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Lecture 6

Author: Kemal Ahmed Instructor: Dr. French Course: Math 2ZZ3 Date: 2013-07-15

Math objects made using MathType; graphs made using Winplot.

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• No bonus points on the test

• 3 hours

Chapter 9.9 – Independence of Path

F is a vector field

In general no relationship between 1

d

C

r

F

and

2 d

C

r

F

.

However, sometimes,

1 2

d d

C C

r r

⋅ = ⋅

F

F

, wheneverC C1, 2starts/ends at the same place.

The vector function, F, is conservative if F can be written as a gradient of some scalar function ϕ (i.e. there exists ϕ, such that∇ =φ F)

If ϕ exists, we say ϕ is the potential function of F.

The integral, d

C

r

F

, is independent of path if

1 2

d d

C C

r r

⋅ = ⋅

F

F

for any two paths, C1, C2, that start / end at the same place.

Fundamental Theorem of Line Integrals

Suppose C is a path in an open region R of the xy-plane, and C is parametrized by

( )

t = x t

( ) ( )

,y t

r ←2D,a≤ ≤t b

If F=P x y

( )

, iˆ+Q x y

( )

, ˆj (2D) is conservative, with potential function,φ

( )

x y, , then

( ) ( )

d d

C C

r r

B A

φ

φ φ

⋅ = ∇ ⋅

= −

F

,

(2)

Where:

( ) ( )

(

)

( ) ( )

(

)

, , A x a y a B x b y b

= =

If conservative => d

C

r

F

is independent of path

Proof

( )

( ) ( )

, , x y r t x t y t φ φ

φ ∂ ∂

= ∇ =

∂ ∂

′ = ′ ′

F

( )

(

)

( ) ( )

(

)

(

)

( ) ( )

d d

d d

d

d d

d

d chain rule d

, by Fundamental Theorem of Calculus (FTC)

C C

b a

b a

b t a t

r r t t

x y

t x t y t

t t x t y t

B A

φ φ

φ

φ

φ φ

= =

⋅ = ⋅

∂ ∂ 

= +

∂ ∂

 

= ←

= ←

= −

F F

Definitions

Region D of2

D is connected if every pair of points in D can be linked by a path lying wholly in D.

D is simply connected if it is connected and every simple closed curve enclosed only points in D.

D is open if it contains no boundary points (i.e. a < x < b), as opposed to closed when it does have boundary points (i.e. a ≤ x b). In an open, connected region, D, d

C

r

F

is independent of path <=> F is conservative

(3)

Proof Using FTLI, Assume d C r

F

is independent of path, =>WTS F is conservative

WT F potential function ofφ

( )

x y, Define

( )

( )

( )

0 0

,

,

, x y d

x y

x y r

φ =

F

ϕ is well-defined because d

C

r

F

is independent of path. Claim ϕ is potential of F.

( )

, ,

( )

, P x y Q x y

x y

φ φ

∂ ==

 

Choose a special path, C, whereC=C1C2

( )

( ) ( ) ( ) ( ) 1 2 , , , ,

0 0 1

d d

, d d

x y x y

x y x y

C C

r r

x y r r

x φ ⋅ ⋅     ∂  = ⋅ +    ∂   ∫ ∫    

F F F F   ( ) ( ) 1 , ,

0 x y d

x y r

x x

φ

∂ ∂

= + ⋅

∂ ∂

F The first part is zero because

1 d C r

F

does not depend on x.

( )

( ) ( )

( )

1 ,

, , d , d

x y

x y P x y x Q x y y

x

= +

∂ 

onC2, dy=0, since C2 is horizontal

( )

( ) ( )

( ) (

)

1 ,

, , d

, FTC

x y

x y P x y x

x x P x y

φ ∂ ∂  =   ∂ ∂ = ←

Similarly, we can show: Q x y

( )

, y

φ

=

(4)

Theorem

In an open, connected region, D, we know this line integral, d

C

r

F

is independent of path <=> d 0

C

r

⋅ =

F

for all C such that C is simple, closed.

Proof

Assume d

C

r

F

is independent of path, show d 0

C

r

⋅ =

F

for C, closed

(

)

1 2

1 2

d d d

d d

0 by independence of path

C C C

C C

r r r

r r

⋅ = ⋅ + ⋅

= ⋅ − ⋅

= ←

F F F

F F

Assume d 0

C

r

⋅ =

F

if C closed WTS

1 2

d d

C C

r r

⋅ = ⋅

F

F

forC C1, 2starting/ending at the same point Let C=C1∪ −C2, C is closed

=> d 0

C

r

⋅ =

F

(by assumption) =

1 2

d d

C C

r r

⋅ + ⋅

F

F

1 2 1 2

d d 0 d d

C C C C

r r r r

⋅ − ⋅ = ⇔ ⋅ = ⋅

F

F

F

F

(5)

F is conservative open, connected region

FTLI

  d

C

r

F

is independent of path

open, connected domain

 

d 0

C

r

⋅ =

F

whenever C closed

open region

s.c. region

P Q

y x

∂ ∂

=

∂ ∂

 

curl F = 0

Conservative

How to tell when F is conservative?

