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

26.5 The curl of a vector field 859
Solutions
1 (a) 3x 2 y,2y 3 z,xz 3
(b) 6xy,6y 2 z,3xz 2
(c) 6xy +6y 2 z +3xz 2
2 (a) 1+2y+2z (b) 3
3 ∇·A=2x+xy
4 (a) 2x+2y+2z
(b) ye xy +2xzcos(xy) +x 3
(c)y−2z
(d)2xy 2 −2y −xy
5 (a) x (b) z 2 (c) −yz (d) 1−y
6 13
7 ∇φ=2xi+2yj−4zk
8 yes
9 (a)z 2 +4xyz +3xz 2
(b)4xz 2 i +8xy 2 zj +4xz 3 k
(c)4z 2 +16xyz +12xz 2
(d)yes
10 yes
11 v = 2xyi +y 2 j −4yzk forexample
26.5 THECURLOFAVECTORFIELD
Athird differential operator isknown ascurl. Itisdefined rather like a vector product.
curlv=∇×v
( ∂
=
∂x , ∂ ∂y ∂z)
, ∂ × (v x
,v y
,v z
)
∣ i jk ∣∣∣∣∣∣∣∣
=
∂ ∂ ∂
∂x ∂y ∂z
∣
v x
v y
v z
This determinant is evaluated in the usual way except that we must regard ∂ ∂x , ∂ ∂y and
∂
as operators, notmultipliers. Thus, forexample,
∂z
∣ ∂ ∂ ∣∣∣∣∣
∂x ∂y
∣ v x
v y
Explicitly wehave
curlv =
means
∂v y
∂x − ∂v x
∂y
( ∂vz
∂y − ∂v ) (
y ∂vx
i +
∂z ∂z − ∂v ) ( ∂vy
z
j +
∂x ∂x − ∂v )
x
k
∂y