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Chapter 14 Vector Calculus Section 14.1 Vector Fields Definition: Vector Field in Three Dimensions Let f, g, and h be defined on a region D in 3 . A vector field in 3 is a function F
that assigns a vector to each
point in its domain, D a vector, written as
, , , , , , , ,
, , , , , , , ,
F x y z f x y z g x y z h x y z
f x y z g x y z h x y z
i j
A vector field , ,F f g h
is continuous and differentiable on a region D of 3 if f, g, and h are continuous or
differentiable on D, respectively. Note, a vector field in 2 is a special case where there is no z component, i.e. , , 0h x y z .
Example: 22 tan / 2yzF x y e x i j k
.
Example: Graph ,F x y y x i j
Applications: - Velocity vector fields
o Flow of water, wind - Force fields - Gravitational fields - Electrical fields - Magnetic fields
Definition: Radial Vector Fields in 3 Let , ,r x y z . A vector field in the form
, , p
rF x y z
r
Where p is a real number. Radial vector fields have the property that their vectors point directly toward or away from the origin at all points.
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Gradient Fields The gradient field of a differentiable function
(the potential function) is the field of gradient vectors
, ,Fx y z x y z
i j k
The gradient field gives direction for which f is increasing most rapidly.
Section 14.2 Line Integrals Definition: Scalar Line Integral in the Plane, Arc Length Parameter
Suppose the scalar-valued function f is defined on a smooth curve C: ,r s x s y s, parameterized by the arc
length s. The line integral of f over C is
0
, lim ,k ksC
f x s y s ds f x s y s s
Provided the limit exists over all partitions of C. When the limit exists, f is said to be integrable on C.
Theorem 14.1 Evaluate Scalar Line Integrals in 2
Let f be continuous on a smooth curve C: ,r t x t y t, for a t b . Then
2 2
, '
, ' '
b
C a
b
a
f ds f x t y t r t dt
f x t y t x t y t dt
Note, the line integral exists if f is continuous at each point of C. Also note, ds is the arc length of a segment of C, so
2 2 2
'dx dy dz
ds dt r t dtdt dt dt
Example: Evaluate 2
C
x zds where cosx t , 2y t , and sinz t for 0 t .
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Subdivision Rule Suppose C is piecewise smooth 1 2 nC C C C where iC is smooth then
1 2 nC C C C
fds fds fds fds
Example: Suppose 1 : , for 3 0C x t y t t and 42 : 2 , for 0 1C x t y t t evaluate
C
xyds where
1 2C C C .
Note suppose we went from 3,3 to 2,1 in a straight line, what would the line integral in the previous example
be? Line Integrals of Vector Fields Definition: Line Integral of a Vector Field Let F
be a vector field that is continuous on a region containing a smooth oriented curve C parameterized by arc
length. Let T
be the unit tangent vector at each point of C consistent with the orientation. The line integral of F
over C is
C
F Tds
Note: 'C C C
F dr F r t dt F Tds
since '
'
r tT
r t and 'ds r t dt .
Let F f g h i j k
be a vector field where f, g and h are continuous functions of , , and x y z and let C be a
piecewise smooth orientable curve with parameterization , ,r t x t y t z t for a t b then
, , ' , ' , 'dr dx dy dz x t dt y t dt z t dt and the line integral of F
along C is
C C
F dr f dx gdy hdz
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Example: Evaluate C
F dr
for , ,F y z x
and 2 , ,t tr t t e e for 0 1t .
Think about , ,C
f x y z ds vs. C C
F dr F Tds
:
The first integral , , 'C C
f x y ds f x t y t r t dt
: Suppose that f is the height of a fence along the curve C,
then the integral is the area of the fence.
For the second integral C C
F dr F Tds
: Recall the scalar projection of u
onto v
cosv
u vcomp u u
v
. Now
think about projecting F
onto T
, then T
F Tcomp F F T
T
since 1T
. So F T
is the height of a picket of
the fence. Comment: The parameterization of a curve C determines its orientation.
- Positive Direction
As t increases from a to b the point ,x t y t moves along the curve from point A to B.
- Opposite Direction
At t increases from a to b the point ,x t y t moves along the curve from point B to A.
