Thomas calculus early transcendentals 13th edition thomas solutions manual

Thomas Calculus Early Transcendentals 13th Edition Thomas Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/thomas-calculus-early-transcendentals-13th-edition-thomas-solutions-manual/

CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS 6.1

VOLUMES USING CROSS-SECTIONS

1.

(diagonal)2 A( x)   2

 x   x    2 x; a  0, b  4; V  b A( x) dx  4 2 x dx   x2  4  16 2

2.

A( x) 

 (diameter)2 4

a

2







  2 x 2  x2  

2





4







0

   1  2 x 2  x 4 ; a  1, b  1;

  2 1 x 2  

0

4

2







1





5 1 b  V   A( x) dx    1  2 x 2  x 4 dx    x  23 x3  x5   2 1  23  51  16 15   1 a 1

2

3.

2





  A( x)  (edge)2   1  x 2    1  x 2     2 1  x 2   4 1  x 2 ; a  1, b  1;      b



1



1





V   A( x) dx   4 1  x 2 dx  4  x  x3   8 1  13  16 3   1 a 1 3



  2  2 1 x   2 1  x2 ; a  1, b  1; 4.   2 1 1 b V   A( x) dx  2 1  x 2  dx  2  x  x3   4 1  13   83 a 1   1  1 x 2   1 x 2    (diagonal)2 A( x)   2 2

2

2

2

3













5. (a) STEP 1) A( x)  12 (side)  (side)  sin 3  12  2 sin x  2 sin x sin 3  3 sin x STEP 2) a  0, b   b





a

0

0

STEP 3) V   A( x) dx  3  sin x dx    3 cos x   3(1  1)  2 3 2



(b) STEP 1) A( x)  (side)  2 sin x

  2 sin x   4 sin x

STEP 2) a  0, b   b



a

0



STEP 3) V   A( x) dx   4 sin x dx   4 cos x 0  8 6. (a) STEP 1) A( x) 



 (diameter)2

 4 (sec x  tan x)2  4 sec2 x  tan 2 x  2sec x tan x

4







 4 sec2 x  sec 2 x  1  2 sin2x  cos x   STEP 2) a   3 , b  3 b

STEP 3) V   A( x) dx   a

 /3 

 /3 4

 2 sec x  1  2

2 sin x cos 2 x

 dx  2 tan x  x  2   

4

        4  2 3  3  2   11    2 3  3  2   11     4 4 3  23     2     2    



Copyright  2014 Pearson Education, Inc.

 /3 1  cos x   /3





431

432

Chapter 6 Applications of Definite Integrals



(b) STEP 1) A( x)  (edge) 2  (sec x  tan x)2  2sec 2 x  1  2 sin2x STEP 2) a   3 , b  3 b

 /3

a

 /3

STEP 3) V   A( x) dx  

 2sec x 1  2

cos x

2sin x cos 2 x



 dx  2  2 3    4 3  

3

2 3

7. (a) STEP 1) A( x)  (length)  (height)  (6  3 x)  (10)  60  30 x STEP 2) a  0, b  2 b

2

2

STEP 3) V   A( x) dx   (60  30 x) dx   60 x  15 x 2   (120  60)  0  60  0 0 a (b) STEP 1) A( x)  (length)  (height)  (6  3x ) 



20 2(63 x ) 2

  (6  3x)(4  3x)  24  6x  9x

2

STEP 2) a  0, b  2 b

2

a

0

STEP 3) V   A( x)dx  

 24  6 x  9 x2  dx  24 x  3x2  3x3 0  (48  12  24)  0  36 2

8. (a) STEP 1) A( x)  12 (base)  (height) 

 x  2x   (6)  6 x  3x

STEP 2) a  0, b  4 6 x1/2  3x  dx   4 x3/2  32 x 2   (32  24)  0  8  0 0

b

STEP 3) V   A( x) dx   a

4

4

2

 diameter   12    2   2  2 x  2x

2

(b) STEP 1) A( x)  12  

x  x3/ 2  14 x 2  8 4

 x  x3/2  14 x2 

STEP 2) a  0, b  4 1 x3    8  64  16   (0)   x  x3/2  14 x 2  dx   12 x 2  25 x5/2  12 5 3 8 15  0 8  0

b

STEP 3) V   A( x) dx  8  a

9.

A( y )  4 (diameter)2  4

4

4

 5 y 2  0   54 y 4 ; 2

d

c  0, d  2; V   A( y ) dy c

 

2 5 4  y dy   54 0 4





2





 y5    5  5    4 2  0  8   0

2

10.

2





 A( y )  12 (leg)(leg)  12  1  y 2    1  y 2    12  2 1  y 2   2 1  y 2 ; c  1, d  1;     





1





d 1 y3   V   A( y ) dy   2 1  y 2 dy  2  y  3   4 1  13  83 1 c   1

Copyright  2014 Pearson Education, Inc.

Section 6.1 Volumes Using Cross-Sections

433

11. The slices perpendicular to the edge labeled 5 are triangles, and by similar triangles we have bh  43  h  34 b.

The equation of the line through (5, 0) and (0, 4) is y   54 x  4, thus the length of the base   45 x  4 and









the height  34  54 x  4   53 x  3. Thus A( x)  12 (base)  (height)  12  54 x  4   53 x  3



5 6 2 12 x  5 x6 0 25

b

6 x 2  12 x  6 and V   25  A( x) dx   5 a



 dx   252 x3  56 x2  6 x0  (10  30  30)  0  10 5

12. The slices parallel to the base are squares. The cross section of the pyramid is a triangle, and by similar

   259 y2  V  cd A( y) dy  05 259 y 2 dy

triangles we have bh  35  b  35 h. Thus A( y )  (base)2  53 y

2

5

3 y 3   15  0  15   25  0

13. (a) It follows from Cavalieri’s Principle that the volume of a column is the same as the volume of a right prism with a square base of side length s and altitude h. Thus, STEP 1) A( x)  (sidelength)2  s 2 ; STEP 2) a  0, b  h; b

h

a

0

STEP 3) V   A( x) dx   s 2 dx  s 2 h (b) From Cavalieri’s Principle we conclude that the volume of the column is the same as the volume of the prism described above, regardless of the number of turns  V  s 2 h The solid and the cone have the same altitude of 12.

The cross sections of the solid are disks of

14. 1) 2)



diameter x  2x  2x . If we place the vertex of the cone at the origin of the coordinate system and make its axis or symmetry coincide with the x-axis then the cone’s cross sections will

 

be circular disks of diameter 4x   4x  2x (see accompanying figure).

The solid and the cone have equal altitudes and identical parallel cross sections. From Cavalier’s Principle we conclude that the solid and the cone have the same volume.

3)

15.

2

2



R( x)  y  1  2x  V     R ( x) dx    1  2x 0 0



2



 dx   02 1  x  x4  dx    x  x2  12x 0 2

2

2

3

2

8  2   2  42  12 3

16.

2

R( y )  x  2  V     R( y ) dy    3y

2

0

 

2 2 3y 2 2 dy   94 y 2 dy    43 y3     43  8  6 2  0 0

0



 

17. R( y )  tan 4 y ; u  4 y  du  4 dy  4 du   dy; y  0  u  0, y  1  u  4 ;

 

2





 /4  /4 1 1  /4 2 1  sec2 u du  4  u  tan u 0 V     R( y )  dy     tan 4 y  dy  4 tan 2 u du  4   0 0 0 0





 4  4  1  0  4  

Copyright  2014 Pearson Education, Inc.

