LECTURE-27-28-LINE INTEGRAL FOR ENGINEERING MATHEMATICS.ppt

Engineering Maths Line Integral

1. DEPARTMENT OF
MATHEMATICS
[YEAR OF ESTABLISHMENT – 1997]
DEPARTMENT OF MATHEMATICS, CVRCE

2. MATHEMATICS - I
FOR BTECH 1
st
SEMESTER COURSE [COMMON TO ALL BRANCHES OF
ENGINEERING]
DEPARTMENT OF MATHEMATICS, CVRCE
TEXT BOOK:
ADVANCED
ENGINEERING
MATHEMATICS
BY ERWIN
KREYSZIG

3. Line Integrals
[Chapters – 9.1,9.2]
10/28/25 LINE INTEGRAL CONTI..........
LECTURES – 27 & 28

4. 10/28/25 LINE INTEGRAL CONTI..........
Content:

►
Introduction to Line Integral

►
Physical Application of Line Integrals

►
Definition and Examples

►
Properties

►
Examples

►
Test Knowledge
► Independent of Path

►
Problem Solved

►
Practice Problems

5. Let C be a curve with a parametric representation

A
C
B
A
B
C
The curve C is called path of integration , A: (a) its initial point
and B: (b) its terminal point of the path of integration C. The
direction from A to B in which t increases is called the positive
direction on C. We can indicate the direction by arrow. If the point
A and B coincide, then C is called a closed path.
INTRODUCTION TO LINE INTEGRAL
ˆˆ ˆ() [ (), (), ()] () () ()r t xt yt zt xti yt j zt k   

r

r


6. 10/28/25 LINE INTEGRAL CONTI..........
INTRODUCTION TO LINE INTEGRAL
The curve C is called a smooth curve , if C has a unique tangent at each
of its points whose direction varies continuously as we move along C.
Mathematically, C has representation

ˆˆ ˆ() [ (), (), ()] () () ()r t xt yt zt xti yt j zt k   

Such that is differentiable and the derivative is continuous
and different from the zero vector at every point of C.

dr
dt

()r t


7. In physics, the line integrals are used, in particular, for
computations of
•mass of a wire;
• center of mass and moments of inertia of a wire;
• work done by a force on an object moving in a vector
field;
• magnetic field around a conductor (Ampere’s Law);
• voltage generated in a loop (Faraday’s Law of magnetic
induction).
10/28/25 LINE INTEGRAL CONTI..........
Physical Applications of Line
Integrals

8. Definition and Evaluation of line Integral:
A line integral of a vector function over a curve C is
defined by
F r

  (1)
b
C a
dr
F r dr F r t dt
dt
    

   
 
 
1 2 3 1 2 3
Alsointermsof cartesianCoordinates with , ,
wehavefrom the above equation
( ) (2)
b
C C a
dr dx dy dz
dx dy dz
F r dr Fdx Fdy Fdz F F F dt
dt dt dt

 
       
 
 
  

 

9. 10/28/25 LINE INTEGRAL CONTI..........
Definition and Evaluation of line Integral:
SOME QUICK REMARKS
 
C
C
If a path of integration is a closed curve, then
may also denote a line integral as instead of .
C
 
Line integral arises naturally in mechanics in computation
of work done by a fource in a displace along a given path.As
such a line integral is also called a int.work egral


10. Definition and Evaluation of line Integral:
Line integral (2) is a definite integral of a function of
taken over the interval on the in the
This definite integral exists if i

directi s
and
on.
is
t at a t b
positive F
continuous C piecewise
xis 



, because this
makes .
smooth
dr
F piecewisecontinous
dt



Line integral (2) with given and is
.
F C independentof
of thechoiceof representationof C


Line integral (2) with given and given and is different along
different paths joing to . That is the line integral (2) dependson .


