Ch. 10 Vector Integral Calculus. Integral Theorems Ch. 10 Vector Integral Calculus. Integral...

Ch. 10 Vector Integral Calculus. Integral Theorems

Ch. 10 Vector Integral Calculus.

Integral Theorems

The vector integral calculus extends integrals from calculus to integrals over curves, surfaces, and solids.

These extended integrals have basic engineering applications in solid mechanics, in fluid flow, and in heat problems.

These different kinds of integrals can be transformed into one another

• in order to simplify evaluations or to gain useful general formulas

• by the powerful formulas of Green, Gauss, Stokes

The formulas involve the divergence and the curl.

Ch. 10 Vector Integral Calculus. Integral Theorems 10.1 Line Integrals

10.1 Line Integrals

Concept of a line integral

: A simple and natural generalization of a definite integral known from calculus

• Line Integral or Curve Integral

: We integrate a given function ( Integrand ) along a curve C in space ( or in the plane ).

• Path of Integration

C : r(t) = [ x(t) , y(t) , z(t) ] = x(t)i + y(t)j + z(t)k ( a t b )

General Assumption

: Every path of integration of a line integral is assumed to be piecewise smooth.

Definition and Evaluation of Line Integrals

A line integral of F(r) over a curve C :

• In term of components

' 'b

C a

dd t t dt

dt

rF r r F r r r

1 2 3 1 2 3' ' 'b

C C a

d F dx F dy F dz F x F y F z dt F r r


Ex.1 Find the value of the line integral when F(r) = [ y , xy ] = yi – xyj and C is the circular arc in the

following figure form A to B.

Represent C by r(t) = [ cost , sint ] = cost i + sint j, where 0 t /2.

Then x(t) = cost, y(t) = sint, and F(r(t)) = y(t)i – x(t)y(t)j = sint i – cost sint j.

By differentiation,

2

0

22 2

0

022

0 1

' sin cos sin cos

sin cos sin sin cos

sin cos sin

1 1 cos 2

2

C

t t, t t t

d t, t t t, t dt

t t t dt

t dt u d

r i j

F r r

10 0.4521

4 3u

10.1 Line Integrals

< Example 1 >


Simple general properties of the line integral

1 2

(5a) constant

(5b)

(5c)

C C

C C C

C C C

k d k d k

d d d

d d d

F r F r

F G r F r G r

F r F r F r

10.1 Line Integrals

< Formula (5c) >

Theorem 1 Direction-Preserving Parametric Transformations

Any representations of C that give the same positive direction on C also yield the same value

of the line integral


Motivation of the Line Integral : Work Done by a Force

• The work W done by a constant force F in the displacement along a straight segment d :

• We define the work W done by a variable force F in the displacement along a curve C : r(t) as

the limit of sums of works done in displacements along small chords of C. ( This definition

amounts to defining W by the line integral.)

Ex.4 Work Done Equals the Gain in Kinetic Energy

Let F be a force and t be time, then dr/dt = v, velocity.

By Newton’s second law,

W F d

b

C a

W d t t dt F r F r v

2 '

2 2

t bb b

t aa a

mm '' t m ' t W m ' t t dt m dt

v vF r v v v v

10.1 Line Integrals


< Proof of Theorem 2 >

Other Forms of Line Integrals : A line integral whose value is a vector rather than a scalar

1 2 3, , b b

C a a

dt t dt F t F t F t dt F r F r r r r

Ex.5 Integrate F(r) = [ xy , yz , z ] along the helix.

22

2 2 2

0 0

1 3cos , 3sin 3 cos , 0, 6 , 6

2 2t dt t t t t t

F r

Ex. Integrate F = [ 0 , xy , 0 ] on the straight segment and the parabola

with 0 t 1, respectively.

1

2

21 1 1 1

42 2 2 2

1'

3

2' 2

5

C

C

t t t d

t t t d

F r r F r r

F r r F r r

1 1 0C : t t, t, r 22 2 0C : t t, t , r

10.1 Line Integrals

Theorem 2 Path Dependence

The line integral generally depends not only on F and on the endpoints A and B of the path, but

also on the path itself which the integral is taken.

Ch. 10 Vector Integral Calculus. Integral Theorems 10.2 Path Independence of Line Integrals

10.2 Path Independence of Line Integrals

Theorem 1 Path Independence

A line integral with continuous in a domain D in space is path independent in D

if and only if is the gradient of some function f in D,

B

1 2 3 1 2 3grad , , A

f f ff F F F F dx F dy F dz f B f A

x y z

F

1 2 3 F , F , F

1 2 3F , F , FF

Ex.1 Path Independence

Show that the integral is path independent in any domain in space and find its value

in the integration from A : ( 0, 0, 0 ) to B : ( 2, 2, 2 ).

Hence the integral is independent of path according to Theorem 1.

