1.3 Solutions 37
Putx=Rcosθ,dx=−Rsinθdθ,y=Rsinθ,dy=Rcosθ,z= 0 , 0 <
θ<∫ 2 π
A.dr=− 2 R^2∫
sin^2 θdθ−R^2∫
cos^2 θdθ
=− 2 πR^2 −πR^2 =− 3 πR^21.12 (a)∇×(∇Φ)=∣ ∣ ∣ ∣ ∣ ∣ ∣
ijk
∂
∂x∂
∂y∂
∂z
∂Φ
∂x∂Φ
∂y∂Φ
∂z∣ ∣ ∣ ∣ ∣ ∣ ∣
=i(
∂^2 Φ
∂y∂z−
∂^2 Φ
∂z∂y)
−j(
∂^2 Φ
∂x∂z−
∂^2 Φ
∂z∂x)
+k(
∂^2 Φ
∂x∂y−
∂^2 Φ
∂y∂x)
= 0
because the order of differentiation is immaterial and terms in brackets
cancel in pairs.
(b) To show∇.(∇×V)= 0
(
ˆi∂
∂x+ˆj∂
∂y+kˆ∂
∂z)
.
∣ ∣ ∣ ∣ ∣ ∣
ijk
∂
∂x∂
∂y∂
∂z
VxVyVz∣ ∣ ∣ ∣ ∣ ∣ =
(
ˆi∂
∂x+ˆj∂
∂y+kˆ∂
∂z)
·
[
ˆi∣
∣
∣
∣
∂
∂y∂
∂z
VyVz∣
∣
∣
∣−
ˆj∣
∣
∣
∣
∂
∂x∂
∂z
VxVz∣
∣
∣
∣+
kˆ∣
∣
∣
∣
∂
∂x∂
∂y
VxVy∣
∣
∣
∣
]
=
∂
∂x∣
∣
∣
∣
∂
∂y∂
∂z
VyVz∣
∣
∣
∣−
∂
∂y∣
∣
∣
∣
∂
∂x∂
∂z
VxVz∣
∣
∣
∣+
∂
∂z∣
∣
∣
∣
∂
∂x∂
∂y
VxVy∣
∣
∣
∣
=
∣ ∣ ∣ ∣ ∣ ∣
∂
∂x∂
∂y∂
∂z
∂
∂x∂
∂y∂
∂z
VxVyVz∣ ∣ ∣ ∣ ∣ ∣
= 0
The value of the determinant is zero because two rows are identical.1.13Φ=x^2 y− 2 xz^3
(a)∇Φ=(
iˆ∂
∂x+ˆj∂
∂y+kˆ∂
∂z)
(x^2 y− 2 xz^3 )=2(xy−z^3 )ˆi+x^2 ˆj+ 6 xz^2 kˆ(b)∇^2 Φ=(
∂^2
∂x^2+
∂^2
∂y^2+
∂^2
∂z^2)
(x^2 y− 2 xz^3 )= 2 y− 12 xz1.14 (a)∇(x^2 y+xz)=(
ˆi∂
∂x+ˆj∂
∂y+kˆ∂
∂z)
(x^2 y+xz)=(2xy+z)ˆi+x^2 ˆj+xˆk
=−ˆi+ˆj+kˆ