Section 15.1 Double Integrals over Rectangles

(1)

Section 15.1 Double Integrals over Rectangles

10. Evaluate the double integral by first identifying it as the volume of a solid.

Z Z

R

(2x + 1)dA, R = {(x, y)|0 ≤ x ≤ 2, 0 ≤ y ≤ 4}

Solution:

526 ¤ CHAPTER 15 MULTIPLE INTEGRALS

10. = 2 + 1 ≥ 0 for 0 ≤  ≤ 2, so we can interpret the integral as the volume of the solid  that lies below the plane  = 2 + 1 and above the rectangle [0 2] × [0 4]. We can picture  as a rectangular solid (with height 1) surmounted by a triangular cylinder; thus



(2 + 1)  = (2)(4)(1) +¹₂(2)(4)(4) = 24

11. = 4 − 2 ≥ 0 for 0 ≤  ≤ 1, so we can interpret the integral as the volume of the solid  that lies below the plane  = 4 − 2 and above the square [0 1] × [0 1]. We can picture  as a rectangular solid (with height 2) surmounted by a triangular cylinder; thus



(4 − 2)  = (1)(1)(2) +¹2(1)(1)(2) = 3

12.Here  =

9 − ², so ²+ ²= 9,  ≥ 0. Thus the integral represents the volume of the top half of the part of the circular cylinder

²+ ²= 9that lies above the rectangle [0 4] × [0 2].

13.2

0( + 3²²)  =

² 2 + 3³

3 ²

=2

=0

=₁

2²+ ³²=2

=0=₁

2(2)²+ (2)³²

−₁

2(0)²+ (0)³²

= 2 + 8²,

3

0( + 3²²)  =



 + 3²³ 3

=3

=0

=

 + ²³=3

=0=

(3) + ²(3)³

−

(0) + ²(0)³

= 3 + 27²

14.2 0 √

 + 2  =

 ·²3( + 2)^32=2

=0=²₃(4)^32−²3(2)^32=¹⁶₃ −⁴3

√2  = ⁴₃(4 −√2 ) ,

3 0 √

 + 2  =

² 2

√ + 2

=3

=0

=¹₂(3)²√

 + 2 −¹2(0)²√

 + 2 = ⁹₂√

 + 2

15.4 1

2

0(6² − 2)   =4 1

3²²− 2=2

=0 =4 1

12²− 4

− (0 − 0)



=4

1(12²− 4)  =

4³− 2²4

1= (256 − 32) − (4 − 2) = 222 16. 1

0

1

0 ( + )²  =1 0

1

0 (²+ 2 + ²)   =1 0

1

3³+ ² + ²=1

=0

=1

0(¹₃ +  + ²)  =1

3 +¹₂²+¹₃³1

0= ¹₃+¹₂+¹₃− 0 = ⁷6

17. 1 0

2

1( + ^−)   =1 0

1

2²+ ^−=2

=1 =1 0

(2 + 2^−) − (¹2+ ^−)



=1

0(³₂ + ^−)  =3

2 − ^−1 0=3

2− ⁻¹

− (0 − 1) =⁵2 − ⁻¹

22. Calculate the iterated integral. R1 0

R2

0 ye^x−ydxdy Solution:

SECTION 15.1 DOUBLE INTEGRALS OVER RECTANGLES ¤ 527 18.6

0

2

0 (sin  + sin )   =6

0 [ sin  − cos ]^=2=0  =6 0



2sin  − 0

− (0 − 1)



=6 0



2sin  + 1

 =

−^2 cos  + 6 0

=

−^2 ·^√2³+^₆

−

−^2 + 0

=

2 3 −^√4³





19.2 0

2

0  sin    =2

0  2

0 sin   [as in Example 5] =

² 2

2 0



− cos 2

0 = (2 − 0)(0 + 1) = 2

20.

 3 1

5 1

ln 

   =

 3 1

1



5 1

ln 

  [by Equation 11]

= [ln ||]³1

₁

2(ln )²5

1 [substitute  = ln  ⇒  = (1) ]

= (ln 3 − 0) ·¹2[(ln 5)²− 0] = ¹2(ln 3)(ln 5)²

21.

