vector

446 Vectors
Along BC, put y = b and dy = 0 in (1) we get F . dr = (x 2 – b 2 ) dx,
where x is from a to 0.
F
BC
. dr =
0
3
0
3
2 2 x 2
– a 2
Along CO, put x = 0 and dx = 0 in (1), we get F . dr 0
( x – b ) dx – bx
ba
a
3 3
...(5)
F
. dr
= 0 ...(6)
CO
Putting the values of integrals (3), (4), (5) and (6) in (2),
we get
F . dr =
C
3 3
2 a 2 2
a
ab – ab 0 2ab
...(7)
3 3
Now we have to evaluate R.H.S. of Stoke’s Theorem i.e.
We have,
F
=
iˆ
ˆj kˆ
x y z
2 2
x – y 2xy
0
S
a
( F).
nds ˆ
(2y 2 y) kˆ
4ykˆ
Also the unit vector normal to the surface S in outward direction is ˆn
k
(z-axis is normal to surface S)
Also in xy-plane ds = dx dy
( F). nˆ
. ds = 4 ˆ. ˆ 4 .
S yk kdxdy ydx dy
R R
Where R be the region of the surface S.
Consider a strip parallel to y-axis. This strip starts on line y = 0 (i.e. x-axis) and end on the line
y = b, We move this strip from x = 0 (y-axis) to x = a to cover complete region R.
( F). nˆ
. ds =
S
From (7) and (8), we get
=
ydy
dx y dx
a b 2
4
a [2 ]
b
0 0 0
0
a 2 2 a 2
b dx b x
0
ab
...(8)
0 2 2 [ ] 2
F . dr = ( ). ˆ
C F nds and hence the Stoke’s theorem is verified.
S
Example 99. Verify Stoke’s Theorem for the function
F = xi
2ˆ – xyj ˆ
integrated round the square in the plane z = 0 and bounded by the lines
x = 0, y = 0, x = a, y = a.
Solution. We have, F = xi
2ˆ – xyj ˆ
F =
iˆ
ˆj kˆ
x y z
2
x = 0
Y
y = b
C
B
(a, b)
x = a
x xy 0
= (0 –0) iˆ
–(0–0) ˆj (– y –0) kˆ – ykˆ
( ˆn to xy plane i.e. ˆk )
O
y = 0
A
X