Section 16.3 The Fundamental Theorem for Line Integrals
648 ¤ CHAPTER 16 VECTOR CALCULUS
F· T ≈
7
= 1
[F(∗ ∗) · T(∗ ∗)] ∆ = [2 + 2 + 2 + 2 + 1 + 1 + 1](2) = 22. Thus, we estimate the work done to be approximately 22 J.
52. Use the orientation pictured in the figure. Then since B is tangent to any circle that lies in the plane perpendicular to the wire, B= |B| T where T is the unit tangent to the circle : = cos , = sin . Thus B = |B| h− sin cos i. Then
B· r =2
0 |B| h− sin cos i · h− sin cos i =2
0 |B| = 2 |B|. (Note that |B| here is the magnitude of the field at a distance from the wire’s center.) But by Ampere’s Law
B· r = 0. Hence |B| = 0(2).
16.3 The Fundamental Theorem for Line Integrals
1. appears to be a smooth curve, and since ∇ is continuous, we know is differentiable. Then Theorem 2 says that the value of
∇ · r is simply the difference of the values of at the terminal and initial points of . From the graph, this is 50 − 10 = 40.
2. is represented by the vector function r() = (2+ 1) i + (3+ ) j, 0 ≤ ≤ 1, so r0() = 2 i + (32+ 1) j. Since 32+ 1 6= 0, we have r0() 6= 0, thus is a smooth curve. ∇ is continuous, and hence is differentiable, so by Theorem 2 we have
∇ · r = (r(1)) − (r(0)) = (2 2) − (1 0) = 9 − 3 = 6.
3. Let ( ) = + 2and ( ) = 2+ 2. Then = + 2 and = 2 + 2. Since 6= , F( ) = i + jis not conservative by Theorem 5.
4. (2− 2) = 2 = (2) and the domain of F is R2which is open and simply-connected, so F is conservative by Theorem 6. Thus, there exists a function such that ∇ = F, that is, ( ) = 2− 2 and ( ) = 2. But
( ) = 2− 2 implies ( ) = 2− 2+ ()and differentiating both sides of this equation with respect to gives
( ) = 2 + 0(). Thus 2 = 2 + 0()so 0() = 0and () = where is a constant. Hence
( ) = 2− 2+ is a potential function for F.
5.
2
= 2· + 2= (2+ 2),
[(1 + )] = (1 + ) · + = + 2+ = (2+ 2).
Since these partial derivatives are equal and the domain of F is R2which is open and simply-connected, F is conservative by Theorem 6. Thus, there exists a function such that ∇ = F, that is, ( ) = 2and ( ) = (1 + ). But
( ) = 2implies ( ) = + ()and differentiating both sides of this equation with respect to gives
( ) = (1 + )+ 0(). Thus (1 + )= (1 + )+ 0()so 0() = 0and () = where is a constant. Hence ( ) = + is a potential function for F.
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SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS ¤ 649
6. () = = (+ )and the domain of F is R2which is open and simply-connected, so F is conservative.
Hence there exists a function such that ∇ = F. Here ( ) = implies ( ) = + ()and then
( ) = + 0(). But ( ) = + so 0() = ⇒ () = + and ( ) = + + is a potential function for F.
7. (+ sin ) = + cos = (+ cos )and the domain of F is R2. Hence F is conservative so there exists a function such that ∇ = F. Then ( ) = + sin implies ( ) = + sin + ()and
( ) = + cos + 0(). But ( ) = + cos so () = and ( ) = + sin + is a potential function for F.
8. (2 + −2) = 2 − 2−3= (2− 2−3)and the domain of F is {( ) | 0} which is open and simply-connected. Hence F is conservative, so there exists a function such that ∇ = F. Then ( ) = 2 + −2 implies ( ) = 2 + −2+ ()and ( ) = 2− 2−3+ 0(). But ( ) = 2− 2−3so
0() = 0 ⇒ () = . Then ( ) = 2 + −2+ is a potential function for F.
