vector calculus - mmedvin.math.ncsu.edu•recall: fundamental theorem of calculus (ftc)...

Vector CalculusBy Dr. Michael Medvinsky, NCSU

online lectures 03-04/2020

Vector Field (definition)

• Definition: Vector Field is a function F that for each (x,y)\(x,y,z) assign a 2\3-dimensional vector, respectively:

• Examples of VF: gradient, direction field of differential equation.

• Vector field vs other functions we learned:

2 3

: function of 1,2,3variables

: vector (of size 1,2,3) valued function, e.g. parametric curve

: parametric surface

: vector field ( 2,3)

n

n

n n

f n

r n

r

n

→ =

→ =

→

→ =F

( ) ( ) ( )

( ) ( ) ( )

, , , ,

( , , ) , , , , , , , ,

x y P x y Q x y

x y z P x y z Q x y z R x y z

=

=

F

F

Vector Field (how to sketch it)

• We draw VF as vectors

starting at points

Examples:

( ) ( ) ( ) ( ) ( ), , , \ , , , , , , , ,P x y Q x y P x y z Q x y z R x y z

( ) ( ), \ , ,x y x y z

x

y

Q

P

General case

Uniform\Constant VF:F=<1,2>Uniform\Constant VF: F=<2,1>

Vector Field (how to sketch it)

• More examples:

( ), , ,f x y xy f y x= = =F ( ), , , ,0y x

x y zz z

= −F( ), ,x y x y=F

Line Integral (the idea)

• Consider smooth curve C be given by:recall: smooth means is continuous and

• Divide [a,b] into subintervals

• Denote a piece of C corresponding to and displacement as .

• Denote a sample point on .

• Consider that function f(x,y) is defined along C.

• What do you think about the followingRieman-like sum

( ) ( ) ( ) , , ,r t x t y t t a b=

tt0 tn

,i

b at a i t t

n

−= + =

1,i it t−is

* *,i ix yis

is

( )* *

1

,n

i i i

i

f x y s=

( )'r t ( )' 0r t

sub−arc 𝑠𝑖

Line Integral (definition)

• Let f be a function defined along a curve C given by

then the line\contour\path\curve integral is defined by

• Recall: Arclength formula:

( ) ( ) ( ) ( ) ( ) ( ) ( ) , , , , ,r t x t y t or r t x t y t z t t a b= =

( ) ( ) ( ) ( )* * * * *

1 1

, lim , , , lim , ,n n

i i i i i i in n

i iC C

f x y ds f x y s or f x y z ds f x y z s→ →

= =

= =

( )2 2 2 2 2 '

b b b

t t t t t

a a a

L x y dt or L x y z dt or L r t dt= + = + + =

Line Integral (2 theorems)

• 1) Let f be continuous function along curve C (defined as before):

regardless of the parameterization as long as the curve traversed exactly once between a and b.

• 2) Let be piecewise smooth curve, then

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

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

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

2 2

2 2 2

, , , or

, , , , , or

' , for either , or , ,

b b

t t

C a a

b b

t t t

C a a

b

C a

dsf x y ds f x t y t dt f x t y t x y dt

dt

dsf x y z ds f x t y t z t dt f x t y t x y z dt

dt

f x ds f r t r t dt x x y x x y z

= = +

= = + +

= = =

1 2 ...C C C=

1 2

...C C C

f ds f ds f ds= + +

Line Integral(examples)

• 1) Let (half circle, r=2), evaluate

Solution: We have the arclength is thus

( ) 2cos ,2sin ,0r t t t t = 2

C

x yds

( ) ( )( ) ( )2 2 2, 2cos ,2sin 4cos 2sin 8cos sinf x y x y f r t f t t t t t t= = = =

( )' 2 sin ,cos 2r t t t= − =

( )( ) ( )

