vector

Vectors 441
Solution.
F =
iˆ
ˆj kˆ
iˆ
ˆj kˆ
x y z
y z x
Obviously ˆn = k ˆ.
Therefore
( F)
nˆ
= (– iˆ
– ˆj – kˆ). kˆ
–1
Hence
( F)
nˆ
ds
= ( 1) dx dy
S
=
S
S
dx dy
= – (1) 2 = – . (Area of circle = r 2 ) Ans.
Example 92. Evaluate Fdr
2 2
by Stoke’s Theorem, where F y iˆ x ˆ j ( x z)
kˆ
and
C
C is the boundary of triangle with vertices at (0, 0, 0), (1, 0, 0) and (1, 1, 0).
(U.P., I Semester, Winter 2000)
Solution. We have, curl F =
=
iˆ
ˆj kˆ
x y z
2 2
F
0. iˆ ˆj 2( x y) kˆ
.
y x ( x z)
We observe that z co-ordinate of each vertex of the triangle is zero.
Therefore, the triangle lies in the xy-plane.
ˆn = ˆk
curl Fn ˆ [ ˆj 2( x yk ) ˆ]. kˆ
2( x y).
In the figure, only xy-plane is considered.
The equation of the line OB is y = x
By Stoke’s theorem, we have
C
Fdr
=
S
(curl Fnˆ
) ds
1 x
= 2( )
2
1
x y dx dy
y
x0 = 2
y 0 xy
dx
0
2 0
2
1 2
= 2 x
2
x
dx
0
2
=
1 x
1
2 dx = 2
0 2
x dx
3
x
=
0
3
x
1
0
= 1 .
3
Example 93. Evaluate Fdr
by Stoke’s Theorem, where
2 2
F ( x y ) iˆ 2 xy ˆj
C and C
is the boundary of the rectangle x = a, y = 0 and y = b. (U.P., I Semester, Winter 2002)
Solution. Since the z co-ordinate of each vertex of the given rectangle is zero, hence the given
rectangle must lie in the xy-plane.
Here, the co-ordinates of A, B, C and D are (a, 0), (a, b), (– a, b) and (– a, 0) respectively.
Curl F =
iˆ
ˆj kˆ
= – 4 y k
x y z
2 2
x y 2xy
0
Ans.