F a b
a b C



11. 1. ( ). ( ) where isaconstant
C C
kF r dr k F r dr k 
    
10/28/25 LINE INTEGRAL CONTI..........
General Properties of Line Integral:
 
1 2 3
1 2 3
1 1 2 2 3 3
1 1 2 2
2. For vector functions , , , , whose line integral
of the form (1) exists and , , , arbitrary constatns
( ) ( ) ( ) ( ) •
( )• ( )
n
n
n n
C
C
G G G G
a a a a
aG r aG r aG r aG r dr
a G r dr a G r
   
 


   


       

   
3 3
• ( )• ( )•
n n
C C C
dr a G r dr a G r dr    
     


12. 10/28/25 LINE INTEGRAL CONTI..........
General Properties of Line Integral:
1
2
1 2 3
3. ( ). ( ). ( ). ( ).
where isthesumofthecurve , , , .
n
C C C C
n
F r dr F r dr F r dr F r dr
C C C C C
      
          


4. (). ().
b a
a b
Frdr Frdr 
    

13. Other Forms of Line Integrals:
 
1 1 1
ˆIf we set and , so that ,
(2) gives another form of line integral given by
(3)
b
C a
f dx
F Fi F f F
dx dt
dt
f r dt f r t dt
 
  
 
   
 
 
  

 

14. 10/28/25 LINE INTEGRAL CONTI..........
Evaluationofintegral ( ) withArclengthasparameter:
C
f r ds

 
 
2 2 2
( ) (), (), ()
4
Where
C C
C
ds
f rds f xt yt zt dt
dt
dr dr
f r t dt
dt dt
ds dx dy dz dr dr
dt dt dt dt dt dt
 

 
 
  
     
    
     
     
 


 

 
 ( ) (), (), ()
b
C a
dr dr
f r ds f xt yt zt dt
dt dt
 
   
 
 
 
 

If limits of t along C are a and b, then

15. 

2 2
Calculate , where ,
and is the straight line segment from 0,0 to 1,4 .
C
F r dr F y x
C
   
 
  
PROBLEM - 1
SOLUTION:
Equation of the straight line segment from 0,0 to 1,4 is 4 .y x
Let 4.x t y t  

ˆ ˆ ˆ ˆHere 4r t xi yj ti tj   

 
d
ˆ ˆ ˆ ˆ 4 4
r d
ti tj i j
dt dt
   


16. 
2
2 2 2
4 , 16 , .F r t t t t t       
  

2 2
Given that , .F y x  
 

 At 0,0 , 0 and at 1,4 , 1.t t 
   
2 2ˆ ˆ ˆ ˆ16 4
dr
F r t t i t j i j
dt
    


2 2 2
16 4 12t t t  

17.  
1
0C
dr
F r dr F r t dt
dt
   

   
Therefore, the required line integral is
1 1
2 2
0 0
12 12tdt tdt  
1
3
0
12
3
t 

 
 
1
12 0 4.
3
 
  
 
 

18. 
 
2 3
Calculate , where cosh , sinh ,
and : , , from 0,0,0 to 2,4,8 .
z
C
F r dr F x y e
C r t t t
  
 
 
 

  

PROBLEM - 2
SOLUTION: 
2 3ˆ ˆˆ ˆ ˆ ˆHere r t xi yj zk ti t j tk     

 
2 3 2d
ˆ ˆˆ ˆ ˆ ˆ 2 3
r d
ti t j tk i tj tk
dt dt
      

  
2 3
At 0,0,0 , , , (0,0,0) 0 tt t t  
 
2 3
At 2,4,8 , , , 2,4,8 2.tt t t  

19. 
3
2
cosh, sinh ,
t
F r t t t e  
 

Given that cosh , sinh , .
z
F x y e 
 

   
3
2 2 ˆ ˆˆ ˆ ˆ ˆcosh sinh 2 3
tdr
F r t ti t j e k i tj t k
dt
      


3
2 2
cosh 2sinh 3
t
t t t te  

20.  
2
0C
dr
F r dr F r t dt
dt
   

   
Therefore, the required line integral is
 
3
2
2 2
0
cosh 2sinh 3
t
t t t te dt  
3
2
2
0
sinh cosh
t
t t e   
 
  
8
sinh2 cosh4 0 1 1e      
 
8
sinh2 cosh4 2e   

21. 
 
2 2
Calculate , where , and is the
quarter - circle from 2,0 to 0,2 with center at 0,0 .
C
F r dr F xy x y C  
 
  
PROBLEM - 3
SOLUTION:
y
x-axis
t=0
t=/2
The radius of the circle
with center at (0,0)
with points (2,0) and
(0,2) on it is 2 .
(0,0)
PATH OF
INTEGRATION
(0,2)
t
y-axis
(2,0)

22.  
d
ˆ ˆ ˆ ˆ 2cos 2sin 2sin 2cos
r d
ti tj ti tj
dt dt
    



Parametric representation of the quarter -circle from
(2,0) to (0,2) with center at 0,0 and radius 2 is
ˆ ˆ =2cos 2sin ,0 .
2
r t ti tj t

