082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

856 Chapter 26 Vector calculus
26.4 THEDIVERGENCEOFAVECTORFIELD
Given a vector field v = v(x,y,z) let us consider what happens when we differentiate
its individual components. If
v=v x
i+v y
j+v z
k
wecantakeeachcomponentinturnanddifferentiateitpartiallyw.r.t.x,yandz,respectively;
thatis, we can evaluate
∂v x
∂x
∂v y
∂y
∂v z
∂z
Ifweaddthecalculatedquantitiestheresultturnsouttobeaveryusefulscalarquantity
known asthedivergenceof v, thatis
divergence ofv = ∂v x
∂x + ∂v y
∂y + ∂v z
∂z
Thisisusuallyabbreviatedtodivv.Alternatively,thenotation ∇ ·visoftenused.Ifwe
use the vector operator notation introduced inthe previous section wehave
( ∂
∇·v=
∂x , ∂ ∂y ∂z)
, ∂ ·v
=
( ∂
∂x , ∂ ∂y , ∂ ∂z)
·(v x
,v y
,v z
)
Interpreting the · as a scalar product wefind
∇·v= ∂v x
∂x + ∂v y
∂y + ∂v z
∂z
( ∂
asbefore,althoughthisisnotascalarproductintheusualsensebecause
∂x , ∂ ∂y ∂z)
, ∂
is a vector operator. We note that the process of finding the divergence is always performed
on a vector field and the resultisalways a scalar field:
divv=∇·v= ∂v x
∂x + ∂v y
∂y + ∂v z
∂z
Example26.5 Ifv =x 2 zi +2y 3 z 2 j +xyz 2 kfind divv.
Solution Partially differentiating the first component ofvw.r.t.xwefind
∂v x
∂x = 2xz
Similarly,
∂v y
∂y = 6y2 z 2
and
Adding these results wefind
∂v z
∂z = 2xyz
divv=∇·v=2xz+6y 2 z 2 +2xyz