1

2 2 2 2

cos'0 0 1

328cos sin 2 16 cos sin 16

3u tr tC f r t

x yds t t dt t tdt u du

=−

= = = =


• 2) Letevaluate

Solution:

( ) ( )1 1 2 28 ,4 ,5 ,0 1 and 1,2, 5 , 1 0C r t t t t t C r t t t= = = = − −

1 2C C

x y zds

+ +

( )( )( )

( )

( ) ( )( )

( )( )( ) ( )

1

2

1 2

11 1 2

0 0 0'

00 0 2

1 1 1'

7 9 88 4 5 8 16 25 7 9 8 7 9 8

2 2

1 25 551 2 5 0 0 25 5 3 5 5 3 5 5 3 5 15

2 2 2 2

C r tf r t

C r tf r t

C C

tx y zds t t t dt tdt

tx y zds t dt tdt t

x y zds

− − −

++ + = + + + + = + = + =

+ + = + − + + = − = − = − − − = + =

+ +

( )1 2

7 9 8 55 7 859

2 2 2C C

x y zds x y zds+

= + + + + + = + = +


• 3) Evaluate where C is a line between (1,2) and (3,-1)

Solution: parameterize

to get

Alternative Solution: parameterize

to get

13C

xyds

( ) ( )1 1,2 3, 1 1 2 ,2 3r t t t t t= − + − = + −

( )( )1 2 2 3

13 13C

t txyds

+ −=

( )( )

4 9

f r t

+

( )

1 1

2

0 0'

12 6

2r t

dt t t dt= + − =

( ) ( )1 3, 1 1,2 3 2 , 1 3r t t t t t= − − + = − − +

( )( )3 2 1 3

13 13C

t txyds

− − +=

( )( )

4 9

f r t

+

( )

1 1

2

0 0'

13 11 6

2r t

dt t t dt= − + − =

Line Integral(theorem + definition)

• In previous example we traversed a curve (line) in two opposite direction and got the same result – it didn’t happen by an accident.

• Theorem: Denote by -C the same curve as C, but with different direction:

• Definition: Denote a line integral with respect to arclength, a line integral with respect to x: ,with respect to y:and analogically in 3D .They often occur together, e.g.

C Cf ds f ds

−=

Cf ds

( ) ( ) ( )( ) ( ), , 'b

C af x y dx f x t y t x t dt=

( ) ( ) ( )( ) ( ), , 'b

C af x y dy f x t y t y t dt=

( ) ( ) ( ), , , , , , , ,C C C

f x y z dx f x y z dy f x y z dz

( ) ( ) ( ) ( ), , , ,C C C

f x y dx g x y dy f x y dx g x y dy+ = +

Line Integral(example)

• Evaluate

Solution:

2 ,4 ,5 ,0 1

t t tt

ydx zdy xdz+

+ +

( )

( )

( )

( )

( ) ( )

( )

( )

1 1 1

2 ,4 ,5 , 0 0 00 1 ' ' '

1 1

0 0

4 2 5 4 2 5

4 20 5 2 29 10 24.5

y zt t tx

t x t y t z t

d d dydx zdy xdz t t dt t t dt t t dt

dt dt dt

t t t dt t dt

+

+ + = + + + +

= + + + = + =

Line Integral of Vector Field

• Reminder:• A work done by variable force f(x) in moving a particle from a to b along the x-

axis is given by .

• A work done by a constant force F in moving object from point P to point Q in space is .

• Unit tangent vector:

• Consider now a variable force F(x,y,z) along a smooth curve C. • Divide C into number of a small enough sub-arcs so that the force is roughly

constant on each sub-arc.