434

Chapter 6 Applications of Definite Integrals

18. R( x)  sin x cos x; R ( x)  0  a  0 and b  2 are the limits of integration; V 

 /2

0

  R( x)  dx    2

 /2

0

(sin x cos x)2 dx   

 /2 (sin 2 x )2

0

4

dx; u  2 x  du  2 dx  du  dx ; 8 4





 2  x  0  u  0, x  2  u     V    18 sin 2 u du  8  u2  14 sin 2u   8  2  0  0  16   0 0 2

19. R( x)  x 2  V     R ( x ) dx 2

0

2

 

0

 x2  dx   02 x4dx    x5 0  325 2

2

5

2

20. R( x)  x3  V     R( x)  dx 2

0

2  x3 dx    x 6 dx    x7   128 7 0  0   0 2

 

2

2

7

3

21. R( x)  9  x 2  V     R ( x)  dx 2

3

9  x 2  dx   9 x  x3  3    3

 

3

3

3

  2  π 18  36π  2 9(3)  27 3 

1

22. R( x)  x  x 2  V     R ( x ) dx 2

0

0 1



2

1





   x  x 2 dx    x 2  2 x3  x 4 dx 0

1



3 4 5    x3  24x  x5    13  12  15   0



 (10  15  6)    30 30

23. R( x)  cos x  V  

 /2

0

 

 /2

0

  R( x)  dx 2

 /2

cos x dx   sin x 0

  (1  0)  

Copyright  2014 Pearson Education, Inc.

Section 6.1 Volumes Using Cross-Sections 24. R( x)  sec x  V  

 /4

 /4



 /4

435

  R ( x ) dx 2

 /4

 /4

sec2 x dx    tan x  /4   [1  (1)]  2

1

1

0

0

25. R( x)  e x  V    [ R ( x)]2 dx    (e x )2 dx 1 2 x

 e 0

1 dx   2 e 2 x   2 (e 2  1) 0

 

 2 1  12  e

 (e2 1) 2e 2

26. R( x)  cot x  V  





 /2

 /6

ln1  ln 12

27. R( x) 

1 2 x

   ln 2

V  

4

1/4

 /2  /2 cos x cot x  dx    cot x dx    dx   [ln(sin x)] /2  /6  /6  /6  /6 sin x

 [ R( x)]2 dx   

 [ R( x)]2 dx   

4

 /2

2

  dx   1

1/4 2 x

2

4 1 4 dx  4 [ln x]1/4  4 4 1/4 x



3

3

3

1

1

1

 ln 4  ln 14   2 ln 4

28. R( x)  e x 1  V    [ R( x)]2 dx    (e x 1 )2 dx    e 2 x 2 dx  2 [e2 x  2 ]13  2 (e4  1)  84.19 29. R( x)  2  sec x tan x  V  

 /4

0

  R( x) dx 2

 2  sec x tan x  dx 0



 /4

2



 /4

 2  2 2 sec x tan x  sec2 x tan 2 x  dx

0

 /4  /4     2 dx  2 2  sec x tan x dx 0 0   /4   (tan x)2 sec2 x dx  0   /4   3  /4  /4     2 x 0  2 2 sec x 0   tan3 x    0   





   2  0  2 2 



  2  2 2  11 3



 2  1  13 (13  0)

Copyright  2014 Pearson Education, Inc.

436

Chapter 6 Applications of Definite Integrals

30. R( x)  2  2sin x  2(1  sin x)  V  

 /2

0



 /2

0

4(1  sin x)2 dx  4 

 /2

0

 4 

 /2

 4 

 /2 3 cos 2 x   2sin x

2

1  sin 2 x  2sin x  dx

1  1 (1  cos 2 x)  2sin x  dx  2 

0 0

  R ( x) dx

2

2



 /2

 4  32 x  sin42 x  2 cos x 





0

 4  34  0  0  (0  0  2)    (3  8)   1

1

31. R( y )  5 y 2  V     R ( y ) dy    5 y 4 dy 2

1

1

1

   y5    [1  (1)]  2   1

2

2

32. R( y )  y 3/2  V     R( y )  dy    y 3 dy 2

0

0

2

 y4     4   4  0

33. R( y )  2sin 2 y  V  

 /2

0



 /2

0

  R( y )  dy 2

 /2

2sin 2 y dy     cos 2 y 0

  [1  (1)]  2

0

y

34. R( y )  cos 4  V     R ( y ) dy 2

2

0

   cos 2

  dy  4 sin   4[0  (1)]  4 y

y 0

4

4

3

2

3 1 dy 0 ( y 1) 2

35. R( y )  y21  V     R( y ) dy  4  2

0

3

 4   y11   4   14  (1)   3  0

Copyright  2014 Pearson Education, Inc.

Section 6.1 Volumes Using Cross-Sections

1

2y

36. R( y ) 

2

y 1

 0

 V     R ( y ) dy 2

0

 dy; [u  y 2  1  du  2 y dy;

1

   2 y y2 1

2

y  0  u  1, y  1  u  2] 2

2

1

1

 V    u 2 du     u1      12  (1)   2



b

37. For the sketch given, a   2 , b  2 ; R ( x)  1, r ( x)  cos x; V     R ( x )   r ( x) a



 /2

 /2

 (1  cos x) dx  2 

 /2

0

 /2

(1  cos x) dx  2  x  sin x 0

d

2



2



 2 2  1   2  2



38. For the sketch given, c  0, d  4 ; R ( y )  1, r ( y )  tan y; V     R( y )    r ( y )  c



 /4

0

2

2



1

0





1

2



2

 dx



1    1  x 2 dx    x  x3     1  13  0   23   0   0 3

1



40. r ( x)  2 x and R( x)  2  V     R( x)    r ( x) 0

1

1

2



2

 dx



   (4  4 x) dx  4  x  x2   4 1  12  2 0   0 2

41. r ( x)  x 2  1 and R( x)  x  3 2



 V     R( x)    r ( x) 1

2



2

 dx



2 2      ( x  3) 2  x 2  1  dx 1  



 



2     x 2  6 x  9  x 4  2 x 2  1  dx 1  

 x 4  x 2  6 x  8  dx     x5  x3  6 2x  8 x  1    1



2

2

 dy

1  tan 2 y  dy   0 /4  2  sec2 y  dy   2 y  tan y 0 /4    2  1  2  

39. r ( x)  x and R( x)  1  V     R ( x)   r ( x) 

5

3

2

 dx

2



       33  3  28  3  8     530533   1175  5     32  8  24  16  15  13  62  8   5 3 2 

Copyright  2014 Pearson Education, Inc.

437

438

Chapter 6 Applications of Definite Integrals

42. r ( x)  2  x and R( x)  4  x 2



2

 V     R( x)    r ( x) 1

2

2

 dx

  2    16  8 x 2  x 4    4  4 x  x 2   dx 1   2 2    12  4 x  9 x 2  x 4  dx   12 x  2 x 2  3 x3  x5  1   1 2 2       4  x 2  (2  x)2  dx 1  

5



 







   24  8  24  32  12  2  3  15    15  33  1085  5 5   43. r ( x)  sec x and R( x)  2

V  

 /4

 /4



 /4

 /4





  R ( x)   r ( x)  2

2

 dx

 2  sec2 x  dx   2x  tan x/4/4  



   2  1   2  1    (  2)  



1

44. R( x)  sec x and r ( x)  tan x  V     R ( x)    r ( x) 0

1





1

2

2

 dx

   sec2 x  tan 2 x dx    1 dx    x 0   0

1

0

2

 dy

46. R( y )  1 and r ( y )  1  y  V     R( y )   r ( y ) 

 dy

1



45. r ( y )  1 and R( y )  1  y  V     R ( y )   r ( y )  0

2





1 1    (1  y )2  1 dy    1  2 y  y 2  1 dy  0 0





1





1 y3      2 y  y 2 dy    y 2  3    1  13  43 0  0

1

0



2



2



1 1    1  (1  y ) 2  dy    1  1  2 y  y 2  dy   0 0 





1





1 y3      2 y  y 2 dy    y 2  3    1  13  23 0  0

Copyright  2014 Pearson Education, Inc.

