vector

Vectors 429
Vector normal to the surface = =
iˆ
ˆj kˆ
x y z
ˆ 2 2 2 2
ˆ
=
ˆ
ˆ
i j k ( x y z a ) 2xiˆ 2yj ˆ 2zk
x y z
ˆn
2xiˆ 2yj ˆ 2zkˆ
xiˆ yj ˆ zkˆ
=
| | 2 2 2 2 2 2
4x 4y 4z x y z
xiˆ
yj ˆ zkˆ
=
[ x 2 + y 2 + z 2 = a 2 ]
a
Here, F = yz iˆ
zxj ˆ xy kˆ
ˆ ˆ ˆ 3
Fn ˆ = ( ˆ ˆ ˆ
xi yj zk xyz
yz i zxj xy k)
a
a
Now, F
n ˆ ds
2 2
dx dy a a x 3xyz dx dy
S
= ( Fn ˆ)
S
| kˆ
. nˆ
|
00
z
a
a
a
2
x
2
2
a a
2
x
2
a y
= 3 xy dy dx 3 x dx
0
0 0 2
=
0
2 2 4
a
4 4 4
3 a 2 2 3 a x x 3 a a 3a
2
x a x dx 0
2
2 4
2 2 4
8
0
( ) .
3
Example 77. Show that Fnds ˆ ,
S
where
2
F = 4 xz î – y 2 ĵ + yz ˆk
and S is the surface of the cube bounded by the planes,
x= 0, x = 1, y = 0, y = 1, z = 0, z = 1.
Solution.
=
ˆ
Fnds ˆ
S
Fnds
OABC
Fnds ˆ
Fnds
ˆ
DEFG
OAGF
Fnds ˆ
Fnds
ˆ
BCED
Fnds
OCEF
ABDG
ˆ
...(1)
Now,
DEFG
Fnds
OABC
2
= (4 ˆ ˆ ˆ
xzi y j yz k)( k)
dx dy = 1 1
OABC
2
(4 xziˆ
y ˆj yz kˆ)
kˆ
dx dy
=
DEFG
2
1
1 y
1
= dx [ x]
0
0
0
2 ˆ
1 1
yz dx dy y (1) dxdy
0 0
1 1
2 2 2
(4 ˆ ˆ
xz i y j yzk) ( j)
dxdz =
OAGF
0 0
Ans.
S.No. Surface Outward ds
normal
1 OABC – k dx dy z = 0
2 DEFG k dx dy z = 1
3 OAGF – j dx dz y = 0
4 BCED j dx dz y = 1
5 ABDG i dy dz x = 1
6 OCEF – i dy dz x = 0
yz dx dy 0
(as z = 0)
2
y dxdz 0
(as y = 0)
OAGF