• The displacement vector becomes unit tangent (T) times displacement ( ):

( )b

aW f x dx=

W PQ= F𝐅 is ~c𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑜𝑛 𝑠𝑖

( ) ( ) ( )( ) * * * *

1, , , ,j i i i i i iPQ s x t y t z t t t t−= T

is

( )( )

( )

'

'

r tt

r t=T

Line Integral of Vector Field(cont)

• Finally the work of F(x,y,z) along C is given by

• Denote

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

( ) ( )

* * * * * *

1

lim , , , ,

, , , ,

n

i i i i i i jn

i

C C

W x t y t z t x t y t z t s

x y z x y z ds ds

→=

=

= =

F T

F T F T

( ) ( )' or 'd r t dr r t= =r

( )( )( )

( )

'

'C

r tds r t

r t = F T F ( )'r t

( )( ) ( )( )( ) '

'

b b b

drr ta a a

dt r t r t dt d= F

F F r

Line Integral of Vector(example)

• Let VF be given by and

• the curve C byEvaluate .

Solution:

, ,x x y x y z= + + +F

Cd F r

( ) sin ,cos ,sin cos ,0 2r t t t t t t = +

( )

( )( ) ( )

( ) ( )

( )

22

'

2 2

2cos2sin

2

0

sin ,cos sin ,2 sin cos cos , sin ,cos sin

32cos 3sin 1 cos 2 1 cos 2

2

3 sin 2 31 cos 2 1 cos 2

2 2 2

r t dtr t

tt

C

d t t t t t t t t t dt

t t dt t t dt

td t t dt t

=

==

= + + − −

= − = + − −

= + − − = + −

F

F r

F r

2

0

sin 2

2

tt

− = −

Line integral of Vector vs Scalar fields

• Letandthen

( ) ( ) ( )( , , ) , , , , , , , ,x y z P x y z Q x y z R x y z=F

( )( ) ( )

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

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

( ) ( ) ( )

'

, , , , , , , , ' , ' , '

, , ' , , ' , , '

, , , , , ,

b

C a

b

a

b

a

b

a

d r t r t dt

P x y z Q x y z R x y z x t y t z t dt

P x y z x t Q x y z y t R x y z z t dt

P x y z dx Q x y z dy R x y z dz

=

=

= + +

= + +

F r F

( ) ( ) ( ) ( ), , ,r t x t y t z t a t b=

Fundamental Theorem for Line Integrals

• Recall: Fundamental Theorem of Calculus (FTC)

• Definition: A vector field F is called a conservative vector field if there exist a potential, a function f, such that .

• Theorem: Let C be a smooth curve given by . Let F be a continuous conservative vector field, and f is a differentiable function satisfying . Then, .

f= F

( ) ,r t a t b

f= F

( )( ) ( )( )C C

d f d f r b f r a = = − F r r

( ) ( ) ( )'b

aF x dx F b F a= −

Fundamental Theorem for Line Integrals(cont)

• Proof:

( )( ) ( )

( )( )

( )( ) ( )( )

' , , , ,

b b

x y z

C a a

b b

x y z

a a

dx dy dzf d f r t r t dt f f f dt

dt dt dt

dx dy dz df f f dt f r t dt

dt dt dt dt

f r b f r a

= =

= + + =

= −

r


• Definition: Let F be continuous on domain D. The integralis called independent of path in D if for any two curves C1, C2 with the same initial and end points, we have

• Corollary: A line integral of a conservative vector field is independent of path.

• Definition: A curve C is called closed if its terminal points coincides.

2 1C Cd d = F r F r

𝐶1

𝐶2

𝐴

𝐵

Cd F r


• Theorem: The integral is independent of path in D if and only if on any closed curve C.

Proof: (➔) Let is independent of path in D. Let C be arbitraryclosed curve. Choose any two points on C, A and B. Let C1 be the curve from A to B, and C2 from B to A, so that , then

() Let on any closed curve C in D. Choose A,B∈D and let C1, C2 be arbitrary paths from A to B. is closed curve, thus

Cd F r

1 2C C C=

1 2 1 2

0C C C C C

d d d d d−

= + = − = F r F r F r F r F r

0C

d = F r

Cd F r

0C

d = F r

1 2C C C= −

1 2 1 2 1 2

0C C C C C C C

d d d d d d d−

= = + = − + = F r F r F r F r F r F r F r


• Theorem: Suppose F=<P,Q> is continuous vector field on an open connected region D. If is independent of path in D, then F is conservative vector field in D, that is there is f such that .