IfF

( )

x y, =P x y

( )

, ˆi+Q x y

( )

, ˆjis conservative on an open region, D, and P,Q are continuous, with continuous 1st order partials, then P Q

y x

∂ ∂

=

∂ ∂

Conversely, if D is a simply connected region and P Q

y x

=

∂ ∂ , F is conservative.

3D

In 3D space, everything is the same, except of the following: F is conservative↔curlF=0

e.g. 1)

( )

3

, 5 4 , 4 8 x y = x+ y xy F

i)

Show d

C

r

F

is independent of path.

F is conservative <=> Q P

x y

=

∂ ∂

4, 4

Q P

x y

==

∂ ∂

ii) Evaluate

( )

( )0,0

1,1 dr

− ⋅

F

You could choose a specific curve and compute as usual. Instead, find potential ϕ and compute

( ) (

0, 0 1,1

)

φ − −φ

Findφ

( )

x y, , such that 5x 4 &y 4x 8y3

x y

φ φ

= +=

∂ ∂

(6)

( )

(

(

)

)

( )

2

5 4

, 5 4 d Partial integration treat like a constant 5

5 2

x y x

x y x y x y

x xy g y

φ

φ ∂

= +

⇒ = + ←

= + +

( )

( )

3

2

4 8 5

4 2 4

x y y

x xy g y y y

x g y φ

φ

=

= ∂  + +

 

∂ ∂  

= +

( )

( )

(

)

3

3

4

8 8 d

2 0

g y y

g y y y

y C C

⇒ = −

⇒ = −

= − + ← =

( )

( )

( )

( ) (

)

2 4

0,0

1,1

5

, 4 2

2

d 0, 0 1,1

5

0 4 2

2 7 2

x y x xy y

r

φ

φ φ

= + −

⋅ = − −

 

= − − −

 

=

F

9.10 – Double Integrals

( )

,

f x y =z

Divide D into n rectangles, fitting within D.

(7)

LetAibe area of ith rectangle, choose sample point

(

x yi*, *i

)

inside ith rectangle, product

(

* *

)

,

i i i

f x yA is a volume of rectangular prism lying under the graph of f x y

( )

, , over ith rectangle

Form Riemann sum:

(

* *

)

1

,

n

i i i

i

f x y A =

Definition:

( )

(

* *

)

1

, d lim ,

n

i i i

n i D

f x y A f x y A

→∞ =

=

∫∫

Double integral of f over D. Volume of solid under f, over D.

Type I Region

( )

( )

( )

{

, , 1 2

}

D= x y a≤ ≤x b g x ≤ ≤y g x

( )

2( )( )

( )

1

, d b g x , d d

a g x D

f x y A= f x y y x

∫∫

∫ ∫

(8)

Iterated integral: evaluate form inside-out, using partial integration

e.g. 2)

3 2

d

R

x y A

∫∫

, R is enclosed byy=x x, =1,y=0

( )

{

, | 0 , 0 1

}

R= x y ≤ ≤y x ≤ ≤x

1

3 2

0 0

1

3 3

0

0

1 6

0

1 7

0

d d 1

d 3

1 d 3

1 1

21 21

x

y x

y

x

x

x y y x

x y x

x x

x = =

= =

=

=

=

= =

∫ ∫

Type II Region

( ) ( )

( )

{

, | 1 2 ,

}

D= x y h y ≤ ≤x h x c≤ ≤y d

(9)

( )

2( )( )

( )

1

, d d h y , d d

c h y D

f x y A= f x y x y

∫∫

∫ ∫

e.g. 3)

3 2

d

R

x y A

∫∫

( )

{

, | 1, 0 1

}

R= x y y≤ ≤x ≤ ≤y

1 1

3 2

0

1 d d

21

yx y x y

=

∫ ∫

=

Changing Order of Integration

It is sometimes useful to change the order of integration fromd dx y→d dy x, or vice versa

e.g. 4)

( )

2

sin d ,

R

x A

∫∫

(10)

, ,

n=x n= π

Point

(

π π,

)

Type II:

{

( )

x y, |y≤ ≤x π, 0< ≤y π

}

( )

2

0 y sin x d dx y

π π

=

∫ ∫

 cannot do! So let’s put it back.

( )

{

, | 0 , 0

}

R= x y ≤ ≤y x ≤ ≤x π

( )

( )

( )

2

0 0

2

0 0

2

0

sin d d

sin d

sin d 2

x

y x y

x y x

y x x

x x x

π π

π

= = =

= =

∫ ∫

Applications of x2 Integration

Area

Area of region, 2

D⊆ is:

( )

1 d

D

A

∫∫

Volume

Volume under graph of f x y

( )

, =z, over region D

e.g. 5)

Volume of solid bounded by:

2 2

2 2

4 4 x y y z

 + =

 

+ =



(11)

With DOUBLE INTEGRATION! 1/8th of region

Region under graph ofy2+z2 =4, over region, D.

