Section 15.2 Double Integrals over General Regions

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Section 15.2 Double Integrals over General Regions

532 ¤ CHAPTER 15 MULTIPLE INTEGRALS

15.2 Double Integrals over General Regions

1. 5 1



0(8 − 2)   =5 1

8 − ²=

=0 =5

1[8() − ()²− 8(0) + (0)²]

=5

1 7² = ⁷₃³5

1= ⁷₃(125 − 1) =⁸⁶⁸3

2. 2 0

²

0 ²   =2 0

1

3³=²

=0  =2 0 1 3

(²)³− (0)³



=2 0 1

3⁷ =¹₃1 8⁸2

0=¹₃(32 − 0) =³²3

3. 1 0



0 ^³  =1 0

1

2²^³=

=0 =1 0 1 2^³

()²− (0)²



=¹₂1

0 ²^³ =¹₂

1 3^³1

0=¹₂ ·¹3

¹− ⁰

=¹₆( − 1) 4. 2

0



0  sin    =2

0 [(− cos )]^==0 =2

0 (− cos  + )  =2

0 ( −  cos ) 

=₁

2²− ( sin  + cos )2

0 (by integrating by parts in the second term)

=

1

2 ·^4² −^2 − 0

− (0 − 0 − 1) = ^8²−^2 + 1

5. 1 0

²

0 cos(³)   =1 0

 cos(³)=²

=0  =1

0 ²cos(³)  = ¹₃sin(³)1

0=¹₃(sin 1 − sin 0) =¹3sin 1 6. 1

0

^ 0

√1 + ^  =1 0

√

1 + ^=^

=0  =1 0 ^√

1 + ^ = ²₃(1 + ^)^321 0

=²₃(1 + )^32−²3(1 + 1)^32=²₃(1 + )^32−⁴3

√2

7. 





²+ 1 =

 4 0

 ^√ 0



²+ 1  =

 4 0

 1

²+ 1·² 2

^=^√^

=0

 =1 2

 4 0



²+ 1

=¹₂

1

2ln²+ 14 0= ¹₄

ln

²+ 1 ⁴

0=¹₄(ln 17 − ln 1) = ¹4ln 17 8. 

(2 + )  =2 1

1

−1(2 + )   =2 1

²+ =1

=−1 =2 1

1 +  − ( − 1)²− ( − 1)



=2

1(−2²+ 4)  =

−²3³+ 2²2 1=

−¹⁶3 + 8

−

−²3 + 2

=⁴₃ 9. 

^−² =3 0



0 ^−²  =3 0



^−²=

=0 =3 0



^−²− 0

 =3

0 ^−²

= −¹2^−²3 0= −¹2

⁻⁹− ⁰

=¹₂

1 − ⁻⁹

10. 



²− ² =2 0

 0 

²− ²  =2 0

−¹3(²− ²)^32=

=0 =2 0

0 +¹₃(²)^32



=2 0

1

3³ = ¹₃·¹4⁴2

0= ₁₂¹ (16 − 0) =⁴3

11. (a) At the right we sketch an example of a region  that can be described as lying between the graphs of two continuous functions of  (a type I region) but not as lying between graphs of two continuous functions of  (a type II region). The regions shown in Figures 6 and 8 in the text are additional examples.

15.2 Double Integrals over General Regions

1. 5 1



0(8 − 2)   =5 1

8 − ²=

=0 =5

1[8() − ()²− 8(0) + (0)²]

=5

1 7² = ⁷₃³5

1= ⁷₃(125 − 1) =⁸⁶⁸3

2. 2 0

²

0 ²   =2 0

1

3³=²

=0  =2 0 1 3

(²)³− (0)³



=2 0 1

3⁷ =¹₃1 8⁸2

0=¹₃(32 − 0) =³²3

3. 1 0



0 ^³  =1 0

1

2²^³=

=0 =1 0 1 2^³

()²− (0)²



=¹₂1

0 ²^³ =¹₂

1 3^³1

0=¹₂ ·¹3

¹− ⁰

=¹₆( − 1)

