Mathematical Physics Vol 1

Chapter 4. Field theory

154

Solution a) 4, b) − 15, c) 1, d) 6. Exercise 92

d 2 f d r 2

2 r

d f d r

a) Prove that ∇ 2 f ( r )=

.

+

b) Find f ( r ) so that ∇ 2 f ( r )= 0.

Solution b) f ( r )= A + B / r , where A and B are constants.

Exercise 93 Prove that the vector field

A =( 2 x 2 + 8 xy 2 z ) i +( 3 x 3 y − 3 xy ) j − ( 4 y 2 z 2 + 2 x 3 z ) k

is not a solenoid field, whereas B = xyz 2 A is.

4.6.6 Integrals, integral theorems

Exercise 94 Find the area of the ellipse.

Solution The area P of any flat surface S is equal to P = x S d x d y . Let the surface S be bounded by a closed curve C , then according to Stokes’ theorem 21 . By adding these two integrals we obtain P = x S d x d y = 1 2 I C x d y − y d x . By switching to polar coordinates x = a cos θ , y = b sin θ , we obtain

2 π Z 0 ( a cos θ )( b cos θ ) d θ − ( b sin θ )( − a sin θ ) d θ =

1 2 I C

1 2

P =

xdy − ydx =

2 π Z 0

2 π Z 0

1 2

ab ( cos 2 θ + sin 2 θ ) d θ = 1 2

abd θ = π ab .