16.3

Conservative Vector Fields

Review:

Work C

dF r = '( )C

t dtF r C

Mdx Ndy Pdz if M N P F i j k

Outward flux across a simple closed

curve C in the plane is C

dsF n if C

Mdy Ndx M N F i j

is called conservative (or a gradient vector field) if

The function is called the of .

f

f potential

F F

F

a) if and only if is path independent:C

f dr

Fundamental theorem for line integrals :

F F1 2

= C C

dr dr F F

= if C is a path from to .

B

C A

dr dr A B F F

b) If , then ( ) ( ) C

f dr f B f A F F

is also called if represents a velocity vector field.C

d circulation F r F

=C

dsF T

When is a vector field conservative? Question : F

Need: f f

M N fx y

F = i j i j or: and

f fM N

x y

a necessary condition: M N

y x

In 3 dimensions: f f f

M N P fx y z

F = i j k i j k

need: , and f f f

M N Px y z

necessary conditions: , , M N M P N P

y x z x z y

Recall:

curl x y z

P N M P N M

y z z x x yM N P

i j k

F i + j k

: If and then curl 0.M N P f Theorem F = i j k F F

necessary conditions: , , M N M P N P

y x z x z y

This condition is "almost" sufficient as well.

But it is always a good test.

In 3 dimensions: f f f

M N P fx y z

F = i j k i j k

Another way to express this, which is easier to remember:

Some terminology for curves C :A region R is called if every closed curve in R

can be shrunk to a point by curves staying in R.

simply connected Definition :

The plane minus the origin is not simply connected.

3-space minus the origin is simply connected.

3-space minus the z-axis is not simply connected.

Important examples :

: If is defined in a connected and

simply connected region, then if and only if

M N

M Nf

y x

Theorem F = i j

F

: If is defined in a connected and

simply connected region, then if and only if

, , or ( ) 0

M N P

f

M N M P N P

y x z x z y

Theorem F = i j k

F

curl F

: To find the potential , integrate one at a time

, (and if in 3-space)

f

f f fM N P

x y z

Hint

A differential is called if

for some function .

Mdx Ndy Pdz exact

Mdx Ndy Pdz df f

Definition :

This is equivalent to saying that (can use same theorems)M N P f i j k

2 4

so is a gradient field.

y y

y y

y x

M x e N y xe

M e N e

F

2, 2 y yf x y Mdx x e dx x xe G y

2,0

2,0

2,02 2

2,0

,

2

4 2 4 2 4

C

y

Work dr f x y

x xe y

F

Find the work done by the force

, 2 4 along the indicated curve.y yx y x e y xe

Example :

F i j

, ,yyf x y xe G y N x y

4y yxe G y y xe

4G y y

22G y y C

2 2, 2yf x y x xe y C

need: and f f

M Nx y

2 2 2 22 3 2 2 3 2xy xz x y y x z z F i j j

2 2

2

2

2 3

2 2

3 2

M xy xz

N x y y

P x z z

2 22 3f Mdx xy xz dx 2 2 2 23

2x y x z ,G y z

2 22 2 2 ( , )y yx y y N f x y G y z or ( , ) 2yG y z y

2 2 2 2 2 232the potential is , ,f x y z x y x z y z C

So is a gradient field, . Now let's find .f f F F

Determine wether the given vector field is a gradient field.

If so, find a potential function.

Example :

2 2 2 2

curl 0 6 6 4 4

2 3 2 2 3 2

x y zxz xz xy xy

xy xz x y y x z z

i j k

F i + j k = 0

2 23 2 3 '( )zx z z P f x z H z

2 2 2 2 2 232

or ( , ) ( ) hence ( )G y z y H z f x y x z y H z

2 or '( ) 2 hence ( )H z z H z z

need: , and f f f

M N Px y z

: is conservative if and only if for every closed curve C:

0C

dr

Theorem F

F

Indeed, if and goes from to and

then ( ) ( ) 0C

f C A B A B

dr f B f A

F

F

: To indicate that the line integral is over a closed curve,

we often write C C

dr dr

Note

F F

1 2

Conversely, assume 0 for any closed curve

and let and be two curves from to with

C

dr C

C C A B A B

F

1 2 1 2

Then 0

C C C C

dr dr dr

F F F

1 2

and hence

C C

dr dr F F

2 2 2 2Compute the work performed by the force

as a particle travels around a circle of radius counter clockwise.