Section 6.1 Volumes Using Cross-Sections



4

y  V     R( y )    r ( y )

47. R( y )  2 and r ( y ) 

0

2

2

 dy

4

4 y2      (4  y ) dy    4 y  2    (16  8)  8 0  0

48. R( y )  3 and r ( y )  3  y 2 3

V  

0



3

0



  R( y )    r ( y )  2



2

 dy



3 3  3  y 2  dy   3 y 2 dy    y    0    3 



1

3

 3

0

 dy

49. R( y )  2 and r ( y )  1  y  V     R( y )    r ( y ) 0





2



2



2 1      4  1  y  dy      1  2 y  y dy 0   1



    3  43  12     18683   76

1 y2      3  2 y  y dy   3 y  43 y 3/2  2  0  0

1



50. R( y )  2  y1/3 and r ( y )  1  V     R ( y )   r ( y )  0





2

2



 dy



2 1 1      2  y1/3  1 dy    4  4 y1/3  y 2/3  1 dy 0 0 



1



1 3 y 5/3      3  4 y1/3  y 2/3 dy   3 y  3 y 4/3  5  0  0





  3  3  35  35 51. (a) r ( x)  x and R ( x )  2 4



 V     R ( x)   r ( x) 0

4

2

2

 dx 4

   (4  x) dx    4 x  x2    (16  8)  8 0   0 2

2



(b) r ( y )  0 and R( y )  y 2  V     R( y )    r ( y ) 0

4

2



2



2

2  y5  dy    y 4 dy    5   325 0  0

(c) r ( x)  0 and R( x)  2  x  V     R( x)   r ( x)  0



4

0

2

2

 dx     2  x  dx 4

0

4

 4  4 x  x  dx   4 x  8 x3  x2 0   16  643  162   83 3/ 2

2

Copyright  2014 Pearson Education, Inc.

2

439

440

Chapter 6 Applications of Definite Integrals



2

(d) r ( y )  4  y 2 and R ( y )  4  V     R ( y )   r ( y )  0





2



2

 dy    16   4  y   dy 2 2

2

0

2



2 2 y5      16  16  8 y 2  y 4 dy    8 y 2  y 4 dy    83 y3  5    0 0  0

 643  325   22415

y

52. (a) r ( y )  0 and R( y )  1  2

 dy       1   dy    1  y   dy   2

 V     R ( y )   r ( y ) 2

2

0

y 2 2

2

0

y2 4

2

0

2





y2 y3   8  2    y  2  12    2  42  12 3  0 y

(b) r ( y )  1 and R ( y )  2  2 2



 V     R( y )    r ( y ) 0

2

2

 dy     2   1 dy     4  2 y  1 dy y 2 2

2

0

2

y2 4

2

0



 



2 y2  y3   8   2  2  8     3  2 y  4  dy   3 y  y 2  12    6  4  12 3 3 0   0



1

53. (a) r ( x)  0 and R( x)  1  x 2  V     R( x)   r ( x)  1

2

2

 dx

1 1  x 2  dx    1  2 x 2  x 4  dx 1  1



2

1

1









3 5 103  16    x  23x  x5   2 1  32  51  2 1515 15   1

 R( x)  r ( x)  dx     2  x  1 dx

1

2

(b) r ( x)  1 and R ( x)  2  x 2  V    1

2

2 2

1

1

1 4  4 x 2  x 4  1 dx     3  4 x 2  x 4  dx   3x  43 x3  x5   2  3  43  15     1 1 1



1

5

  215 (45  20  3)  56 15 1

(c) r ( x)  1  x 2 and R( x)  2  V    1

1

 R( x)  r( x)  dx    4  1  x   dx 2

2

2 2

1

1

1 4  1  2 x 2  x 4  dx     3  2 x 2  x 4  dx   3x  23 x3  x5   2  3  23  15    1 1  1



1

5

  215 (45  10  3)  64 15

54. (a) r ( x)  0 and R( x)   bh x  h b



 V     R ( x)   r ( x) 0



b

0

2

2

 dx

  bh x  h  dx   0b  bh x2  2bh x  h2  dx 2

2

2

2

b





3 2 2   h2  x 2  xb  x    h 2 b3  b  b   h3 b  3b  0

Copyright  2014 Pearson Education, Inc.

1

Section 6.1 Volumes Using Cross-Sections

 

 R( y)  r ( y)  dy   b  1   dy

h

y

441

2

(b) r ( y )  0 and R( y )  b 1  h  V    0

h

y 2 h

2 h 0

2





2 h 2 y y2  y2 y3     b 2   1  h  2  dy   b2  y  h  2    b2 h  h  h3   b3 h 0 3 h  h  0

55. R( y )  b  a 2  y 2 and r ( y )  b  a 2  y 2 a



 V     R( y )    r ( y )  a

2

2

 dy

2 2 b  a 2  y 2    b  a 2  y 2   dy   a      

 

a  a

a

4b a 2  y 2 dy  4b 

a

a 2  y 2 dy

a

2

 4b  area of semicircle of radius a  4b   2a  2a 2 b 2 5

5

56. (a) A cross section has radius r  2 y and area  r 2  2 y. The volume is  2 ydy    y 2   25 .  0 0 (b) V (h)   A(h)dh, so dV  A(h). Therefore dV  dV  dh  A(h)  dh , so dh  A(1h)  dV . dh dt dh dt dt dt dt 3

3

 81  3 units  83  units .

For h  4, the area is 2 (4)  8 , so dh sec sec dt

57. (a)

R( y )  a 2  y 2  V   

h a

a











ha

3  ( h  a )3 y3       a 2 h  a3  3  a3  a3  a 2  y 2 dy    a 2 y  3    a  







2

3 3  h (3a  h)    a 2 h  13 h3  3h 2 a  3ha 2  a3  a3    a 2 h  h3  h 2 a  ha 2  3  

(b) Given dV  0.2 m3 /sec and a  5 m, find dh dt dt

h 4

. From part (a), V (h) 

 dV  10 h   h 2  dV  dV  dh   h(10  h) dh  dh dh dt dh dt dt dt

h4

 h 2 (15 h ) 3

 5 h2   3h

3

0.2 1  4 (10   1 m/sec.  4) (20 )(6) 120

58. Suppose the solid is produced by revolving y  2  x about the y -axis. Cast a shadow of the solid on a plane parallel to the xy -plane.

Use an approximation such as the Trapezoid Rule, n

2

dˆ b 2 to estimate    R( y )  dy     2k  y. a  k 1 

59. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a disk of radius h has been removed. Thus its area is A1   R 2   h 2   ( R 2  h2 ). The cross section of the 2





R 2  h2 . Therefore its area is A2    R 2  h 2    R 2  h 2 .   We can see that A1  A2 . The altitudes of both solids are R. Applying Cavalieri’s Principle we find hemisphere is a disk of radius





 

Volume of Hemisphere  (Volume of Cylinder)  (Volume of Cone)   R 2 R  13  R 2 R  32  R3 .

Copyright  2014 Pearson Education, Inc.

442 60.

Chapter 6 Applications of Definite Integrals 6

6





 

  60536   365 cm3 .

6





2 4 x 36  x 2  V  x  R( x)  12 36  x 2 dx  144    R( x) dx    144  36 x  x dx



6

2

0

2

0





 12 x3  x    12  63  6   6 12  36  196  144 5  0 144 5 144 5 144  5

5

3

0

 

The plumb bob will weigh about W  (8.5) 365  192 gm, to the nearest gram.

61.

R( y )  256  y 2  V  

7

16

2

7

3

   256(16  7)    1053 cm  3308 cm



3 3    (256)(7)  73  (256)(16)  163     

62. (a)

7

y   256  y 2  dy    256 y  3  16    16

  R( y ) dy   

73 3

163 3

3

3

 R( x)2 dx   0 (c  sin x)2 dx   0  c 2  2c sin x  sin 2 x  dx 0

R( x) | c  sin x |, so V   









 c2  2c sin x  1cos2 2 x  dx   0  c2  12  2c sin x  cos22 x  dx    c2  12  x  2c cos x  sin42 x 0    c 2  2  2c  0   (0  2c  0)     c 2  2  4c  . Let V (c)    c 2  2  4c  . We find the   



0

       2  4   2  4; Evaluate V at the endpoints: V (0)  2 and V (1)    32   4   2  (4   ) .

extreme values of V (c) : dV   (2c  4)  0  c  2 is a critical point, and V 2   4  2  8 dc 2

2

2

2

Now we see that the function’s absolute minimum value is 2  4, taken on at the critical point c  2 .

