vector

444 Vectors
2 2
= [(2 x – y) dx – yz dy – y z dz]
C
= (2 x – y) dx,
C
since on C, z = 0 and 2z = 0
2
dx 2
= (2cos t –sin t) dt (2cos t –sin t)(–sin t)
dt
0 dt
0
2 2
2 1–cos 2t
= (– sin 2t sin t) dt – sin 2t dt
0
0
2
2
cos 2t t sin 2t
1 1
= – –
2 2 4
0 2 2
iˆ
ˆj kˆ
...(1)
Curl F =
x y z = (– 2yz 2 yz) iˆ
(0–0) ˆj (01)
kˆ kˆ
2 x – y
2
– yz
2
– y z
Z
Curl F . nˆ
= kˆ. nˆ
nk ˆ.
ˆ
Curl F.
nds ˆ = ˆ. ˆ ˆ. ˆ dx dy
. .
S nkds nk
S R ˆ ˆ
Where R is the projection of S on xy-plane.
O
Y
=
2
1 1– x
dx dy =
2
–1 – 1– x
1 2 1 2
2 1– x dx 4 1– x dx
–1 0
1
2 –1
x
1 1
= 4 1– x sin x 4 .
2 2
0 2 2
From (1) and (2), we have
F . dr = Curl F . nds ˆ which is the Stoke's theorem.
C
Example 97. Verify Stoke’s Theorem for 2 ˆ ˆ
2
F ( x y 4) i 3 xyj (2 xz z ) kˆ
over the surface of hemisphere x 2 + y 2 + z 2 = 16 above the xy-plane.
Solution. Fdr , where c is the boundary of the circle x 2 + y 2 + z 2 = 16
c
(bounding the hemispherical surface)
2 2 ˆ
= [( x y 4) iˆ 3 xyj ˆ (2 xz z ) k]( idx ˆ ˆjdy)
Putting
c
2
= [( x y 4) dx 3 xy dy)]
c
x = 4 cos , y = 4 sin , dx = – 4 sin d , dy = 4 cos d
=
2
2 2
0
[(16 cos 4sin 4) ( 4sin d ) (192 sin cos d )]
2 2 2 2
= 16 [ 4cos sin sin sin 12 sin cos ] d
0
2
2 2
= 16 (8 sin cos sin sin ) d
=
=
0
2
16 sin d
0
2
2 2
=
0
16 4 sin d
To evaluate surface integral
F
=
1
64 = – 16 .
22
iˆ
ˆj kˆ
x y z
X
2 2
x y 4 3xy 2 xz z
2
0
2
0
C
...(2)
Ans.
n
sin cos d0
n
cos sin d
0