Multivariate Calculus - Bruce E. Shapiro

178 LECTURE 22. LINE INTEGRALS
Therefore the line integral is
∫
∫ 10
F · dr = 2tdt +
ABC
0
∫ 18
10
= t 2∣ ( )∣
∣ 10 3 ∣∣∣
18
0 + 2 t2 − 24t
(3t − 24)dt (22.15)
10
(22.16)
= 10 − 0 + 3 × 182 − (24)(18) − 3 × 102 + 240
2
2
(22.17)
= 10 + 486 − 432 + 240 = 304 (22.18)
Definition 22.2 The work done on an object when it is moved along a path C
and subjected to a force F is
∫
W = F · r (22.19)
Example 22.2 Find the work required to lift a satellite of mass 1000 kg from the
earth’s surface (r = 6300 km) to a geostationary orbit (r = 42, 000 km) under the
influence of the Earth’s gravity
F = − µm
r 2 k
where r is the distance of the object from the center of the Earth; m is its mass; and
µ = GM = 3.986 × 10 14 meter 3 /second 2 is the product of the universal constant of
gravity and the mass of the Earth. (Work is measured in units of joules, where one
joule equals one kilogram meter 2 /second 2 .)
Solution.The work is the line integral
W =
C
∫ 42×10 6
6.3×10 6 F · dr
where we have converted the distances from meters to kilometers. Substituting the
equation for F and assuming the trajectory moves completely in the z directions,
∫ 42×10 6
W = −µm
6.3×10 6 1
r 2 dz
∣ 42×106
6.3×10 6
= µm r
(
= (3.986 × 10 14 ) × (1000)
= −5.4 × 10 10 Joules
1
42 × 10 6 − 1
6.3 × 10 6 )
Joules
The work is negative because the line integral measured the work done by the force
field on the satellite, i.e., work must be done by the satellite against the field to
get it into orbit. Converting to day-to-day units, since one watt is the same as one
Joule/second, one kilowatt-hour is the energy used by expending 1000 Joules per
second (one kilowatt) for one hour (3600 seconds), namely 3.6×10 6 Joules.Thus the
Revised December 6, 2006. Math 250, Fall 2006