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Section 16.4 Green’s Theorem
SECTION 16.4 GREEN’S THEOREM ¤ 655
3. (a) 1: = ⇒ = , = 0 ⇒ = 0 , 0 ≤ ≤ 1.
2: = 1 ⇒ = 0 , = ⇒ = , 0 ≤ ≤ 2.
3: = 1− ⇒ = −, = 2− 2 ⇒ = −2 , 0 ≤ ≤ 1.
Thus + 23 =
1 +2 +3
+ 23
= 1
00 +
2
03 +
1
0
−(1− )(2− 2)− 2(1− )2(2− 2)3
= 0 +
14420
+ 1
0
−2(1− )2 − 16(1− )5
= 4 +
23(1− )3 + 8
3(1− )6
10
= 4 + 0− 103
= 23
(b) + 23 =
(23)−
() =
1
0
2
0(23 − )
= 1
0
124 −
=2
=0 =
1
0(85 − 22) = 4
3− 2
3= 2
3
4. (a) 1: = ⇒ = , = 2 ⇒ = 2 , 0 ≤ ≤ 1
2: = 1− ⇒ = −, = 1 ⇒ = 0 , 0 ≤ ≤ 1
3: = 0 ⇒ = 0 , = 1− ⇒ = −, 0 ≤ ≤ 1
Thus22 + =
1+2+3
22 +
= 1
0
2(2)2 + (2)(2 )
+ 1
0
(1− )2(1)2(−) + (1− )(1)(0 )
+ 1
0
(0)2(1− )2(0 ) + (0)(1− )(−)
= 1
0
6 + 24
+
1
0
−1 + 2− 2 +
1
00
=
177 + 2
5510+− + 2 − 1
3310+ 0 =
17
+ 25
+−1 + 1− 1
3
= 22
105
(b)22 + =
()−
(22) =
1
0
1
2( − 22)
= 1
0
122 − 22
=1
=2 =
1
0
12− 2 − 1
24 + 6
=
12− 1
33 − 1
105 + 1
7710
= 12− 1
3− 1
10+ 1
7= 22
105
5. The region enclosed by is [0 3]× [0 4], so
+ 2 =
(2)−
() =
3
0
4
0(2 − )
= 3
0
4
0 =
30
40
= (3 − 0)(4− 0) = 4(3 − 1)
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
656 ¤ CHAPTER 16 VECTOR CALCULUS
6. The region enclosed by is given by {( ) | 0 ≤ ≤ 1 0 ≤ ≤ 2}, so(2 + 2) + (2 − 2) =
(2 − 2)−
(2 + 2)
= 1
0
2
0(2− 2)
= 1
0
2 − 2
=2
=0
= 1
0(42 − 42) =
1
00 = 0
7.
+
√+ (2+ cos 2) =
(2+ cos 2)−
+
√
= 1
0
√2
(2− 1) = 1
0(√− 2) =
2332 − 1
3310
= 13
8.4 + 23 =
(23)−
(4) =
(23 − 43)
= −2
3 = 0
because ( ) = 3 is an odd function with respect to and is symmetric about the -axis.
9.3 − 3 =
(−3)−
(3) =
(−32 − 32) = 2
0
2
0(−32)
= −3 2
0 2
03 = −3
20
14420
= −3(2)(4) = −24
10.(1− 3) + (3 +
2
) =
(3 + 2
)−
(1− 3) =
(32 + 32)
= 2
0
3
2(32) = 3
2
0 3
23
= 320
14432
= 3(2) · 14(81− 16) = 195
2
11. F( ) = h cos− sin + cosi and the region enclosed by is given by
{( ) | 0 ≤ ≤ 2 0 ≤ ≤ 4− 2}. is traversed clockwise, so − gives the positive orientation.
