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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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Text of Section 16.4 Green’s mathcal/download/108/HW/16.4.pdf Section 16.4 Green’s Theorem...

  • 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