4. 2 0



0  sin    =2

0 [(− cos )]^==0 =2

0 (− cos  + )  =2

0 ( −  cos ) 

=₁

2²− ( sin  + cos )2

0 (by integrating by parts in the second term)

=

1

2 ·^4² −^2 − 0

− (0 − 0 − 1) = ^8²−^2 + 1

5. 1 0

²

0 cos(³)   =1 0

 cos(³)=²

=0  =1

0 ²cos(³)  = ¹₃sin(³)1

0=¹₃(sin 1 − sin 0) =¹3sin 1 6. 1

0

^ 0

√1 + ^  =1 0

√

1 + ^=^

=0  =1 0 ^√

1 + ^ = ²₃(1 + ^)^321 0

=²₃(1 + )^32−²3(1 + 1)^32=²₃(1 + )^32−⁴3

√2

7.







²+ 1 =

 4 0

 ^√ 0



²+ 1  =

 4 0

 1

²+ 1·² 2

=√



=0

 =1 2

 4 0



²+ 1

=¹₂

1

2ln²+ 14 0= ¹₄

ln

²+ 1 ⁴

0=¹₄(ln 17 − ln 1) = ¹4ln 17 8. 

(2 + )  =2 1

1

−1(2 + )   =2 1

²+ =1

=−1 =2 1

1 +  − ( − 1)²− ( − 1)



=2

1(−2²+ 4)  =

−²3³+ 2²2 1=

−¹⁶3 + 8

−

−²3 + 2

=⁴₃ 9. 

^−² =3 0



0 ^−²  =3 0

^−²=

=0 =3 0

^−²− 0

 =3

0 ^−²

= −¹2^−²3 0= −¹2

⁻⁹− ⁰

=¹₂

1 − ⁻⁹

10. 



²− ² =2 0

 0 

²− ²  =2 0



−¹3(²− ²)^32=

=0 =2 0



0 +¹₃(²)^32



=2 0

1

3³ = ¹₃·¹4⁴2

0= ₁₂¹ (16 − 0) =⁴3

11. (a) At the right we sketch an example of a region  that can be described as lying between the graphs of two continuous functions of  (a type I region) but not as lying between graphs of two continuous functions of  (a type II region). The regions shown in Figures 6 and 8 in the text are additional examples.



  =3 0

3

²    =3 0

 ·¹2² = 3

 = ² =¹₂3

0 (9²− ⁴)  = ¹₂3

0(9³− ⁵) 

=¹₂

9 ·¹4⁴−¹6⁶3 0= ¹₂₉

4· 81 −¹6 · 729

− 0

= ²⁴³₈

or 

  =9 0

^√

3   =9 0

₁

2² = √

 = 3 = ¹₂9 0

 −¹9²

  = ¹₂9 0

²−¹9³



=¹₂₁

3³−¹9 ·¹4⁴9 0=¹₂₁

3 · 729 −36¹ · 6561

− 0

= ²⁴³₈

15. The curves  =  − 2 or  =  + 2 and  = ²intersect when  + 2 = ² ⇔

²−  − 2 = 0 ⇔ ( − 2)( + 1) = 0 ⇔  = −1,  = 2, so the points of intersection are (1 −1) and (4 2). If we describe  as a type I region, the upper boundary curve is  = √ but the lower boundary curve consists of two parts,

 = −√for 0 ≤  ≤ 1 and  =  − 2 for 1 ≤  ≤ 4.

Thus  = {( ) | 0 ≤  ≤ 1, −√

 ≤  ≤√

 } ∪ {( ) | 1 ≤  ≤ 4,  − 2 ≤  ≤√

 } and



  =1 0

^√

−√

   +4 1

^√

−2  . If we describe  as a type II region,  is enclosed by the left boundary

 = ²and the right boundary  =  + 2 for −1 ≤  ≤ 2, so  =

( ) | −1 ≤  ≤ 2, ²≤  ≤  + 2and



  =2

−1

+2

²   . In either case, the resulting iterated integrals are not difﬁcult to evaluate but the region  is more simply described as a type II region, giving one iterated integral rather than a sum of two, so we evaluate the latter integral:



  =2

−1

+2

²    =2

−1

 = +2

 = ²  =2

−1( + 2 − ²)  =2

−1(²+ 2 − ³) 

=₁

3³+ ²−¹4⁴2

−1=₈

3+ 4 − 4

−

−¹3+ 1 −¹4

=⁹₄

16. As a type I region,  = {( ) | 0 ≤  ≤ 4,  ≤  ≤ 4} and



²^ =4 0

4

²^ . As a type II region,

 = {( ) | 0 ≤  ≤ 4, 0 ≤  ≤ } and

²^ =4 0



0 ²^ .

Evaluating

²^requires integration by parts whereas

²^does not, so the iterated integral corresponding to  as a type II region appears easier to evaluate.



²^ =4 0



0 ²^  =4 0

^=

=0 =4 0



^²− 



=

1

2^²−¹2²4 0=₁

2¹⁶− 8

−₁

2− 0

= ¹₂¹⁶−¹⁷2

17. 1

0

²

0  cos    =1 0

 sin  = ²

 = 0  =1

0  sin ²

= −¹2cos ²1

0= −¹2(cos 1 − cos 0) =¹2(1 − cos 1)

SECTION 15.2 DOUBLE INTEGRALS OVER GENERAL REGIONS ¤ 535

18. 

(²+ 2)  =1 0



³(²+ 2)   =1 0

² + ²=

=³ 

=1

0(³+ ²− ⁵− ⁶)  =1

4⁴+¹₃³−¹6⁶−¹7⁷1 0

=¹₄+¹₃−¹6−¹7 =²³₈₄

19. 

² =2 1

7−3

−1 ²  =2 1

²=7−3

=−1 

=2

1 [(7 − 3) − ( − 1)] ² =2

1(8²− 4³) 

=8

3³− ⁴2

1= ⁶⁴₃ − 16 −⁸3+ 1 = ¹¹₃

20. 

   =1 0

 √1−²

0   

=1 0

1

2²=√

1−²

=0  =1

0 1

2(1 − ²) 

= ¹₂1

0( − ³)  = ¹₂1

2²−¹4⁴1 0

= ¹₂₁

2−¹4− 0

= ¹₈

21.  2

−2

 √4−²

−√

4−²(2 − )  

=

 2

−2



2 −¹2²=√

4−²

=−√

4−²



=2

−2

2√

4 − ²−¹2

4 − ² + 2√

4 − ²+¹₂ 4 − ²



=2

−24√

4 − ² = −⁴3

4 − ²322

−2= 0 [Or, note that 4√

4 − ²is an odd function, so2

−24√

4 − ² = 0.]

22. 

   =1 0

4−3

   

=1

0 []^=4_=^−3 =1

0(4 − 3²− ²) 

=1

0(4 − 4²)  =

2²−⁴3³1

0= 2 −⁴3 − 0 = ²3

23.

D

 =1 0

^√

² (3 + 2)   =1 0

3 + ²=√



=² 

=1 0

(3√

 + ) − (3³+ ⁴)

 =1

0(3^32+  − 3³− ⁴) 

=

3 ·²5^52+¹₂²−³4⁴−¹5⁵1

0=⁶₅ +¹₂ −³4 −¹5 − 0 = ³4

24.  =2

−2

4

²(1 + ²²)  

=2

−2

 +¹₃³²=4

=²  =2

−2(4 +⁶¹₃²−¹3⁸) 

=

4 +⁶¹₉³−27¹⁹2

−2= 8 +⁴⁸⁸₉ −⁵¹²27 + 8 +⁴⁸⁸₉ −⁵¹²27 = ²³³⁶₂₇

25.  =2

1

7− 3

1    =2 1

1

2² = 7− 3

 = 1 

=¹₂2 1 

(7 − 3)²− 1

 =¹₂2

1(48 − 42²+ 9³) 