  


2 2
4sin cos, 16sin cosF r t t t t t  
 

2 2
Given that , .F xy x y 
 


23.    
2 2ˆ ˆ ˆ ˆ4sin cos 16sin cos 2sin 2cos
dr
F r t t ti t tj ti tj
dt
      


2 2 3
8sin cos 32sin cost t t t 
 
2
0C
dr
F r dr F r t dt
dt
   

   
Therefore, the required line integral is
 
2
2 2 3
0
8sin cos 32sin cost t t tdt

  
2 2
2 2 3
0 0
8 sin cos 32 sin cost tdt t tdt
 
  

24. 3 2
2
2 2
00
sin
8 32 sin (1 sin )cos
3
t
t t tdt


 
  
 
 

3 5
2 2
0 0
8 sin sin
32 32
3 3 5
t t
 
   
  
   
   
2 2
2 4
0 0
1
8 0 32 sin cos 32 sin cos
3
t tdt t tdt
 
 
   
 
 
 
2
0
8 1 1 8 32 32 32 8
32 0 32 0 8
3 3 5 3 3 5 5 5

   
          
   
   

25. 2 2
1. [( ),( )] : 1,1 4Q F x y y x C xy x     
TEST YOUR KNOWLEDGE
.2 [ , . ], : ( ) [2cos , ,2sin ] (2,0,0) (2,2 ,0)Q F x y y zz x C r t t t t from to     
2 2
3 3
3
2
.3: [ , ], :
y x
Q F e e C y x 
10/28/25 LINE INTEGRAL CONTI..........

26. 
2 2
Evaluate the integral , where
and : 3 from (0,0) to (2,6).
C
f r ds f x y
C y x
 



PROBLEM - 4
Solution:
Given path is the line : 3 from (0,0) to (2,6).C y x
Let . So, 3 3.x t y x t  

Therefore, the parametric representation of the
ˆ ˆ ˆ ˆgiven straight line is 3 .r t xi yj ti tj   


27.  
d
ˆ ˆ ˆ ˆ 3 3
r d
ti tj i j
dt dt
   

  
ˆ ˆ ˆ ˆ 3 3 1 9 10
dr dr
i j i j
dt dt
       
 
 At 0,0, 0 and at 2,6, 2.t t 
2 2
Given that .f x y 

2 2
2
3 10 .f r t t t t   


28. 10/28/25 LINE INTEGRAL CONTI..........

2
2
0
10 10t dt
Therefore, the required line integral is
 
2
0
( )
C C
dr dr dr dr
f r ds f r t dt f r t dt
dt dt dt dt
     
   
  
2
2
0
10 10t dt
2
3
0
10 10
3
t 

 
 

3
2
10 10 0
3
 
  
 
 
8 80 10
10 10 .
3 3
 
 
 
 

29. 
 
2 2 2
Evaluate the integral , where
and : cos , sin , 2 , 0 4 .
C
f r ds f x y z
C r t t t t 
  
  



PROBLEM - 5
Solution:
 
The parametric representation of the
given path is cos , sin , 2 .C r t t t t

 
d
ˆ ˆˆ ˆ ˆ ˆ cos sin 2 sin cos 2 .
r d
ti tj tk ti tj k
dt dt
      


30.   
2 2
ˆ ˆˆ ˆ ˆ ˆ sin cos 2 sin cos 2
sin cos 4 5
dr dr
ti tj k ti tj k
dt dt
t t
        
   
 
2 2 2
Given that .f x y z  

2 2 2
2 2 2 2
cos sin 2
cos sin 4 4 1.
f r t t t t
t t t t
   
    


31.  
4
2
0
4 1 5t dt

 
Therefore, the required line integral is
 
4
0
( )
C C
dr dr dr dr
f r ds f r t dt f r t dt
dt dt dt dt

     
   
  
 
4
2
0
5 4 1t dt

 
4
3
0
4
5
3
t t

 
 
 
 

34
5 4 4 0
3
 
 
  
 
 
3256
5 4 5
3
  

32. 10/28/25 LINE INTEGRAL CONTI..........
 
1
2
3
3 3
Evaluate the integral , where
and is the hypocycloid cos , sin ,0 .
C
f r ds f x xy
C r t t t 
 
   
 



PROBLEM - 6
Solution:
3 3
The parametric representation of the
given path is cos , sin .C r t t 
 

 
3 3 2 2d
ˆ ˆ ˆ ˆ cos sin 3cos sin 3sin cos .
r d
ti tj t ti t tj
dt dt
    


33. 10/28/25 LINE INTEGRAL CONTI..........
  
