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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.

2

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

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,F y x

, ,F y z x

With(plots

;

x

x

s):

,F x y

, 2,F y x

/

x

,sinF y x

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