vector

Vectors 399
Vx
= Vx
dydz Vx
dx dy dz
x
Vx
= dx dy dz
(Minus sign shows decrease)
x
Vy
Similarly, the decrease in mass of fluid to the flow along y-axis = dx dy dz
y
Vz
and the decrease in mass of fluid to the flow along z-axis = dx dy dz
z
V V
x y Vz
Total decrease of the amount of fluid per unit time = dx dy dz
x y z
V
V
x y Vz
Thus the rate of loss of fluid per unit volume =
x y z
= i j k
.( iV
x jV y kV
z)
= . V div V
x y z
If the fluid is compressible, there can be no gain or loss in the volume element. Hence
div V = 0 ...(1)
and V is called a Solenoidal vector function.
Equation (1) is also called the equation of continuity or conservation of mass.
xi yj
zk
Example 34. If v
,
2 2 2
x y z
find the value of div v .
(U.P., I Semester, Winter 2000)
Solution. We have, v =
xi yj
zk
2 2 2
x y z
div v =
xi yj zk
. v = i j k .
x y z 2 2 2 1/2
( x y z )
=
x y z
2 2 2 1/2 2 2 2 1/2 2 2 2 1/2
x ( x y z ) y ( x y z ) z
( x y z )
1
2 2 2 1/2 1
2 2 2
( x y z ) x. ( x y z )
2
.2x
2
=
2 2 2
( x y z )
1 1
2 2 2 2
1
2 2 2 1/2 1 2 2 2 1/2
2 2 2
( x y z ) y. ( x y z ) 2 2y
( x y z ) z. ( x y z ) .2z
2
2
2 2 2
2 2 2
( x y z )
( x y z )
=
=
2 2 2 2 2 2 2 2 2 2 2 2
( x y z ) x ( x y z ) y ( x y z ) z
2 2 2 3/2 2 2 2 3/2 2 2 2 3/2
( x y z ) ( x y z ) ( x y z )
2 2 2 2 2 2
y z x z x y
2 2 2 3/2
( x y z )
=
2 2 2
2( x y z ) 2
2 2 2 3/2
( x y z ) 2 2 2
( x y z )
Example 35. If u = x 2 + y 2 + z 2 , and r xi y j zk,
then find div ( ur ) in terms of u.
(A.M.I.E.T.E., Summer 2004)
Ans.