B
A
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Now looking at our line integrals, because ds is always positive then C C
fds fds
. However,
C C
F Tds F Tds
(the force acts in the opposite directions).
For example: Consider the line from 1,2 to 9,8 . Let 1 : 1 8 , 2 6C x t y t for 0 1t then
1
1
0
1 8 2 6 10 290C
xyds t t dt and 1
70C
ydx xdy . Let 2 : 9 4 , 8 3C x t y t for 0 2t then
2
2
0
9 4 8 3 5 290C
xyds t t dt and 2
2 2
0 0
8 3 4 9 4 3 70C
ydx xdy t dt t dt .
Work
Recall that cosD
W F D F D comp F D
now the displacement D
is the unit tangent vector, T
so the
work done by a force F f g h i j k
in moving an object over a smooth curve r t
from t a to t b is t b
t a
W F Tds
Different Ways to Write the Work Integral t b
t a
t b
t a
b
a
b
a
b
a
W F Tds
F dr
drF dt
dt
dx dy dzf g h dt
dt dt dt
f dx gdy hdz
Example: Determine the work done by the force field 2,F x y x xy i j
moving the particle along the quarter of
the circle cos ,sinr t t t for 0 / 2t .
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If F
is a velocity field of fluid flowing through a region in space, then the following integral gives the fluid’s flow along the curve. Definitions: If r t
is a smooth curve in the domain of a continuous velocity field F
, the flow along the curve from t a to
t b is b
a
Flow F Tds
The integral in this case is called a flow integral. If the curve is a closed loop, the flow is called the circulation around the curve. To find the rate at which fluid is entering or leaving the region enclosed by a smooth curve C in the xy-plane. Definition: If C is a smooth closed curve in the domain of a continuous vector field , ,F f x y g x y i j
in the plane and if
n
is the outward-pointing unit normal vector on C, the flux of F
across C is
Flux of F across C
C F nds
Note:
- Flux is the integral of the scalar component of F
in the direction of the outward normal. - Circulation is the integral of the scalar component of F
in the direction of the unit tangent vector.
For a curve in the plane, the outward normal vector is n T k
if the curve is traversed counterclockwise as t
increases. (The outward normal vector is n k T
if the curve is traversed clockwise as t increases.)
Now dx dy dy dx
nds ds ds dx
T×k i j k i j
and dy dx
F n f gds ds
.
Calculating Flux Across a Smooth Closed Plane Curve
Flux of across C
F f g C fdy gdx i j
Example: Find the flux of the field 2 , 3F x y
across the circle cos , sinr t a t a t for 0 2t .
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Section 14.3 Conservative Vector Fields Definition: Simple and Closed Curves Suppose a curve C is described parametrically by r t
, where a t b . Then C is a simple curve if 1 2r t r t
for all 1t and 2t , with 1 2a t t b ; that is, C never intersects itself between its endpoints. The curve C is closed if
r a r b ; that is, the initial and terminal points of C are the same.
Definition: Connected and Simply Connected Regions
An open region R in 2 3 is connected if it is possible to connect any two points of R by a continuous curve
lying in R. An open region R is simple connected if every closed simple curve in R can be deformed and contracted to a point in R. Definition: Conservative Vector Field A vector field F
is said to be conservative on a region if there exists a scalar function such that F
on that
region. Theorem 14.3 Test for Conservative Vector Fields Let , , , , , ,F f x y z g x y z h x y z i j k
be a field on a connected and simply connected domain whose
component functions have continuous first partial derivatives. Then F
is conservative if and only if
, , andf g f h g h
y x z x z y
Example: Show that the field cos sinx xF e y e y z i j k
is conservative and find a potential function for it.
Example: Show that the field F z y z y x i j k
is not conservative.
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Example: Show the vector field 2 2 2 2 22 2 , 4 2 ,2 3xy z yz xy x xz y x y zx xy z is conservative and find
its potential function. Theorem 14.4: Fundamental Theorem of Line Integrals Let C be a smooth curve joining the points A to B in the plane or in space and given by r t
. Let be a
differentiable function with continuous gradient vector F
on a domain containing C. Then
C C
F dr dr B A
Proof: Since F
then
'b
C C a
b
a
b
a
F dr dr r t r t dt
dx dy dzdt
x dt y dt z dt
dr t dt
dt
r b r a B A
Example: Find the work done by 2 2 zF x y y x ze i j k
over the following paths from 1,0,0 to 1,0,1 .