9. (2cos + cos ) = 2 cos − sin = (2 sin − sin ) and the domain of F is R2which is open and simply connected. Hence F is conservative so there exists a function such that ∇ = F. Then ( ) = 2cos + cos implies
( ) = 2sin + cos + ()and ( ) = 2 sin − sin + 0(). But ( ) = 2 sin − sin so
0() = 0 ⇒ () = and ( ) = 2sin + cos + is a potential function for F.
10. (ln + ) = 1 + 1 = (ln + )and the domain of F is {( ) | 0 0} which is open and simply connected. Hence F is conservative so there exists a function such that ∇ = F. Then ( ) = ln +
implies ( ) = ln + ln + () and ( ) = + ln + 0(). But ( ) = ln + so 0() = 0 ⇒
() = and ( ) = ln + ln + is a potential function for F.
11. (a) F has continuous first-order partial derivatives and
(2) = 2 =
(2)on R2, which is open and
simply-connected. Thus, F is conservative by Theorem 6. Then we know that the line integral of F is independent of path;
in particular, the value of
F· r depends only on the endpoints of . Since all three curves have the same initial and terminal points,
F· r will have the same value for each curve.
(b) We first find a potential function , so that ∇ = F. We know ( ) = 2and ( ) = 2. Integrating
( )with respect to , we have ( ) = 2 + (). Differentiating both sides with respect to gives
( ) = 2+ 0(), so we must have 2+ 0() = 2 ⇒ 0() = 0 ⇒ () = , a constant.
Thus ( ) = 2 + , and we can take = 0. All three curves start at (1 2) and end at (3 2), so by Theorem 2,
F· r = (3 2) − (1 2) = 18 − 2 = 16 for each curve.
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SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS ¤ 651
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).
(b) r(0) = h0 0 0i, r(2) = h1 2 i so
F· r = (1 2 ) − (0 0 0) = 1 −2 − 0 = 1 −2. 19. The functions 2−and 2 − 2− have continuous first-order derivatives on R2and
2−
= −2−=
2 − 2−, so F( ) = 2−i+
2 − 2−
jis a conservative vector field by Theorem 6 and hence the line integral is independent of path. Thus a potential function exists, and ( ) = 2−
implies ( ) = 2−+ ()and ( ) = −2−+ 0(). But ( ) = 2 − 2−so
0() = 2 ⇒ () = 2+ . We can take = 0, so ( ) = 2−+ 2. Then
2− + (2 − 2−) = (2 1) − (1 0) = 4−1+ 1 − 1 = 4.
20. The functions sin and cos − sin have continuous first-order derivatives on R2and
(sin ) = cos =
( cos − sin ), so F( ) = sin i + ( cos − sin ) j is a conservative vector field by Theorem 6 and hence the line integral is independent of path. Thus a potential function exists, and ( ) = sin implies
( ) = sin + ()and ( ) = cos + 0(). But ( ) = cos − sin so
0() = − sin ⇒ () = cos + . We can take = 0, so ( ) = sin + cos . Then
sin + ( cos − sin ) = (1 ) − (2 0) = −1 − 1 = −2.
21. If F is conservative, then
F· r is independent of path. This means that the work done along all piecewise-smooth curves that have the described initial and terminal points is the same. Your reply: It doesn’t matter which curve is chosen.
22. The curves 1and 2connect the same two points but
1F· r 6=
2F· r. Thus F is not independent of path, and therefore is not conservative.
23. F( ) = 3i+ 3j, =
F· r. Since (3) = 0 = (3), there exists a function such that ∇ = F. In fact, ( ) = 3 ⇒ ( ) = 144+ () ⇒ ( ) = 0 + 0(). But ( ) = 3so
0() = 3 ⇒ () = 144+ . We can take = 0 ⇒ ( ) = 144+144. Thus
=
F· r = (2 2) − (1 0) = (4 + 4) −1 4 + 0
= 314. 24. F( ) = (2 + ) i + j, =
F· r. Since (2 + ) = 1 = (), there exists a function such that
∇ = F. In fact, ( ) = 2 + ⇒ ( ) = 2+ + () ⇒ ( ) = + 0(). But ( ) = so 0() = 0 ⇒ () = . We can take = 0 ⇒ ( ) = 2+ . Thus
=
F· r = (4 3) − (1 1) = (16 + 12) − (1 + 1) = 26.