=¹₂

24²− 14³+⁹₄⁴2 1=³¹₈

26.  =2

0

2−

0 (²+ ²+ 1)   =2 0

² +¹₃³+ =2−

=0 

=2 0

²(2 − ) +¹3(2 − )³+ (2 − ) − 0



=2 0

−⁴3³+ 4²− 5 +¹⁴3

 =

−¹3⁴+⁴₃³−⁵2²+¹⁴₃2 0

= −¹⁶3 +³²₃ − 10 +²⁸3 − 0 = ¹⁴3

27.  =2

0

4−2

0 (4 − 2 − )   =2 0

4 − 2 −¹2²=4−2

=0 

=2 0

4(4 − 2) − 2(4 − 2) −¹2(4 − 2)²− 0



=2 0

2²− 8 + 8

 =₂

3³− 4²+ 82

0= ¹⁶₃ − 16 + 16 − 0 =¹⁶3

28.  =1

0

2−

   

=1 0

=2−

=  =1

0(2 − 2²) 

=

²−²3³1 0=¹₃

29.  =2

−2

4

²² 

=2

−2

²=4

=²  =2

−2(4²− ⁴) 

=4

3³−¹5⁵2

−2=³²₃ −³²5 +³²₃ −³²5 =¹²⁸₁₅

SECTION 15.2 DOUBLE INTEGRALS OVER GENERAL REGIONS ¤ 539 38. The region of integration is the portion of the ﬁrst quadrant bounded by the axes and the curve  =√

4 − ². The solid lies under the graph of  =  +  and above the graph of  = , so its volume is

 =2 0

 √4−²

0 ( + )   −2 0

 √4−²

0    =2 0

 √4−²

0 ( +  − )  

=2 0

 +¹₂²−¹2²=√

4−²

=0  =2

0

√

4 − ²+¹₂(4 − ²) −¹2(4 − ²) − 0



=2 0

√

4 − ²+ 2 −¹2²− 2 +¹2³

 =

−¹3(4 − ²)^32+ 2 − ¹6³− ²+¹₈⁴2 0

=

4 −⁴3− 4 + 2

−

−¹3· 4^32

=²₃+⁸₃ =¹⁰₃

39. The solid lies below the plane  = 1 −  −  or  +  +  = 1 and above the region

 = {( ) | 0 ≤  ≤ 1 0 ≤  ≤ 1 − }

in the -plane. The solid is a tetrahedron.

40. The solid lies below the plane  = 1 −  and above the region

 =

( ) | 0 ≤  ≤ 1 0 ≤  ≤ 1 − ² in the -plane.

41. The two bounding curves  = ³−  and  = ²+ intersect at the origin and at  = 2, with ²+   ³−  on (0 2).

Using a CAS, we ﬁnd that the volume of the solid is

 =

2 0

²+ 

³− 

(³⁴+ ²)   = 13,984,735,616 14,549,535

42. For || ≤ 1 and || ≤ 1, 2²+ ² 8 − ²− 2². Also, the cylinder is described by the inequalities −1 ≤  ≤ 1,

−√

1 − ²≤  ≤√

1 − ². So the volume is given by

 =

 1

−1

 √1− ²

−√

1− ²

(8 − ²− 2²) − (2²+ ²)

  = 13

2 [using a CAS]

43. The two surfaces intersect in the circle ²+ ²= 1,  = 0 and the region of integration is the disk : ²+ ²≤ 1.

Using a CAS, the volume is



(1 − ²− ²)  =

 1

−1

 √1−²

−√

1−²(1 − ²− ²)   =  2.

1

(2)

50. Because the region of integration is

 =

( ) | arctan  ≤  ≤^4, 0 ≤  ≤ 1

=

( ) | 0 ≤  ≤ tan , 0 ≤  ≤ ^4

 we have

 1 0

 4 arctan 

 ( )   =





 ( )  =

4 0

 tan  0

 ( )  

51.

 1 0

 3 3

^²  =

 3 0

 3 0

^²  =

 3 0



^²=3

=0



=

 3 0

  3



^² =¹6 ^²3

0=⁹− 1 6

52.  1

0

 1

²

√ sin    =

1 0

^√ 0

√ sin    =

 1 0

√ sin  []^=√=0 

=

1 0

(√

 sin ) (√

 − 0)  =

1 0

 sin  

= − cos ]¹0+1 0 cos  

[by integrating by parts with  = ,  = sin  ]

= [− cos  + sin ]¹0= − cos 1 + sin 1 − 0 = sin 1 − cos 1

53.

