vector

460 Vectors
4 2
x
2
y
2
(4z xz 3) dzdydx =
3
2 xz
2z 3z
dy dx
3
4
x
2
y
2
64x
2 2 2 2 3/2 2 2
= 32 12 {2( x y ) xx ( y ) 3 x y } dydx
3
64x
2 2 2 2 3/2 2 2
= 44 2( x y ) xx ( y ) 3
x y dydx
3
Putting x = r cos and y = r sin , we have
64r
cos 2 3
= 44 2r r cos r 3r
rddr
3
Limits of r are 0 to 4.
and limits of are 0 to 2
=
=
=
=
=
2
2 4 64r
cos 3 5 2
44 2 cos 3
0
0
r r r r ddr
3
3 4 6
2
2 64 r cos r r
3
0
2
0
2
0
2
0
22r cos r d
9 2 6
3 4 6
2 64 (4) cos (4) (4)
3
22(4) cos (4)
d
9 2 6
6
64 64 (4)
352 cos 128 cos 64
d
9 6
6
64 64 (4)
160 cos d
9 6
2
6
64 64 (4)
6
64 64 (4)
= 160
sin
9 6 = 160 (2 ) sin 2
9 6
0
= 320 Ans.
Example 117. The vector field 2
F x iˆ
zj ˆ yzkˆ
is defined over the volume of the cuboid
given by 0 x a, 0 y b, 0 z c, enclosing the surface S. Evaluate the surface integral
F . ds
(U.P., I Semester, Winter 2001)
S
Solution. By Divergence Theorem, we have
2 ˆ
2
( x iˆ zj ˆ yz k). ds div ( x iˆ zj ˆ yz kˆ) dv,
S
v
where V is the volume of the cuboid enclosing the surface S.
ˆ
2
ˆ
=
ˆ
ˆ
i j k .( x iˆ z ˆ
j yzk)
dv
v
x y z
2
= ( x ) ( z) ( yz)
dx dy dz
v
x y z
=
a b c
=
(2 x
y ) dx dy dz
0 0 0
x y z
a b a b
c
= 0
0 0 0 0
dx [2 xz yz] dy dx (2 xc yc)
dy
4
0
a b c
dx dy (2 x y ) dz
0 0 0