Q # 93: Evaluate I   ( 3x 2 y 2 dx  2x 3 ydy ) between O(0,0) and A( 2,4)

a) along c 1 i.e. y  x 2
b) along c 2 i.e. y  2x
c) along c 3 i.e. x  0 from (0,0) to (0,4) and y  4 from (0,4) to ( 2,4)

Figure # 120
2 2 3
I   ( 3x y dx  2x ydy )
C

The path c 1 is y  x 2
 dy  2xdx
2
I 1   ( 3x 2 y 2 dx  2x 3 ydy )
0
2
I 1   ( 3x 2 ( x 2 ) 2 dx  2x 3  x 2  2xdx)
0
2
I 1   ( 3x 6  4x 6 )dx
0
2 2
6  x7 
I 1   7x dx  7 
0  7 0
I 1  128
a) In (b), the path of integration changes to c 2 i.e. y  2x

Figure # 121

1 Dr. A.N.M. Rezaul Karim/Associate Professor/CSE/IIUC

 I   ( 3x 2 y 2 dx  2x 3 ydy )
C
The path c 2 is y  2x
 dy  2dx
2
I 2   ( 3x 2 y 2 dx  2x 3 ydy )
0
2
I 2   ( 3x 2 ( 2x) 2 dx  2x 3  2x  2dx )
0
2
I 2   (12x 4  8x 4 )dx
0
2
I 2   20x 4 dx
0
2
 x5 
I 2   20 
 5 0
I 2  128
c) In the third case, the path c 3 is split
i. x  0 from (0,0) to (0,4)
ii. y  4 from (0,4) to ( 2,4)
Sketch the diagram and determine I 3

Figure # 122
i. From (0,0) to (0,4) x  0  dx  0
I   ( 3x 2 y 2 dx  2x 3 ydy )
C
The path c 3 is x  0
 dx  0
0
I 3   ( 3x 2 y 2 dx  2x 3 ydy )
0

2 Dr. A.N.M. Rezaul Karim/Associate Professor/CSE/IIUC

 4
I 3   ( 3.(0) 2 y 2 (0)  2(0) 3  ydy )
0

I3  0
ii. From (0,4) to ( 2,4) y  4  dy  0
I   ( 3x 2 y 2 dx  2x 3 ydy )
C
The path c 3 is y  4  dy  0
2
I 3   ( 3x 2 .4 2 dx  2x 3 .4.0)
0
2
I 3   48x 2 dx
0
I 3  128
In the above example, we have just worked through, we took three different paths and in
each case , the line integral produced the same result. It appears, therefore, that in this
case, the value of the integral is independent of the path of integration taken.

Figure # 123
We have been dealing with I   ( 3x y 2 dx  2x 3 ydy )
2

On reflection, we see that the integrand 3x 2 y 2 dx  2x 3 ydy is of the form

Pdx  Qdy which we have met before and that it is, in fact, an exact differential of the
function z  x 3 y 2 , for
z z
 3x 2 y 2 and  2x 3 y
x y
This always happens. If the integrand of the given integral is seen to be an exact
differential, then the value of the line integral is independent of the path taken and
depends only on the coordinates of the two end points.

3 Dr. A.N.M. Rezaul Karim/Associate Professor/CSE/IIUC

 Figure # 124
If I   ( Pdx  Qdy ) and ( Pdx  Qdy ) is an exact differential, then I C1  I C2
C

Figure # 125
If we reverse the direction of c 2 , then I C1   I C2  I C1  I C 2  0
Hence the integration taken round a closed curve is zero, provided ( Pdx  Qdy ) is an
exact differential.
 If ( Pdx  Qdy ) is an exact differential, I   ( Pdx  Qdy )  0
C

Q# 94: Evaluate I   {3ydx  ( 3x  2y )dy } from A(1,2) to B( 3,5) .

C
Now path is given, so the integrand is doubtless an exact differential of some
function z  f ( x, y ) .
Here, ( Pdx  Qdy )  3ydx  ( 3x  2y )dy
 P  3y and Q  3x  2y
P Q
In fact  3 and 3
y x
We have already dealt with the integration of exact differentials, so there is no difficulty.
Compare with I   {Pdx  Qdy } .
C
z
P  3y  z   3ydx  3xy  f ( y ) ----------------------------(i)
x

4 Dr. A.N.M. Rezaul Karim/Associate Professor/CSE/IIUC

 z
Q  3 x  2y  z   ( 3x  2y )dy  3xy  y 2  F( x) -------------(ii)
y
For (i) and (ii) to agree
f ( y )  y 2 and F( x )  0
Hence z  3xy  y 2
I   {3ydx  ( 3x  2y )dy }
C