F · r = − −( cos− sin) + ( + cos) = −
( + cos)−
( cos− sin)
= −
( − sin+ cos− cos+ sin) = − 2
0
4−2
0
= − 2
0
122=4−2
=0 = − 2
0
12(4− 2)2 = − 2
0(8− 8+ 22) = − 8− 42 + 2
3320
= − 16− 16 + 163− 0
= − 163
12. F( ) =− + 2 − + 2
and the region enclosed by is given by {( ) | −2 ≤ ≤ 2 0 ≤ ≤ cos}.
is traversed clockwise, so − gives the positive orientation.
F · r = − − − + 2+
− + 2
= −
− + 2
−
− + 2
= − 2−2 cos
0(2− 2) = − 2−2
2 − 2
=cos
=0
= − 2−2(2 cos− cos2 ) = − 2−22 cos− 1
2(1 + cos 2)
= − 2 sin+ 2cos− 12
+ 1
2sin 2
2−2 [integrate by parts in the first term]
= − − 14 − − 1
4
= 12
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
656 ¤ CHAPTER 16 VECTOR CALCULUS
8. The region enclosed by is [0 5]× [0 2], so
cos + 2 sin =
(2 sin )−
(cos ) =
5
0
2
0[2 sin − (− sin )]
= 5
0(2+ 1)
2
0sin =
2 +
50
− cos 20
= 30(1− cos 2)
9.3 − 3 =
(−3)−
(3) =
(−32 − 32) = 2
0
2
0(−32)
= −3 2
0 2
03 = −3
20
14420
= −3(2)(4) = −24
10.−2 + (4 + 222) =
(4 + 222)−
(−2) =
(43 + 42 − 0)
= 4
(2 + 2) = 4
2
0
2
1( cos )(2)
= 4 2
0cos
2
14 = 4
sin
20
15521
= 0
11. F( ) = h cos− sin + cosi and the region enclosed by is given by
{( ) | 0 ≤ ≤ 2 0 ≤ ≤ 4− 2}. is traversed clockwise, so − gives the positive orientation.
F · r = − −( cos− sin) + ( + cos) = −
( + cos)−
( cos− sin)
= −
( − sin+ cos− cos+ sin) = − 2
0
4−2
0
= − 2
0
122=4−2
=0 = − 2
0
12(4− 2)2 = − 2
0(8− 8+ 22) = − 8− 42 + 2
3320
= − 16− 16 + 163− 0
= − 163
12. F( ) =− + 2 − + 2
and the region enclosed by is given by {( ) | −2 ≤ ≤ 2 0 ≤ ≤ cos}.
is traversed clockwise, so − gives the positive orientation.
F · r = − − − + 2+
− + 2
= −
− + 2
−
− + 2
= − 2−2 cos
0(2− 2) = − 2−2
2 − 2
=cos
=0
= − 2−2(2 cos− cos2 ) = − 2−22 cos− 1
2(1 + cos 2)
= − 2 sin+ 2cos− 12
+ 1
2sin 2
2−2 [integrate by parts in the first term]
= − − 14 − − 1
4
= 12
13. F( ) = h − cos sin i and the region enclosed by is the disk with radius 2 centered at (3−4).
is traversed clockwise, so − gives the positive orientation.
F · r = − −( − cos ) + ( sin ) = −
( sin )−
( − cos )
= −
(sin − 1− sin ) =
= area of = (2)2 = 4
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
656 ¤ CHAPTER 16 VECTOR CALCULUS
6. The region enclosed by is given by {( ) | 0 ≤ ≤ 1 0 ≤ ≤ 2}, so(2 + 2) + (2 − 2) =
(2 − 2)−
(2 + 2)
= 1
0
2
0(2− 2)
= 1
0
2 − 2
=2
=0
= 1
0(42 − 42) =
1
00 = 0
7.
+
√+ (2+ cos 2) =
(2+ cos 2)−
+
√
= 1
0
√2
(2− 1) = 1
0(√− 2) =
2332 − 1
3310
= 13
8.4 + 23 =
(23)−
(4) =
(23 − 43)
= −2
3 = 0
because ( ) = 3 is an odd function with respect to and is symmetric about the -axis.
