vector calculus - nc state university · vector calculus part ii by dr. michael medvinsky, ncsu...

Vector Calculuspart II

By Dr. Michael Medvinsky, NCSU

online lectures 03/2020

Fundamental Theorem for Line Integrals(cont)

• Theorem: Suppose F=<P,Q> is a conservative vector field and P,Q has continuous first order partial derivatives on domain D, then

Proof: Let f be the potential, i.e. , therefore

P Q

y x

=

, ,x yP Q f f f= = =F

xy yx

P Qf f

y x

= = =


• Definitions:1) A simply connected curve is a curve that doesn’t intersect itself between endpoints. 2) A simple closed curve is a curve with butfor any .

3) A simply connected region: is a region D in which every simple closed curve encloses only points from D. In other words D consist of one piece and has no hole.

( ) ( )r a r b= ( ) ( )1 2r t r t

1 2a t t b


• Theorem: Let be a vector field on an open simply connected region D. If P,Q have continuous first order partial derivatives on domain D and , then F is conservative.

• Example: Determine whether is conservative.Solution: Not conservative, since

,P Q=F

P Q

y x

=

( ), sin , sinx y x y y x=F

( ) ( )sin cos cos siny xy xP x y x y y x y x Q= = = =


• Example: Show that and find the potential.

Solution: , indeed F is conservative.

• To find the potential start with

note that the constant of integration can be function of y.

• To find g differentiate and compare to Q:

to get

• Finally, since any potential works, set const=0 to get

( ), , ,x y P Q x y x y= = + −F

( ) ( ) ( )2

, ,2

x

xf x y f x y dx x ydx yx g y= = + = + +

( ) ( )1y x xyP x y x y Q= + = = − =

( )'yf x g y x y= + = −

( )2 2

,2 2

x yf x y yx= + −

( ) ( )2

2'

yg y g y dx ydx const= = − = − +

Green’s Theorem

• Definition: A simple closed curve is said to be positive oriented if it traversed counterclockwise.

Counterclockwise – positively oriented Clockwise – negatively oriented

Green’s Theorem(the theorem)

• Green’s Theorem: Let C be positively oriented piecewise-smooth, simple closed curve in the plane and let D be the region bounded by C. If P and Q have continuous partial derivatives on an open region that contains D, then

• Note: The circle on the line integral ( ) is sometime related to the positive oriented curve and sometime even drawn with an arrow on the circle:

Q Px y

C D

Pdx Qdy dA

+ = −

Green’s Theorem(cont)

• One views the Green’s theorem as a counterpart of Fundamental Theorem of Calculus

• Green’s theorem

• FTC theorem

• Notice that in both, the left side is on the domain while the right one is at the boundary of the domain.

Q Px y

D D

dA Pdx Qdy

− = +

( ) ( ) ( )'b

aF x dx F b F a= −

Green’s Theorem(proof)

Proof:

• Formulate D as domain of type I and show that

thus, let and let , as depicted

which is the same as

• Similarly, one formulate D as domain of type II to show thatQ

x

D D

Qdy dA

=

Py

D D

Pdx dA

= − ( ) ( ) 1 2,D a x b g x y g x=

2 3 41C C CC C=

C1:y=g1(x)

-C3:y=g2(x)C2

C4

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( )( )

, 2,, 1 1 ,1 , 1 12 ,2

1,

x g ax g b g b t b g b t g a t a gx a txD

b b

a b

Pdx Pdx Pdx Pdx Pdx

P dx x x Pdxg

−− + +

= + − +

= +

( )( )2,

b a

a a

xP x xx dg Pd− +

( )

( )

( )

( )

( )( ) ( )( )2 1

1 2

1 2, ,

g x g xb b b

P P Py y y

D a g x a g x a

dA dydx dydx P x g x P x g x dx

− = − = = −

Green’s Theorem(cont)

• Example: Let D be square . EvaluateSolution: Using Green’s theorem,

• Verify

( ) ( )2 3 2 2D

x xy dx y xy dy

− + − 0,2 0,2

( ) ( ) ( ) ( )

( )

2 3 2 2 2 3

222 2 2 22 22

2 2 2 2 3

00 0 0 00 0

2 2

22 3 2 3 4 3 2 2 8

2 2

D D DP Q

Q Px xy dx y xy dy dA y xy x xy dA

x y x y

xy xy dxdy xy y dy y y dy y y

− + − = − = − − −

= − + = − + = − + = − + =

( ) ( )2 3 2

2 2 2 2 2 22 2

2 2 2 3 2 2 2

0 00 0

0,,2

0 0 0 0

2,,0

2

4 2 8 4 4 2 16 8 8

yxyD x

x xy dx y xy dy

x dx y ydy x xdx y dy xdx ydy x y

− + − = + − − =

= + − − − − = − = − = − =

Green’s Theorem(extensions)

• How to use Green’s theorem beyond its original formulation? • In the case when the curve C is not closed (but its line integral isn’t “nice”):

• Connect the endpoints of C with any simple curve 𝐶1 to get 𝐶2 = 𝐶 ∪ 𝐶1

• Now, can conveniently(?) be evaluated using Green’s theorem and

• Hint: The best choice of 𝐶1 will make easy.