(See also the accompanying graph.)

2

(b) From the discussion in part (a) we conclude that the function’s absolute maximum value is 2 , taken on at the endpoint c  0. (c) The graph of the solid’s volume as a function of c for 0  c  1 is given at the right. As c moves away from [0, 1] the volume of the solid increases without bound. If we approximate the solid as a set of solid disks, we can see that the radius of a typical disk increases without bounds as c moves away from [0, 1].

63. Volume of the solid generated by rotating the region bounded by the x-axis and y  f ( x ) from x  a to b

x  b about the x-axis is V    [ f ( x)]2 dx  4 , and the volume of the solid generated by rotating the same a

b

b

a

a

b

region about the line y  1 is V    [ f ( x)  1]2 dx  8 . Thus    f ( x)  1 dx     f ( x) dx  8  4



b

a

2

2

a

 f ( x)  2 f ( x)  1   f ( x)  dx  4   (2 f ( x)  1) dx  4  2 f ( x) dx   dx  4 2

b

b

b

a

a

a

2

b

b

a

a

  f ( x) dx  12 (b  a)  2   f ( x) dx  4b2 a 64. Volume of the solid generated by rotating the region bounded by the x-axis and y  f ( x ) from x  a to x  b b

about the x-axis is V     f ( x) dx  6 , and the volume of the solid generated by rotating the same 2

a

b

region about the line y  2 is V     f ( x)  2 dx  10 . Thus 2

a

Copyright  2014 Pearson Education, Inc.

Section 6.2 Volumes Using Cylindrical Shells

a   f ( x)  2 dx  a   f ( x) dx  10  6   a  f ( x)  4 f ( x)  4   f ( x)  dx  4 b

b

2

b

2

2

2

b

b

b

b

b

a

a

a

a

a

  (4 f ( x)  4) dx  4  4 f ( x) dx  4 dx  4   f ( x) dx  (b  a )  1   f ( x) dx  1  b  a 6.2

VOLUMES USING CYLINDRICAL SHELLS

1. For the sketch given, a  0, b  2; b

V   2 a



shell radius



 dx   2 x 1   dx  2   x   dx  2     2    2

shell height

2

x2 4

0

x3 4

0

2 x4 16 0

x2 2

 2  3  6

4 2

16 16

2. For the sketch given, a  0, b  2; b

V   2 a





shell radius

 dx   2 x  2   dx  2   2 x   dx  2  x    2  4  1  6 2

shell height

2

x2 4

0

x3 4

0

2

2 x4 16 0

3. For the sketch given, c  0, d  2; d

V   2 c



shell radius



shell height

2

 dy   2 y   y  dy  2  y dy  2    2 2

y4 4

2 3

2

0

0

0

4. For the sketch given, c  0, d  3; d

V   2 c



shell radius



shell height

3

3 3

2

0

0

5. For the sketch given, a  0, b  3; b

V   2 a



shell radius



shell height

 dx   2 x   x  1  dx; 3

2

0

u  x 2  1  du  2 x dx; x  0  u  1, x  3  u  4   



4

4

  23  (8  1)  143

 V    u1/2 du    23 u 3/2   23 43/2  1   1 1 6. For the sketch given, a  0, b  3; b

V   2 a



shell radius



shell height

 dx   2 x 

  dx;

3

9x

0

x3  9 

[u  x3  9  du  3x 2 dx  3 du  9 x 2 dx; x  0  u  9, x  3  u  36] 36

36

3u 1/2 du  6  2u1/2   12  9 9

 V  2 

 36  9   36

7. a  0, b  2; b

V   2 a

2



shell radius



3

 dy   2 y  3  3  y  dy  2  y dy  2    

shell height

 dx   2 x  x     dx

2

2

2

x 2

0

  2 x 2  32 dx    3 x 2 dx    x3   8  0 0 0

Copyright  2014 Pearson Education, Inc.

y4 4

0

9 2

443

444

Chapter 6 Applications of Definite Integrals

8. a  0, b  1; V   2



1

2

b

a

shell height

 dx   2 x  2 x   dx 1

0

  dx    3x dx    x   

   2 32x 0



shell radius

1

x 2

3 1 0

2

0

9. a  0, b  1; b

V   2 a



shell radius





shell height

 dx   2 x (2  x)  x  dx 1

2

0



1

3 4 1  2  2 x  x 2  x3 dx  2  x 2  x3  x4  0   0









  5  2 1  13  14  2 121243  10 12 6

10. a  0, b  1;

  dx   2 x  2  x   x  dx  2  x  2  2 x  dx  4   x  x  dx b

V   2 a



shell radius

1

1

shell height

1

2

0

2

2

0

3

0

1

 4  x2  x4   4   0 2

4

 12  14   

11. a  0, b  1; b

V   2 a

1



shell radius





shell height

 dx   2 x  x  (2 x 1) dx 1

0



1

 2  x3/2  2 x 2  x dx  2  25 x5/2  23 x3  12 x 2   0 0  2

 25  23  12   2  12302015   715

12. a  1, b  4;

 dx   2 x  x  dx  3  x dx  3  x   2  4  1   b

V   2 a



shell radius



4 1/2 1

4

shell height 2 3

1

3/2 4 1

3 2

1/2

3/2

 2 (8  1)  14

13. (a)

 x  sin x , 0  x   sin x, 0  x   x ; since sin 0  0 we have x f ( x)    x f ( x)   x0 x0  0,  x, sin x, 0  x    x f ( x)  sin x, 0  x   x f ( x)   x0 sin x,

Copyright  2014 Pearson Education, Inc.

Section 6.2 Volumes Using Cylindrical Shells b

(b) V   2 a



shell radius



shell height

 dx    2 x  f ( x) dx and x  f ( x)  sin x, 0  x   by part (a) 0





 V  2  sin x dx  2   cos x 0  2 ( cos   cos 0)  4 0

14. (a)

 x  tan 2 x , 0  x   4 x x g ( x)   x0  x  0,

 tan 2 x, 0  x   /4  x g ( x)   ; since tan 0  0 we have x0  0,

 tan 2 x, 0  x   /4  x g ( x)  tan 2 x, 0  x   /4 x g ( x)   2 x0  tan x,

 dx    2 x  g ( x) dx and x  g ( x)  tan x, 0  x   /4 by part (a)     V  2  tan x dx  2   sec x  1 dx  2  tan x  x   2 1       b

(b) V   2 a



shell radius



/4

2

/4

shell height

2

0

/4

0

/4

2

0

0

15. c  0, d  2;

  dy   2 y  y  ( y) dy    2   y  y  dy  2      d

V   2 c

2



shell radius

3/2

2

shell height

0

5/ 2

y3 3

2y 5

2

0

2

0

 2   23   2  8 52  83   16  52  13   3 2 5  16  15 

  2  52 

5

3

16. c  0, d  2;

  dy   2 y  y  ( y) dy    2   y  y  dy  2     16      d

V   2 c

2



shell radius

3

2

2

0

y4 4

0

 16

2

shell height

y3 3

 56   403

2

2 4

0

1 3

17. c  0, d  2;

   dy   2 y  2 y  y  dy    2   2 y  y  dy  2     2      d

V   2 c

2



shell radius

2

3

0





2

shell height

2

0

2 y3 3

y4 4

2

0

16 3

16 4

  8  32 13  14  32 12 3

Copyright  2014 Pearson Education, Inc.