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Does path matter? Consider the following paths for the previous example: a. The line segment 1, 0, 0 1x y z .
b. The helix cos ,sin , , 0 22
tr t t t t
c. The x-axis from 1,0,0 to 0,0,0 followed by the parabola 2 , 0z x y from 0,0,0 to 1,0,1 .
Theorem 14.5 Line Integrals on Closed Curves
Let R in 2 (or in 3 ) be an open region. Then F
is a conservative vector field on R if and only if 0C
F dr
on all simple closed smooth oriented curves C in R
Section 14.4 Green’s Theorem in the Plane Definition: The k-component of the curl (circulation density) of a vector field F f g i j
at the point ,x y is the scalar
curl g f
Fx y
k
Theorem 14.6: Green’s Theorem Circulation Form Let C be a simple closed smooth curve, oriented counterclockwise, that encloses a connected and simply connected region R in the plane. Let F f g i j
be a vector field with f and g continuous first partial derivatives in an open
region containing R. Then the counterclockwise circulation of F
around C equals the double integral of
curl F k
over R enclosed by C.
C C R
g fF Tds fdx gdy dxdy
x y
Example: Apply Green’s Theorem to evaluate 6 2C
y x dx y x dx over C the circle 2 32 3 4x y .
Definition: The divergence (flux density) of a vector field F f g i j
at the point ,x y is
f g
div Fx y
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Theorem 14.7: Green’s Theorem (Flux Form) Let C be a simple closed smooth curve, oriented counterclockwise, that encloses a connected and simply connected region R in the plane. Let F f g i j
be a vector field with f and g continuous first partial derivatives in an open
region containing R. Then the outward flux of F
across C equals the double integral of div F
over the region R enclosed by C.
C C R
f gF nds fdy gdx dxdy
x y
Example: Evaluate 2 22C
x dy y dx and C is the upper half of the unit circle and the line segment 1 1x .
Example: Evaluate 2 3
C
xydx x y dy over the triangle with vertices 0,0 , 1,0 and 1,2 .
Look at more general regions, textbook, see examples 5 and 6. Note these are simple connected regions in polar coordinates.. Area of a Region R
1 1 1
2 2 2R R C C C
A dA dA xdy ydx xdy ydx
Example: Find the area of the region enclosed by the ellipse cos sinr t a t b t i j
for 0 2t .
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Section 14.5 Divergence and Curl
Definition: Divergence of a Vector Field The divergence of a vector field , ,F f g h
that is differentiable on a region of 3 is
f g hdiv F F
x y z
If 0F
the vector field is source free Definition: Curl of a Vector Field The curl of a vector field , ,F f g h
that is differentiable on a region of 3 is
F curl F
h g f h g fi j k
y z z x x y
If 0F
, the vector field is irrotational. Theorem 14.9 Curl of a Conservative Vector Field Suppose that F
is a conservative vector field on an open region D of 3 . Let F
, where is a potential
function with continuous second partial derivatives on D. Then 0F
, that is the curl of the gradient
is the zero vector and F
is irrotational. Theorem 14.10 Divergence of the Curl
Suppose that , ,F f g h
, where f, g and h have continuous second partial derivatives. Then 0F
. The
divergence of the curl is zero.
Properties Let F
be a
connected 1. There
2. C
F d
3. C
F d
4. F
Section 14 Definition:The area o
Is
Example: F
Example: I
of a Conservata conservative region in D in exists a potent
dr B A
0dr on all si
0
at all poin
.6 Surface Inte
: of the smooth s
Find the surfac
Integrate the fu
tive Vector Fievector field wh
3 . Then F
tial function
A for all poin
imple smooth c
nts of D.
egrals
surface
,r u v f u
ce area of the c
unction ,G x y
eld hose compone has the follow
such that F
nts A and B in D
closed oriented
, ,u v g u vi
A
ap cut from the
,y z z over th
nts have continwing equivalent
.