2 2 2 2

ˆ ˆ ˆ ˆ3cos sin 3sin cos 3cos sin 3sin cos
dr dr
dt dt
t ti t tj t ti t tj

     
 

1
2
3Given that .f x xy 
 
1
2
3 3 3
3
cos cos sinf r t t t t  

  
2 2
2 2
3cos sin 3sin cost t t t  
4 2 4 2
9cos sin 9sin cost t t t 
2 2
9cos sint t

34. 6
cos cossint t t 
 
6
0
cos cos sin 3cos sint t t t tdt

 
Therefore, the required line integral is
 
0
( )
C C
dr dr dr dr
f r ds f r t dt f r t dt
dt dt dt dt

     
   
  
7 2 2
0 0
3 cos sin 3 cos sint tdt t tdt
 
  

35.  
8
2
00
cos 3
3 2cos sin
8 4
t
t t dt


 
  
 
 

2
0
1 1 3
3 sin 2
8 8 4
tdt

    
    
    
    

0
3 1 cos4
0
4 2
t
dt

 
 
 
 

0
3 1
sin4
8 4
t t

 
 
 
 
 
3 1 1
0 0 0
8 4 4

    
   
    
    
3
8


36. 10/28/25 LINE INTEGRAL CONTI..........
6.Evaluate the integral ,
C
f rds

where

37. 8)10(8
4
cos
32
2
0
4









t
ANS
10/28/25 LINE INTEGRAL CONTI..........

38. TEST YOUR KNOWLEDGE
( )f rdswithArclengthasparameter
3
1. , : [2cos ,2sin ].0
2
Q f x y C r t t t

   
2 2
2. 16 81 , : ( ) [3cos ,2sin ],0Q f x y C r t t t t     
2
.3 1 sinh , : [,cosh ],0 2Q f xC r t t t    
10/28/25 LINE INTEGRAL CONTI..........

39. LINE INTEGRAL INDEPENDENT OF
PATH
A line integral of the type
is said to be independent of path in a domain D
in space if for every pair of end points A and B
in D, in the above integral has the same value
for all paths in D the begin at A and end at B.
10/28/25 LINE INTEGRAL CONTI..........
 
1 2 3 1 2 3
( )
B
C C A
dx dy dz
F r dr Fdx Fdy Fdz F F F dt
dt dt dt
 
      
 
 
  
 

40. LINE INTEGRAL INDEPENDENT OF
PATH
  
1 2 3 1 2 3
( )
B
C C A
F r dr Fdx Fdy Fdz Fdx Fdy Fdz        
 
A line integral of the type
with continuous F
1, F
2, F
3 in a domain D in space is
independent of path in D if and only if
is the gradient of some function f in D. Moreover,
 
1 2 3
, ,F F F F

THEOREM - 1
  
1 2 3 1 2 3
( )
B
C C A
F r dr Fdx Fdy Fdz Fdx Fdy Fdz f B f A          
 

41. LINE INTEGRAL INDEPENDENT OF PATH
EXAMPLE – 1:
 
2 2
Show that the integral • 3 2 is
independent of path in any domain in space and find its value
if has the initial point : (0,1,2) and the terminal point :(1, 1,7).
C C
F dr x dx yzdy y dz
C A B
  

 

Let us see whether there exists a scalar function f such that
SOLUTION:
2 2 ˆˆ ˆGiven that 3 2F xi yzj yk  

F gradf


42. LINE INTEGRAL INDEPENDENT OF PATH
2 2 ˆ ˆˆ ˆ ˆ ˆ3 2
f f f
xi yzj yk i j k
x y z
  
     
  
2
2
3 (1)
2 (2)
(3)
f
x
x
f
yz
y
f
y
z

 


  


 


3
Integrating 1 w.r.t. , we get ( , , ) ( , ) (4)
( , ) is a function of and .
x f x y z x k y z
k y z y z
    

43. LINE INTEGRAL INDEPENDENT OF PATH
(5)
f k
y y
 
   
 
 

2
Integrating 6 w.r.t. y, we get ( , ) (7)
where is a function of .
k y z y z z
z z


    
  
3
Differentiating 4 w.r.t. , we get ( , )
f
y x k y z
y y
 
 
 