I   {3ydx  3xdy  2ydy }

C

I   {3xdy  3ydx  2ydy }

C

I   {d( 3xy)  d( y 2 )}
C
( 3,5 ) 2
I (1,2) d(3xy  y )


I  3xy  y 2  ( 3,5 )
( 1, 2 )

I  3  3  5  5 2
 3  1  2  22 
I  45  25  6  4  70  10  60

Exact differentials in three independent variables

A line integral in space naturally involves three independent variables, but the method is
very much like that for two independent variables.
dz  Pdx  Qdy  Rdw is an exact differential of z  f ( x, y , w )
P Q P R R Q
if  ;  ;  ;
y x w x y w
If the test is successful, then
a) I   ( Pdx  Qdy  Rdw ) is independent of the path of integration
C

b) I   ( Pdx  Qdy  Rdw ) is zero

C
Green’s Theorem:
Let P and Q be two functions of x and y that are finite and continuous inside and on the
boundary c of a region R in the xy-plane.

Figure # 126

5 Dr. A.N.M. Rezaul Karim/Associate Professor/CSE/IIUC

If the first partial derivatives are continuous within the region and on the boundary, then
Green’s theorem states that
 P Q 
  y  x dxdy    (Pdx  Qdy )
R
That is, a double integral over the plane region R cane be transformed into a line integral
over the boundary c of a region and the action is reversible.
Green’s Theorem
Green’s Theorem enables an integral over a plane area to be expressed in terms of a line
integral round its boundary curve.
If P and Q are two single-valued functions of x and y, continuous over a plane surface S,
and c is its boundary curve, then
P Q
  ( Pdx  Qdy )    (  )dxdy ------------------------------(i)
c R
y x
Q P
  ( Pdx  Qdy )   (  )dxdy -------------------------------(ii)
c R
x y
Where the line integral is taken round c in an anticlockwise manner. In vector terms, this
becomes:
S is a two-dimensional space enclosed by a simple closed curve c.

dS  dS  dxdy
  
dS   dS  k dxdy
  
If F  P i  Q j Where P  P( x, y ) and Q  Q( x, y ) then
  
i j k
            
Curl F    F  ( i  j k )  ( P i  Q j) 
x y z x y z
P Q 0

 
      
(0)  (Q )]  j[ (0)  ( P )]  k[ (Q ) 
 i[ ( P )]
y z x z x y
    
 i [0]  j[0  0]  k[ (Q )  ( P )]
x y
  
 k[ (Q )  ( P )]
x y
Q P
Since P and Q are functions of x and y, that is in the xy-plane  0
z z
     
So  curl F . dS   curl F .  dS and in the xy plane   k

6 Dr. A.N.M. Rezaul Karim/Associate Professor/CSE/IIUC

    Q P     Q P    Q P   

 curl F . dS   
S  x y 
k  .  dS   
S  x y 
k  . k dS    
S  x y dxdy [ k . k  1]
S

   Q P 
  curl F . dS     dxdy --------------------------------------(i)
S
S  x y 
We have, from Stoke’s theorem:
   
  curl F . dS   F . dr
S c
        
  ( P i  Q j).(dx i  dy j dz k ) [ r  x i  y j z k ]
c

  ( Pdx  Qdy ) -----------------------------------------(ii)

c
From (i) and (ii)
   Q P 
  curl F . dS     dxdy   ( Pdx  Qdy )
S
S 
x y  c

 Q P 
    dxdy   ( Pdx  Qdy )
S  x y  c

Q# 95: Evaluate I   {( 2x  y )dx  (3x  2y )dy } taken in an anticlockwise manner

C

round the triangle with vertices at O(0,0), A(1,0), B(1,2)

Figure # 127
I   {( 2x  y )dx  (3x  2y )dy }
C

a) There are clearly three stages with c1 , c 2 , c 3 . Work through the complete evaluation to
determine the value of I . It will be good revision.
i) The equation of OA is:

7 Dr. A.N.M. Rezaul Karim/Associate Professor/CSE/IIUC

 y  y1 x  x1 y0 x0 y x
       y  x.0  y  0
y 1  y 2 x1  x 2 0 0 01 0 1
c1 is y  0  dy  0
I   {( 2x  y )dx  (3x  2y )dy }
C

