Section 16.4 Green’s Theorem

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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 

  + ²³ = 

₁+ ₂+ ₃

  + ²³

=1

0 0  +2

0 ³ +1 0

−(1 − )(2 − 2) − 2(1 − )²(2 − 2)³



= 0 +₁

4⁴2 0+1

0

−2(1 − )²− 16(1 − )⁵



= 4 +2

3(1 − )³+⁸₃(1 − )⁶1

0= 4 + 0 −¹⁰3 =²₃ (b)

  + ²³ =





(²³) −^ ()

 =1 0

2

0 (2³− )  

=1 0

1

2⁴− =2

=0  =1

0(8⁵− 2²)  = ⁴₃−²3 = ²₃

4. (a) 1:  =  ⇒  = ,  = ² ⇒  = 2 , 0 ≤  ≤ 1

2:  = 1 −  ⇒  = −,  = 1 ⇒  = 0 , 0 ≤  ≤ 1

3:  = 0 ⇒  = 0 ,  = 1 −  ⇒  = −, 0 ≤  ≤ 1

Thus



²² +   = 

₁+₂+₃

²² +  

=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+ 2⁴

 +1 0

−1 + 2 − ²

 +1 0 0 

=₁

7⁷+²₅⁵1 0+

− + ²−¹3³1

0+ 0 =₁

7+²₅ +

−1 + 1 −¹3

=₁₀₅²²

(b)

²² +   =





() −^ (²²)

 =1 0

1

²( − 2²)  

=1 0

₁

2²− ²²=1

=² =1 0

₁

2 − ²−¹2⁴+ ⁶



=1

2 −¹3³−10¹⁵+¹₇⁷1

0= ¹₂−¹3−10¹ +¹₇ = ₁₀₅²² 5.The region  enclosed by  is [0 3] × [0 4], so



^ + 2^ =





(2^) − ^ (^)

 =3 0

4

0 (2^− ^)  

=3

0 ^4 0  =

^3 0

4

0= (³− ⁰)(4 − 0) = 4(³− 1)

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(4²− 4²)  =1

0 0  = 0 7. 





 + ^√^

 + (2 + cos ²)  =





(2 + cos ²) −^



 + ^√^



=1 0

^√

² (2 − 1)   =1 0(√

 − ²)  =

2

3^32−¹3³1 0= ¹₃ 8. 

 ⁴ + 2³ =





(2³) −^ (⁴)

 =

(2³− 4³) 

= −2

³ = 0

because ( ) = ³is an odd function with respect to  and  is symmetric about the -axis.

9. 

³ − ³ =





(−³) −^ (³)

 =

(−3²− 3²)  =2

0

2

0(−3²)   

= −32

0 2

0 ³ = −3

2

0

1 4⁴2

0= −3(2)(4) = −24

10. 

(1 − ³)  + (³+ ^²)  =





(³+ ^²) −^ (1 − ³)

 =

(3²+ 3²) 

=2

0

3

2 (3²)    = 32

0 3 2 ³

= 3

2

0

₁

4⁴3

2= 3(2) ·¹4(81 − 16) = ¹⁹⁵2  11. F( ) = h cos  −  sin   +  cos i 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

₁

2²=4−2

=0  = −2 0

1

2(4 − 2)² = −2

0(8 − 8 + 2²)  = −

8 − 4²+²₃³2 0

= −

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

−2

2 cos  −¹2(1 + cos 2)



= −

2 sin  + 2 cos  −¹2

 +¹₂sin 22

−2 [integrate by parts in the ﬁrst term]

= −

 −¹4 −  −¹4

= ¹₂

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(4²− 4²)  =1

0 0  = 0 7. 





 + ^√^

 + (2 + cos ²)  =





(2 + cos ²) −^



 + ^√^



=1 0

^√

² (2 − 1)   =1 0(√

 − ²)  =

2

3^32−¹3³1 0= ¹₃ 8. 

 ⁴ + 2³ =





(2³) −^ (⁴)

 =

(2³− 4³) 

= −2

³ = 0

because ( ) = ³is an odd function with respect to  and  is symmetric about the -axis.

9. 

³ − ³ =





(−³) −^ (³)

 =

(−3²− 3²)  =2

0

2

0(−3²)   

= −32

0 2

0 ³ = −3

2

0

₁

4⁴2

0= −3(2)(4) = −24

10. 