From (2) and (5), we have 2 (6)
k
yz
y

   

 
3 2
So from 4 we have ( , , ) (8)f x y z x y z z     

44. LINE INTEGRAL INDEPENDENT OF PATH
  
3 2
Differentiating 8 w.r.t. , we get
f
z x y z z
z z

 
  
 
2
(9)
f d
y
z dz

   

2 2
From (3) and (9), we have
d
y y
dz

  0
d
dz

 
 On integration w.r.t. , we get sayz z Aconstant c 

3 2
So from 8 we have ( , , ) (10)f x y z x y z c     

45. 3 2
So we find that there exists a function
( , , ) such that .f x y z x y z c F gradf   

LINE INTEGRAL INDEPENDENT OF PATH
 
  
2 2
2 2
Therefore, the integral • 3 2 is
independent of path and its value is given by
• 3 2 1, 1,7 0,1,2 .
C C
C C
F dr x dx yzdy y dz
F dr x dx yzdy y dz f f
  
     
 
 


  
3 2 3 2
at 1, 1,7 at 0,1,2
-x y z c x y z c

       
   
 
3 2 3 2
1 1 7 - 0 1 2 8 2 6.c c
   
        
      

46. LINE INTEGRAL INDEPENDENT OF
PATH
  
1 2 3 1 2 3
( )
B
C C A
F r dr Fdx Fdy Fdz Fdx Fdy Fdz        
 
A line integral of the type
is independent of path in a domain D if and only if
its value around every closed path in D is zero.
THEOREM - 2

47. LINE INTEGRAL INDEPENDENT OF PATH
1 2 3
1 2 3
Let ( , , ), ( , , ), ( , , )becontinuousfunctionshaving
continuousfirstpartialderivativesinadomain inspace.
is called in if it is the differential
of a
exact
F x y z F x y z F x y z
D
Fdx Fdy Fdz D
f f f
df dx dy dz
x y z
 
  
  
  
 
1 2 3
differentiable function , , everywhere in , i.e., if we have
.
f x y z D
Fdx Fdy Fdz df  
1 2 3
1 2 3 1 2 3
From
ˆˆ ˆwe have , , , . . .
f f f
Fdx Fdy Fdz df dx dy dz
x y z
f f f
F F F ie F Fi F j Fk gradf
x y z
  
     
  
  
      
  


48. LINE INTEGRAL INDEPENDENT OF PATH
A domain D is called a simply connected domain if
every closed curve in D can be continuously shrunk
to any point in D without leaving D.
SIMPLY CONNECTED DOMAINS
NOT SIMPLY CONNECTED DOMAINS

49. LINE INTEGRAL INDEPENDENT OF
PATH
 
1 2 3
1 2 3
1 2 3
Let ( , , ), ( , , ), ( , , )becontinuousfunctionhaving
continuousfirstpartialderivativesinadomain inspace.If theline
integral isindependentof pathin ,
Then is exact then i
C
F x y z F x y z F x y z
D
Fdx Fdy Fdz D
Fdx Fdy Fdz
 
 

 1 2 3
3 3 2 1 2 1
n
ˆˆ ˆ 0
. . , ,
D
curl F curl Fi F j Fk
F FF F F F
ie
y z z x x y
   
    
  
     

THEOREM – 3:

50. LINE INTEGRAL INDEPENDENT OF PATH
 
 
1 2 3
1 2 3
1 2 3
Let ( , , ), ( , , ), ( , , )becontinuousfunction
havingcontinuousfirstpartialderivativesinadomain in
ˆˆ ˆspace.If 0
and is simply connected then the integral
C
F x y z F x y z F x y z
D
curlF curl Fi F j Fk
D
Fdx Fdy Fdz
   
 

isindependentofpathin .D
THEOREM – 4:

51. LINE INTEGRAL INDEPENDENT OF PATH
Example – 1:
Show that the form under integral sign is exact in
the plane and evaluate the integral.

 
 
 
 

3π
2,
2
x
0,π
e cosydx-sinydy
Solution:
 
1 2 3
1 2 3
We know that a differential form
ˆˆ ˆis exact if 0.
Fdx Fdy Fdz
CurlF Curl Fi F j Fk
 
   

1 2 3
ˆ ˆˆ ˆ ˆ ˆSet cos sin 0
x x
F Fi F j Fk e yi e yj k     


52. 10/28/25 LINE INTEGRAL CONTI..........

