vector

426 Vectors
Example 73. A vector field is given by F (sin y) iˆ x(1 cos y) ˆj.
Evaluate the line integral
over a circular path x 2 + y 2 = a 2 , z = 0. `(Nagpur University, Winter 2001)
Solution. We have,
F
C
Work done = . dr
F
C
= [(sin ) ˆ (1 cos ) ˆ].[ ˆ ˆ
y i x y j dxi dyj]
( z = 0 hence dz = 0)
C
. d r = sin ydx x (1 cos y ) dy (sin y dx x cos y dy xdy )
C
C
= d ( xsin y)
x dy
C
(where d is differential operator).
The parametric equations of given path
x 2 + y 2 = a 2 are x = a cos , y = a sin ,
Where varies form 0 to 2
F
C
. d r =
=
C
2
2
d [ a cos sin ( a sin )] a cos . a cos d
0 0
2
2
2 2
d [ a cos sin ( a sin )] a cos .
d
0 0
= [ a cos sin ( asin )] a cos d
2
2
2 2
0 0
2
2 21cos 2 a sin 2
= 0 a
d
0
2 2
2
0
2
a
2
= .2a
2
Example 74. Determine whether the line integral
Ans.
2 2 2 2
(2 xyz ) dx ( x z z cos yz) dy (2x yz y cos yz)
dz is independent of the path of
integration ? If so, then evaluate it from (1, 0, 1) to
0, ,1 .
2
2 2 2 2
Solution. (2 xy z ) dx ( x z z cos yz) dy (2x yz y cos yz)
dz
=
c
c
2 2 2 2
[(2 xy ziˆ) ( x z z cos yz) ˆj (2x yz y cos yz) kˆ].( idx ˆ ˆjdy kdz ˆ )
= Fdr
c
This integral is independent of path of integration if
F
= 0
iˆ
ˆj kˆ
F =
F
x y z
2 2 2 2
2xyz x z z cos yz 2x yz y cos yz
= (2x 2 z + cos yz – yz sin yz – 2x 2 z – cos yz + yz sin yz) = iˆ
–(4 xyz – 4 xyz) ˆj (2 xz –2 xz ) k
= 0
Hence, the line integral is independent of path.
d = dx
dy
dz
x y z
(Total differentiation)
2
2 2 ˆ