Section 15.2 Double Integrals over General Regions

Section 15.2 Double Integrals over General Regions

9. (a) Express the double integralRR

Df (x, y)dA as an iterated integral for the given function f and region D.

(b) Evaluate the iterated integral.

SECTION 15.2 Double Integrals over General Regions 1059

10

If m < f sx, yd < M for all sx, yd in D, then

m  AsDd <

y

D

y ^f sx, yd dA < M  AsDd

Figure 20 illustrates Property 10 for the case m . 0. The volume of the solid below the graph of z − fsx, yd and above D is between the volumes of the cylinders with base D and heights m and M. (Compare to Figure 5.2.17, which illustrates the analogous property for single integrals.)

EXAMPLE 6 Use Property 10 to estimate the integral yy

D

e

sin x cos y dA, where D is the

disk with center the origin and radius 2.

SOLUTION

Since 21 < sin x < 1 and 21 < cos y < 1, we have

21 < sin x cos y < 1 and, because the natural exponential function is increasing, we have

e²¹

<

esin x cos y

<

e¹

− e

Thus, using m − e

²¹

− 1ye, M − e, and AsDd − s2d

²

in Property 10, we obtain 4

e

< y

D

y ^e

sin x cos y dA <

4e

^■

y z

z=M

z=m z=f(x, y)

x

D

FIGURE 20

15.2 Exercises

1–6 Evaluate the iterated integral.

1.

y

1⁵

y

0^x s8x 2 2yd dy dx 2.

y

0²

y

0^y2 x²y dx dy

3.

y

0¹

y

0^y xe^y3 dx dy 4.

y

0^y2

y

0^x x sin y dy dx

5.

y

0¹

y

0^s2 cosss³d dt ds 6.

y

0¹

y

0^ev s1 1 e^v dw dv

7–10

(a) Express the double integral yyD fsx, yd dA as an iterated integral for the given function f and region D.

(b) Evaluate the iterated integral.

7. fsx, yd − 2y 8. fsx, yd − x 1 y

y=3x-≈

y=x (2, 2) y

0 x

D (1, 1)

0 y

x D

2

9. fsx, yd − xy 10. fsx, yd − x

y

0 x

D y=œ„x

y=x-2

0 y

x D

y=6-x

y=≈

11–14 Evaluate the double integral.

11.

y

D

y

_x^2y₁₁ ^{dA, D −}

^h

^{sx, yd}

|

0 < x < 4, 0 < y <sx

j

12.

y

D

y

s2x 1 yd dA, D − hsx, yd

|

1 < y < 2, y 2 1 < x < 1j 13.

y

D

y

^{e2y2 dA, D −hsx, yd}

|

0 < y < 3, 0 < x < yj 14.

y

D

y

^{ysx22y2 dA, D −hsx, yd}

|

0 < x < 2, 0 < y < xj 15. Draw an example of a region that is

(a) type I but not type II (b) type II but not type I

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Solution:

SECTION 15.2 DOUBLE INTEGRALS OVER GENERAL REGIONS ¤ 1501 To find , we first use Fubini’s Theorem to find that





  ( )   =





  ( )  , and then use the Fundamental Theorem twice, as above, to get =  ( ). So = =  ( ).

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. (a) We express the iterated integral as a Type I:

 2 0

 3−²



2  . A Type II would require the sum of two integrals.

(b)

2 0

 3−²



2   =

 2 0

²=3−²

=  =

 2 0

(3 − ²)²− ()²

 =

 2 0

8²− 6³+ ⁴



=₈

3³−³2⁴+¹₅⁵2

0=⁶⁴₃ − 24 +³²5 =⁵⁶₁₅ 8. (a) We express the iterated integral as a Type II:1

0

2−

 ( + )  . A Type I would require the sum of two integrals.

(b)

1 0

 2−



( + )   =

 1 0

² 2 + 

=2−

=

 =

 1 0

(2 − )²

2 + (2 − )



−

² 2 + ²





=1

0 (2 − 2²)  =

2 −²3³1

0= 2 −²3 =⁴₃

9. (a) We express the iterated integral as a Type II. A Type I would require the sum of two integrals. The curves intersect when

√ =  − 2 ⇒  = ²− 4 + 4 ⇔ 0 = ²− 5 + 4 ⇔ ( − 4)( − 1) = 0 ⇔  = 1 or  = 4. The

point for  = 1 is not in . Thus, the point of intersection of the curves is (4 2) and the integral is2 0

+2

²   .

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1502 ¤ CHAPTER 15 MULTIPLE INTEGRALS (b)

 2 0

 +2

²

   =

 2 0



² 2

=+2

=²

 = 1 2

 2 0



( + 2)²− (²)²

 = 1 2

 2 0

[³+ 4²+ 4 − ⁵] 

=¹₂₁

4⁴+⁴₃³+ 2²−¹6⁶2 0=¹₂

4 +³²₃ + 8 −³²3

= 6

10. (a) We express the iterated integral as a Type I. A Type II would require the sum of two integrals. The curves intersect when 6 −  = ² ⇔ ²+  − 6 = 0 ⇔ ( + 3)( − 2) = 0 ⇔  = −3 or  = 2. The point for  = −3 is not in . Thus, the point of intersection of the two curves is (2 4) and the integral is

 2 0

 6−

²

  .

(b)

 2 0

 6−

²

   =

2 0



=6−

=²  =

2 0

[(6 − ) − ²]  =

2 0

[6 − ²− ³] 

=



3²−³ 3 −⁴

4

²

0

= 12 −8

3− 4 = 16 3

11.







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 12. 

(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

13. 

^−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

14. 



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

15. (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.

(b) Now we sketch an example of a region  that can be described as lying between the graphs of two continuous functions of  but not as lying between graphs of two continuous functions of . The first region shown in Figure 7 is another example.

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21. Set up iterated integrals for both orders of integration. Then evaluate the double integral using the easier order and explain why its easier.

Z Z

D

sin²xdA, D is bounded by y = cos x, 0 ≤ x ≤ π

2, y = 0, x = 0 Solution:

1504 ¤ CHAPTER 15 MULTIPLE INTEGRALS

19. 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

20. 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

21.

D

If we describe  as a type I region,  = {( ) | 0 ≤  ≤ 2, 0 ≤  ≤ cos }

and

sin²  =2 0

cos 

0 sin²  . As a type II region,

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



sin²  =

 1 0

cos⁻¹ 0

sin²  . Evaluating

 cos⁻¹ 0

sin² will result in a very difficult integral. Therefore, we evaluate the iterated integral that describes  as a type I region because integrating sin²with respect to  is easy.

 2 0

 cos  0

sin²   =

 2 0

sin²

=cos 

=0  =

 2 0

cos  sin² 

=

 1 0

²

 = sin 

 = cos  



=

³ 3

1 0

=1 3

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25. Evaluate the double integral.

Z Z

D

y²dA, D is the triangular region with vertices (0, 1), (1, 2), (4, 1) Solution:

SECTION 15.2 DOUBLE INTEGRALS OVER GENERAL REGIONS ¤ 535

18. 



² =

 1

−1

 √1−² 0

² 

=1

−1²1 2²=√

1−²

=0  =¹₂1

−1²(1 − ²) 

=¹₂1

−1(²− ⁴)  = ¹₂1

3³−¹5⁵1

−1

=¹₂₁

3 −¹5 +¹₃ −¹5

=₁₅²

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

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64. Evaluate the integral by reversing the order of integration. R2 0

R1

y/2y cos(x³− 1)dxdy Solution:

SECTION 15.2 DOUBLE INTEGRALS OVER GENERAL REGIONS ¤ 541

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

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