 
^ ^ ^ ^ ^ ^
1 2 3
ˆˆ ˆcos sin 0
cos sin 0
x x
x x
curl F
curl e yi e yj k
i j k i j k
x y z x y z
F F F e y e y

  
     
 
     


LINE INTEGRAL INDEPENDENT OF PATH
   
  
ˆ ˆ0 sin cos 0
ˆ ˆˆ ˆsin cos 0 0 0 0
x x
x x
e y i e y j
y z z z
e y e y k i j k
x z
      
    
   
      
  
      
 
  


53. 10/28/25 LINE INTEGRAL CONTI..........
LINE INTEGRAL INDEPENDENT OF PATH
 
1 2 3
So the given differential form
cos -sin is exact.
x
Fdx Fdy Fdz e ydx ydy  
 
 
 
1 2 3
So there exists function, say , , such that
cos -sin .
.. cos -sin
x
x
f x y z
df Fdx Fdy Fdz e ydx ydy
f f f
ie dx dy dz e ydx ydy
x y z
   
  
  
  

54. 10/28/25 LINE INTEGRAL CONTI..........



cos 1
sin 2
0 3
x
x
f
e y
x
f
e y
y
f
z

 


 


 

On comparing the coefficients of dx, dy and dz we get
LINE INTEGRAL INDEPENDENT OF PATH

55. 10/28/25 LINE INTEGRAL CONTI..........
LINE INTEGRAL INDEPENDENT OF PATH
On integrating (1) partially w. r. t. x we get
 cos , cos , (4)
( , ) is a function of and .
x x
f e ydx y z e y k y z
k y z y z
    
Differentiating (4) partially w.r.t. y we get
 cos , sin (5)
x xf k
e y k y z e y
y y y
  
     
  
From (2) and (5), we get

sin sin 0
.. is a function of z alone, say , ( )
x x k
e y e y
y
ie k k y z c z

   



56. LINE INTEGRAL INDEPENDENT OF PATH
From (4) we have
cos (6).
x
f e y c z  
Differentiating (6) partially w. r. t. z we get
 cos (7)
xf dc
e y c z
z z dz
 
   
 
From (3) and (7), we get
 0 A constant = say
dc
c z C
dz
  
From (6) we have
cos (8)
x
f e y C  

57. Hence the exact differential is
 cos sin cos
x x x
e ydx e ydy df d e y c   

   

3 3 3
2, 2, 2,
2 2 2
0, 0, 0,
cos sin cos
x x
e ydx ydy df d e y c
  
  
     
     
     
      
LINE INTEGRAL INDEPENDENT OF PATH

58.  

3
(2, )
2
(0, )
2 0
cos
3
cos cos
2
0 1 1
x
e y c
e c e c
c c




  
 
 
   
 
 
    
ANS
10/28/25 LINE INTEGRAL CONTI..........
LINE INTEGRAL INDEPENDENT OF PATH

59. 
 
^ ^ ^ ^ ^ ^
1 2 3
ˆˆ ˆ3 3 2
3 3 2
i j k i j k
curl F
x y z x y zcurl yi xj zk
F F F y x z

     
 
       

EXAMPLE – 2:
Here
ˆ ˆ ˆ
F=3yi+3xj+2zk


60.   
ˆˆ ˆ2 3 3 2 3 3z x i y z j x y k
y z z x x y
          
     
     
          
To find the exact differential
Suppose that there a function f(x,y,z) in space such
that 3ydx+3xdy+2zdz=df.
10/28/25 LINE INTEGRAL CONTI..........

ˆˆ ˆ0 0 0 0 3 3 0i j k      

LINE INTEGRAL INDEPENDENT OF PATH

61. 10/28/25
LINE INTEGRAL CONTI..........



3 1
3 2
2 3
f
y
x
f
x
y
f
z
z

 


 


 

LINE INTEGRAL INDEPENDENT OF PATH

62. Integrating 1 w.r.t. , we get ( , , ) 3 ( , ) (4)
( , ) is a function of and .
x f x y z xy k y z
k y z y z
    
3 (5)
f k
x
y y
 
    
 

k
Since ( , ) 0 and =0, integrating w.r.t. ,
y
we get ( , ) function of z-alone = z say
k y z y
k y z A 




  Differentiating 4 w.r.t. , we get 3 ( , )
f
y xy k y z
y y
 
 
 
From (2) and (5), we have 3 3 0
k k
x x
y y
 
   
 
