Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro 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
LECTURE 22. LINE INTEGRALS 179 energy required to lift 1000 kg is (5.4×10 10 )/(3.6×10 6 ) = 15, 000 kW-hours. That’s equivalent to leaving a 60-Watt light on for 28 years; running a typical 1000-Watt hair dryer continuously for about 20 months; or running a typical home dryer (2800 Watts) for about 7 months. Definition 22.3 A closed curve is a curve that follows a path from a point back to itself, such as a circle. Definition 22.4 Let C be a closed curve, and F be a vector field. Then the circulation of the vector field is ∮ F · dr C where the special integral symbol ∮ indicates that the path of integration is closed. The circulation of a vector field is a measure of the “circularity” of its flow. For example consider the two vector fields in figure 22.4. Suppose we follow the illustrated circular paths in a counter-clockwise fashion in both cases, such as r(t) = (cos t, sin t), 0 ≤ t ≤ 2π In the vector field on the left, the magnitude of the field is the same everywhere, Figure 22.4: Illustration of the same path through two different vector fields. but is direction is circular, giving a flow around the origin. F(x, y) = 1 √ (−y, x) x 2 + y2 Math 250, Fall 2006 Revised December 6, 2006.
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LECTURE 22. LINE INTEGRALS 179<br />
energy required to lift 1000 kg is (5.4×10 10 )/(3.6×10 6 ) = 15, 000 kW-hours. That’s<br />
equivalent to leaving a 60-Watt light on for 28 years; running a typical 1000-Watt<br />
hair dryer continuously for about 20 months; or running a typical home dryer (2800<br />
Watts) for about 7 months. <br />
Definition 22.3 A closed curve is a curve that follows a path from a point back<br />
to itself, such as a circle.<br />
Definition 22.4 Let C be a closed curve, and F be a vector field. Then the circulation<br />
of the vector field is ∮<br />
F · dr<br />
C<br />
where the special integral symbol ∮ indicates that the path of integration is closed.<br />
The circulation of a vector field is a measure of the “circularity” of its flow.<br />
For example consider the two vector fields in figure 22.4. Suppose we follow the<br />
illustrated circular paths in a counter-clockwise fashion in both cases, such as<br />
r(t) = (cos t, sin t),<br />
0 ≤ t ≤ 2π<br />
In the vector field on the left, the magnitude of the field is the same everywhere,<br />
Figure 22.4: Illustration of the same path through two different vector fields.<br />
but is direction is circular, giving a flow around the origin.<br />
F(x, y) =<br />
1<br />
√ (−y, x)<br />
x 2 + y2 Math 250, Fall 2006 Revised December 6, 2006.
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