Answers and Solutions to Section 13.2 Homework Problems 1"27 ...

For the curve C3 we have
Z
z 2 dx z dy + 2y dz =
Thus Z
C
C3
Z 1
0
4 dt = 4.
z 2 dx z dy + 2y dz = 1 25 77
+ + 4 =
2 3 6 .
13. This is di¤erent from the number 13 in the book.
(a) The curve C1 is
Thus
Therefore Z
C1
r (t) = 3i + (6t 3) j
0 t 1.
F dr =
r 0 (t) = 6j.
Z 1
0
F (r (t)) 6j dt.
Since the vector 6j makes an acute angle with any of the vectors
F (r (t)), 0 t 1, the dot product F (r (t)) 6j is positive for all
t such that 0 t 1, meaning that the integrand is a positive–
valued function of t. Therefore
Z
F dr > 0.
(b) The curve C2 is
Thus
Therefore
Z
C2
F dr =
C1
r (t) = 3 cos (t) i + 3 sin (t) j
0 t 2 .
r 0 (t) = 3 sin (t) i + 3 cos (t) j.
Z 1
0
F (r (t)) ( 3 sin (t) i + 3 cos (t)) dt.
4