Proof: Let (a,b)∈D be arbitrary fixed point. Define .

Due to independency of path we can choose path C from (a,b) to (x,y) that crosses (x1,y)∈D, x1 is const.

Similarly,

Cd F r

f= F

( )( )

( ),

,

,

x y

a b

f x y d= F r

( )( )

( )

( )

( )1, ,

, ,

,

x y x y

x

a b a b

d df x y d d

dx dx= = F r F r

( )

( )

1

0, no ,

,

xx y

x y

d dd Pdx Qdy

dx dx

=

+ = + F r

( )

( )

1 1

, 0

,

x y xdy

FTCx y x

dPdx P

dx

=

= =

( )( )

( )

( )

( )

( )

( )1

1

, , ,

, , ,

,

x y x y x y

y

a b a b x y

d d df x y d d d

dy dy dy= = + F r F r F r

0, no y

dPdx

dy

=

=( )

( )

1 1

, 0

,

x y ydx

FTCx y y

dQdy Qdy Q

dy

=

+ = =

(x,y1)

(x1,y) (x,y)

(a,b)


• 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


• Recall: , where 𝐓 is the unit tangent vector.

• Green’s theorem in vector form (Tangential Component): Let𝐅 = 𝑃 𝑥, 𝑦 , 𝑄 𝑥, 𝑦 , 0 , then

• Proof:

( )

0 0

1

curl , , , ,0 0 , 0,

curlD D D D D

Q P Q P Q PP Q

x y z z z x y x y

Q P Q PdA dA dA Pdx Qdy d

x y x y

= =

=

= = − − − = −

= − = − = + =

F k

F k k k F r

( )curlD D

dA d

= F k F r

D D

d ds

= F r F T


• Recall: unit normal vector is given by

• Green’s theorem in vector form (Normal Component): Let F=<P,Q> be a vector field and , then

• Proof:

( ) ( ) ( ),r t x t y t=

( )div ,C D

ds x y dA = F n F

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

( )

( ) ( )

( )

( ) ( )

( )

' , ' ' '

' '

' '

'

, , ,y t x t Py t Qx t

r t r t

Py t Qd t

r tC

r t P x t y t Q x t y t

ds

− −

+

= =

=

F n

F n ( )'r t

( )( )

( )'

div ,

b b

a a

b

Green sa D D D

dt Pdy Qdx

QP P QQ dx Pdy dA dA x y dA

x y x y

= −

− = − + = − = + =

F

( )( )

( )

( ) ( )

( )

in 2 ,

verify!' , ''

' '

D

y t x tt

t r tt

−= =

T

Tn


• Example: Evaluate , where D is unit disk.

• Solution: Previously we would evaluate it directly using a trigonometric identity

• Now we can do something else, let and , then , and

• Furthermore, notice that F is conservative VF (since ), therefore vanishes on every closed curve, i.e. no integration required here.

( )( ) ( )

2 2

2 2 2 2 2 2 2 2

0 0'

cos sin ,cos sin ,0 sin , cos ,0 cos sin cos sin 0D D

r tr t

y x dA d t t t t t dt t t t tdt

− = = − = − = F

F r

2 2, ,0x y xy=F

2 2

D

y x dA−

( )1 22 1 2 1 4

2 2 2 2 2 3

00 0 0 0 0

sin 2 1sin cos cos 2 0 0

4 2 4D

ry x dA r rdrd r drd

− = − = − = − = − =

( )2 2 curly x− = F k( ) cos ,sinr t t t=

2 2x y= F


• Example: Evaluate , where C is a circle of radius 2.