y x

x y x y

r

Example : F i j

parametrize the circle: cos( ), sin( )x r t y r t sin( ), cos( )dx r t dy r t

2 2 2 2 C C

y xdr dx dy

x y x y

F i j i j

2 2sin( ) cos( )

= ( sin( )) ( cos( )C

r t r tr t r t dt

r r

i j i j

2

2 2 2 2

2

0

1= sin ( ) cos ( )r t r t dt

r

2

0

= 1 2dt

is conservative since with arctan (check it!)y

f fx

Puzzle 2 : F F

is not defined at (0,0), and the plane minus the origin is not simply connected !Solution : F

(for any radius !)r

2 2 2 2

2 22 2 2 2 2 2

is conservative since

1 2x y x xx x y

x x y x y x y

Puzzle 1 : F

2 2 2 2

2 22 2 2 2 2 2

1 2 and

x y y yy x y

y x y x y x y

Gravitational vector fieldExample :

2

GmM

rF

| r | | r |11 2 2

is the vector from the center of the sun to the planet

is the mass of the sun

is the mass of the planet

is the gravitational constant

6.674 10 G = (from 1798 )

M

m

G

N m kg

r

3/23 2 2 2

x y z

x y z

r i j kF

| r |

1/2

2 2 2 is conservative with potential . . ( , , ) f i e f x y z

x y z

Claim : F

| r |

1/2

2 2 2 Indeed: ( , , ) f x y z x y z

and similarly for and f f

y z

The work necessary to "escape" the force field from a point :pExample :

= ( ) ( ) | |

C p

dr dr f f pp

F F

3/22 2 2

3/22 2 2

hence 22

f xx x y z

x x y z

GmM

Newton: force = mass x acceleration ''mExample : F r

What is the work performed? Question :

= '( )dr t dt F F r ' ' '

2m dt

r r1 '' '( )

2m t dt r r

since ' ' ' '' ' ' '' 2 '' ' r r r r r r r r

' ' ' hence =

2C

dr m dt

r r

F

If is also conservative, (sign is physics convention)f F F

work performed is gain in kinetic energy

2' '2 2

m mdt dt v v | v |

2 is the kinetic energy:2

mk | v |

is constant along any path ( )

(conservation of kinetic energy + potential energy)

k f tClaim : r

0

= ( ( )) ' ( ( )) ( (0))

t

C

dr k t dt k t k F r r r

since = ( ( )) ( (0)) and ( ( )) ( (0))C

dr fdr f t f dr k t k F r r F r r hence ( ( )) ( ( )) ( (0)) ( (0))k t f t k f r r r r

16.4

Greens Theorem

Closed Curve Line Integral C

Pdx Qdy

Closed Curve Orientation:

Counter-clockwise

Clockwise

C

Pdx Qdy

C

Pdx QdyC

Pdx Qdy

Green’s Theorem (in the plane = 2 dim.)

Suppose that is a simple piecewise smooth closed curve.C

C is the boundary of a region .R If , , , and are continuous on , then y xP Q P Q R

x yR C

Q P dA Pdx Qdy

2 , 3P y Q xy

3 , 2x yQ y P y

2 3GThm

C R

y dx xydy ydA 2

0 1

sinr rdrd

2

2

0 1

sin d r dr

14

3

2

3

0

1

cos3

r

8 1

cos cos03

7

1 13

2Compute the closed line intgeral 3

where C is the indicated curve.

C

y dx xydyExample : x yR CQ P dA Pdx Qdy

x yC R

Pdx Qdy Q P dA

If 1, then x yQ P

R

dA R

The area of the

interior region

If you have the parametrization of a closed curve and want to find the enclosed

area then you can use this consequence of Green's Theorem to set up the line integral.

1 1choose: , 2

2 2x yP y Q x Q P

We can use Green's theorem to compute areas:

1( )

2R

area R xdy ydx

stands for the boundary

of the region

R

R

2R

dA x yC R

Pdx Qdy Q P dA

1 1

2 2Pdx Qdy xdy ydx

aa

b

b

2 2

2 2The parametrization of the ellipse 1:

x y

a b

cosx a t

siny b t

1Area

2C

xdy ydx 0 2t

2

0

c1

cos sin2

os sina t bb t a tdtdt t

sindx a tdt

cosdy b tdt

2

2 2

0

1cos sin

2ab t ab t dt

2

2 2

0

cos sin 2

abt t dt

2

02

abdt

2ab

2 ab

2 2

2 2Find the area enclosed by the ellipse 1.

x y

a b Example :

positively oriented