9.3 − 3 =
(−3)−
(3) =
(−32 − 32) = 2
0
2
0(−32)
= −3 2
0 2
03 = −3
20
14420
= −3(2)(4) = −24
10.(1− 3) + (3 +
2
) =
(3 + 2
)−
(1− 3) =
(32 + 32)
= 2
0
3
2(32) = 3
2
0 3
23
= 320
14432
= 3(2) · 14(81− 16) = 195
2
11. F( ) = h cos− sin + cosi and the region enclosed by is given by
{( ) | 0 ≤ ≤ 2 0 ≤ ≤ 4− 2}. is traversed clockwise, so − gives the positive orientation.
F · r = − −( cos− sin) + ( + cos) = −
( + cos)−
( cos− sin)
= −
( − sin+ cos− cos+ sin) = − 2
0
4−2
0
= − 2
0
122=4−2
=0 = − 2
0
12(4− 2)2 = − 2
0(8− 8+ 22) = − 8− 42 + 2
3320
= − 16− 16 + 163− 0
= − 163
12. F( ) =− + 2 − + 2
and the region enclosed by is given by {( ) | −2 ≤ ≤ 2 0 ≤ ≤ cos}.
is traversed clockwise, so − gives the positive orientation.
F · r = − − − + 2+
− + 2
= −
− + 2
−
− + 2
= − 2−2 cos
0(2− 2) = − 2−2
2 − 2
=cos
=0
= − 2−2(2 cos− cos2 ) = − 2−22 cos− 1
2(1 + cos 2)
= − 2 sin+ 2cos− 12
+ 1
2sin 2
2−2 [integrate by parts in the first term]
= − − 14 − − 1
4
= 12
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
1
SECTION 16.4 GREEN’S THEOREM ¤ 657
13. F( ) = h − cos sin i and the region enclosed by is the disk with radius 2 centered at (3−4).
is traversed clockwise, so − gives the positive orientation.
F · r = − −( − cos ) + ( sin ) = −
( sin )−
( − cos )
= −
(sin − 1− sin ) =
= area of = (2)2 = 4
14. F( ) =√
2 + 1 tan−1 and the region enclosed by is given by {( ) | 0 ≤ ≤ 1 ≤ ≤ 1}.
is oriented positively, so
F · r =
√2 + 1 + tan−1 =
tan
−1−
(2 + 1)
=
1
0
1
1
1 + 2− 0
=
1
0
1
1 + 2
=1
= =
1
0
1
1 + 2(1− )
=
1
0
1
1 + 2−
1 + 2
=
tan
−1− 1
2ln(1 +
2)
10
=
4− 1
2ln 2
15. Here = 1 +2 where
1 can be parametrized as = , = 0, −2 ≤ ≤ 2, and
2 is given by = −, = cos , −2 ≤ ≤ 2.
Then the line integral is1+2
34 + 54 = 2−2(0 + 0) +
2−2[(−)3(cos )4(−1) + (−)5(cos )4(− sin )]
= 0 + 2−2(
3 cos4 + 5 cos4 sin ) = 1154 − 4144
11252 + 7,578,368
253,125 ≈ 00779
according to a CAS. The double integral is
−
=
2
−2
cos
0
(544 − 4
33) = 1
15
4 − 41441125
2+ 7,578,368
253,125 ≈ 00779, verifying Green’s
Theorem in this case.
16. We can parametrize as = cos , = 2 sin , 0 ≤ ≤ 2. Then the line integral is + =
2
0
2 cos − (cos )3(2 sin )5
(− sin ) +
2
0(cos )3(2 sin )8 · 2 cos
= 2
0(−2 cos sin + 32 cos3 sin6 + 512 cos4 sin8 ) = 7,
according to a CAS. The double integral is
−
=
1
−1
√4− 42
−√
4− 42(3
28+ 5
34) = 7.