4

4  2

2

445

446

Chapter 6 Applications of Definite Integrals

18. c  0, d  1;

  dy   2 y  2 y  y  y  dy  2  y  y  y  dy  2   y  y  dy d

V   2 c



shell radius

1

1

shell height

2

0

1

2

0

2

3

0

1





 y3 y 4   2  3  4   2 13  14  6  0 19. c  0, d  1; d

V   2 c



shell radius



shell height

 dy  2  y  y  ( y) dy 1

0

3 1

1

 2  2 y 2 dy  43  y   43  0 0

20. c  0, d  2; d

V   2 c



shell radius



shell height

 dy   2 y  y   dy 2

y 2

0

2 2 y2   y 3   8 dy  3  0 3 0 2

 2 

21. c  0, d  2; d

V   2 c



shell radius



shell height

 dy   2 y (2  y)  y  dy 2

2

0

2

y y   2 y  y 2  y 3  dy  2  y 2  3  4  0  0

 2 

3

2



4



 2 4  83  16  6 (48  32  48)  163 4 22. c  0, d  1; d

V   2 c



shell radius





shell height

 dy   2 y (2  y)  y  dy 1

2

0

1



1 y3 y 4    2  2 y  y 2  y 3 dy  2  y 2  3  4  0  0





 2 1  13  14  6 (12  4  3)  56

 (b) V   2  b

23. (a) V   2 a

b

a

 6



8  83

shell radius shell radius

 

  32

shell height shell height

 dx   2 x (3x)dx  6  x dx  2  x   16  dx   2 (4  x) (3x)dx  6   4 x  x  dx  6 2 x  x  2

2 2

0

0

3 2 0

2

2

0

0

2

Copyright  2014 Pearson Education, Inc.

2

2 1 3 3 0

Section 6.2 Volumes Using Cylindrical Shells b

(c) V   2 a





shell radius





shell height

 dx   2 ( x  1) (3x)dx  6   x  x  dx  6  x  x 

shell height

 dy   2 y  2  y  dy  2   2 y  y  dy  2  y  y 

2

2

0

0

2

2 1 2 2 0

1 3 3

 6 83  2  28 d

(d) V   2 c



shell radius



6

6

1 3

0

1 2 3

0

6 1 3 9 0

2

 2 (36  24)  24 d

(e) V   2 c



shell radius



 dy   2 (7  y)  2  y  dy  2  14  y  y  dy 6

shell height

6

1 3

0

1 2 3

13 3

0

6

y 2  19 y 3   2 (84  78  24)  60  2 14 y  13 6  0 d

(f ) V   2 c



shell radius



 dy   2 ( y  2)  2  y  dy  2   4  y  y  dy 6

shell height

6

1 3

0

1 2 3

4 3

0

6

 2  4 y  23 y 2  19 y 3   2 (24  24  24)  48  0 b

24. (a) V   2 a





shell radius





shell height

 2 16  32  965 5 b

(b) V   2 a



shell radius



shell height

 dx   2 x 8  x  dx  2  8x  x  dx  2 4 x  x  2

2

3

0

4

2 1 5 5 0

2

0

 dx   2 (3  x) 8  x  dx  2   24  8x  3x  x  dx 2

2

3

0



2

3

4

0



  2  24 x  4 x 2  34 x 4  15 x5   2 48  16  12  32  264 5 5  0 b

(c) V   2 a



shell radius



shell height

 dx   2 ( x  2) 8  x  dx  2  16  8x  2 x  x  dx 2

2

3

0



2

3

4

0



  2 16 x  4 x 2  12 x 4  15 x5   2 32  16  8  32  336 5 5  0

 (e) V   2  d

(d) V   2 c

d

c

 2



d

c



shell radius

96  384 7

(f ) V   2

 2π

shell radius



b

a





 dy   2 y  y dy  2  y dy    y    (128)    dy   2 (8  y) y dy  2  8 y  y  dy  2 6 y  y  8

shell height

8 4/3

1/3

0

8

shell height

8

1/3

0

7/3 8 0

6 7

0

1/3

6 7

4/3

4/3

0



8

8

1/3

0

4/3

1/3

3 7

0

b



7/3

 936π 7

shell radius



shell height

 dx   2 (2  x)  x  2  x  dx  2   4  3x  x  dx 2

2

2

1

2

3

1



2

a

8 3 7/3 7 0

 dy   2 ( y  1) y dx  2   y  y  dy  2  y  y 

shell height



 2  4 x  x3  14 x 4   2 (8  8  4)  2 4  1  14  272   1 (b) V   2

768 7

  576 7

shell radius

384  12 7

25. (a) V   2



 

shell radius



shell height 2

 dx   2 ( x  1)  x  2  x  dx  2   2  3x  x  dx 2

2

2

1

1





 2  2 x  32 x 2  14 x 4   2 (4  6  4)  2 2  32  14  272   1

Copyright  2014 Pearson Education, Inc.

3

3 4

4/3 8 0

447

448

Chapter 6 Applications of Definite Integrals



 dy   2 y  y    y  dy   2 y  y  ( y  2)  dy  4  y dy  2   y  y  2 y  dy    y   2  y  y  y      d

(c) V   2 c



shell radius

shell height

1 3/2

4

0

1

 85 (1)  2



3/2

1

4

0

1

2

5/2 1 0

8 5

 645  643  16  2  52  13  1  725

2 4 1

1 3 3

 dy   2 (4  y)  y    y  dy   2 (4  y)  y  ( y  2)  dy  4   4 y  y  dy  2   y  y  6 y  4 y  8 dy d

(d) V   2 c

shell radius

1



2 5/2 5

shell height

1

4

0

1

4

3/2

0

2

3/2

1

1

4

 4  83 y 3/2  52 y5/2   2  13 y 3  52 y 5/2  3 y 2  83 y 3/2  8 y   0  1





 643  645  48  643  32  2  13  25  3  83  8  1085  .

shell radius



 4 83  52  2 b

26. (a) V   2 a



shell height

 dx   2 (1  x)  4  3x  x  dx  2   x  x  3x  3x  4 x  4 dx 1

2

1

d

c



shell radius



shell height

1 4  4 y 5/4 dy  4 y 0 3 1





1

5

4

3

2

1

 2  16 x 6  51 x5  34 x 4  x3  2 x 2  4 x   2   1 (b) V   2

1

4

1

 16  15  34  1  2  4  2  16  15  34  1  2  4  565

 dy   2 y  y    y  dy   2 y  1

4

4

4

0

1

4 y  4 y      3   dy 3  



4  ydy [u  4  y  y  4  u  du   du; y  1  u  3, y  4  u  0]

0

3





3

 169  y 9/4   4  (4  u ) u du  169 (1)  4  4 u  u 3/2 du  169  4  83 u 3/2  52 u 5/2   0 0 3 3 3 0 3 3



27. (a) V   2





  169  4 8 3  18 3  169  885  872 5 45

d

c

 24

shell radius



shell height

 14  15   2420  65

d

(b) V   2 c



shell radius



shell height











 dy   2 (1  y) 12  y  y  dy  24  (1  y)  y  y  dy 1

2

1

3

0

1



3

 13  12  15   24  301   45

 dy   2   y  12  y  y  dy  24    y   y  y  dy    24   y  y  y  dy  24  y  y    24       d

(c) V   2 c

shell radius

1 8 0 5

2



2

0

1  y3 y 4 y5   24  y 2  2 y 3  y 4 dy  24  3  2  5   24 0  0



1



1 1  y 4 y5  dy   2 y 12 y 2  y 3 dy  24  y 3  y 4 dy  24  4  5  0 0  0

1

shell height

13 3 5

0

4

8 5

8 3 15

13 20

2

3

4

y5 5

1 8 0 5

1

8 15

0

2

13 20

3

1 5

 (32  39  12)  24  2  24 60 12

(d) V   2



 dy   2  y   12  y  y  dy  24   y    y  y  dy    24   y  y  y  y  dy  24   y  y  y  dy  24  y  y     1

3

d

c

shell radius



4

2 5

0

 24



2  3 1 15 20 5



1

shell height 2

0

2 3 5

2

2 5

1

3

0

1 2 0 5

2

3 3 5

4

 (8  9  12)  24  2  24 60 12

Copyright  2014 Pearson Education, Inc.

