Mathematical Physics Vol 1

4.6 Examples

159

on the curve c , respectively. It follows that

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

Z c

v · d r =

1 Z 0 1 Z 0

9 t 2 d t − 28 t 6 d t + 60 t 9 d t

=

( 9 t 2 − 28 t 6 t + 60 t 9 ) d t

=

1 0 = 5 .

= 3 t 3 − 4 t 7 + 6 t 10

b) Given that

Z c

= Z c 1

+ Z c 2

+ Z c 3

where c = c 1 ∪ c 2 ∪ c 3 , we will divide the line OBCA as follows - segment c 1 , connecting points O ( 0 , 0 , 0 ) and B ( 1 , 0 , 0 ) , and lying on the x axis, which means that y = 0, z = 0, d y = 0, d z = 0, while x takes the values from 0 to 1; - segment c 2 , connecting points B ( 1 , 0 , 0 ) and C ( 1 , 1 , 0 ) , and lying in the xy plane, parallel to the y axis, which means that x = 1, z = 0, d x = 0, d z = 0, while y takes the values from 0 to 1; - segment c 3 , connecting points C ( 1 , 1 , 0 ) i A ( 1 , 1 , 1 ) , and lying in a plane parallel to the plane yz and parallel to the z axis, which means that x = 1, y = 1, d x = 0, d y = 0, while z takes the values from 0 to 1.

Figure 4.28