Chap. 10 Vector Integral Calculus. Integral Theorems

중앙대학교 건설환경플랜트공학과 교수

김 진 홍

- 9주차 강의 내용 -

Chap. 10 Vector Integral Calculus. Integral Theorems

10.1 Line Integrals

 The concept of a line integral is a simple and natural generalization of a definite integral

(1)



a^b f (x)dx known from calculus.

 (1) means that the integrand f(x) is integrated from x=a to x=b.

But, in a line integral f(x) is integrated along a curve

C

in space (or in the plane). Hence curve integral would be a better name.

 We represent the curve

C

by a parametric representation (2) r(t)[x(t),y(t),z(t)]x(t)iy(t)jz(t)k

 The curve

C

is called the path of integration.

A; r(a) ; initial point B; r(b) ; terminal point

) (at b

Oriented curve

General assumption

Every path of integration of a line integral is assumed to be piecewise smooth; that is, it consists of finitely many smooth curves.

Definition and Evaluation of Line Integrals

 A line integral of a vector function F(r) over a curve

C

:r(t) is defined by

(3*)



^{ }



^{  }



^{b  }

a

C C

dt z F y F x F dz

F dy F dx F dr

r

F( ) ( _{1 2 3} ) ( ₁ ' ₂ ' ₃ ')

 If the path of integration

C

in (3) is a closed curve, then instead of we also write



C



C

(3)

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

C

b

a









^{( )}



^{( ( )) '( )) '}

In terms of components, with and , (3) becomesdr[dx,dy,dz] dt

 d '

Ex. 1) Evaluation of a Line Integral in the Plane

Find the value of line integral (3) when F(r) = and

C

is the circular arc from A to B.

] , [y xy

Chap. 10 Vector Integral Calculus. Integral Theorems

Sol.)

C

is represented by r(t) = [cost,sint] costisintj

t y

t

x  cos ,  sin

tj t ti

t t t

j t y t x i t y t

r F

sin cos sin

] sin cos , sin [ ) ( ) ( ) ( ))

( (



















tj ti

t t

t

r ' ( )  [  sin , cos ]   sin  cos

by (3)

) cos ( tu

dt t t t

dt t t t

t t

dr r

C

F ( ) [ sin , cos sin ] [ sin , cos ]

^/2

(sin cos

²

sin )

0 2 2

/

0

     





  

^



^

4521 . 3 0 0 1 ) 4

( )

2 cos 1 2( 1

2 / 0

0 1

2     









^{ t dt}



^{u du}

^

) 2 / 0

(  t

Ex. 2) Line Integral in Space ; Find the line integral of (3) yk

xj zi y x z r

F( )[ , , ]   and

C

is the helix.

(4) r(t)[cos t,sin t,3t] cos ti sin tj 3tk

) 2 / 0

(  t  when

Sol.) x cost, y sint, z 3t

Example 1

Example 2

) 3 cos sin

( ) sin cos

3 ( ) ( ' )) (

(r t r t ti tj tk ti tj k

F        

t t

t

t( sin ) cos 3sin

3   ² 



Chap. 10 Vector Integral Calculus. Integral Theorems



^{ }



^{     }

 C F(r) dr ^/2( 3tsin t cos t 3sin t)dt 6 0

0

2

 



99 . 21 7 

  Ex. 3) Evaluation of Line Integral

zxk yzj

xyi zx

yz xy r

F( )[ , , ]  

and

C

is the twisted cubic given by r(t)  [t,t²,t³]ti  t² j  t³k ) 1 0

(  t  when

Sol.) x t, y t², z t³

6 3 2 4

5

3 ) ( 2 3 ) 5

( ) ( ' )) (

(r t r t t i t j t k i tj t k t t

F         



^{ }



^{ }

 C F(r) dr ¹(t 5t⁶)dt 27 /28

0 3



C F(r)^dr

Chap. 10 Vector Integral Calculus. Integral Theorems

Simple general properties of the line integral (5a)

(5b)



CkF^dr ^k



CF^dr (k ; constant)



C(F ^G)^dr ^



CF^dr^



CG^dr (5c)



C

F ^ dr ^ 

C

F

1

^ dr ^ 

C

F

2

^ dr

Ex. 4) Work Done by a Variable Force

If F in Ex. 1 is a force, the work done by F is the displacement along the quarter-circle is 0.4521, measured in suitable unit, newton-meters(nt·m, also called joules, abbreviation J).

Ex. 5) Find the work done by a force F(r)= [x²,xy]in moving a particle along the quarter-circle r(t)[cos t,sin t](0 t



/2)

Sol.)

x  cos t , y  sin t

t t t

r t r

F ( ( ))  ' ( )   2 cos

²

sin

j t t i

t t

r

F( ( ))cos² cos sin

r ' ( t )   sin t i  cos tj

Therefore the work done is dt t t dr

r

CF( ) ^/2( 2cos sin )

0



^{ }



^{ } 2

^{2 cos} _{3 3} ²

2 /

0

3

  



 

t



Chap. 10 Vector Integral Calculus. Integral Theorems

(6) ^



C ^{ }



^

b

a F r t v t dt dr

F

W ( ( )) ( )

) ( ' )

( t mv t r

m

F   

by Newton's second law

b t

a t b

a m v

vdt mv W





^{ } 

 ²

' 2 Path Dependence

Theorem 2

The line integral (3) generally depends not only on F and on the endpoints A and B of the path, but also on the path itself along which the integral is taken.