2 5

2

2 3 15

3 20

3

4

y5 5

1 0

Section 6.2 Volumes Using Cylindrical Shells d

28. (a) V   2 c





shell radius

d

c





shell radius

y2 2

2

0

y4 4

y2 2

2

2

0

    32     32     32   

2

(b) V   2

 dy   2 y      dy   2 y  y   dy  2   y   dy

shell height

 y4 y6   2  4  24   2  0

24 4

26 24

1 4

4 24

1 4

1 6

2 24

y4 4

2

3

0

8 3

 dy   2 (2  y)      dy   2 (2  y)  y   dy y2 2

2

shell height

0

y4 4

y2 2

2

y4 4

2

0

2





2 y4 y5   2 y3 y5 y 4 y 6   2   2 y 2  2  y 3  4  dy  2  3  10  4  24   2 16  32  16  64  85 3 10 4 24 0   0

d

(c) V   2 c





shell radius

 dy   2 (5  y)      dy   2 (5  y)  y   dy y2 2

2

shell height

0

y4 4

y2 2

2

2 y5   5 y3 5 y5 y 4 y 6   2   5 y 2  54 y 4  y 3  4  dy  2  3  20  4  24   2 0   0

d

c





shell radius

y4 4

2

0

2

(d) V   2

64  8  16  24  403  160  20 4

 dy   2  y        dy   2  y    y   dy 2

shell height

5 8

0

y2 2

y4 4

y2 2

2

2

d

29. (a) About x-axis: V   2 c



shell radius



shell height

2

5 8

0

4 6 3 5 2 y5 5 y 4  dy  2  y  y  5 y  5 y   2  2   y 3  4  58 y 2  32    4 24 24 160 0   0

y4 4

 4  164  6424  4024  160 160 

 dy

1 y  y  dy  2   y 3/2  y 2  dy  0 0 1  2  52 y 5/2  13 y3   2  52  13   215  0 1

  2 y

b

About y -axis: V   2



1

a





shell radius 1



shell height



 dx



  2 x x  x 2 dx  2  x 2  x3 dx 0

0

1





 2  x3  x4   2 13  14  6   0 3

4

b

(b) About x-axis: R( x)  x and r ( x)  x 2  V    a

1



 R( x)  r ( x)  dx     x  x  dx 2

5

About y -axis: R( y )  1

 y 2 y3    2  3    0 b

30. (a) V    a

d

y and r ( y )  y  V    c

2

 R( x)  r ( x)  dx      2  x  dx

4

4

 R( y)  r ( y)  dy     y  y  dy

4

2

0

x 2

2

2

 3 x 2  2 x  4  dx     x4  x 2  4 x  0 4   0



2



 12  13   6

2

1

2

0

   x3  x5    13  15  215   0 3

449

3

4

  16  16  16   16

Copyright  2014 Pearson Education, Inc.

2

1

0

2

y5 4

450

Chapter 6 Applications of Definite Integrals b

(b) V   2 a





shell radius

shell height

4

 dx   2 x   2  x  dx   2 x  2   dx  2   2x   dx 4



4

x 2

0

4

x 2

0

x2 2

0



 2  x 2  x6   2 16  64  323 6   0 3

b

(c) V   2 a





shell radius

shell height

 dx   2 (4  x)   2  x  dx   2 (4  x)  2   dx  2  8  4 x   dx 4

0

4

4

x 2



4

x 2

0

x2 2

0



 2 8 x  2 x 2  x6   2 32  32  64  643 6   0 3



b

(d) V     R( x)    r ( x) a

2

2



4 3 2 x  10 x  28 0 4



d

31. (a) V   2 c

2

 2 

1







shell radius



 dx    (8  x)   6    dx     64 16 x  x   36  6 x   dx 4

2

0

x 2

2

4

2

0

x2 4

 dx    x4  5x2  28x 0   [16  (5) (16)  (7) (16)]   (3) (16)  48 4

3

shell height

 dy   2 y( y 1) dy 2

1

2



 y3 y 2  y 2  y dy  2  3  2   1

    2  73  2  12   3 (14  12  3)  53

 2  83  42  13  12   

b

(b) V   2 a





shell radius



shell height

 



 dx   2 x(2  x) dx  2   2 x  x  dx  2  x   2

2

1

1

2

2



2 x3 3 1

    34  32   43 b shell shell (c) V   2  radius  height  dx  12 2  103  x  (2  x) dx  2 12  203  163 x  x2  dx a 2 x  83 x 2  13 x3   2  40  2  20  32  8  20  8  1   2  33   2 3 1  3 3 3   3 3 3   2  4  83  1  13   2  1238  331   2    

d

(d) V   2 c

d

32. (a) V   2 c



shell radius



shell height

 dy   2 ( y  1)( y 1) dy  2  ( y  1)  2 



shell radius



shell height

 dy   2 y  y  0 dy

b



shell radius



4



shell height



4

1

2

   8 24 4



4

0

 2 x  x3/2  dx 5

 2  x 2  52 x5/2   2 16  252  0





2 ( y 1)3   23 3  1

 dx

  2 x 2  x dx  2  0

1

2

0

2

a

2

2

2  y4   2  y 3 dy  2  4   2 0  0

(b) V   2

2



 2 16  64  25 (80  64)  325 5 b

(c) V   2 a



shell radius



shell height

 dx   2 (4  x)  2  x  dx  2  8  4x  2 x  x  dx 4

0 4  2 8 x  83 x3/2  x 2  25 x5/2   2  0

4

1/2

3/2

0

32  643  16  645   215 (240  320  192)  215 (112)  22415

Copyright  2014 Pearson Education, Inc.

Section 6.2 Volumes Using Cylindrical Shells

c





16  16 3 4

d

(d) V   2

 2

shell radius





shell height

 dy   2 (2  y)  y  dy  2   2 y  y  dy  2  y   2

2

2

0

2

3

y4 4

2 3 3

0

451

2 0

 (4  3)  8  32 12 3

  dy   2 y  y  y  dy     2  y  y  dy  2     2        (b) V   2    dy   2 (1  y)  y  y  dy    2   y  y  y  y  dy  2        d

33. (a) V   2 c

1



shell radius

2

4

1

shell height

y3 3

0

d

c

shell radius

1

2

y5 5

 2



d

34. (a) V   2 c



shell radius







1 3

0

4 15

1 5

3

0

y2 2

4

0

1111 2 3 4 5

1

1

shell height

3

3

0

y3 3

y4 4

y5 5

1

0

 260 (30  20  15  12)  730

shell height

 dy   2 y 1   y  y  dy 1

3

0

1



1  y 2 y3 y5   2  y  y 2  y 4 dy  2  2  3  5  0  0

 2

 12  13  15   230 (15  10  6)  1115

(b) Use the washer method: d



V     R ( y )   r ( y )  c

2

2

 dy    1   y  y   dy    1  y  y  2 y  dy 1

3 2

2

1

0

1



2

6

4

0



y3 y7 2 y5    (105  35  15  42)  97    y  3  7  5    1  13  17  52  105 105  0 (c) Use the washer method: 2 2 1  1 d  2 2 V     R( y )    r ( y ) dy     1  y  y 3   0  dy    1  2 y  y 3  y  y 3  dy 0 0 c    













 



1



1 y3 y7 y 4 2 y5      1  y 2  y 6  2 y  2 y3  2 y 4 dy    y  3  7  y 2  2  5  0  0





 (70  30  105  2  42)  121   1  13  17  1  12  52  210 210

 dy   2 (1  y) 1   y  y  dy  2  (1  y) 1  y  y  dy    2  1  y  y  y  y  y  dy  2  1  2 y  y  y  y  dy  2  y  y       d

(d) V   2 c

1



shell radius



3

 2 1  1  13  14  15

1

3

0

2

0



1

shell height

4

1

2

3

4

0



3

0

  260 (20  15  12)  23 30

Copyright  2014 Pearson Education, Inc.