17. By Green’s Theorem, =
F · r =(+ ) + 2 =
(2 − ) where is the path described in the
question and is the triangle bounded by . So
= 1
0
1−0
(2 − ) = 1
0
133 −
= 1−= 0
= 1
0
13(1− )3 − (1− )
=− 1
12(1− )4 − 1
22 + 1
3310
=− 1
2+ 1
3
− − 112
= − 1
12
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
658 ¤ CHAPTER 16 VECTOR CALCULUS
18. By Green’s Theorem, =
F · r =
sin +sin + 2 + 1
33 =
(2 + 2 − 0) , where
is the region (a quarter-disk) bounded by . Converting to polar coordinates, we have
= 20
5
02 · =
20
14450
= 12
6254
= 625
8.
19. Let 1 be the arch of the cycloid from (0 0) to (2 0), which corresponds to 0 ≤ ≤ 2, and let 2 be the segment from
(2 0) to (0 0), so 2 is given by = 2 − , = 0, 0 ≤ ≤ 2. Then = 1 ∪ 2 is traversed clockwise, so − is
oriented positively. Thus − encloses the area under one arch of the cycloid and from (5) we have
= − − =1
+2
= 2
0(1− cos )(1− cos ) +
2
00 (−)
= 2
0(1− 2 cos + cos2 ) + 0 =
− 2 sin + 1
2 + 1
4sin 2
20
= 3
20. = =
2
0(5 cos − cos 5)(5 cos − 5 cos 5)
= 2
0(25 cos2 − 30 cos cos 5+ 5cos2 5)
=25
12+ 1
4sin 2
− 30
18
sin 4 + 112
sin 6
+ 5
12+ 1
20sin 10
20
[Use Formula 80 in the Table of Integrals]
= 30
21. (a) Using Equation 16.2.8, we write parametric equations of the line segment as = (1− )1 + 2, = (1− )1 + 2,
0 ≤ ≤ 1. Then = (2 − 1) and = (2 − 1) , so − =
1
0[(1− )1 + 2](2 − 1) + [(1− )1 + 2](2 − 1)
= 1
0(1(2 − 1)− 1(2 − 1) + [(2 − 1)(2 − 1)− (2 − 1)(2 − 1)])
= 1
0(12 − 21) = 12 − 21
(b) We apply Green’s Theorem to the path = 1 ∪ 2 ∪ · · · ∪ , where is the line segment that joins ( ) to
(+1 +1) for = 1, 2, , − 1, and is the line segment that joins ( ) to (1 1). From (5),
12
− =
, where is the polygon bounded by . Therefore
area of polygon= () =
= 1
2
−
= 12
1
− +2
− + · · ·+ −1
− +
−
To evaluate these integrals we use the formula from (a) to get
() = 12[(12 − 21) + (23 − 32) + · · ·+ (−1 − −1) + (1 − 1)].
(c) = 12[(0 · 1− 2 · 0) + (2 · 3− 1 · 1) + (1 · 2− 0 · 3) + (0 · 1− (−1) · 2) + (−1 · 0− 0 · 1)]
= 12(0 + 5 + 2 + 2) = 9
2
22. By Green’s Theorem, 12
2 = 1
2
2 = 1
= and
− 12
2 = − 1
2
(−2) = 1
= .
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
658 ¤ CHAPTER 16 VECTOR CALCULUS
18. By Green’s Theorem, =
F · r =
sin +sin + 2 + 1
33 =
(2 + 2 − 0) , where
is the region (a quarter-disk) bounded by . Converting to polar coordinates, we have
= 20
5
02 · =
20
14450
= 12
6254
= 625
8.
