vector

440 Vectors
Solution.
v =
iˆ
ˆj kˆ
iˆ
ˆj kˆ
x y z
y z x
Obviously ˆn = k ˆ.
Therefore
( v)
nˆ
= (– iˆ
– ˆj – kˆ). kˆ
–1
Hence
( v)
nds ˆ = ( 1) dx dy
S
=
S
S
dx dy
= – (1) 2 = – . (Area of circle = r 2 ) Ans.
Example 90. Use Stoke’s Theorem to evaluate
v ,
c dr
2
where v y iˆ xyj ˆ xzkˆ
, and c
is the bounding curve of the hemisphere x 2 + y 2 + z 2 = 9, z > 0, oriented in the positive
direction.
Solution. By Stoke’s theorem
( v)
nds ˆ =
S
v
c
dr = (curl v) nˆ ds ( v)
nds
ˆ
S S
iˆ
ˆj kˆ
ˆ ˆ
ˆ
(0 0) i ( z 0) j ( y 2 y)
k
v =
x y z zj ˆ ykˆ
2
y xy xz
2 2 2
i j k ( x y z 9)
ˆn = | = x y z
|
| |
=
2 xiˆ 2 yj ˆ 2 zkˆ xiˆ yj ˆ zkˆ xiˆ yj ˆ zkˆ
2 2 2 2 2 2
4x 4y 4 z x y z 3
ˆ ˆ ˆ
( v)
nˆ
2
= ( ˆ ˆ xi yj zk yz yz yz
zj yk)
3 3 3
ˆ ˆ ˆ
ˆn
kds ˆ = dx dy
xi yj zk . k ˆ
z
dx = dx dy ds = dx dy
3
3
ds = 3 dx dy
z
2yz
3
dx dy = 2 ydxdy
3 z
=
= 2r sin r d dr
3
3
2 r
0
0
2
3
2 sin d r dr
0 0
= 2( cos )
= – 2 (– 1 + 1) 9 = 0 Ans.
3
Example 91. Evaluate the surface integral curl F . nˆ
dS by transforming it into a line
S
integral, S being that part of the surface of the paraboloid z = 1 – x 2 – y 2 for which
z0and
F yiˆ
zj ˆ xkˆ
. (K. University, Dec. 2008)
2