• In the case the region D has a hole, i.e. is not a simply connected.

• Rewrite D as union of simply connected regions (see example)

• Use the version of Green’s theorem for Union of Domains(TBD on next slide)

1C2 1C C C

= − 2C

D1

D2


• Theorem: Let D be a domain. Rewrite D as union of 2subdomains, e.g., let and , such that and ,

then 1 2D D D=

1 2D C C = 3 1 2C D D= 1 1 3D C C = ( )2 2 3D C C = −

( )1 3 2 3

1 2 1 2

Q Px y

D C C C C

C C C C

dA Pdx Qdy Pdx Qdy

Pdx Qdy Pdx Qdy Pdx Qdy

−

− = + + +

= + + + = +


• Example: Evaluate .

• Solution: For a smart use of Green’s Theorem: choose any P and Q,

such that .

• For example , gives

DA dA=

1Q P

x y

− =

0,P Q x= =

D DA dA xdy

= =


• Let C: . Evaluate:

• Solution: reformulate the curve as or which is a

half circle, or in polar coordinates . Connect the

ends of the half circle with a line along x-axis, from 0 to 1.

( ) 2, , 0,1r t t t t t= −

( )

1 1 ,0

cos2

0 0

1 1 1

6 2 3

1cos cos 2 2 2 sin

6

x yC C C C R x

x x

x yR R R

Pdx Qdy Pdx Qdy Pdx Qdy Q P dA Pdx Qdy

Q P dA e y e y y dA ydA r rdrd

Pdx Qdy

+ = + − + = − − + = − = −

− = − − = = =

+

( )1

20 1

2

,0 00

1sin 0 0

2 2

t

x

d te t tdt

dt

=

= − + = =

( ) ( )2 2sin cos cosx x

C

P Q

e y y x dx e y y dy− + + −

cos ,0 / 2r =

2y x x= −2 2y x x+ =


• Example: Let C be a ring with radiuses 1 and 2 centered at the origin.

1 2

3 3 2 2 2 2

22 2 43

0 1 1

3 3 3 3

453 3 2

4 2

x xy y

C D DQ Q QP P P

y dx x dy x y dA x y dA

rr drd

+ = − − + − −

= − = − = −

Curl and Divergence

• Let F=<P,Q,R> be a vector field on . Assume that all partial derivatives of P,Q,R exists, then• the curl of F is defined as

• the divergence of F is defined as

curl det , ,

i j k

R Q P R Q P

x y z y z z x x y

P Q R

= = = − − −

F F

3

div , , , ,P Q R

P Q Rx y z x y z

= = = + +

F F

Curl and Divergence(cont)

• Example: Let . Then ,

curl 0 since

cos sin ,

cos ,and

cos

and

R Qx yz xyz yz

y z

P Ry yz

z x

Q Pz yz

x y

=

= − =

= =

= =

F

sin , cos , cosf yz xz yz xy yz= =F

( ) ( ) ( )2 2 2 2

div , , sin , cos , cos

0 sin sin sin

yz xz yz xy yzx y z

xz yz xy yz x y z yz

=

= + − + − = − +

F

( ), , sinf x y z x yz=

Curl and Divergence (cont)

• Theorem: Suppose f(x,y,z) has continuous second-order partial derivatives, then

• Theorem: If F is vector field defined on whose component functions have continuous partial derivatives and , then F is a conservative vector field.

• Theorem: Suppose F=<P,Q,R> is a vector field on and has continuous second-order partial derivatives, then

curl 0=F

( )curl , , , , 0x y z zy yz xz zx yx xyf curl f f f f f f f f f = = − − − =

3

3

2 22 22 2

divcurl div , , 0R Q P R Q P Q Q

y z z x x y x

P P

y z z

R R

x y x yxyz z

= − − − = − + − + − =

F

vector calculus - nc state university · vector calculus part ii by dr. michael medvinsky, ncsu...

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