surface of the cube
oriented outward,
z
1
1
1
x
52
y
and let
F (x, y, z) = 2xi + 3yj + z 2k,
Using the divergence theorem, the flux is thus
given by
Z Z
σ
∂f ∂g ∂h
+
+
dV
∂x ∂y ∂z
2 + 3 + 2z dV =
F · n dS =
Z Z Z
=
Z Z Z
=
Z 1Z 1Z 1
G
G
0
0
Z Z Z
5 + 2z dV
G
0
(5 + 2z) dzdydx = 6.
Example : Let σ be the surface of the solid G
enclosed by the circular cylinder x2 +y 2 = 9 and
the planes z = 0 and z =
2, oriented outward.
z
2
y
3
x
and let F (x, y, z) = x3i + y 3j + z 2k,
53
Using the divergence theorem, the flux is thus
given by
Z Z
F · n dS =
σ
∂f ∂g ∂h
+
+
dV
∂x ∂y ∂z
3x2 + 3y 2 + 2z dV
Z Z Z
=
Z Z G
Z
=
Z 2π Z 3 Z 2
G
=
=
=
=
=
0
0
0
Z 2π Z 3 Z 2
0 0
0
Z 2π Z 3 (3r2 + 2z) rdzdrdθ
(3r3 + 2rz) dzdrdθ
3
3r z + rz
0
0
Z 2π Z 3 0
0
Z 2π
0
2
z=0
drdθ
6r + 4r drdθ
4

Z 2π


0
3
2
3
3r
+ 2r2
dθ
2
r=0
279
dθ = 279π.
2
Divergence viewed as flux density
Suppose we have a very small sphere G of surface σ centered at (x0, y0, z0).
Φ(G) =
Z Z
σ
F · n dS =
Z Z Z
G
∇ · F dV
≈ ∇ · F (x0, y0, z0)
Z Z Z
dV
G
= ∇ · F (x0, y0, z0) × (volume of G).
54
Thus, we have
∇ · F (x0, y0, z0) =
lim
vol(G)→0
1
vol(G)
Z
Z
σ(G)
F · n dS
This limit, called the flux density of F at
the point (x0, y0, z0), is sometimes taken as
the definition of divergence. Note that we do not
require the introduction of a coordinate system
in order to define it this way.
Incompressible Fluid Flows
If ∇ · F > 0, then, Φ(G) > 0 for a sufficiently
small sphere G, thus, net flow out of G is pos-
itive. We call points with ∇ · F > 0 sources.
Similarly, if ∇ · F < 0, then net flow out of G is
negative. We call points with ∇ · F < 0 sinks.
Hence, for incompressible fluid flows without sources and sinks, we
must have ∇ · F = 0 at every point. In hydrodynamics, this is called
the continuity equation for incompressible fluids.
55
Gauss’ Law for inverse-square fields
Theorem
If
F (r) =
c
r
||r||3
is an inverse square field in 3-space, and if σ is a
closed orientable surface that surrounds the origin and has outward orientation, then the flux
Φ of F across σ is
Φ =
Z Z
σ
F · n dS = 4 π c.
Example : Electrostatic field
Q
r
4πǫ0||r||3
F (r) =
Thus, the flux is given by
Φ =
Z Z
σ
F · n dS
Q 
Q
=
= 4π 
.
4πǫ0
ǫ0


56
Stokes’ Theorem
Theorem
Let σ be a piecewise smooth orientable surface
that is bounded by a simple, closed, piecewise
smooth curve C with positive orientation. If the
components of the vector field
F (x, y, z) = f (x, y, z)i + g(x, y, z)j + h(x, y, z)k
are continuous and have continuous first partial
derivative on some open set containing σ, and
if T is the unit tangent vector to C, then
I
C
F · T ds =
Z Z
σ
(∇ × F ) · n dS.
Note that dd rs = T . If the surface σ can be
described by G(x, y, z) = 0, then
I
C
F · dr =
Z Z
R
(∇ × F ) · ∇ G dA.
57
Note that in this theorem, if we choose
F = f (x, y)i + g(x, y)j + 0k,
and let σ be a flat surface over region R, then
it can be shown that the above relation degenerates into
I
C
f (x, y) dx + g(x, y) dy =
Z Z
R
∂g ∂f

dA.
−
∂x ∂y


Thus, the Green’s Theorem is just a special case
of the Stokes’ Theorem.
Example : Suppose
F (x, y, z) = 2zi + 3xj + 5yk,
and let σ be the portion of the paraboloid z =
4 − x2 − y 2 for which z ≥ 0, (oriented upward),
and let C be the circle x2 + y 2 = 4 that forms
the boundary of σ in the xy-plane.
Thus, parametrization of the circle is given by
x = 2 cos(t),
y = 2 sin(t),
58
z=0
(0 ≤ t ≤ 2π)
I
C
=
=
Z 2π
0
Z 2π
0
F dr =
I
C
2z dx + 3x dy + 5y dz
[2(0)(−2 sin(t)) + (6 cos(t))(2 cos(t)) + 5(2 sin(t))(0)] dt
2
12 cos (t) dt = 12π.
On the other hand, the curl of F is given by
∂
∂x
i
∇×F =
2z
j
∂
∂y
∂ ∂ z k
= 5i + 2j + 3k.
3x 5y
Thus, if we let G = z − (4 − x2 − y 2), then
Z Z
R
(∇ × F ) · ∇ G dA =
=
=
Z Z
Z R
Z
Z Z
R
∂z
∂z
j + k dA
(∇ × F ) · − i −
∂x
∂y
∂z
∂z
(5i + 2j + 3k) · − i −
j + k dA
∂x
∂y
(5i + 2j + 3k) · (2xi + 2yj + k) dA

=
Z R
Z
=
Z 2π Z 2
(10x + 4y + 3) dA
R
0


0
(10r cos(θ) + 4r sin(θ) + 3) rdrdθ
= 12π.
59

Example : Let C be the rectangle in the plane
z = y oriented as
z
3
3
y
1
x
and let
F (x, y, z) = x2i + 4xy 3j + y 2xk.
Now, for the curl of F , we have
∂
∂x
2
i
∇×F =
x
j
∂
∂y
∂ ∂ z 2 k
4xy 3 y x
= 2xyi − y 2j + 4y 3k.
Thus, if we let G = z − (y), then
I
C
F dr =
Z Z
(∇ × F ) · ∇ G dA
=
Z Z
∂z
∂z
j − k dA
(∇ × F ) · + i +
∂x
∂y
R
R

60

Z Z
=
∂z
∂z
(2xyi − y j + 4y k) · + i +
j − k dA
∂x
∂y
(2xyi − y 2j + 4y 3k) · (0i + j − k) dA
2
=
Z R
Z
=
Z 1Z 3
R
0
0

3

(−y 2 − 4y 3) dydx = −90.
Circulation
Suppose we have a very small closed oriented
curve disk-shape region with boundary Ca centered at P0 and with radius a and area σa. The
term
I
Ca
F · T ds
would be the circulation of F around Ca. Now,
I
Ca
F · T ds =
Z Z
σa
(∇ × F ) · n dS
≈ ((∇ × F )(P0)) · n
Z Z
dS
σa
= ((∇ × F )(P0)) · n × area of σa.
Hence
1
((∇ × F )(P0)) · n = lim
a→0 Area(σa )
I
Ca
F · T ds
We can thus, consider ((∇ × F )(P0)) · n as the
circulation density of F at P0 in the direction
of n.
61