2 2 4 2, 2, 2 0, 0, 0 4 4 8 16C

xdx ydy zdz f B f A f f

2 2 4C

xdx ydy zdz

2 2 21 2 32 2 , 4z grad 2 , 2 , 4 2

f f fx, y f x F y F z F f x y z

x y z

F

Ch. 10 Vector Integral Calculus. Integral Theorems 10.2 Path Independence of Line Integrals

Theorem 2 Path Independence

The integral is path independent in a domain D if and only if its value around every closed path

in D is zero.

Theorem 3* Path Independence

The integral is path independent in a domain D in space if and only if the differential form

has continuous coefficient functions and is exact in D.1 2 3 F , F , F1 2 3d F dx F dy F dz F r

is exact if and only if there is f in D such that F = grad f.1 2 3d F dx F dy F dz F r


Theorem 3 Criterion for Exactness and Path Independence

Let in the line integral, be continuous and have

continuous first partial derivatives in a domain D in space. Then

• If the differential form is exact in D – and thus line integral is path

independent -, then in D, curl F = 0 ; in components .

• If curl F = 0 holds in D and D is simply connected, then is exact in D

– and thus line integral is path independent .

1 2 3 F , F , F 1 2 3

C C

d F dx F dy F dz F r r

1 2 3d F dx F dy F dz F r

3 32 1 2 1F FF F F F, ,

y z z x x y

1 2 3d F dx F dy F dz F r



Ex.3 Exactness and Independence of Path. Determination of a Potential

Show that the differential form under the integral sign of

is exact, so that we have independence of path in any domain, and find the value of I from A : ( 0, 0, 1 )

to B : ( 1, /4, 2 ).

Exactness :

To find f

2 23 2 1 3 2 12 cos sin , 4 , 2

y z z x x yF x z yz yz yz F F xyz F F xz F

2 2 2 22

2 21

2 23

cos sin ,

2 2 0

2 cos ' 2 cos ' 0 const

x x x

z

f F dy x z z yz dy x yz yz g x z

f xyz g F xyz g g h z

f x yz y yz h F x yz y yz h h

2 2 sin , 1 4 sin 0 14 2

f x yz yz f B f A

2 2 2 22 cos 2 cosC

I xyz dx x z z yz dy x yz y yz dz


Ch. 10 Vector Integral Calculus. Integral Theorems 10.3 Calculus Review : Double Integrals

Double integral

: Volume of the region between the surface defined by the function and the plane

Definition of the double integral

• We subdivide the region R by drawing parallels to the x- and y-axes.

• We number the rectangles that are entirely within R from 1 to n.

• In each such rectangle we choose a point in the kth rectangle,

whose area we denote by .

The length of the maximum diagonal of the rectangles approaches zero as n approaches infinity.

• We form the sum .

• Assuming that f ( x , y ) is continuous in R and R is bounded by finitely many smooth curves, one

can show that this sequence converges and its limit is independent of the choice of

subdivisions and corresponding points .

10.3 Calculus Review : Double Integrals

k kx , y

kΔA

1

n

n k k kk

J f x , y ΔA

1 2, , n nJ J

k kx , y

< Subdivision of a region R >


Properties of double integrals

Mean Value Theorem

R is simply connected, then there exists at least one point in R such that we have

where A is the area of R.

1 2

constant

Figure

R R

R R R

R R R

kf dxdy k f dxdy k

f g dxdy f dxdy gdxdy

f dxdy f dxdy f dxdy

0 0, ,R

f x y dxdy f x y A

0 0,x y


< Formula >


Evaluation of Double Integrals by Two Successive Integrations

•

•

, ,h xb

R a g x

f x y dxdy f x y dy dx

, ,q yd

R c p y

f x y dxdy f x y dx dy


< Evaluation of a double integral >

< Evaluation of a double integral >

Ch. 10 Vector Integral Calculus. Integral Theorems 10.4 Green’s Theorem in the Plane

10.4 Green’s Theorem in the Plane

Theorem 1 Green’s Theorem in the Plane

Let R be a closed bounded region in the xy-plane whose boundary C consists of finitely many

smooth curves. Let and be functions that are continuous and have continuous

partial derivatives and everywhere in some domain containing R. Then

Here we integrate along the entire boundary C of R in such a sense that R is on the left as we

advance in the direction of integration.

1 2 1 2, curlR C

F F F F dxdy d F i j F k F r

1 2, ,F x y F x y

1 2 F F

y x

2 11 2

R C

F Fdxdy F dx F dy

x y

• Vectorial form

< Region R whose boundary C consists of two parts >


Ex. 1 Verification of Green’s Theorem in the Plane

Green’s theorem in the plane will be quite important in our further work. Let us get used to it by verifying it

for and C the circle .