2

y3 3

y4 4

y5 5

1 0

452

Chapter 6 Applications of Definite Integrals d

35. (a) V   2 c



shell radius



shell height

 dy   2 y  8 y  y  dy 1

2

0

2

y   2 2 y 3/2  y 3  dy  2  4 5 2 y 5/2  4  0  0 4

2

 2 

 4 2  2 5  24   2  2   5 4    





423  44 5 4





 2  4 85  1  85 (8  5)  245 b

(b) V   2 a

 2



5

22 5





shell radius 4



shell height

4  2  32

4

0

   6

3/2

0

2 32

7

9

160

160

x3 8

2 5/2 5

 2 (32  20)   2 3   24 3  48

8

2 5

4

x2 8

5

5

  dx   2 x  2 x  x   x  dx  2  x  x  x  dx  2   x  x  dx b

36. (a) V   2 a



 dx   2 x  x   dx  2   x   dx  2  x  

shell radius

1

1

shell height

2

0

1

2

0

2

3

0

1





 2  x3  x4   2 13  14  6   0 3

4

b

(b) V   2 a





shell radius



shell height

 dx   2 1  x   2 x  x   x dx  2  1  x   x  x  dx 1

2

0



1

2 4 1  2  x  2 x 2  x3 dx  2  x2  23 x3  x4   2 0   0



b

a

2

 dx     x

1/2





1

37. (a) V     R( x)    r ( x)

2

1/16

1

1

 12  23  14   212 (6  8  3)  6



 1 dx

1     2 x1/2  x    (2  1)  2  14  16  1/16  





7  9   1  16 16

d

(b) V   2 c

 2 

2

1







shell radius



shell height

 dy   2 y    dy 1

1 y4

0



1 16

2

y y2   y 3  16 dy  2   12 y 2  32   1

 



1   2  2     18   12  32  

   321 

 2 (8  1)  9 d



38. (a) V     R ( y )   r ( y )  c

2

2

2



2

0

 dy       dy 2

1

 

1 y4

1 16



y 1 1  1  1      13 y 3  16      24 8 3 16    1

 ( 2  6  16  3)  11  48 48

Copyright  2014 Pearson Education, Inc.

4 x4 32 0

Section 6.2 Volumes Using Cylindrical Shells b

(b) V   2 a





shell radius



shell height

 

453

 dx   2  1 dx  2   x  x  dx  2  x   1



1    2  23  12  23  18  32  

1

1 x

1/4

1/2

1 x2 2 1/4

2 3/2 3

1/4

 34  1  61  161   48 (4 16  48  8  3)  1148

39. (a) Disk: V  V1  V2 b

b

V1   1   R1 ( x) dx and V2   2   R2 ( x)  dx with R1 ( x)  2

a1

2

a2

x 2 3

and R2 ( x)  x ,

a1  2, b1  1; a2  0, b2  1  two integrals are required (b) Washer: V  V1  V2

 dx with R ( x)   V     R ( x)    r ( x)  dx with R ( x)  b

V1   1   R1 ( x)   r1 ( x) 2

2

1

a1

b2

2

2

2

a2

2

2

 two integrals are required



d shell (c) Shell: V   2 radius c



shell height

x 2 3

x 2 3

2

 dy   2 y  d

c

and r1 ( x )  0; a1  2 and b1  0;

shell height

and r2 ( x)  x ; a2  0 and b2  1

 dy where shell height  y  3 y  2  2  2 y ; 2

2

2

c  0 and d  1. Only one integral is required. It is, therefore preferable to use the shell method.

However, whichever method you use, you will get V   .

40. (a) Disk: V  V1  V2  V3 d

Vi   i   Ri ( y ) dy, i  1, 2, 3 with R1 ( y )  1 and c1  1, d1  1; R2 ( y )  2

y and c2  0 and d 2  1;

ci

R3 ( y )  ( y )1/4 and c3  1, d3  0  three integrals are required (b) Washer: V  V1  V2



d

Vi   i   Ri ( y )    ri ( y )  ci

2

2

 dy, i  1, 2 with R ( y)  1, r ( y)  y , c  0 and d  1; 1

1

1

1

R2 ( y )  1, r2 ( y )  ( y )1/4 , c2  1 and d 2  0  two integrals are required



b shell (c) Shell: V   2 radius a



shell height

 dx   2 x  b

a

shell height

 dx , where shell height  x    x   x  x , a  0 2

4

2

4

and b  1  only one integral is required. It is, therefore preferable to use the shell method.

However, whichever method you use, you will get V  56 . b



41. (a) V     R( x)    r ( x) a

2

4

2

 dx     25  x   (3)  dx     25  x  9 dx    16  x  dx 

4

2

2

2

4



4

2

4

 



   16 x  13 x3    64  64   64  64  256 3 3 3   4

  Volume of portion removed  500  256  244 (b) Volume of sphere  43  (5)3  500 3 3 3 3



b shell 42. V   2 radius a



shell height

 dx    2 x sin  x  1 dx; [u  x 1  du  2 x dx; 1

2

2

1





x  1  u  0, x  1    u   ]    sin u du    cos u 0   (1  1)  2 0

Copyright  2014 Pearson Education, Inc.

4

4

2

454

Chapter 6 Applications of Definite Integrals



b shell 43. V   2 radius a



2

2





shell height

 dx   2 x   x  h  dx  2    x  h x  dx  2  x  x  r

0

r

h r

0

h 2 r

h 3 3r

r h 2 2 0

2

 2  r 3h  r 2h  13  r h

c



2

2

d shell 44. V   2 radius



shell height

 dy   2 y  r  y    r  y  dy  4  y r  y dy r

2

2

2

r

2

0

2

2

0

0

2

[u  r  y  du  2 y dy; y  0  u  r , y  r  u  0]  2  2 r

2 r 2 1/2 3 2 r  4  u du  2 u du  3 u  0 0



 43 r 3 45. W (a)  

f (a) f ( a)

a

 [( f 1 ( y ))2  a 2 ]dy  0   2 x[ f (a)  f ( x)]dx  S (a); a

W (t )   [ f 1 ( f (t )))2  a 2 ] f (t )   (t 2  a 2 ) f (t ); also t

t

t

a

a

a

S (t )  2 f (t )  x dx  2  xf ( x) dx  [ f (t )t 2   f (t )a 2 ]  2  xf ( x)dx  S (t )   t 2 f (t )  2 tf (t )   a 2 f (t )  2 tf (t )   (t 2  a 2 ) f (t )  W (t )  S (t ). Therefore, W(t) = S(t) for all t  [a, b].

46. V   

 /3

0

[22  (sec y )2 ]dy   [4 y  tan y ]0 /3  

 43  3 

2 2 1 b 1  shell  shell  47. V   2  dx   2 xe x dx   e x   (e1  e0 )   1  1e   a 0 0  radius  height 



48. Use washer cross sections. A washer has inner radius r = 1, outer radius R  e x /2 , and area

 ( R 2  r 2 )   (e x  1). The volume is V 

ln 3

0

 (e x  1)dx   [e x  x]

ln 3 0

  (3  ln 3  1)   (2  ln 3)

6.3

1.

ARC LENGTH

 x2  2   2 x   x 2  2   x 3 3  L   1   x 2  2  x 2 dx   1  2 x 2  x 4 dx 0 0 3 2 3 3   1  x 2  dx   1  x 2  dx   x  x3  0 0   0 dy  13  32 dx

1/2

3

 3  27  12 3

Copyright  2014 Pearson Education, Inc.



