1000 Solved Problems in Modern Physics

36 1 Mathematical Physics

+

∫ 4

0

(

y^2

8

+y

)

ydy

4

+

(

y^2

8

−y

)

dy (alongy^2 = 8 x)

=+

16

3

1.8 (a) It is sufficient to show that CurlF= 0

∇×F=

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

ijk

∂

∂x

∂

∂y

∂

∂z

2 xy+z^2 x^22 xz

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

=ˆi. 0 −ˆj(2z− 2 z)+kˆ(2x− 2 x)= 0

(b) dΦ=F.dr=((2xy+z^2 )ˆi+x^2 ˆj+ 2 xzkˆ)).(ˆidx+ˆjdy+kˆdz)

=(2xy+z^2 )dx+x^2 dy+ 2 xzdz

=(2xydx+x^2 dy)+(z^2 dx+ 2 xzdz)

=d(x^2 y)+d(z^2 x)=d(x^2 y+xz^2 )

ThereforeΦ=x^2 y+xz^2 +constant

(c) Work done=Φ 2 −Φ 1 = 5. 0

1.9 LetU=x+y;V=x−y

∂U

∂x

=1;

∂V

∂y

=− 1

The curvesy=x^2 andy^2 = 8 xintersect at (0, 0) and (2, 4).

∫∫(

∂U

∂x

−

∂V

∂x

)

dxdy=

∫∫

S

(1−(−1))dxdy= 2

∫ 2

x= 0

∫ 2 √ 2 x

y=x^2

dxdy

= 2

∫ 2

0

[∫ 2 √ 2 x

x^2

dy

]

dx= 2

∫ 2

0

(2

√

2

√

x−x^2 )dx= 2

[

4

√

2

3

x^3 /^2 −

x^3

3

] 2

0

=

16

3

This is in agreement with the value obtained in Problem 1.7 for the line inte-

gral.

1.10 Use the divergence theorem
∫∫
A.ds=

∫∫∫

∇.Adν

But∇.A=

∂

∂x

x^3 +

∂

∂y

y^3 +

∂

∂z

z^3

= 3 x^2 + 3 y^2 + 3 z^2 =3(x^2 +y^2 +z^2 )= 3 R^2

∫∫

A.ds=

∫∫∫

3 R^2 dν=

∫

(3R^2 )(4πR^2 dR)

= 12 π

∫

R^4 dR=

12

5

πR^5

1.11

∫

c

A.dr=

∫

(2yˆi− 3 xˆj+zkˆ).(dxiˆ+dyˆj+dzkˆ)

=

∫

(2ydx− 3 xdy+zdz)