vector

464 Vectors
Fnds
OCDG
=
ˆ =
=
b
2
2 yc
ac
dy
2
0
=
OCDG
2 2
2 yc
acy
4
2 2 2 ˆ
b
0
=
2 2
2 b c
a bc ...(5)
4
{( x yz) iˆ ( y zx) ˆj ( z x y) k} ( iˆ
) dydz
b c
x 2 yz dydz
=
0 0 ( )
2
b yc
2 2
y c
= dy = =
0 2 4 0
Adding (1), (2), (3), (4), (5) and (6), we get
b
b
c
=
dy ( yz ) dz
0 0
2 2
b c
4
2
b yz
dy
0
2
c
0
...(6)
Fnds
ˆ
2 2 2 2 2 2 2 2
2 2
=
a b abc a b a c ab c
a c
4 4 4 4
2 2 2 2
b c
2 b c
a bc
4 4
= abc 2 + ab 2 c + a 2 bc
= abc (a + b + c) ...(B)
From (A) and (B), Gauss divergence Theorem is verified.
Verified.
2
Example 120. Verify Divergence Theorem, given that F 4 xziˆ
– y ˆj yz kˆ
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.
Volume Integral =
F
ˆ
2 ˆ
=
ˆ
ˆ
i j k (4 zxiˆ y ˆj yzk)
x y z
= 4 z – 2 y + y
= 4 z – y
Fdv
= (4 z y)
dx dy dz
=
1 1 1
dx dy (4 z y ) dz
0 0 0
=
=
1 dx 1 dy 2 1
(2 z yz )
0 0
0
=
2
1 y
dx
2 y
0 2 =
1
0
1 1
dx dy (2 y)
0 0
1 1
dx 2
0
2
= 3 1 3 ( ) 1
0
0
2 dx x = 3 2 2
...(1)
To evaluate Fnds ˆ , where S consists of six plane surfaces.
S
Over the face OABC , z = 0, dz = 0, ˆn = – ˆk , ds = dx dy
1 1 2
F. nds ˆ (– y ˆj) (– kˆ
) dxdy 0
0 0
Over the face BCDE, y = 1, dy = 0