19. Let 1 be the arch of the cycloid from (0 0) to (2 0), which corresponds to 0 ≤ ≤ 2, and let 2 be the segment from
(2 0) to (0 0), so 2 is given by = 2 − , = 0, 0 ≤ ≤ 2. Then = 1 ∪ 2 is traversed clockwise, so − is
oriented positively. Thus − encloses the area under one arch of the cycloid and from (5) we have
= − − =1
+2
= 2
0(1− cos )(1− cos ) +
2
00 (−)
= 2
0(1− 2 cos + cos2 ) + 0 =
− 2 sin + 1
2 + 1
4sin 2
20
= 3
20. = =
2
0(5 cos − cos 5)(5 cos − 5 cos 5)
= 2
0(25 cos2 − 30 cos cos 5+ 5cos2 5)
=25
12+ 1
4sin 2
− 30
18
sin 4 + 112
sin 6
+ 5
12+ 1
20sin 10
20
[Use Formula 80 in the Table of Integrals]
= 30
21. (a) Using Equation 16.2.8, we write parametric equations of the line segment as = (1− )1 + 2, = (1− )1 + 2,
0 ≤ ≤ 1. Then = (2 − 1) and = (2 − 1) , so − =
1
0[(1− )1 + 2](2 − 1) + [(1− )1 + 2](2 − 1)
= 1
0(1(2 − 1)− 1(2 − 1) + [(2 − 1)(2 − 1)− (2 − 1)(2 − 1)])
= 1
0(12 − 21) = 12 − 21
(b) We apply Green’s Theorem to the path = 1 ∪ 2 ∪ · · · ∪ , where is the line segment that joins ( ) to
(+1 +1) for = 1, 2, , − 1, and is the line segment that joins ( ) to (1 1). From (5),
12
− =
, where is the polygon bounded by . Therefore
area of polygon= () =
= 1
2
−
= 12
1
− +2
− + · · ·+ −1
− +
−
To evaluate these integrals we use the formula from (a) to get
() = 12[(12 − 21) + (23 − 32) + · · ·+ (−1 − −1) + (1 − 1)].
(c) = 12[(0 · 1− 2 · 0) + (2 · 3− 1 · 1) + (1 · 2− 0 · 3) + (0 · 1− (−1) · 2) + (−1 · 0− 0 · 1)]
= 12(0 + 5 + 2 + 2) = 9
2
22. By Green’s Theorem, 12
2 = 1
2
2 = 1
= and
− 12
2 = − 1
2
(−2) = 1
= .
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
660 ¤ CHAPTER 16 VECTOR CALCULUS
+ +
−0
+ =
−
=
0 = 0
and
F · r =0 F · r. We parametrize 0 as r() = cos i + sin j, 0 ≤ ≤ 2. Then
F · r =
0
F · r =
2
0
2 ( cos ) ( sin ) i +2 sin2 − 2 cos2
j
2 cos2 + 2 sin2 2 ·
− sin i + cos j
=1
2
0
− cos sin2− cos
3 =
1
2
0
− cos sin2− cos
1− sin
2
= −1
2
0
cos = −1
sin
20
= 0
28. and have continuous partial derivatives on R2, so by Green’s Theorem we have
F · r =
−
=
(3− 1) = 2
= 2 ·() = 2 · 6 = 12
29. Since is a simple closed path which doesn’t pass through or enclose the origin, there exists an open region that doesn’t
contain the origin but does contain. Thus = −(2 + 2) and = (2 + 2) have continuous partial derivatives on
this open region containing and we can apply Green’s Theorem. But by Exercise 16.3.35(a), = , soF · r =
0 = 0.
30. We express as a type II region: = {( ) | 1() ≤ ≤ 2(), ≤ ≤ } where 1 and 2 are continuous functions.
Then
=
2()
1()
=
[(2() )−(1() )] by the Fundamental Theorem of
Calculus. But referring to the figure, =
1 +2 +3 +4
.
Then1
= (1() ) ,
2
=4
= 0,
and3
= (2() ) . Hence
=
[(2() )−(1() ) ] =
() .
