Mathematical Physics Vol 1

Chapter 4. Field theory

160

Let us now calculate the integrals along this segments, namely c 1 , c 2 and c 3 .

1

1 Z 0 ( 3 x 2 + 6 ( 0 )) d x − 14 ( 0 )( 0 ) 0 + 20 x ( 0 ) 2 0 = 1 Z 0 1 Z 0 ( 3 ( 1 ) 2 + 6 ( y )) 0 − 14 y ( 0 ) d y + 20 ( 1 )( 0 ) 2 0 = 0 , 1 Z 0 ( 3 ( 1 ) 2 + 6 ( y )) 0 − 14 ( 1 )( 0 ) 0 + 20 ( 1 )( z ) 2 d z = 1 Z 0

I 1 = Z c 1 I 2 = Z c 2 I 3 = Z c 3

3 x 2 d x = x 3

= 1 ,

=

0

=

1

20 z 3 3

20 3

20 z 2 d z =

=

=

.

0

Finally, adding the three integrals we obtain I = I 1 + I 2 + I 3 = Z c

20 3

23 3

v · d r = 1 + 0 +

=

.

c) The straight line, passing through points O ( 0 , 0 , 0 ) and A ( 1 , 1 , 1 ) , can be repre sented in parametric form as x = y = z = t . Parameter values t = 0 and t = 1 correspond to points O and A , respectively, and thus it follows Z c v · d r = 1 Z 0 ( 3 t 2 + 6 t ) d t − 14 ( t )( t ) d t + 20 ( t )( t ) 2 d t

1 Z 0 ( 3 t 2 + 6 t − 14 t 2 + 20 t 3 ) d t = 1 Z 0

13 3

( 6 t − 11 t 2 + 20 t 3 ) d t =

=

.

Note that these examples show that the values of the curvilinear integrals depend on the path (line) along which the integrals are calculated, and which passes through the given points.

Exercise 98 Observe the force F = 3 xy i − y 2 j . Calculate the work of this force R c F · d r along the curve c in the xy plane, given by the equation y = 2 x 2 , from point O ( 0 , 0 ) to point A ( 1 , 2 ) .

Solution Given that the integration is performed in the xy plane, it follows that d r = d x i + d y j ,