vector

Vectors 449
c
L.H.S. of (1) =
Fdr
ABC
=
Fdr F dr F dr
= 1 + 36 – 16 = 21 ...(2)
AB BC CA
Curl F = F =
ˆ
ˆ
ˆ
i j k [( x y) iˆ (2 x z) ˆj ( y z) k]
x y z
iˆ
ˆj kˆ
=
x y z
x y 2x z y z
x y z
Equation of the plane of ABC is = 1
2 3 6
Normal to the plane ABC is
(1 1) iˆ(0 0) ˆj (2 1) kˆ
2iˆ
kˆ
ˆ
=
ˆ
ˆ
x y z
i j k 1
=
x y z2 3 6
iˆ
ˆj kˆ
Unit Normal Vector =
2 3 6
1 1 1
4 9 36
1
ˆn = (3iˆ
2 ˆj kˆ
)
14
i ˆ ˆ ˆ
j
k
2 3 6
R.H.S. of (1) = curl Fnds
= ˆ ˆ 1
ˆ ˆ ˆ dx dy
(2 ) (3 2 )
s i k i j k
s
4 1
(3iˆ
2 ˆj kˆ).
kˆ
14
(6 1) dx dy
= = 7
s
14 1 dx dy = 7 Area of OAB
14
1
= 7
23
= 21 ...(3)
2
with the help of (2) and (3) we find (1) is true and so Stoke’s Theorem is verified.
Example 101. Verify Stoke’s Theorem for
F
= (y – z + 2) î + (yz + 4) ĵ – (xz) ˆk
over the surface of a cube x = 0, y = 0, z = 0, x = 2, y = 2, z = 2 above the XOY plane
(open the bottom).
Solution. Consider the surface of the cube as shown in the figure. Bounding path is OABCO
shown by arrows.
Fdr = [( y z 2) iˆ ( yz 4) ˆj ( xz) kˆ] ( idx ˆ ˆjdy kdz ˆ )
= ( y z 2) dx ( yz 4) dy xzdz
c
Fdr = Fdr Fdr Fdr Fdr
c
OA AB BC CO
(1) Along OA, y = 0, dy = 0, z = 0, dz = 0
...(1)