(1 − ³)  + (³+ ^²)  =





(³+ ^²) −^ (1 − ³)

 =

(3²+ 3²) 

=2

0

3

2 (3²)    = 32

0 3 2 ³

= 3

2

0

1 4⁴3

2= 3(2) ·¹4(81 − 16) = ¹⁹⁵2  11. F( ) = h cos  −  sin   +  cos i 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

1

2²=4−2

=0  = −2 0

1

2(4 − 2)² = −2

0(8 − 8 + 2²)  = −

8 − 4²+²₃³2 0

= −

16 − 16 +¹⁶3 − 0

= −¹⁶3

12. F( ) =

^−+ ² ^−+ ²

and the region  enclosed by  is given by {( ) | −2 ≤  ≤ 2 0 ≤  ≤ cos }.



F· r = −

−

^−+ ²

 +

^−+ ²

 = −







^−+ ²

−^

^−+ ²



= −2

−2

cos 

0 (2 − 2)   = −2

−2

2 − ²=cos 

=0 

= −2

−2(2 cos  − cos²)  = −2

−2

2 cos  −¹2(1 + cos 2)



= −

2 sin  + 2 cos  −¹2

 +¹₂sin 22

−2 [integrate by parts in the ﬁrst term]

= −

 −¹4 −  −¹4

= ¹₂

1

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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).



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)

1 0

=  4 −1

2ln 2 15. Here  = 1+ 2where

1can be parametrized as  = ,  = 0, −2 ≤  ≤ 2, and

2is given by  = −,  = cos , −2 ≤  ≤ 2.

Then the line integral is



₁+₂

³⁴ + ⁵⁴ =2

−2(0 + 0)  +2

−2[(−)³(cos )⁴(−1) + (−)⁵(cos )⁴(− sin )] 

= 0 +2

−2(³cos⁴ + ⁵cos⁴ sin )  = ₁₅¹⁴−⁴¹⁴⁴1125²+^7,578,368_253,125 ≈ 00779 according to a CAS. The double integral is







 −





 =

 2

−2

 cos  0

(5⁴⁴− 4³³)   = ₁₅¹⁴−⁴¹⁴⁴1125²+^7,578,368_253,125 ≈ 00779, 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− 4²

−√

4− 4²

(3²⁸+ 5³⁴)   = 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

₁

3³−  = 1−

 = 0  =1 0

₁

3(1 − )³− (1 − )



=

−12¹(1 − )⁴−¹2²+¹₃³1 0=

−¹2+¹₃

−

−12¹

= −12¹

658 ¤ CHAPTER 16 VECTOR CALCULUS 18. By Green’s Theorem,  =

F· r =

sin   +

sin  + ²+¹₃³

 =

(²+ ²− 0) , where

is the region (a quarter-disk) bounded by . Converting to polar coordinates, we have

 =2 0

5

0 ²·    =

2 0

₁

4⁴5

0=¹₂₆₂₅

4

=⁶²⁵₈ .

19. Let 1be the arch of the cycloid from (0 0) to (2 0), which corresponds to 0 ≤  ≤ 2, and let ²be the segment from (2 0)to (0 0), so 2is given by  = 2 − ,  = 0, 0 ≤  ≤ 2. Then  = ¹∪ ²is 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

0 0 (−)

=2

0 (1 − 2 cos  + cos²)  + 0 =

 − 2 sin  +¹2 +¹₄sin 22

0 = 3

20.  =

  =2

0 (5 cos  − cos 5)(5 cos  − 5 cos 5) 

=2

0 (25 cos² − 30 cos  cos 5 + 5 cos²5) 

= 25₁

2 +¹₄sin 2

− 30₁

8sin 4 +₁₂¹ sin 6 + 5₁

2 +₂₀¹ sin 102

0

[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 (12− ²1)  = 12− ²1

(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),

1 2



  −   =

, where  is the polygon bounded by . Therefore area of polygon = () =

  = ¹₂

  −  

=¹₂

₁  −   +

₂  −   + · · · +

_−1  −   +

_  −   To evaluate these integrals we use the formula from (a) to get

() =¹₂[(12− ²1) + (23− ³2) + · · · + (^−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) =⁹₂ 22. By Green’s Theorem, _2¹ 

² = _2¹ 

2  = _¹ 

  = and

−2¹



² = −2¹



(−2)  = ¹



  = .

658 ¤ CHAPTER 16 VECTOR CALCULUS 18. By Green’s Theorem,  =

F· r =

sin   +

sin  + ²+¹₃³

 =

(²+ ²− 0) , where

is the region (a quarter-disk) bounded by . Converting to polar coordinates, we have

 =2 0

5

0 ²·    =

2 0

₁

4⁴5

0=¹₂₆₂₅

4

=⁶²⁵₈ .