 4 1

2 1



 +





  =

 4 1



 ln || +1

·1 2²

=2

=1

 =

 4 1



 ln 2 + 3 2



 =₁

2²ln 2 +³₂ln ||4 1

=

8 ln 2 +³₂ln 4

−1

2ln 2 + 0

= ¹⁵₂ ln 2 +³₂ln 4 or ¹⁵₂ ln 2 + 3 ln(4^12) = ²¹₂ ln 2

22. 1 0

2

0 ^^−  =1 0

2

0 ^^−  =2 0 ^1

0 ^− [by Equation 11]

= [^]²₀ 

(− − 1)^−1

0 [by integrating by parts]

=

²− ⁰

−2⁻¹−

−⁰

= (²− 1)(1 − 2⁻¹) or ²− 2 + 2⁻¹− 1

23. 3 0

2

0 ²sin³   =2

0 sin³ 3

0 ² [by Equation 11] =2

0 (1 − cos²) sin  3 0 ²

=1

3cos³ − cos 2 0

1 3³3

0=

(0 − 0) −1 3− 1

·¹3(27 − 0) = ²3(9) = 6

24. 1 0

2 1

^

   =

 1 0

^

 2 1

1

 [as in Example 5]

=

^− ^1 0

ln ||2

1 [by integrating by parts]

= [( − ) − (0 − 1)](ln 2 − 0) = ln 2

25. 2 0



0  sin²   =2 0  

0 sin²  [as in Example 5] =2 0  

0 1

2(1 − cos 2) 

=1 2²2

0·¹2

 −¹2sin 2

0 = (2 − 0) ·¹2

 −¹2sin 2

−

0 −¹2sin 0

= 2 · ¹2[( − 0) − (0 − 0)] = 

26. 1 0

3

0 ^{ + 3}  =1 0

3

0 ^^3  =3 0 ^1

0 ^3 = [^]³₀1 3^31

0

=

³− ⁰

·¹3

³− ⁰

=¹₃(³− 1)²or ¹₃(⁶− 2³+ 1)

27.

 sec²  =2 0

4

0  sec²   =2

0  4

0 sec²  =

1 2²2

0



tan 4 0

= (2 − 0) (tan^4 − tan 0) = 2(1 − 0) = 2

34. Calculate the double integral. RR

R 1

1+x+ydA, R = [1, 3] × [1, 2].

Solution:

528 ¤ CHAPTER 15 MULTIPLE INTEGRALS 28.

( + ⁻²)  =2 1

2

0( + ⁻²)   =2 1

 +¹₂²⁻²=2

=0  =2 1

2 + 2⁻²



=

²− 2⁻¹2

1= (4 − 1) − (1 − 2) = 4

29.





²

²+ 1 =

1 0

 3

−3

²

²+ 1  =

 1 0



²+ 1

 3

−3

² =

1

2ln(²+ 1)1 0

1 3³3

−3

=¹₂(ln 2 − ln 1) ·¹3(27 + 27) = 9 ln 2

30.





tan 

√1 − ²  =

 12 0

 3 0

tan 

√1 − ²  =

 12 0

√ 1

1 − ²

 3 0

tan   =

sin⁻¹12 0



ln |sec |3 0

=

sin^{−1 1}₂− sin⁻¹0 

lnsec^₃

 − ln |sec 0|

=_

6 − 0

(ln 2 − ln 1) =^6ln 2

31.6 0

3

0  sin( + )  

=6 0

− cos( + ) = 3

 = 0  =6 0

 cos  −  cos

 +^₃



= 

sin  − sin

 +^₃6 0 −6

0

sin  − sin

 +^₃

 [by integrating by parts separately for each term]

=^₆₁

2 − 1

−

− cos  + cos

 +^₃6

0 = −12^ −

−^√2³+ 0 −

−1 +¹2

=^√³₂⁻¹−12^

32.