 I   {( 2x  0)dx  ( 3x  2.0).0}
C
1 1
 2x 2 
 I1   2xdx    1
0  2 0
ii) The equation of AB is:
y  y1 x  x1 y 0 x1 y x1
      2 x  2  0  2 x  2  x  1
y 1  y 2 x1  x 2 0 2 11 2 0
c 2 is x  1  dx  0
I   {( 2x  y )dx  ( 3x  2y )dy }
C

 I   {( 2.1  y ).0  ( 3.1  2y )dy }

C
2 2
 2y 2 
 I 2   (3  2y )dy   3y   2
0  2 0
iii) The equation of OB or BO is:
y  y1 x  x1 y0 x0 y x
       y  2 x  y  2x
y 1  y 2 x1  x 2 0 2 01  2 1
c 3 is y  2x
Then, Given y  2 x
dy d
 (2 x)
dx dx
dy
2
dx
 dy  2dx
I   {( 2x  y )dx  ( 3x  2y )dy }
C

I   {( 2x  2x)dx  ( 3x  2  2x)2dx}
C

 I   {4xdx  ( 3x  4x)2dx}
C

 I   {4x  6x  8x}dx
C
0
0  2x 2 
 I 3   2 xdx     1
1  2 1

8 Dr. A.N.M. Rezaul Karim/Associate Professor/CSE/IIUC

 Finally I  I 1  I 2  I 3  1  2  ( 1)  2
b) By Green’s theorem
I   {( 2x  y )dx  ( 3x  2y )dy }
C
Here, P  2x  y
P
 1
y
and
Q  3x  2y
Q
 3
x

We have, Green’s theorem

P Q
 ( Pdx  Qdy )    ( y  x )dxdy
c R

Now,
P Q
I    (  )dxdy
R
y x
I    (1  3)dxdy
R

I     2dxdy
R

I  2 dxdy  2A ----------------------------------(i)

R

[  dxdy  A ]
R

1
Now, The area of the triangle: A   base  height
2
1
A   OA  AB [Figure 122]
2
1
A  1 2  1
2
From (i),
I  2  A  2  1  2 [ A  1]
L.H .S  R .H .S (Proved)

Q# 96: Evaluate the line Integral I   {xydx  ( 2x  y )dy } round the region bounded by
C
2 2
the curves y  x and x  y by the use of Green’s theorem.

9 Dr. A.N.M. Rezaul Karim/Associate Professor/CSE/IIUC

 Figure # 128
Answer: Points of intersection are O(0,0) and A(1,1) . P and Q are known, so there is no
difficulty.
Now, By Green’s theorem
I   {xydx  ( 2x  y )dy }
C
Here, P  xy
P
 x
y
and Q  2x  y
Q
 2
x

Figure # 129
P Q
We have,  ( y  x )dxdy    ( Pdx  Qdy )
R c

P Q
   (  )dxdy   ( Pdx  Qdy )
R
y x c

P Q
  ( Pdx  Qdy )    (  )dxdy
c R
y x
P Q
Now, I    (  )dxdy
R
 y  x
I    ( x  2)dxdy
R

10 Dr. A.N.M. Rezaul Karim/Associate Professor/CSE/IIUC

 1 y x
I    (x  2)dydx [Here upper limit of y is y  x and the lower limit is y  x 2 ]
0 y  x2
1
I    ( x  2)y x 2 dx
x

0
1
 I    ( x  2) x  x 2 dx  
0
1
3 1
 I    (x 2
 x 3  2x 2
 2x 2 )dx
0
1
2 5 1 4 3 2  31
 I   x 2  x4  x 2  x3  
5 4 3 3  0 60
Now, L.H.S.  Pdx  Qdy

  xydx  ( 2x  y )dy

1
I 1   x.x 2 dx  ( 2x  x 2 )2xdx
0
1
  x 3 dx  4x 2  2x 3 dx
0
1
  (  x 3 dx  4x 2 )dx
0
1
  x4 x3 
 4 
 4 3 0
  1 4  13
  
 4 3  12
and  Pdx  Qdy

  xydx  ( 2x  y )dy

0
I 2   xydx  ( 2x  y )dy
1
0
1 1
I 2   x. xdx  ( 2x  x ) x 2 dx
1
2
0
3
2
1
2
1
I 2   (x x  )dx
1
2

11 Dr. A.N.M. Rezaul Karim/Associate Professor/CSE/IIUC

 0
 5 3

x 2 x 2 1 
I2     x
 5 3 2 
 2 2  1
 2 2 1 17
I 2  0      
 5 3 2 30
13 17 31
Finally I  I 1  I 2   
12 30 60

L.H .S  R .H .S (Proved)

12 Dr. A.N.M. Rezaul Karim/Associate Professor/CSE/IIUC