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652 ¤ CHAPTER 16 VECTOR CALCULUS
25. We know that if the vector field (call it F) is conservative, then around any closed path ,
F· r = 0. But take to be a circle centered at the origin, oriented counterclockwise. All of the field vectors that start on are roughly in the direction of motion along , so the integral around will be positive. Therefore the field is not conservative.
26. If a vector field F is conservative, then around any closed path ,
F· r = 0. For any closed path we draw in the field, it appears that some vectors on the curve point in approximately the same direction as the curve and a similar number point in roughly the opposite direction. (Some appear perpendicular to the curve as well.) Therefore it is plausible that
F· r = 0 for every closed curve which means F is conservative.
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. = {( ) | 0 3} consists of those points between, but not on, the horizontal lines = 0 and = 3.
(a) Since 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.”)
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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. =
( ) | 2consists 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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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+ 12+ 12)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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1
SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS ¤ 653 (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. = {( ) | 1 || 2} consists of those points between, but not on, the vertical lines = 1 and = 2, together with the points between the vertical lines = −1 and = −2.
(a) The region does not include any of its boundary points, so it is open.
(b) consists of two separate pieces, so it is not connected. [For instance, both the points (−15 0) and (15 0) lie in but they cannot be joined by a path that lies entirely in .]
(c) Because is not connected, it’s not simply-connected.
33. =
( ) | 1 ≤ 2+ 2≤ 4 ≥ 0
is the semiannular region in the upper half-plane between circles centered at the origin of radii 1 and 2 (including all boundary points).
(a) includes boundary points, so it is not open. [Note that at any boundary point, (1 0) for instance, any disk centered there cannot lie entirely in .]
(b) The region consists of one piece, so it’s connected.
(c) is connected and has no holes, so it’s simply-connected.
34. = {( ) | ( ) 6= (2 3)} consists of all points in the -plane except for (2 3).
(a) has only one boundary point, namely (2 3), which is not included, so the region is open.
(b) is connected, as it consists of only one piece.
(c) is not simply-connected, as it has a hole at (2 3). Thus any simple closed curve that encloses (2 3) lies in but includes a point that is 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.
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654 ¤ CHAPTER 16 VECTOR CALCULUS
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+ 12)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.
16.4 Green's Theorem
1. (a) 1: = ⇒ = = 0 ⇒ = 0 0 ≤ ≤ 5.
2: = 5 ⇒ = 0 = ⇒ = 0 ≤ ≤ 4.
3: = 5 − ⇒ = − = 4 ⇒ = 0 0 ≤ ≤ 5.
4: = 0 ⇒ = 0 = 4 − ⇒ = − 0 ≤ ≤ 4
Thus
2 + 2 =
1+ 2+ 3+ 4
2 + 2 =5
0 0 +4
0 25 +5
0(−16 + 0) +4 0 0
= 0 +25 224
0+ [−16]50+ 0 = 200 + (−80) = 120
(b) Note that as given in part (a) is a positively oriented, piecewise-smooth, simple closed curve. Then by Green’s Theorem,
2 + 2 =
(2) − (2)
=5 0
4
0(2 − 2) =5 0
2− 2=4
=0
=5
0(16 − 16) =
82− 165
0= 200 − 80 = 120
2. (a) Parametric equations for are = 4 cos , = 4 sin , 0 ≤ ≤ 2. Then = −4 sin , = 4 cos and
− =2
0 [(4 sin )(−4 sin ) − (4 cos )(4 cos )]
= −162
0 (sin2 + cos2) = −162
0 1 = −16(2) = −32
(b) Note that as given in part (a) is a positively oriented, smooth, simple closed curve. Then by Green’s Theorem,
− =
(−) − ()
=
(−1 − 1) = −2
= −2(area of ) = −2 · (4)2= −32
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