vector

Vectors 461
=
=
a b a b
2 a
2
y b
cdx(2 x y) dy c2xy dx c 2bx
dx
2
2
0 0 0 0 0
2 2
a
2
2
2
bx b x ab b
c c ab abc a
2 2 2 2
0
Example 118. Verify the divergence Theorem for the function F = 2 x2 yi – y 2 j + 4 x z 2 k
taken over the region in the first octant bounded by y 2 + z 2 = 9 and x = 2.
Solution.
FdV
=
V
= (4 xy 2y 8 xz)
dxdydz
=
=
ˆ
ˆ
ˆ 2 ˆ 2ˆ
2
i j k (2x yi y j 4 xz k)
dV
x y z
=
2 3 9 y
2
dx dy (4 xy 2y 8 xz)
dz
0 0 0
2 3 9
2
2 y
dx (4 2 4 )
0 dy xyz yz xz
0
0
2 3 2 2 2
dx [4 xy 9 y 2y 9 y 4 x(9 y )] dy
0 0
3
2 4x
2 2 3/2 2 2 3/2 4xy
= dx (9 y ) (9 y ) 36
xy
0
2 3 3 3 0
2
2
= (0 0 108 x 36 x 36 x 18) dx
2
= (108 x 18) dx
2
x
0 = 108 18 x
0
2 0
= 216 – 36 = 180 ...(1)
Here Fnds
ˆ = Fnˆ ds Fnˆ ds Fnˆ ds Fnˆ ds F
nˆ
ds
S
OABC OCE OADE ABD BDEC
F nˆ
2 2 2
ds (2 x yiˆ
y ˆj 4 xzkˆ
). nds ˆ
=
BDEC
BDEC
Normal vector
= = ˆ
i ˆ
j kˆ
x y z (y2 + z 2 – 9)
= 2 yj ˆ 2 zkˆ
2 yj ˆ 2zkˆ
yj ˆ zkˆ
Unit normal vector = ˆn =
=
2 2 2 2
4y
4z
y z
yj ˆ zkˆ
yj ˆ zkˆ
= =
9 3
ˆ
2 ˆ 2ˆ
2 yj zk 1
(2x yi y j 4 xzk)
ds
3 3
= ( 4 )
3
BDEC
3
y xz ds
BDEC
yj ˆ zkˆ
( ˆ ) ˆ z dxdy
dx dy ds nk ds k ds or ds
3 3 z
3
3
1
3 3 dxdy
2 3
y 2
= ( y 4 xz )
3
= dx 4 xz dy
z
0
0
z
BDEC
3
3
2
27 sin
2
2
= dx
4 x (9cos )
0
0
3cos
3
Ans.
y
3sin ,
z
3cos