Section 16.3 The Fundamental Theorem for Line Integrals
22. (a) Find a function f such that F = ∇f and (b) use part (a) to evaluateR
CF·dr along the given curve C.
F(x, y, z) = (y2z + 2xz2) i + 2xyz j + (xy2+ 2x2z) k, C : x =√
t, y = t + 1, z = t2, 0 ≤ t ≤ 1.
Solution:
650 ¤ CHAPTER 16 VECTOR CALCULUS
(b) is a smooth curve with initial point (1 1) and terminal point
414, so by Theorem 2
F· r =
∇ · r = 414
− (1 1) = (12 + 1) − (3 + 1) = 9.
13. (a) If F = ∇ then ( ) = 23and ( ) = 32.
( ) = 23implies ( ) = 1333+ ()and ( ) = 32+ 0(). But ( ) = 32so 0() = 0 ⇒
() = , a constant. We can take = 0, so ( ) = 1333.
(b) is a smooth curve with initial point r(0) = (0 0) and terminal point r(1) = (−1 3), so by Theorem 2
F· r =
∇ · r = (−1 3) − (0 0) = −9 − 0 = −9.
14. (a) ( ) = 2implies ( ) = 133+ ()and ( ) = 0 + 0(). But ( ) = 2so
0() = 2 ⇒ () = 133+ . We can take = 0, so ( ) = 133+133. (b)
F· r = (2 8) − (−1 2) =8 3+ 5123
−
−13+ 83= 171.
15. (a) ( ) = implies ( ) = + ( ) and so ( ) = + ( ). But ( ) = so
( ) = 0 ⇒ ( ) = (). Thus ( ) = + () and ( ) = + 0(). But
( ) = + 2, so 0() = 2 ⇒ () = 2+ . Hence ( ) = + 2(taking = 0).
(b)
F· r = (4 6 3) − (1 0 −2) = 81 − 4 = 77.
16. (a) ( ) = 2 + 22implies ( ) = 2 + 22+ ( )and so ( ) = 2 + ( ). But
( ) = 2so ( ) = 0 ⇒ ( ) = (). Thus ( ) = 2 + 22+ ()and
( ) = 2+ 22 + 0(). But ( ) = 2+ 22, so 0() = 0 ⇒ () = . Hence
( ) = 2 + 22(taking = 0).
(b) = 0 corresponds to the point (0 1 0) and = 1 corresponds to (1 2 1), so
F· r = (1 2 1) − (0 1 0) = 5 − 0 = 5.
17. (a) ( ) = implies ( ) = + ( )and so ( ) = + ( ). But ( ) = so
( ) = 0 ⇒ ( ) = (). Thus ( ) = + ()and ( ) = + 0(). But
( ) = , so 0() = 0 ⇒ () = . Hence ( ) = (taking = 0).
(b) r(0) = h1 −1 0i, r(2) = h5 3 0i so
F· r = (5 3 0) − (1 −1 0) = 30+ 0 = 4.
18. (a) ( ) = sin implies ( ) = sin + ( ) and so ( ) = cos + ( ). But
( ) = cos + cos so ( ) = cos ⇒ ( ) = cos + (). Thus
( ) = sin + cos + ()and ( ) = − sin + 0(). But ( ) = − sin , so 0() = 0 ⇒
() = . Hence ( ) = sin + cos (taking = 0).
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34. Let F = ∇f, where f (x, y) = sin(x − 2y). Find curves C1 and C2that are not closed and satisfy the equation.
(a)R
C1F·dr = 0 (b)R
C2F·dr = 1 Solution:
652 ¤ CHAPTER 16 VECTOR CALCULUS
27. From the graph, it appears that F is conservative, since around all closed paths, the number and size of the field vectors pointing in directions similar to that of the path seem to be roughly the same as the number and size of the vectors pointing in the opposite direction. To check, we calculate
(sin ) = cos =
(1 + cos ). Thus F is conservative, by Theorem 6.
28. ∇( ) = cos( − 2) i − 2 cos( − 2) j (a) We use Theorem 2:
1F· r =
1∇ · r = (r()) − (r()) where 1starts at = and ends at = . So because (0 0) = sin 0 = 0 and ( ) = sin( − 2) = 0, one possible curve 1is the straight line from (0 0) to ( ); that is, r() = i + j, 0 ≤ ≤ 1.
(b) From (a),
2F· r = (r()) − (r()). So because (0 0) = sin 0 = 0 and
2 0= 1, one possible curve 2is r() = 2 i, 0 ≤ ≤ 1, the straight line from (0 0) to
2 0.