 1 0

 1

√

³+ 1   =

 1 0

 ² 0

³+ 1   =

 1 0

³+ 1 []^=_=0² 

=

 1 0

²

³+ 1  = ²₉

³+ 1321 0

= ²₉

2^32− 1^32

= ²₉ 2√

2 − 1

54.

 2 0

1

2

 cos(³− 1)   =

 1 0

 2

0

 cos(³− 1)  

=

 1 0

cos(³− 1)₁

2²=2

=0 

=

 1 0

2²cos(³− 1)  = ²3sin(³− 1)1 0

= ²₃[0 − sin(−1)] = −²3sin(−1) =²3sin 1

55.  1

0

 2 arcsin 

cos 

1 + cos²  

=2 0

sin  0 cos √

1 + cos²  

=2 0 cos √

1 + cos²

=sin 

=0 

=2 0 cos √

1 + cos² sin  

Let  = cos ,  = − sin  ,

 = (− sin )



=0 1 −√

1 + ² = −¹3

1 + ²320 1

=¹₃√

8 − 1

=¹₃ 2√

2 − 1

56. 8

0

 2

√3

^⁴  =

 2 0

 ³ 0

^⁴ 

=

 2 0

^⁴

=³

=0  =

2 0

³^⁴

= ¹₄^⁴2

0= ¹₄(¹⁶− 1)

57.  = {( ) | 0 ≤  ≤ 1, −  + 1 ≤  ≤ 1} ∪ {( ) | −1 ≤  ≤ 0,  + 1 ≤  ≤ 1}

∪ {( ) | 0 ≤  ≤ 1, − 1 ≤  ≤  − 1} ∪ {( ) | −1 ≤  ≤ 0, − 1 ≤  ≤ − − 1}, all type I.





² =

 1 0

 1 1− 

²  +

 0

−1

 1

 + 1

²  +

1 0

− 1

−1

²  +

 0

−1

_{− − 1}

−1

² 

= 4

 1 0

 1 1− 

²  [by symmetry of the regions and because ( ) = ²≥ 0]

= 41

0 ³ = 4₁

4⁴1 0= 1 58.  =

( ) | −1 ≤  ≤ 0, − 1 ≤  ≤  − ³

∪

( ) | 0 ≤  ≤ 1,√

 − 1 ≤  ≤  − ³, both type II.





  =

 0

−1

 −³

−1

   +

1 0

 −³

√−1

   =

 0

−1

 = −³

 =−1  +

1 0

=−³

=√−1

=0

−1(²− ⁴+ )  +1

0(²− ⁴− ^32+ ) 

=₁

3³−¹5⁵+¹₂²0

−1+

1

3³−¹5⁵−²5^52+¹₂²1 0

= (0 −¹¹30) + (₃₀⁷ − 0) = −15²

59. Since ²+ ²≤ 1 on , we must have 0 ≤ ²≤ 1 and 0 ≤ ²≤ 1, so 0 ≤ ²²≤ 1 ⇒ 3 ≤ 4 − ²²≤ 4 ⇒

√3 ≤

4 − ²²≤ 2. Here we have () =¹2(1)²=^₂, so by Property 11,

√3() ≤



4 − ²² ≤ 2() ⇒ ^√2³ ≤



4 − ²² ≤  or we can say

2720 



4 − ²²  3142. (We have rounded the lower bound down and the upper bound up to preserve the

inequalities.)

60. is the triangle with vertices (0 0), (1 0), and (1 2) so ( ) = ¹₂(1)(2) = 1. We have 0 ≤ sin⁴( + ) ≤ 1 for all , , and Property 11 gives 0 · () ≤

sin⁴( + ) ≤ 1 · ( ) ⇒ 0 ≤

sin⁴( + ) ≤ 1.