1. ( Circular disk R has area . )

2. We must orient C counterclockwise

2 2 21 27 , 2 2 1F y y F xy x x y

2 1 2 2 2 7 9 9R R R

F Fdxdy y y dxdy dxdy

x y

2 21 2

22

1 2

0

23 2 2 2

0

7 sin 7sin 2 2 2cos sin 2cos

sin 7sin sin 2 cos sin cos cos

sin 7sin 2cos sin 2cos

π

C

π

F y y t t, F xy x t t t

F x' F y' dt t t t t t t t dt

t t t t t dt

0 7 0 2 9 π - π π

cos sin , sin cos t t , t ' t t , t r r



Ex. 3 Area of a Plane Region in Polar Coordinates

Polar coordinates :

21 1 1cos sin cos sin cos sin

2 2 2C C C

A xdy ydx r dr r d r dr r d r dθ

cos , sin cos sin , sin cosx r y r dx dr r d dy dr r d

Ex. 2 Area of a Plane Region as a Line Integral Over the Boundary

Some Applications of Green’s Theorem

1 2

1 2

0,

, 0

R C

R C

F F x dxdy xdy

F y F dxdy ydx

1

2 C

A xdy ydx


Ch. 10 Vector Integral Calculus. Integral Theorems 10.7 Triple Integrals. Divergence Theorem of Gauss

10.7 Triple Integrals. Divergence Theorem of Gauss

Triple integral

: An integral of a function f ( x,y,z ) taken over a closed bounded region T in space

Definition of the triple integral

• We subdivide T by planes parallel to the coordinate planes.

• We consider those boxes of the subdivision that lie entirely inside T,

and number them from 1 to n.

• In each such box we choose an arbitrary point, say, in box k.

The maximum length of all edges of those n boxes approaches zero as n approaches infinity.

• The volume of box k we denote by . We now form the sum .

• Assuming that f ( x,y,z ) is continuous in T and T is bounded by finitely many smooth surfaces,

one can show that this sequence converges and its limit is independent of the

choice of subdivisions and corresponding points .

,k k kx , y z

kΔV 1

,n

n k k k kk

J f x , y z ΔV

1 2, , n nJ J

,k k kx , y z


Theorem 1 Divergence Theorem of Gauss

Let T be a closed bounded region in space whose boundary is a piecewise smooth orientable

surface S. Let F (x,y,z) be a vector function that is continuous and has continuous first partial

derivatives in some domain containing T. Then

In components of and of the outer unit normal vector

of S, formula becomes

31 21 2 3 1 2 3cos cos cos

T S S

FF Fdxdydz F F F dA F dydz F dzdx F dxdy

x y z

1 2 3, , F F FF cos , cos , cos n

divT S

dV dA F F n



Ex. 1 Evaluation of a Surface Integral by the Divergence Theorem

Before we prove the theorem, let us show a typical application. Evaluate

where S is the closed surface consisting of the cylinder and the circular disks z = 0

and z = b

Polar coordinates ( dxdydz = rdrddz )

3 2 2

S

I x dydz x ydzdx x zdxdy 2 2 2 0x y a z b

2 2 2x y a

3 2 2 2 2 2 21 2 3, , div 3 5F x F x y F x z x x x x F

2

2 2 2

0 0 0

2 4 42 4

0 0 0

5 5 cos

5 5 cos 5

4 4 4

b a

T z r

b b

z z

I x dxdydz r rdrd dz

a ad dz dz a b


< Surface S in Example 1 >

Ch. 10 Vector Integral Calculus. Integral Theorems 10.9 Stokes’s Theorem

10.9 Stokes’s Theorem

Theorem 1 Stokes’s Theorem

Let S be a piecewise smooth oriented surface in space and let the boundary of S be a piecewise

smooth simple closed curve C. Let F(x,y,z) be a continuous vector function that has continuous

first partial derivatives in a domain in space containing S. Then

Here n is a unit normal vector of S and, depending on n, the integration around C is taken in the

sense shown. Furthermore, is the unit tangent vector and s the arc length of C.

In components, formula becomes

Here, and R is the region with

boundary curve is the uv-plane corresponding to S represented by r(u,v).

curl 'S C

dA s ds F n F r

3 32 1 2 11 2 3 1 2 3

R C

F FF F F FN N N dudv F dx F dy F dz

y z z x x y

' /d dsr r

1 2 3 1 2 3[ , , ], [ , , ], , ' [ , , ]F F F N N N dA dudv ds dx dy dz F N n N r

C


Ex. 1 Verification of Stokes’s Theorem

Let us first get used to it by verifying it for F = [ y,z,x ] and S the paraboloid

Case 1. The curve C is the circle .

Its unit tangent vector :

The function F on C :

Case 2. The surface integral.

A normal vector of S :

2 2

0 0

' sin sin 0 0C

d s s ds s s ds

F r F r r

2 1 2

0 0 0

curl curl 2 2 1

2 1 1 2 cos sin 1 cos sin 0 0 2

3 2 2

S R R

r

dA dxdy x y dxdy

r rdrd d

F n F N

2 2, 1 , 0z f x y x y z

cos , sin , 0s s sr

' sin , cos , 0s s s r

sin , 0, coss s sF r

1 2 3 1 2 3, , curl curl , , curl , , 1, 1, 1F y F z F x F F F y z x F

grad , 2 , 2 , 1z f x y x y N

curl 2 2 1x y F N

< Surface S in Example 1 >

10.9 Stokes’s Theorem

Ch. 10 Vector Integral Calculus. Integral Theorems Ch. 10 Vector Integral Calculus. Integral...

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