1 + ² 1 + ²  =

1 0

 1 0

1 + ²

1 + ²   =

 1 0

(1 + ²) 

 1 0

1

1 + ² =

 +¹₃³1 0

tan⁻¹¹

0

=

1 +¹₃− 0 4 − 0

=^₃

33.

sin( − )  =2 0

2

0 sin( − )   =2

0 [cos( − )]^=2=0  =2 0

cos( −^2) − cos 



=

sin( −^2) − sin 2

0 = sin 0 − sin^2 −

sin(−^2) − sin 0

= 0 − 1 − (−1 − 0) = 0

34.





1

1 +  +  =

3 1

 2 1

1

1 +  +   =3

1 [ln(1 +  + )]^=2_=1 =3

1 [ln( + 3) − ln( + 2)] 

=

( + 3) ln( + 3) − ( + 3)

−

( + 2) ln( + 2) − ( + 2) 3 1

[by integrating by parts separately for each term]

= (6 ln 6 − 6 − 5 ln 5 + 5) − (4 ln 4 − 4 − 3 ln 3 + 3) = 6 ln 6 − 5 ln 5 − 4 ln 4 + 3 ln 3

35. =  ( ) = 4 −  − 2 ≥ 0 for 0 ≤  ≤ 1 and 0 ≤  ≤ 1. So the solid is the region in the ﬁrst octant which lies below the plane  = 4 −  − 2

and above [0 1] × [0 1].

54. Find the average value of f over the given rectangle.

f (x, y) = e^y√

x + e^y, R = [0, 4] × [0, 1]

Solution:

SECTION 15.1 DOUBLE INTEGRALS OVER RECTANGLES ¤ 531

48. () = 4 · 1 = 4, so

ave= 1

()





 ( )  =¹₄

 4 0

 1 0

^√

 + ^  = ¹₄

 4 0

2

3( + ^)^32=1

=0

=¹₄ ·²3

4

0[( + )^32− ( + 1)^32]  = ¹₆

2

5( + )^52−²5( + 1)^524 0

=¹₆ ·²5[(4 + )^52− 5^52− ^52+ 1] = ₁₅¹[(4 + )^52− ^52− 5^52+ 1] ≈ 3327

49.







1 + ⁴ =

 1

−1

 1 0



1 + ⁴  =

 1

−1

 1 + ⁴

 1 0

  [by Equation 11] but () = 

1 + ⁴ is an odd function so

 1

−1

 ()  = 0(by Theorem 4.5.6 [ET 5.5.7]). Thus







1 + ⁴ = 0 ·

 1 0

  = 0.

50. 

(1 + ²sin  + ²sin )  =

1  +

²sin   +

²sin  

= () +

−



−²sin    +

−



−²sin   

= (2)(2) +

−²

−sin   +

−sin  

−²

But sin  is an odd function, so

−sin   =

−sin   = 0(by Theorem 4.5.6 [ET 5.5.7]) and



(1 + ²sin  + ²sin )  = 4²+ 0 + 0 = 4². 51. Let ( ) =  − 

( + )³. Then a CAS gives1 0

1

0  ( )   = ¹₂and1 0

1

0  ( )   = −¹2.

To explain the seeming violation of Fubini’s Theorem, note that  has an inﬁnite discontinuity at (0 0) and thus does not satisfy the conditions of Fubini’s Theorem. In fact, both iterated integrals involve improper integrals which diverge at their lower limits of integration.

52. (a) Loosely speaking, Fubini’s Theorem says that the order of integration of a function of two variables does not affect the value of the double integral, while Clairaut’s Theorem says that the order of differentiation of such a function does not affect the value of the second-order derivative. Also, both theorems require continuity (though Fubini’s allows a ﬁnite number of smooth curves to contain discontinuities).

(b) To ﬁnd , we ﬁrst hold  constant and use the single-variable Fundamental Theorem of Calculus, Part 1:

= 

( ) = 



 



 



 ( ) 



 =

 



 ( ) . Now we use the Fundamental Theorem again:

= 



 



 ( )  =  ( ).

To find , we first use Fubini’s Theorem to find that





  ( )   =





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

1