Lecture 6
Author: Kemal Ahmed Instructor: Dr. French Course: Math 2ZZ3 Date: 2013-07-15Math objects made using MathType; graphs made using Winplot.
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Chapter 9.9 – Independence of Path
F is a vector fieldIn general no relationship between 1
d
C
r
⋅
∫
F
and2 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 if1 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 tr ←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∫
,Where:
( ) ( )
(
)
( ) ( )
(
)
, , A x a y a B x b y b
= =
If conservative => d
C
r
⋅
∫
F
is independent of pathProof
( )
( ) ( )
, , 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 of2
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 conservativeProof Using FTLI, Assume d C r ⋅
∫
F
is independent of path, =>WTS F is conservativeWT 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 yx y
φ φ
∂ = ∂ =
∂ ∂
Choose a special path, C, whereC=C1∪C2
( )
( ) ( ) ( ) ( ) 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 because1 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
( )
, yφ
∂ =
∂
Theorem
In an open, connected region, D, we know this line integral, d
C
r
⋅
∫
F
is independent of path <=> d 0C
r
⋅ =
∫
F
for all C such that C is simple, closed.Proof
Assume d
C
r
⋅
∫
F
is independent of path, show d 0C
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 WTS1 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
F is conservative open, connected region
FTLI
d
C
r
⋅
∫
F
is independent of pathopen, connected domain
d 0
C
r
⋅ =
∫
F
whenever C closedopen 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 Qy 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 x− y 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
− ⋅
∫
FYou 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 8y3x y
φ φ
∂ = + ∂ = −
∂ ∂
( )
(
(
)
)
( )
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
−
= + −
⋅ = − −
= − − −
=
∫
F9.10 – Double Integrals
( )
,f x y =z
Divide D into n rectangles, fitting within D.
LetAibe area of ith rectangle, choose sample point
(
x yi*, *i)
inside ith rectangle, product(
* *)
,
i i i
f x y ∆A is a volume of rectangular prism lying under the graph of f x y
( )
, , over ith rectangleForm 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
∫∫
∫ ∫
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
( )
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)
( )
2sin d ,
R
x A
∫∫
, ,
n=x n= π
Point
(
π π,)
Type II:
{
( )
x y, |y≤ ≤x π, 0< ≤y π}
( )
20 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 dD
A
∫∫
Volume
Volume under graph of f x y
( )
, =z, over region De.g. 5)
Volume of solid bounded by:
2 2
2 2
4 4 x y y z
+ =
+ =
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?
Mass of lamina
( )
, dR
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-axis2 2
d , d
x y
R R
I =
∫∫
y ρ A I =∫∫
x ρ A←2nd moments of inertia9.11 – 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
∫∫
∫ ∫
Theorem:
( )
, d(
cos , sin)
d dR 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= −x −y ⇔ x +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
V − x 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 −
− +
∫ ∫
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.
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 dR
R=
∫∫
AIf 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 1x y
∂ ∂
− =
∂ ∂
Some choices:
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
−
= =
= = +
= −
= +
=
=