29. Since F is conservative, there exists a function such that F = ∇, that is, = , = , and = . Since ,
, and have continuous first order partial derivatives, Clairaut’s Theorem says that = = = ,
= = = , and = = = .
30. Here F( ) = i + j + k. Then using the notation of Exercise 29, = 0 while = . Since these aren’t equal, F is not conservative. Thus by Theorem 4, the line integral of F is not independent of path.
31. =
( ) | 2
consists of those points beneath, but not on, the parabola = 2.
(a) Because does not include any of its boundary points, it is open.
More formally, at any point in there is a disk centered at that point that lies entirely in .
(b) Any two points chosen in can always be joined by a path that lies entirely in , so is connected. ( consists of just one “piece.”)
(c) is connected and it has no holes, so it’s simply-connected. (Every simple closed curve in encloses only points that are in .)
32. = {( ) | + 6= 1} consists of those points not on the line
= 1 − .
(a) The region does not include any of its boundary points, so it is open.
(b) consists of two separate pieces (each piece shaded slighly differently), so it is not connected.
(c) Because is not connected, it is not simply-connected.
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42. (a) Suppose that F is an inverse square force field, that is, F(r) = |r|cr3, for some constant c, where r = x i + y j + z k.
Find the work done by F in moving an object from a point P1 along a path to a point P2 in terms of the distances d1and d2 from these points to the origin.
(b) An example of an inverse square field is the gravitational field F = −(mMG)r/|r|3 discussed in Example 16.1.4.
Use part (a) to find the work done by the gravitational field when the earth moves from aphelion (at a maximum distance of 1.52 × 108 km from the sun) to perihelion (at a minimum distance of 1.47 × 108 km). (Use the values m = 5.97 × 1024kg, M = 1.99 × 1030kg, and G = 6.67 × 10−11N · m2/kg2.)
(c) Another example of an inverse square field is the electric force field F = εqQr/|r|3discussed in Example 16.1.5.
Suppose that an electron with a charge of −1.6 × 10−19C is located at the origin. A positive unit charge is positioned a distance 10−12m from the electron and moves to a position half that distance from the electron. Use part (a) to find the work done by the electric force field. (Use the value ε = 8.985 × 109.)
Solution:
1
SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS ¤ 653
33. = {( ) | ≥ 0 ≥ 0 + 1} consists of those points in the triangle formed by the axes and the line = 1 − . It includes the axes as boundary points, but not the line.
(a) includes some boundary points, so it is not open. [Note that at any boundary point, (0 0) for instance, any disk centered there cannot lie entirely in .]
(b) The region consists of one piece, so it is connected.
(c) Because is connected and has no holes, it is simply-connected.
34. =
( ) | 2+ 2 1consists of those points that are outside, and not on, the circle 2+ 2= 1.
(a) The region does not include any of its boundary points, so it is open.
(b) is connected because it consists of only one piece.
(c) is not simply-connected because it has a hole of radius 1. Thus, any simple closed curve that lies in , but goes around the circle 2+ 2= 1, includes points that are not in .
35. (a) = −
2+ 2,
= 2− 2
(2+ 2)2 and =
2+ 2,
= 2− 2
(2+ 2)2. Thus
=
. (b) 1: = cos , = sin , 0 ≤ ≤ , 2: = cos , = sin , = 2 to = . Then
1
F· r =
0
(− sin )(− sin ) + (cos )(cos ) cos2 + sin2 =
0
= and
2
F· r =
2
= −
Since these aren’t equal, the line integral of F isn’t independent of path. (Or notice that
3F· r =2
0 = 2where
3is the circle 2+ 2= 1, and apply the contrapositive of Theorem 3.) This doesn’t contradict Theorem 6, since the domain of F, which is R2except the origin, isn’t simply-connected.
36. (a) Here F(r) = r|r|3and r = i + j + k. Then (r) = −|r| is a potential function for F, that is, ∇ = F.
(See the discussion of gradient fields in Section 16.1.) Hence F is conservative and its line integral is independent of path.
Let 1= (1 1 1)and 2= (2 2 2).
=
F· r = (2) − (1) = −
(22+ 22+ 22)12 +
(21+ 21+ 21)12 =
1
1 − 1
2
.
(b) In this case, = −() ⇒
= −
1
152 × 1011 − 1 147 × 1011
= −(597 × 1024)(199 × 1030)(667 × 10−11)(−22377 × 10−13) ≈ 177 × 1032J (c) In this case, = ⇒
=
1
10−12− 1 5 × 10−13
=
8985 × 109 (1)
−16 × 10−19
−1012
≈ 1400 J.
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