• Solution 1:

• Solution 2: Notice that normal to C is and

thus

( )22 2 4

2 2 2

0 0 0

1 1 1div , 2

2 2 2 4C D D

rds x y dA y x dA r rdrd

= = + = = =

F n F

( ),

,

x yt

x y=n

2 2

2 2

x y

x yC

ds+

2 2

2 2 2 2

2 2

2cos 2sin

2cos 2sinx y

x y t tC

t tds+ +

= ( )'r t ( )2 2 2

22 2 2

00 0 0

12 8

4cos sin sin 2 sin 4tdt t tdt tdt t

= = = − =

2 22 2

2 2

,

,

1,

2

x yx y

x yx yxy x y

=

+=

F n

Surface Integral

• Definition: Let 𝑆 be surface . Let 𝑃𝑖𝑗∗ be a

sample point on a patch 𝑆𝑖𝑗 which area is given by .• 𝑆𝑖𝑗 is defined by , consequently

The surface integral is given by

( ) ( ) ( ) ( ) ( ), , , , , , , ,r u v x u v y u v z u v u v D=

ij u vS r r u v =

1 1, ,i i j ju u v v− −

( ) ( ) ( ) * * * * * * * * *

1 1 1, , , , , , , , ,ij i j i j i j i i j j jP x u v y u v z u v u u u v v v− − =

( ) ( )*

,1 1

, , limm n

ij ijm n

i jS

f x y z dS f P S→

= =

=

Surface Integral(cont)

• Theorem:

• Note that the area of the surface S is

• Reminder:

• Note: The relationship of surface integral and surface area are analogical to the relationship between the lineintegral and arclength.

• Example:

( ) ( )( ), , , u vD

S

f x y z dS f r u v r r dA=

( ) ( )( ) ( ), , '

b

C a

f x y z ds f r t r t dt=

( ) 1 u vD

S

A S dS r r dA= =

( )( )

( )( ) ( ) ( )22

: ,

, , , , , 1x y

D z g x y D

f x y z dS f x y g x y g g dA =

= + +


• Example: Let S be a unit sphere. The parametric representation is given by

( )

2 2 3 22

2 2

2 3 2

0 0 0 0

2

3

0 0

1 cos 2cos sin sin sin cos

2

1 1 1 4sin 2 cos cos

2 2 3

sin cos sin

3

cosS D D

r r dAx dS

d d d d

dA

= =

+= = −

= + − + =

( ), sin cos ,sin sin ,cos ,0 ,0 2r =

2 2

sin sin ,sin cos ,0

sin cos ,sin sin ,cos sin sin

cos cos ,cos sin , sinr r

=

= −

=

−

Surface Integral (cont)

• Definition: A two-sided surface S is called oriented if the unit normal vector n is defined at every point (except the boundary points). The orientation is chosen by direction (positive or negative) of the unit normal, in other words by choosing the side.

• Example: A Mobius strip is one-sided surface, therefore non orientable.

• A torus is two-sided, so it can be oriented inward or outward.

• Definition: A closed surface is a boundary of solid region. A closed surface is considered positive oriented if the unit normal points outward.


• Definition: Flux of F across surface S is defined by

• Notice the difference between and , and the similarity with .

• Evaluation:

S S

d dS = F S F n

dS dS dr

( )( )( ) ( ),

u v

u v

u v

u v

r r

r r

S S S

r r

u v u vr r

D D

d dS dS

r u v r r dA r r dA

= =

= =

F S F n F

F F


•

• Rate of flow of a fluid with density 𝜌 and velocity field Ԧ𝑣:

• Electric flux of an electric field E through the surface S: .

• Gauss’s Law: a net charge enclosed by a closed surface S:

where 𝜀0 is permittivity of free space.

• Let K be a conductivity constant of a substance and let u(x,y,z) denote

temperature of a body. The of heat flow is given by −𝐾∇𝑢 and the rate

of heat flow by .

