vector

Vectors 445
= (0 – 0) iˆ – (2 z – 0) ˆj (3 y – 1) kˆ – 2 zj ˆ (3 y – 1) kˆ
ˆ ˆ ˆ 2 2 2
i j k ( x y z 16)
ˆn = | = x y z
|
| |
2xiˆ
2yj ˆ 2zkˆ
xiˆ
yj ˆ zkˆ
xiˆ
yj ˆ zkˆ
=
=
=
2 2 2 2 2 2
4x 4y 4z
x y z 4
ˆ ˆ ˆ
( F)
nˆ
= [– 2 ˆ (3 – 1) ˆ xi yj zk
zj y k]
2 yz (3y 1) z
=
4
4
ˆk n
xiˆ
yj ˆ zkˆ
z
ds = dx dy
. k ds = dx dy ds = dx dy
4
4
ds = 4 dx dy
z
2 (3 1) 4
( F)
nˆ
ds yz y z dx dy
4 z
= [ 2 y (3 y 1)] dx dy = ( y 1) dxdy
On putting x = r cos , y = r sin , dx dy = r d dr, we get
=
=
= ( r sin 1) r d dr
2
3 2
4
2
d ( r sin r)
dr
= r
r
d
sin
3 2 = 64
d
sin 8
0
3
0 0
2
64 64 64
= cos 8
= 16
= – 16
3 3 3
0
The line integral is equal to the surface integral, hence Stoke’s Theorem is verified. Proved.
2 2
Example 98. Verify Stoke’s theorem for a vector field defined by F ( x – y ) iˆ 2xy ˆj
in
the rectangular in xy-plane bounded by lines x = 0, x = a, y = 0, y = b.
(Nagpur University, Summer 2000)
Solution. Here we have to verify Stoke’s theorem F
. dr
= ( ). ˆ
C F
nds
S
Where ‘C’ be the boundary of rectangle (ABCD) and S be the surface enclosed by curve C.
2
F 2 2
= ( x – y ) iˆ (2 xy)
ˆj
F . dr =
2 2
[( x – y ) iˆ 2 xyj ˆ].[ idx ˆ ˆj dy]
F . dr = (x
2
+ y 2 ) dx + 2xy dy ...(1)
Now, F
. dr
= F . dr F . dr F . dr F . dr
C
...(2)
OA AB BC CO
Along OA, put y = 0 so that k dy = 0 in (1) and F . dr = x 2 dx,
Where x is from 0 to a.
3
a
a
2
F
. dr
3
x dx
x a
OA =
0 3 3
Along AB, put x = a so that dx = 0 in (1), we get F .
Where y is from 0 to b.
b
0
0
d r
= 2ay dy
...(3)
2 2
F . dr 2aydy
= [ ay ] b
0
ab
...(4)
AB