61. The average value of a function  of two variables deﬁned on a rectangle  was deﬁned in Section 15.1 as ave=_()¹ 

 ( ). Extending this deﬁnition to general regions , we have ave= _()¹ 

 ( ).

Here  = {( ) | 0 ≤  ≤ 1 0 ≤  ≤ 3}, so () =¹2(1)(3) =³₂ and

ave= _()¹ 

 ( ) = _32¹ 1 0

3

0   

= ²₃1 0

₁

2²=3

=0  =¹₃1

0 9³ = ³₄⁴1 0= ³₄ 62. Here  =

( ) | 0 ≤  ≤ 1 0 ≤  ≤ ², so

() =1

0 ² = ¹₃³1

0= ¹₃and

ave=_()¹ 

 ( ) = _13¹ 1 0

²

0  sin   

= 31 0

− cos =²

=0 

= 31 0

 −  cos(²)

 = 31

2²−¹2sin(²)1 0

= 31

2−¹2sin 1 − 0

=³₂(1 − sin 1) 63. Since  ≤ ( ) ≤ ,

  ≤

 ( )  ≤

 by (8) ⇒



1  ≤

 ( )  ≤ 

1 by (7) ⇒ () ≤

 ( )  ≤ () by (10).

64.





 ( )  =

 1 0

 2

0

 ( )   +

 3 1

 3−

0

 ( )  

=

 2 0

 3−

2

 ( )  

65. First we can write

( + 2)  =

  +

2 . But ( ) =  is an odd function with respect to  [that is, (− ) = −( )] and  is symmetric with respect to . Consequently, the volume above  and below the graph of  is the same as the volume below  and above the graph of , so



  = 0. Also,

2  = 2 · () = 2 ·¹2(3)²= 9since  is a half disk of radius 3. Thus

( + 2)  = 0 + 9 = 9.

66. The graph of ( ) =

²− ²− ²is the top half of the sphere ²+ ²+ ²= ², centered at the origin with radius

, and  is the disk in the -plane also centered at the origin with radius . Thus



²− ²− ²represents the

volume of a half ball of radius  which is¹₂·⁴3³= ²₃³.

544 ¤ CHAPTER 15 MULTIPLE INTEGRALS 67. We can write

(2 + 3)  =

2  +

3 .

2 represents the volume of the solid lying under the plane  = 2 and above the rectangle . This solid region is a triangular cylinder with length  and whose cross-section is a triangle with width  and height 2. (See the ﬁrst ﬁgure.)

Thus its volume is¹₂·  · 2 ·  = ². Similarly,

3 represents the volume of a triangular cylinder with length , triangular cross-section with width  and height 3, and volume¹₂·  · 3 ·  = ³2². (See the second ﬁgure.) Thus



(2 + 3)  = ² +³₂²

68. In the ﬁrst quadrant,  and  are positive and the boundary of  is  +  = 1. But  is symmetric with respect to both axes because of the absolute values, so the region of integration is the square shown at the left. To evaluate the double integral, we ﬁrst write



(2 + ²³− ²sin )  =

2  +

²³ −

²sin  .

Now ( ) = ²³is odd with respect to  [that is, ( −) = −( )]

and  is symmetric with respect to , so

²³ = 0.

Similarly, ( ) = ²sin is odd with respect to  [since (− ) = −( )] and  is symmetric with respect to , so

²sin   = 0.  is a square with side length√ 2, so

2  = 2 · () = 2√

2² = 4, and



(2 + ²³− ²sin )  = 4 + 0 + 0 = 4.

69. 



³+ ³+√

²− ²

 =

³ +

³ +



√²− ². Now ³is odd with respect

to  and ³is odd with respect to , and the region of integration is symmetric with respect to both  and , so

³ =

³ = 0.





√²− ²represents the volume of the solid region under the

graph of  =√

²− ²and above the rectangle , namely a half circular cylinder with radius  and length 2 (see the ﬁgure) whose volume is

1

2· ² =¹₂²(2) = ². Thus





³+ ³+√

²− ²

 = 0 + 0 + ² = ².