34 1 Mathematical Physics

1.2∇(xy^2 +xz)=

(

ˆi∂

∂x

+ˆj

∂

∂y

+ˆk

∂

∂z

)

(xy^2 +xz)

=(y^2 +z)ˆi+(2xy)ˆj+xkˆ

= 2 ˆi− 2 ˆj−kˆ,at(− 1 , 1 ,1)

A unit vector normal to the surface is obtained by dividing the above vector

by its magnitude. Hence the unit vector is

(2iˆ− 2 ˆj−kˆ)[(2)^2 +(−2)^2 +(−1)^2 ]−^1 /^2 =

2

3

ˆi−^2

3

ˆj−^1

3

kˆ

1.3 F∝ 1 /r^2

∇.(r−^3 r)=r−^3 ∇.r+r.∇r−^3

But∇.r=

(

iˆ∂

∂x

+ˆj

∂

∂y

+kˆ

∂

∂z

)

.

(

ˆix+ˆjy+kzˆ

)

=

∂x

∂x

+

∂y

∂y

+

∂z

∂z

= 3

r.∇r−^3 =(xiˆ+yˆj+zkˆ).

(

ˆi∂

∂x

+ˆj

∂

∂y

+ˆk

∂

∂z

)

(x^2 +y^2 +z^2 )−^3 /^2

=(xˆi+yˆj+zkˆ).

(

−

3

2

)

.(2xˆi+ 2 yˆj+ 2 zkˆ)(x^2 +y^2 +z^2 )−^5 /^2

=−3(x^2 +y^2 +z^2 )(x^2 +y^2 +z^2 )−

5

(^2) =− 3 r−^3
Thus∇.(r−^3 r)= 3 r−^3 − 3 r−^3 = 0
1.4 By problem∇×A= 0 and∇×B=0, it follows that
B.(∇×A)= 0
A.(∇×B)= 0
Subtracting,B.(∇×A)−A.(∇×B)= 0
Now∇.(A×B)=B.(∇×A)−A.(∇×B)
Therefore∇.(A×B)=0, so that (A×B) is solenoidal.
1.5 (a) Curl{rf(r)}=∇×{rf(r)}=∇×{xf(r)ˆi+yf(r)ˆj+zf(r)kˆ}

∣∣
∣∣
∣∣
∣∣
ˆi ˆj kˆ
∂
∂x
∂
∂y
∂
∂z
xf(r)yf(r)zf(r)
∣∣
∣∣
∣∣
∣∣

(
z
∂f
∂y
−y
∂f
∂z
)
iˆ+
(
x
∂f
∂z
−z
∂f
∂x
)
ˆj+
(
y
∂f
∂x
−x
∂f
∂y
)
kˆ
But∂∂xf=

(

∂f

∂r

)(

∂r

∂x

)

=∂∂fr∂(x

(^2) +y (^2) +z (^2) ) 1 / 2
∂x =
xf′
r
Similarly∂∂yf=yf
′
r and
∂f
∂z=
zf′
r, where prime means differentiation with
respect tor.
Thus,
curl{rf(r)}=
(
zy f′
r
−
yz f′
r
)
ˆi+
(
xzf′
r
−
zx f′
r
)
ˆj+
(
yx f′
r
−
xyf′
r
)
kˆ= 0