19. Let 1be the arch of the cycloid from (0 0) to (2 0), which corresponds to 0 ≤  ≤ 2, and let ²be the segment from (2 0)to (0 0), so 2is given by  = 2 − ,  = 0, 0 ≤  ≤ 2. Then  = ¹∪ ²is 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

0 0 (−)

=2

0 (1 − 2 cos  + cos²)  + 0 =

 − 2 sin  +¹2 +¹₄sin 22

0 = 3

20.  =

  =2

0 (5 cos  − cos 5)(5 cos  − 5 cos 5) 

=2

0 (25 cos² − 30 cos  cos 5 + 5 cos²5) 

= 25₁

2 +¹₄sin 2

− 30₁

8sin 4 +₁₂¹ sin 6 + 5₁

2 +₂₀¹ sin 102

0

[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 (12− ²1)  = 12− ²1

(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),

1 2



  −   =

, where  is the polygon bounded by . Therefore area of polygon = () =

  = ¹₂

  −  

=¹₂

₁  −   +

₂  −   + · · · +

_−1  −   +

_  −   To evaluate these integrals we use the formula from (a) to get

() =¹₂[(12− ²1) + (23− ³2) + · · · + (^−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) =⁹₂ 22. By Green’s Theorem, _2¹ 

² = _2¹ 

2  = _¹ 

  = and

−2¹



² = −2¹



(−2)  = ¹



  = .





  +   +



−⁰

  +   =







 −





 =





0  = 0 and

F· r =

0F· r. We parametrize ⁰as r() =  cos  i +  sin  j, 0 ≤  ≤ 2. Then





F· r =



⁰

F· r =

 2

0

2 ( cos ) ( sin ) i +

²sin² − ²cos²

 j

²cos² + ²sin²2 ·



−  sin  i +  cos  j





= 1



 2

0

− cos  sin² − cos³

 =1



2

0

− cos  sin² − cos 

1 − sin²



= −1



 2

0

cos   = −1

sin 

2

0

= 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),  = , so



F· r =

0  = 0.

30. We express  as a type II region:  = {( ) | ¹() ≤  ≤ ²(),  ≤  ≤ } where ¹and 2are continuous functions.

Then







  =

 



 ₂()

₁()



   =

 



[(2() ) − (¹() )] by the Fundamental Theorem of

Calculus. But referring to the ﬁgure,

  = 

₁+ ₂+ ₃+ ₄

 .

Then

₁  =

(1() ) ,

₂  =

₄  = 0, and

₃  =

 (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] = ±



()

() 





  +   +



−⁰

  +   =







 −





 =





0  = 0 and

F· r =

⁰F· r. We parametrize ⁰as r() =  cos  i +  sin  j, 0 ≤  ≤ 2. Then





F· r =



0

F· r =

 2

0

2 ( cos ) ( sin ) i +

²sin² − ²cos²

 j

²cos² + ²sin²2 ·



−  sin  i +  cos  j





= 1



 2

0

− cos  sin² − cos³

 =1



2

0

− cos  sin² − cos 

1 − sin²



= −1



 2

0 cos   = −1

sin 

2

0

= 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),  = , so



F· r =

0  = 0.

30. We express  as a type II region:  = {( ) | ¹() ≤  ≤ ²(),  ≤  ≤ } where ¹and 2are continuous functions.

Then







  =

 



 ₂()

₁()



   =

 



[(2() ) − (¹() )] by the Fundamental Theorem of Calculus. But referring to the ﬁgure,

  = 

1+ 2+ 3+ 4

 .

Then

₁  =

(1() ) ,

₂  =

₄  = 0, and

3  =

 (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] = ±



()

() 

2

(3)

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) curl F = ∇ × F =





i j k

  

²² ²² ²²





=

 

(²²) − 

(²²)

 i−

 

(²²) − 

(²²)

 j+

 

(²²) − 

(²²)

 k

= (2² − 2²) i − (2² − 2²) j + (2²− 2²) k = 0

(b) div F = ∇ · F = 

(²²) + 

(²²) + 

(²²) = ²²+ ²²+ ²²

2. (a) curl F = ∇ × F =





i j k

  

0 ³² ⁴³





=

 

(⁴³) − 

(³²)

 i−

 

(⁴³) − 

(0)

 j+

 

(³²) − 

(0)

 k

= (4³³− 2³) i − (0 − 0) j + (3²²− 0) k = (4³³− 2³) i + 3²²k

(b) div F = ∇ · F = 

(0) + 

(³²) + 

(⁴³) = 0 + ³²+ 3⁴²= ³²+ 3⁴²

3. (a) curl F = ∇ × F =





i j k

  

^ 0 ^





= (^− 0) i − (^− ^) j + (0 − ^) k

= ^i+ (^− ^) j − ^k

(b) div F = ∇ · F = 

(^) + 

(0) + 

(^) = ^+ 0 + ^= (^+ ^)

4. (a) curl F = ∇ × 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) div F = ∇ · F = 

(sin ) + 

(sin ) + 

(sin ) = 0 + 0 + 0 = 0