Mathematical Physics Vol 1

Chapter 4. Field theory

176

Exercise 117

Calculate s S

F · n d S ,where F = 4 xz i − y 2 j + yz k , and S is the surface of a cube bounded

by x = 0, x = 1, y = 0, y = 1, z = 0 and z = 1.

Solution Using Gauss’s theorem (see p. 102, relation (4.92)) we obtain y V ∇ · F d V = y V ∂ ∂ x ( 4 xz )+ ∂ ∂ y ( − y 2 )+ ∂ ∂ z ( yz ) d V = = y V ( 4 z − y ) d V = 1 Z x = 0 1 Z y = 0 1 Z z = 0 ( 4 z − y ) d z d y d x =

1

1 Z x = 0 1 Z x = 0

1 Z y = 0 1 Z y = 0

( 2 z 2 − yz )

d y d x =

=

z = 0

2 3

( 2 − y ) d y d x =

=

.

Exercise 118

Calculate

I = I c

A · d r , where A =( 2 x − y ) i − yz 2 j − y 2 z k , and c is the boundary of the surface S , the upper half of the sphere x 2 + y 2 + z 2 = 1: a) directly, b) using Stokes’ theorem. Solution a) The boundary c of the surface S is a circle in the xy plane, with a radius of one, and center in the coordinate origin. If x = cos t , y = sin t , z = 0, 0 ≤ t ≤ 2 π are parametric equation of the curve c , then I = I c A · d r = I c [( 2 x − y ) d x − yz 2 d y − y 2 z d z ]=

2 π Z 0 ( 2cos t − sin t )( − sin t ) d t = π .

=

b) According to Stokes’ theorem I c

A · d r = x S

( ∇ × A ) · d S = I .