Section 16.4 Green’s Theorem

Section 16.4 Green’s Theorem

10. Use Greens Theorem to evaluate the line integral along the given positively oriented curve. R_Cy⁴dx + 2xy³dy, C is the ellipse x²+ 2y²= 2.

Solution:

1638 ¤ CHAPTER 16 VECTOR CALCULUS

8. 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 9. 



 + ^√

 + (2 + cos ²)  =





(2 + cos ²) −^

 + ^√



=1 0

^√

² (2 − 1)   =1 0(√

 − ²)  =

2

3^32− ¹3³1 0= ¹₃ 10. 

 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.

11. 

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

12. 

(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 

13. The region  enclosed by  is given by {( ) | 1 ≤  ≤ 2, 0 ≤  ≤ 2} (in polar coordinates),which is traversed counterclockwise, so  has positive orientation. Thus,





3 + ^2

 +

tan⁻¹ + 3²

 =



 

(tan⁻¹ + 3²) −^(3 + ^2)



=

6  = 62 0

2

1  cos     

Switching to polar coordinates

= 62

0 cos  2

1 ² = 6

sin =2

=0

_3

3

=2

=1

= 6(1 − 0)₈

3−¹3

= 14

14. The region  enclosed by  is given by {( ) | ²≤  ≤ 4, 0 ≤  ≤ 2}.  is traversed clockwise, so − gives the positive orientation. Then

−

−(^23+ ²)  + (^43− ²)  = −



 

(^43− ²) −^(^23+ ²)



= −2 0

4

²(−2 − 2)   =2 0

4

²(2 + 2)  

=2 0

²+ 2=4

=² =2

0(16 + 8 − ⁴− 2³) 

=

16 + 4²−^5⁵ −^2⁴

2

0= 16(2) + 4(2²) −²5⁵ −²2⁴ − 0 = ¹⁶⁸5

° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

18. Use Greens Theorem to evaluate R_CF·dr. (Check the orientation of the curve before applying the theorem.) F(x, y) =√

x²+ 1, tan⁻¹x, C is the triangle from (0, 0) to (1, 1) to (0, 1) to (0, 0).

Solution: SECTION 16.4 GREEN’S THEOREM ¤ 657

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

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 .

° 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.c

25. (a) If C is the line segment connecting the point (x1, y1) to the point (x2, y2), show that Z

C

xdy − ydx = x1y²− x2y1

(b) If the vertices of a polygon, in counterclockwise order, are (x1, y1), (x2, y2),. . ., (xn, yn) show that the area of the polygon is

A =1

2[(x₁y₂− x2y₁) + (x₂y₃− x3y₂) + · · · + (x_n−1y_n− xny_n−1) + (x_ny₁− x1y_n)]

(c) Find the area of the pentagon with vertices (0, 0), (2, 1), (1, 3), (0, 2), and (−1, 1).

Solution:

1

658 ¤ CHAPTER 16 VECTOR CALCULUS

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



  = .

° 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.c

31. Use the method of Example 5 to calculate R

CF·dr, where

F(x, y) = 2xy i + (y²− x²) j (x²+ y²)²

and C is any positively oriented simple closed curve that encloses the origin.

Solution:

SECTION 16.4 GREEN’S THEOREM ¤ 659

23. We orient the quarter-circular region as shown in the figure.

 = ¹₄²so  = 1

²2





²and  = − 1

²2





².

Here  = 1+ 2+ 3where 1:  = ,  = 0, 0 ≤  ≤ ;

2:  =  cos ,  =  sin , 0 ≤  ≤ ^2; and

3:  = 0,  =  − , 0 ≤  ≤ . Then



² =

₁² +

₂² +

₃² =

0 0  +2

0 ( cos )²( cos )  + 0 0 

=2

0 ³cos³  = ³2

0 (1 − sin²) cos   = ³

sin  −¹3sin³2 0 = ²₃³ so  = 1

²2





² = 4

3.



² =

₁² +

₂² +

₃² =

0 0  +2

0 ( sin )²(− sin )  + 0 0 

=2

0 (−³sin³)  = −³2

0 (1 − cos²) sin   = −³1

3cos³ − cos 2

0 = −²3³, so  = − 1

²2





² = 4

3. Thus ( ) =

4

34

3

 .

24. Here  = ¹₂and  = 1+ 2+ 3, where 1:  = ,  = 0, 0 ≤  ≤ ;

2:  = ,  = , 0 ≤  ≤ ; and ³:  = ,  = _^,  =  to  = 0. Then



² =

₁² +

₂² +

₃² = 0 +

0 ² +0

(²)



= ² +^_1 3³0

= ² −¹3² = ²₃²

Similarly,

² =

₁² +

₂² +

₃² = 0 + 0 +0





²

 = _^22 ·¹3³0

= −¹3². Thus

 = _2¹ 

² = _¹ · ²3² = ²₃and  = −2¹



² = −¹

−¹3²

= ¹₃, so ( ) =2 3¹₃. 25. By Green’s Theorem, −¹₃

³ = −¹3

(−3²)  =

²  = and

1 3

³ = ¹₃

(3²)  =

²  = .

26. By symmetry the moments of inertia about any two diameters are equal. Centering the disk at the origin, the moment of inertia about a diameter equals

= ¹₃

³ = ¹₃2

0 (⁴cos⁴)  = ¹₃⁴2

0

3

8+¹₂cos 2 +¹₈cos 4

 =¹₃⁴ · ^3(2)8 = ¹₄⁴ 27. As in Example 5, let ⁰be a counterclockwise-oriented circle with center the origin and radius , where  is chosen to

be small enough so that ⁰lies inside , and  the region bounded by  and ⁰. Here

 = 2

(²+ ²)² ⇒ 

 = 2(²+ ²)²− 2 · 2(²+ ²) · 2

(²+ ²)⁴ = 2³− 6² (²+ ²)³ and

 = ²− ²

(²+ ²)² ⇒ 

 = −2(²+ ²)²− (²− ²) · 2(²+ ²) · 2

(²+ ²)⁴ = 2³− 6²

(²+ ²)³. Thus, as in the example,

° 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.c

660 ¤ CHAPTER 16 VECTOR CALCULUS





  +   +



−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 =







 −





 =



(7 − 3)  = 4



 = 4 · () = 4 · 5 = 20

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] = ±



()

() 

° 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.c

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