( ): ,

, , , ,1x y x y

D z g x y D D

d P Q R g g dA Pg Qg RdA =

= − − = − − + F S

SdS v n

Sd E S

0S

Q d= E S

S

K u d− S


• Example: Let F=<x,y,z> and

( )( ) ( ) ( )2

0

0 0

, sin sin 2 cos 4C D D

d r dA d dr r A d

= = = = − = F S F

( ), sin cos ,sin sin ,cos ,0 ,0 2r =

2 2

sin sin ,sin cos ,0

sin cos ,sin sin ,cos sin

cos cos ,cos sin , sinr r

−

=

= −

( )( ) ( ) ( )2 2 2 2 2

,

sin sin sin cos

sin

cos cos cos sin

r r r r r r

=

=

+

=

−

F

Stokes’ Theorem

• Let S be an oriented piecewise-smooth surface that is bounded by a simple closed piecewise-smooth boundary curve C with positive orientation. Let F be a vector field whole components have continuous partial derivatives on an open region in that contains S. Then,

3

curlS S

d d

= F r F S

Stokes’ Theorem(cont)

• Stokes’ Theorem says that the line integral around the boundary curve of S of the tangential component of F is equal to the surface integral over of the normal component of the curl of F.

• One of the important uses of Stoke’s Theorem is in calculating surface integrals over “non convenient” surface using surface integral over more convenient surface with the same boundary:

1 1 2 2

curl curlS S S S S

d d d= =

= = F S F r F S


• One see Stokes’ Theorem as a sort of higher dimensional version of Green’s theorem. Really, if S is flat and lies in xy plane, then n=k and therefore

which is a vector form of Green’s theorem.

• Thus, Green’s theorem is a private case of Stokes Theorem.

( )curl curlS S S

d d dS

= = F r F S F k


• Proof (of the light version): We restrict our proof only for the case of S given as z=g(x,y).

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )'

, , , , , ,

b b

x yS

a a

b

x y y xGreen s

a D

x z x x z x

d P x y z dx Q x y z dy R x y z dz Pdx Qdy R z dx z dy

P Rz dx Q Rz dy Q Rz P Rz dAx y

Q Q z R R z

= + + = + + +

= + + + = + − +

= + + +

F r

( ) y yxz Rz+( ) y z y y z yP P z R R z− + + +( ) x xyz Rz+( )

( ) ( ) ( )

( )( ): ,

, , , ,1 curl

D

y z x z x y x y

D

x y u v

D D D z g x y

R Q P R Q P

y z z x x y

dA

R Q z P R z Q P dA

z z dA r r dA d =

− − −

= − − − − + −

= − − = =

F F S

( ) ( )( ),

' 'x y

f x g x y x

f g g y

= = +


• Example: Verify Stokes’ Theorem for over S:

• From the other side we have, , therefore

, ,yz xz xy=F 2 2 1z x y= +

( )2

cos ,sin ,1 0

2

0

2 2

2 2

00 0

cos ,sin ,1 sin ,cos ,0

sin ,cos ,cos sin sin ,cos ,0

sin 2sin cos cos 2 0

2

t t

d t t t t dt

t t t t t t dt

tt tdt tdt

= −

= −

= − + = = =

F r F

xyz= F curl 0 0S S

d d = = F S S


• Example: Evaluate

• Solution: It is clear that a direct evaluation of the line integral is

awkward. Therefore, denote , and

use Stokes’ Theorem. We also need . Finally,

( )2

, , 2 2 6 6 6Stokes

disc discz

I x y z dS dA A disc ==

= − + = = = n kn

𝐼 = ර

cos𝑡,sin𝑡,2

𝑒− Τ𝑥2 2 − 𝑦𝑧 𝑑𝑥 + 𝑒− Τ𝑦2 2 + 𝑥𝑧 + 2𝑥 𝑑𝑦 + 𝑒− Τ𝑧2 2 + 5 𝑑𝑧

curl , , 2 2x y z= − +F

2 2 2/2 /2 /2, 2 , 5x y ze yz e xz x e− − −= − + + +F


• Example: Let and C be an intersection between cylinderand . Evaluate .