31. Using the first part of (5), we have that
= () =
. But = ( ), and =
+
,
and we orient by taking the positive direction to be that which corresponds, under the mapping, to the positive direction
along , so
=
( )
+
=
( )
+ ( )
= ±
( )
−
( )
[using Green’s Theorem in the -plane]
= ±
+ ( ) 2
−
− ( ) 2
[using the Chain Rule]
= ±
−
[by the equality of mixed partials] = ±
()
()
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
660 ¤ CHAPTER 16 VECTOR CALCULUS
+ +
−0
+ =
−
=
0 = 0
and
F · r =0 F · r. We parametrize 0 as r() = cos i + sin j, 0 ≤ ≤ 2. Then
F · r =
0
F · r =
2
0
2 ( cos ) ( sin ) i +2 sin2 − 2 cos2
j
2 cos2 + 2 sin2 2 ·
− sin i + cos j
=1
2
0
− cos sin2− cos
3 =
1
2
0
− cos sin2− cos
1− sin
2
= −1
2
0
cos = −1
sin
20
= 0
28. and have continuous partial derivatives on R2, so by Green’s Theorem we have
F · r =
−
=
(3− 1) = 2
= 2 ·() = 2 · 6 = 12
29. Since is a simple closed path which doesn’t pass through or enclose the origin, there exists an open region that doesn’t
contain the origin but does contain. Thus = −(2 + 2) and = (2 + 2) have continuous partial derivatives on
this open region containing and we can apply Green’s Theorem. But by Exercise 16.3.35(a), = , soF · r =
0 = 0.
30. We express as a type II region: = {( ) | 1() ≤ ≤ 2(), ≤ ≤ } where 1 and 2 are continuous functions.
Then
=
2()
1()
=
[(2() )−(1() )] by the Fundamental Theorem of
Calculus. But referring to the figure, =
1 +2 +3 +4
.
Then1
= (1() ) ,
2
=4
= 0,
and3
= (2() ) . Hence
=
[(2() )−(1() ) ] =
() .
31. Using the first part of (5), we have that
= () =
. But = ( ), and =
+
,
and we orient by taking the positive direction to be that which corresponds, under the mapping, to the positive direction
along , so
=
( )
+
=
( )
+ ( )
= ±
( )
−
( )
[using Green’s Theorem in the -plane]
= ±
+ ( ) 2
−
− ( ) 2
[using the Chain Rule]
= ±
−
[by the equality of mixed partials] = ±
()
()
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
2
SECTION 16.5 CURL AND DIVERGENCE ¤ 661
The sign is chosen to be positive if the orientation that we gave to corresponds to the usual positive orientation, and it is
negative otherwise. In either case, since () is positive, the sign chosen must be the same as the sign of ( )
( ).
Therefore () =
=
( )
( )
.16.5 Curl and Divergence
1. (a) curlF = ∇×F =
i j k
22 22 22
=
(22)−
(22)
i−
(22)−
(22)
j +
(22)−
(22)
k
= (22 − 22) i− (22 − 22) j + (22 − 22)k = 0
(b) divF = ∇ · F =
(22) +
(22) +
(22) = 22 + 22 + 22
2. (a) curlF = ∇×F =
i j k
0 32 43
=
(43)−
(32)
i−
(43)−
(0)
j +
(32)−
(0)
k
= (433 − 23) i− (0− 0) j + (322 − 0)k = (433 − 23) i + 322 k
(b) divF = ∇ · F =
(0) +
(32) +
(43) = 0 + 32 + 342 = 32 + 342
3. (a) curlF = ∇×F =
i j k
0
= ( − 0) i− ( − ) j + (0− )k
= i + ( − ) j− k
(b) divF = ∇ · F =
() +
(0) +
() = + 0 + = ( + )
4. (a) curlF = ∇×F =
i j k
sin sin sin
= ( cos − cos ) i− ( cos − cos ) j + ( cos − cos )k
= (cos − cos ) i + (cos − cos) j + (cos − cos )k
(b) divF = ∇ · F =
(sin ) +
(sin ) +
(sin) = 0 + 0 + 0 = 0
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3
