vector

Vectors 465
Fnds ˆ =
1 1
0 0
(4 xziˆ ˆj zkˆ
) ( ˆj)
dx dz
ˆn = ˆ, j ds dxdz =
=
1 1
dx dz =
0 0
1 1
0 0
dxdz
1 1
0 z 0
( x) ( ) = – (1) (1) = – 1
Over the face DEFG, z = 1, dz = 0, ˆn = ˆk , ds = dx dy
Fnds ˆ =
=
1 1 2
0 0
[4 x (1) y ˆj y(1) kˆ]()
kˆ
dx dy
1 1
ydxdy =
0 0
1 1
dx ydy = ( x)
0 0
Over the face OCDG, x = 0, dx = 0, ˆn = – iˆ,
1
0
ds = dy dz
Fnds ˆ = 1 1 ˆ 2 ˆ ˆ ˆ
(0 i y j yzk ) ( i ) dydz = 0
0 0
Over the face AOGF, y = 0, dy = 0, ˆn = – ĵ ,
Fnds ˆ =
1 1
(4 ˆ) ( ˆ
xzi j)
dxdz = 0
0 0
Over the face ABEF, x = 1, dx = 0, ˆn = î ,
Fnds ˆ =
1 1 2
0 0
[(4 zi ˆ y ˆ j yzk ˆ)()]
i ˆ dydz
=
1 dy
1
4
0 0
1 2 1
dy (2 z )
0
0
1
2
y
2
0
ds = dx dz
ds = dy dz
1
0
= 1 2
1 1
4 zdydz
0 0
= zdz = = 2 dy = 2( y) 2
On adding we see that over the whole surface
Fnds
1
ˆ = 01 0 0
2
2 = 3 2
From (1) and (2), we have
1. Use Divergence Theorem to evaluate
V
Fdv
Fnds
S
=
ˆ
EXERCISE 5.15
2 2 2 2 2
( y z i ˆ z x ˆ j x y 2ˆ k ). ds ,
s
1
0
...(2)
Verified.
where S is the upper part of the sphere x 2 + y 2 + z 2 = 9 above xy- plane. Ans. 243
8
2. Evaluate ( F). ds,
where S is the surface of the paraboloid x 2 + y 2 + z = 4 above the xy-plane and
S
2 ˆ ˆ
2
F ( x y 4) i 3 xyj (2 xz z ) kˆ
.
Ans. – 4
2 2 3 2
3. Evaluate [ xz dy dz ( x y z ) dzdx (2 xy y z) dxdy],
where S is the surface enclosing a
s
region bounded by hemisphere x 2 + y 2 + z 2 = 4 above XY-plane.
4. Verify Divergence Theorem for 2
F x iˆ zj ˆ yzkˆ
, taken over the cube bounded by
x = 0, x = 1, y = 0, y = 1, z = 0, z = 1.
2
5. Evaluate (2 ˆ ˆ ˆ
xyi yz j xz k)
ds
over the surface of the region bounded by
S
x = 0, y = 0, y = 3, z = 0 and x + 2 z = 6 Ans. 351
2