• Thus direct integration won’t be nice; therefore we try Stokes. Let the surface S be elliptical region on plane .

2 2, ,y x z= −F2 2 1x y+ = 2y z+ =

( )32

2 2

1

0

: 2 : 1 curl

22 1 2 2

00 0 0 03

1 2

2 3 3

1 12 sin

2 2

curl 0,0,1 2 , ,1 1 2

1 2 sin 2 sin cos

x y

C S z y DD x y

rr

d d y g g dA ydA

r rdrd d d

= − + =

+

= = + − − = +

= + = = + = − =

F n

F r F S

C

d F r

( )

( )( ) ( )

( )

22

23 2

cos ,sin ,2 sin

sin ,cos , 2 sin sin ,cos , cos

sin cos 2 sin cos

r t t t t

r t d t t t t t t

t t t t

= −

= − − − −

= + − −

F r

2y z+ =

Divergence Theorem

• Let E be a simple solid region and let S be the boundary surface of E, given with positive (outward) orientation. Let F be a vector field whose component functions have continuous partial derivatives on an open region that contains E. Then

• Note the similarity with Normal Component Green’s Theorem:

( )div ,C D

ds x y dA = F n F

divS E

d dV = F S F

Divergence Theorem(cont)

• Proof: Let F=<P,Q,R>, we want to show

• Consider , then

• are proved in a similar manner using the expressions for E as a type II or type III region, respectively.

div zyE

xE E SE Ss Show

S S

P dV P dQ dV Q dSdV R dV R dSS

dS d

= + + = + +

=

=

iF k n

F n F S

nn j

( )

( )

( )( ) ( )( )

1 3

2

1

,

1

,

2 ,, , ,, ,

u x y

zE

D u x

S

y D

S

R x y u x y

R

R dV RdzdA R x y u x y

d SS

dA

R d

= = −

+ =

k nk n2

3either 0 or

S S

S

R dS R dS = =

− = k n

k n k n

( ) ( ) ( ) ( ) 1 2type I

, , : , , , ,E x y z x y D u x y z u x y=

,S E SE

yx Q dVP dV QP S dSd == j ni n


• Example: Let and S be defined by

on top, on sides and at the bottom.

212

, ,xy y z= −F

0z =

2 24 3 3 ,1 4z x y z= − −

2 2 1,0 1x y z+ =

divS E

d dV y y = = − F S F

( )

2

22 1 4 3 2 1

4 3

0

0 0 0 0 0

12 1

2 2 4

00 0

1

3 54 3 2 2

4 2

rr

E

dV rdzdrd rz drd

r r drd r r

−−

+ = =

= − = − =


• Example: Let , S spherical solid of radius 2 in first octant.

• Example: Let

3 3 3, ,x y z=F

/2 /2 2

2 2 2 4

0 0 0

2/2 /2 /25/2

0

0 0 00

div 3 3 sin

3 32 48 483 sin sin cos

5 5 2 5 5

S E E

d dV x y z dV d d d

d d d

= = + + = =

= = = − =

F S F

23 cos , , sinzy z x e x y=F

div 0 0S E E

d dV dV = = = F S F

Decision Tree Vector CalculusQuestion

Line Integral Surface Integral

ClosedCurve?

ClosedSurface?

The Divergence Theorem

Parametrize,Solve Directly

yes no

Work/FluxLine/Surface

yes no

2D/3DConservative?

Green’s Theorem Stoke’s Theorem

2D 3D

Fundamental Theorem of Line

Integral

Parametrize,Solve Directly

yes no

vector calculus - mmedvin.math.ncsu.edu•recall: fundamental theorem of calculus (ftc)...

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