2

2

2

2 4 2

0 0

2 4

2

0 0

8 4 d

8 4 d d

8 4 d d

D

x

y

y A

y y x

y x y

− − −

= −

= −

∫∫

∫ ∫

∫ ∫

Moments of Inertia

Lamina: 2D region with a density function, p (x,y) What is the mass of this lamina region?

(12)

Mass of lamina

( )

, d

R

m ρ x y A

= =

∫∫

Centre of mass of lamina:

, y

x M

M

x y

m m

= =

d , d

y x

R R

M =

∫∫

xρ A M =

∫∫

yρ A←1st moments of inertia about x, y-axis

2 2

d , d

x y

R R

I =

∫∫

y ρ A I =

∫∫

x ρ A←2nd moments of inertia

9.11 – Polar Coordinates

Specify any point in 2

 by distance form origin (r), and∠made by its position vector with positive x-axis

2 2 2 cos

,

sin x r x y r

y r θ θ

= 

+ =  =

Rectangles

{

( )

r,θ |a≤ ≤r b c, ≤ ≤θ d

}

Type I

( ) ( )

( )

{

}

( )

2( )( )

( )

1

1 2

, | ,

, d g , d d

g R

r g r g

f r A β θ f r r r

α θ

θ θ θ α θ β

θ θ θ

≤ ≤ ≤ ≤

=

∫∫

∫ ∫

Type II

( ) ( )

( )

{

r,θ | f r1 ≤ ≤θ f2 r ,a≤ ≤r b

}

( )

2( )( )

( )

1

, d b f r , d d

a f r R

f r θ A= f r θ r θ r

∫∫

∫ ∫

(13)

Theorem:

( )

, d

(

cos , sin

)

d d

R R

f x y A= f r θ r θ r r θ

∫∫

∫∫

e.g. 6)

Find volume of solid bound byz= −1 x2−y z2, =0

2 2 2 2

0 1= −xyx +y =1

(

)

2 1 2

0 0 1 d d

V =

∫ ∫

π −r r r θ OR

(

)

2

2

1 1 2 2

1 1 1 d d

x x

Vx y y x

− − −

=

∫ ∫

− −

( )

2 2 2

2 2

1 ,

x y r

z x y f x y

+ =

= − − =

I think she meant to use this, though:

e.g. 7)

(

)

2

5 25

5 0 4 3 d d

x

x y y x

− +

∫ ∫

(14)

What kind of region will we get? R:

{

( )

x y, | 0≤ ≤y 25−x2, 5− ≤ ≤x 5

}

( )

{

, | 0 5, 0

}

R= r θ ≤ ≤r ≤ ≤θ π

( )

( )

(

)

5

0 0 4 cosr 3 sinr r rd d

π

θ + θ θ

∫ ∫

9.12 – Green’s Theorem

Green’s Theorem

The positive orientation of a closed curve, C, is one for which the region enclosed by C is on the left.

Green’s Theorem: Let C be a piecewise-smooth curve bounding a simple connected region, R.

If , ,P Q P, Q y x

∂ ∂

∂ ∂ are all continuous on R, then:

d d d

R

Q P

P x Q y A

x y

∂ ∂ 

+ =

∂ ∂

 

∫∫

Holy Green

Green’s theorem may be extended to regions with holes.

(15)

You can split your region into 2 pieces and apply Green to 2 parts separately, then add them together.

e.g. 8)

Example 1, Chapter 16 from Stewart’s Calculus Early Transcendentals 7E

4

d d

C

x x+xy y

, C is a triangular curve consisting of line segments from:

( ) ( )

( ) ( )

( ) ( )

0, 0 to 1, 0 1, 0 to 0,1 0,1 to 0, 0

We could do it directly, but Green’s theorem is faster

(

)

d d

0 d

C R

R

Q P

r A

x y

y A

∂ ∂ 

⋅ =

∂ ∂

 

= −

∫∫

∫∫

F

Can be expressed as T1 or T2

( )

1 1

0 0

1 d d

6

x

y y x

=

∫ ∫

=

Area of region, R, enclosed by C Area of

( )

1 d

R

R=

∫∫

A

If we can parametrize the boundary of R, then we may use Green’s Theorem. However, that requiresP x y Q x y

( ) ( )

, , , , such that Q P 1

x y

∂ ∂

− =

∂ ∂

Some choices:

(16)

0, d

C

P Q x

A x y

= =

=

, 0

d

C

P y Q

A y x

= − =

= −

2

0

,

2 2

1

d d

2

y x

P Q

A πx y y x

= =

=

e.g. 9)

Find area of ellipse

2 2

2 2 1

x y a +b =

Boundary curve of the ellipse has parametrization

( )

( )

cos

, 0 2 sin

x a t

t

y b t π

=

 ≤ ≤

 = 

The most useful method for this question:

( )

( )

( )

( )

2

2 2

0

2

0

, ,

2 2

1 d d d

1

d d

2 1

cos d sin d 2

1

d 2

R C

C

y x

P x y Q

A A P x Q y

x y y x

ab t t ab t t

ab t

ab

π π

π

= =

= = +

= −

= +

=

=

∫∫

References

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