 So from 4 we have ( , , ) 3 z (6)f x y z xy    
LINE INTEGRAL INDEPENDENT OF PATH

63.   Differentiating 6 w.r.t. , we get 3 .
f
z xy z
z z

 
 
 
(7)
f d
z dz

  

From (3) and (7), we have 2
d
z
dz


2
d
z
dz

 

2
On integration w.r.t. , we get ,
where c is an arbitrary constant.
z z z c 
LINE INTEGRAL INDEPENDENT OF PATH

64. 
2
So from 6 we have ( , , ) 3 .f x y z xy z c  
 




 

4,1,2 4,1,2 4,1,2
2
0,0,0 0,0,0 0,0,0
3 3 2 3ydx xdy zdz df d xy z c        

4,1,2
2
0,0,0
3xy z c   
 
  
2 2
3 4 1 2 3 0 0 0c c
 
     
 
= 16
ANS
 
2 2
4,1,2 0,0,0
3 3
At At
xy z c xy z c        
   
LINE INTEGRAL INDEPENDENT OF PATH

65. 
^ ^ ^ ^ ^ ^
2 2
1 2 3
So
2 2 1
i j k i j k
curl F
x y z x y z
F F F xy x ydy
     
 
     

EXAMPLE – 3:
Here ˆ ˆ ˆ
2 2
F=2xy i+2x yj+k


66.  

2 2
2 2
ˆ ˆ1 2 2 1
ˆ
2 2
x y i xy j
y z z x
x y xy k
x y
      
   
   
      
  
 
 
  
So the given differential form s exact.(path independent)
10/28/25 LINE INTEGRAL CONTI..........
 
ˆ ˆˆ ˆ ˆ ˆ0 0 0 0 4 4 0 0 0 0i j xy xy k i j k         

LINE INTEGRAL INDEPENDENT OF PATH

67. 10/28/25 LINE INTEGRAL CONTI..........



2
2
2 1
2 2
1 3
f
xy
x
f
xy
y
f
z

 


 


 

LINE INTEGRAL INDEPENDENT OF PATH

68. 10/28/25 LINE INTEGRAL CONTI..........

2 2
Integrating 1 w.r.t. , we get ( , , ) ( , ) (4)
( , ) is a function of and .
x f x y z x y k y z
k y z y z
    
2
2 (5)
f k
x y
y y
 
    
 

k
Since ( , ) 0 and =0, integrating w.r.t. ,
y
we get ( , ) function of z-alone = z say
k y z y
k y z A 




  
2 2
Differentiating 4 w.r.t. , we get ( , )
f
y x y k y z
y y
 
 
 
2 2
From (2) and (5), we have 2 2 0
k k
x y x y
y y
 
   
 
 
2 2
So from 4 we have ( , , ) z (6)f x y z x y    
LINE INTEGRAL INDEPENDENT OF PATH

69.   
2 2
Differentiating 6 w.r.t. , we get .
f
z x y z
z z

 
 
 
(7)
f d
z dz

  

From (3) and (7), we have 1
d
dz


1
d
dz

 
On integration w.r.t. , we get ,
where K is an arbitrary constant.
z z z K 
LINE INTEGRAL INDEPENDENT OF PATH

70. 10/28/25

2 2
So from 6 we have ( , , ) .f x y z x y z K  
 




 

, , , , , ,
2 2
0,0,0 0,0,0 0,0,0
3 3 2
a b c a b c a b c
ydx xdy zdz df d x y z K        

, ,
2 2
0,0,0
abc
xy z K   
 
  
2 2 2 2
0 0 0a b c K K
 
     
 
= a
2
b
2
+c
ANS
LINE INTEGRAL INDEPENDENT OF PATH

71. 
^ ^ ^
1 2 3
i j k
curlF
x y z
F F F
  

  

EXAMPLE – 4:
Here ˆ ˆ ˆ
y zx zF= i j+ k

LINE INTEGRAL INDEPENDENT OF PATH

72.  

^ ^ ^
ˆ ˆ
ˆ
i j k
z zx i y z j
y z z x
x y z
zx y k
y zx z x y
      
   
   
         
 
     
  
 
   
So the given differential form s exact.(path dependent)
ANS
10/28/25 LINE INTEGRAL CONTI..........
 
ˆˆ ˆ0 0 0 1x i j z k       0

LINE INTEGRAL INDEPENDENT OF PATH