Ex. 6) Work Done Equals the Gain in Kinetic Energy

Let F be a force, t be a time, so that dr/dt=v, velocity.

Then (3) becomes

* Proof Line Integral F(r) = [0, xy, 0] along the straight segment C₁ : r₁(t)

= [t, t, 0] and the parabola C₂ : r₂(t) = [t, t², 0] with 0 t1 Sol.) F(r₁(t))·r₁’(t) = t², F(r₂(t))·r₂’(t) = 2t⁴



 ^{ }

0¹

^

2

1 / 3

) (

1

dt t dr r

C

F 

^{ }



0^{1 }

4

2F(r) dr 2t dt 2/5

C ⇒ Path Dependence

Chap. 10 Vector Integral Calculus. Integral Theorems

10.2 Path Independence of Line Integrals

The line integral

(1) F(r) dr (F₁dx F₂dy F₃dz) (dr [dx,dy,dz])

C   C   

 

is path independent in a domain D in space if and only if : (

Theorem

1) F = grad f (see Sec. 9.7 for the gradient)

(

Theorem

2) Integration around closed curves

C

in

D

always gives 0.

(

Theorem

3) curl F = 0 (provided

D

is simply connected) Theorem 1

Path Independence

A line integral (1) with continuous F₁, F_2,F₃ in a domain D in space is path independent in D if and only if F = [F₁, F₂, F₃] is the gradient of some function f in D.

(2) F = grad f, thus,

z F f y F f x F f



 



 



  _{2 3}

1 , ,

Chap. 10 Vector Integral Calculus. Integral Theorems

Theorem 3

Path Independence

The integral (1) is path independent in D if and only if curl F = 0 ; in components

y F x

F x

F z

F z

F y

F



 







 







 



 _{3 2 1 3 2 1}

, ,

Theorem 2

Path Independence

The integral (1) is path independent in D if and only if its value around every closed path in D is zero.

Chap. 10 Vector Integral Calculus. Integral Theorems

Ex. 1) Show that

and find its value in the integration from A(0,0,0) to B(2,2,2).

Sol.) F [2x,2y,4z]

) 4 2

2 ( )

(r dr xdx ydy zdz F

C   C  

 

is path independent

= grad f, where f x²y²2z²

z F f y F f x F f



 



 



  _{2 3}

1 , ,

because

Chap. 10 Vector Integral Calculus. Integral Theorems

Hence the integral is independent of path and the line integral is, 16

8 4 4 ) 0 , 0 , 0 ( ) 2 , 2 , 2

(  f    

f

Ex. 2) Evaluate the integral from A(0,1,2) to B(1,-1,7).

Sol.)

z F f y F f x F f



 



 



  _{2 3}

1 , ,



^{ }

C

dz y yzdy dx

x 2 )

3

( ^{2 2}

by showing that F has a potential.

, 3 ²

x x f



  2 ,

y yz f



 

) , ( ,

3 ² f x³ g y z x

x f  



 

z y f



 

2

) ( )

, ( ,

2yz g y z y²z h z y

g y

f   



 





)

2 (

3 y z h z

x

f   

0 ,

)

( ²

2   

 

 y h z y h

z f

z y x f  ³ ²



Thus, the integral is f(1,1,7) f(0,1,2) 17(02)6

C

h Disregarding

C

, k

y yzj i

x

F 3 ² 2  ²

Chap. 10 Vector Integral Calculus. Integral Theorems

Ex. 3) Show that integral

evaluate its value.

Sol.)



^{      }

C

dz xy yz dy

xz z x dx yz

y ) ( 3 ) (9 1)

( ^{3 2}

is independent of the path C between (1,1,1) to (2,1,4) and

1 9

, 3

, ₂ ³ ₃ ²

1 y yz F  x z xz F  yz xy F



Thus, the integral is f(2,1,4) f(1,1,1)1984194 k

xy yz j

xz z x i yz y

F (  ) ( 3 ³ ) (9 ² 1)

,

1 ²

1

x z F y

F



 



 

 ^{1 3},

x y F z F



 

 



y x F z z

F



 



 

 _{2 2 3}

9

curl F = 0 and the integral is independent of the path.

x, yz f

y 

 

 f  xyxyzg(y,z) x 3z³ xz, y

xz g y x

f   







 



) ( 3yz³ h z xyz

xy

f    

1 9

) (

9 ²   ² 



 

 xy yz h z yz xy

z f

z yz xyz xy

f    

 3 ³

) ( 3

,

3z³ g yz³ h z y

g   





C z h z

h( )1,   Disregarding C,

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