# Section 16.4 Green’s mathcal/download/108/HW/16.4.pdf Section 16.4 Green’s Theorem SECTION16.4GREEN STHEOREM¤ 655 3. D F 1 { = wig{ gw | =0g| 0 \$ 1 F 2 { =1 i g{ =0gw | =w i g|

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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 + 8

3 (1− )61

0 = 4 + 0− 10

3 = 2

3

(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 + 2

5 5 1 0 + − + 2 − 1

3 3 1 0 + 0 =

 1 7

+ 2 5

 + −1 + 1− 1

3

 = 22

105

(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 − 1

2 4 + 6

 

= 

1 2− 133 − 1105 + 177

1 0

= 12 − 13 − 110 + 17 = 22105

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)

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 (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 − 1

3 3 1 0

= 1 3

8.   4  + 23  =

 

  

(23)−  

(4)   =

 

(23 − 43) 

= −2   3  = 0

because ( ) = 3 is 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) · 1 4 (81− 16) = 195

2 

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 + 2

3 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+ 2cos− 12 + 12 sin 22−2 [integrate by parts in the first term] = −  − 1

4  −  − 1

4  

= 1 2 

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 0

sin   =  2 + 

5 0

− cos 2 0

= 30(1− cos 2)

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.   −2 + (4 + 222)  =

 

  

(4 + 222)−  

(−2)   =

 

(43 + 42 − 0) 

= 4 

 (2 + 2)  = 4

 2 0

 2 1

( cos )(2)   

= 4  2 0

cos    2 1 4  = 4

 sin 

2 0

 1 5 5 2 1

= 0

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20

= − 16− 16 + 16 3 − 0 = − 16

3

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