D and all smoo
d curves C in D
,h u vj k ,
u v
R
r r dA
e paraboloid z
he cylindrical su
nuous second pt properties
oth oriented cu
D.
,a u b c
2 22z x y b
urface 2 2y z
partial derivativ
urves C from A
c v d
by the cone z
4 , 0z , 1
ves on an open
to B.
2 2x y .
4x .
12
n
Example: I
y a , and
Example: U
2,F y xz
Section 14 Theorem 1Let S be a p
a vector fiecirculation
equals the
So the linethe normal For orientaas the norm
Integrate ,G x
d z a .
Use a paramete
, 1z with the
.7 Stokes’ Theo
4.13: Stokes’piecewise smo
eld whose comn of F
around
integral of cur
e integral arounl component of
ation, suppose ymal vector, then
, ,y z x y
erization of fin
e outward norm
orem
Theorem ooth oriented su
mponents have cC in the direct
rl F n
over S.
nd the boundaryf curl F
.
you are walkinn the surface is
z over the sur
nd S
Flux F
mal (away from
urface having a
continuous firsion counterclo
C C
F Tds
y curve of S of
ng around the bs to your left.
rface of the cub
nd across th
m the z-axis),
a piecewise sm
st partial derivackwise with re
S
F dr curl
f the tangent co
boundary curve
be cut from the
he surface z
mooth boundary
atives on an opespect to the su
l F nd
omponent of F
e with your hea
e first octant by
2 22 x y , 0
y curve C. Let
pen region conturface’s unit no
F
is equal to th
ad pointing in
y the planes x
2z for
t F f g i j
taining S. Thenormal vector n
he surface integ
the same direc
13
a ,
hk be
n the
gral of
ction
14
Example: Use the surface integral in Stokes’ Theorem to calculate the circulation of the field
2 2 2 2 2 2F y z x z x y i j k
around the curve C: the boundary of the triangle cut from the plane
1x y z by the first octant, counterclockwise when viewed from above.
Example: Use the surface integral in Stokes’ Theorem to calculate the flux of the curl of the field
F y z z x x z i j k
across the surface S: 2, cos sin 9r r r r r i j k
for 0 3r and
0 2
An Important Identity 0curl
Theorem 0Curl F
Related to the Closed-Loop Property
If 0F
at every point of a simple connected open region D in space, then on any piecewise smooth closed path C in D
0C
F dr
Section 14.8 Divergence Theorem Theorem 14.15 Divergence Theorem Let F
be a vector field whose components have continuous first partial derivatives in a connected and simply
connected region D enclosed by a smooth oriented surface S. Then’
DS
F ndS FdV
Where n
is the unit outward vector on S.
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Example: Use the Divergence Theorem to evaluate S
F nd
where 2 3 3, ,F x xy x y
and S is the tetrahedron
bounded by 1x y z and the coordinate planes with outward normal vectors.
Example: Use the Divergence Theorem to evaluate S
F nd
where 3 3 3, ,F x y z
and S is the cylindrical can
2 2 1x y from 0z to 2z with outward normal vectors.
Example: (Exercise 22, page 923 see figure) The base of the closed cubelike surface shown is the unit square in the xy-plane. The four sides line in the planes 0x , 1x , 0y and 1y . The top is an arbitrary smooth surface
whose identity is unknown. Let , , 3F x y z
and suppose the outward flux of F
through side A is 1 and
through side B is -3. Can you conclude anything about the outward flux through the top? Example: (Exercise 23)
a. Show that the flux of the position vector field , ,F x y z
outward through a smooth closed surface S is
three times the volume of the region enclosed by the surface. b. Let n
be the outward unit normal vector field on S. Show that it is not possible for F
to be orthogonal to
n
at every point of S.
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Green’s Theorem and Its Generalization to Three Dimensions
Normal form of Green’s Theorem: C R
F nds FdA
Divergence Theorem: S T
F nd FdV
Tangential form of Green’s Theorem: C C R
F Tds F dr F kdA
Stokes’ Theorem: C C S
F Tds F dr F nd