Section 6.3 Arc Length

2.

dy  32 dx

xL

4

1  94 x dx;

0

u  1  9 x  du  9 dx  4 du  dx; 4 4 9  x  0  u  1; x  4  u  10]



10





10



8 10 10  1  L   u1/2 94 du  94  23 u 3/2   27  1 1

3.

  y  

dx  y 2  1  dx dy dy 4 y2

L

3

1

2

4

1 2

1 16 y 4

3 1  y 4  12  1 4 dy   y 4  12  1 4 dy 1

16 y

16 y

2

3  2   y  1 2  dy    y 2  1 2  dy 1  1 4y  4y 



3

3



 



 y 3 y 1  1 11   3  4   27  1  13  14  9  12 3 12 3 4  1 9

4.

( 1 4 3) ( 2)  9  12  53 12 6

   y2   L   1   y  2   dy    y  2   dy dx  1 y1/2  1 y 1/2  dx 2 2 dy dy 9

1 4

1

2

1 4

1 y

9

1 y

1

2

1 4



1 y



9 9     12  y  1  dy  12  y1/2  y 1/2 dy 1 1 y   9

9  y 3/ 2   12  23 y 3/2  2 y1/2    3  y1/2   1  1





3



 33  3  13  1  11  13  32 3

5.

  y  

dx  y 3  1  dx dy dy 4 y3

L

2

1



2

1



2

6

1 2

1 16 y 6

2 1  y 6  12  1 6 dy   y 6  12  1 6 dy 16 y

1

2

16 y

2

2  3 y 3   3 y 3   y 4 y 2   y  4  dy  1  y  4  dy   4  8       1

 

1 11  16  1  14  81  4  32 4 (16)(2) 4 8 18 4  123  12832 32

Copyright  2014 Pearson Education, Inc.

455

456

6.

Chapter 6 Applications of Definite Integrals

   y 2 y 

2

dx  y  1  dx dy dy 2 2 y2

L

3

2

2

4

4

1 4





1  14 y 4  2  y 4 dy

 y 4  2  y 4  dy 2 3 3  12   y 2  y 2  dy  12   y 2  y 2  dy 2 2 

3

1 4

2

3



 



 y3   12  3  y 1   12  27  1  83  12   3 3   2



 



 12 26  83  12  12 6  12  13 3 4 7.

 

2/3 dy dy 2  x1/3  14 x 1/3  dx  x 2/3  12  x16 dx

L

8

1



8

2/3

1  x 2/3  12  x16 dx

x1/3  14 x 1/3  dx 1 

2/3

x 2/3  12  x16 dx  

1

2

8

x1/3  14 x 1/3  dx   34 x 4/3  83 x 2/3   1 1 8  83  2 x 4/3  x 2/3   83  2  24  22   (2  1)   1  



8

8

 83 (32  4  3)  99 8 8.

dy  x 2  2 x  1  4 2  x 2  2 x  1  14 1 2 dx (4 x  4) (1 x )

 (1  x)2  14 L

2

0



2

0

 

2 1  dy 1  (1  x)4  12  2 dx (1 x ) 16(1 x )4

1  (1  x) 4  12 

(1  x)4  12 

(1 x )4 dx 16

(1 x )4 dx 16 2

(1 x )2   (1  x)2  4  dx  0  



2





2 3 (1 x )2    (1  x) 2  4  dx; [u  1  x  du  dx; x  0  u  1, x  2  u  3]  L   u 2  14 u 2 du 1 0  3



 



1  1  1  1081 4 3  106  53   u3  14 u 1   9  12 3 4 12 12 6  1 3

Copyright  2014 Pearson Education, Inc.

Section 6.3 Arc Length 9.

        2

dx  1  x  dx dy x 4 dy 2

L

1

x 2 4

1 x

1 x2 2

1 2

2 1  12  1  x dx  1 x 2 16

y

x2 16

0.5

1  1  x 2 dx  x 2 2 16



y  ln x 

2

1  1x  4x  dx  1  1x  4x  dx  ln x  x8  1  2

2

2

2

 ln 2  48    ln1  18   ln 2  83 10.

 

dy dy 2  x  41x  dx  dx 3

L

1



3

1

1 

1 x

2

0

2

5 4

 12  1 2 dx  16 x

 1

16 x

3

y

3



11.

1

x  41x dx   x2  14 ln x     1

0

  x    x  

2 dy 2 1  dy  x   2 dx dx 4x

L

3

1

1 x

4

2

1 4 x2

2

4

1 2

1 16 x 4

3

4

2

4

1  3

12.



3

3

L

1

1/2

1  x8  12 

4

1 4 x4

2

8

1 2

1 dx  16 x8 1

8

1/2  x  41x  dx   x5  121x  1/2 

sec 4 y  1 

L

 /4

 /4

3

y

x5 1  5 12 x 3

x

y

0.5

3

0

 15  121    1601  23   373 480 dx  dy

2

1

5

4

13.

1

2

4

4

4

x3 1  3 4x

1

8

1

0

1 16 x8

1/2 x  12  161x dx  1/2  x  41x  dx  1

y

2

 9  121    13  13   536

  x    x  

dy dy 2  x 4  1 4  dx  dx 4x

x

3

4

2

x 2  1 2 dx   x3  41x     1 4x

2

10 6

2

1

y

8

 12  1 4 dx  16 x

1 x  12  161x dx  1  x  41x  dx  3

x 2 ln x  2 4

2

2 x  41x dx 

3

2

x

2

y

2

  92  14 ln 3   12  14 ln1  4  14 ln 3 3

1

0.5

 x  41x   x2  21  161x

x 2  12  1 2 dx  

x2 8

0.5

   sec y 1 dx dy

2



4



1  sec4 y  1 dy  

 /4

 /4

sec 2 y dy

 /4

  tan y  /4  1  (1)  2

Copyright  2014 Pearson Education, Inc.

1

1.5

x

457

458

14.

Chapter 6 Applications of Definite Integrals

dy  dx

3x4  1 

dy 2 dx



1

L

   3x  1 4



1  3x 4  1 dx  

2

1

2

3 x 2 dx

1

 3  x3   33  1  (2)3   33 (1  8)  7 3 3   2   3

15. (a)

   L   1    dx dy 2 dx

2

1



2

1

(c)

16. (a)

1  4x 2 dx

L  6.13

 

dy dy 2  sec2 x  dx  sec4 x dx 0

L

 /3

(c)

17. (a)

   cos y



2

(b)

2

1  cos 2 y dy

L  3.82

 

dx   y dx  dy dy 1 y 2

L

1/2

1/2



1/2

1/2

(c)

1  sec4 x dx

dx  cos y  dx dy dy 0

18. (a)

(b)

L  2.06

L (c)

(b)

dy dy 2  2 x   4x2 dx dx

1

1  y 2 

y

2

2

1 y 2

1/2

(b)

y2 1 y

2

dy  

1/2

1/2

1 dy 1 y 2

dy

L  1.05

Copyright  2014 Pearson Education, Inc.

Section 6.3 Arc Length 19. (a) 2 y  2  2 dx  dy 3

L

20. (a)

 



1  x 2 sin 2 x dx

L  4.70

 

 /6

0



cos x

dx  dy

sin 2 x  cos 2 x dx cos 2 x

0

sec2 y  1   /4

 /3



 /4

 /3

23. (a)

 /6

0

L  0.55

L

(c)

1  tan 2 x dx  

 /6 dx  /6  sec x dx

0

22. (a)

(b)

dy dy 2  tan x  dx  tan 2 x dx

L

(c)

(b)

dy dy 2  cos x  cos x  x sin x   x 2 sin 2 x dx dx 0

21. (a)

(b)

2

L  9.29

L (c)

2

dx dy

1  ( y  1)2 dy

1

(c)

   ( y  1)

459

   sec y 1 dx dy

2



(b)

2



1  sec 2 y  1 dy  

 /4

 /3

| sec y | dy

sec y dy

L  2.20

  corresponds to dy 2 dx

1 here, so take dy as 1 . Then y  dx 4x 2 x

x  C and since (1, 1) lies on the curve,

C  0. So y  x from (1, 1) to (4, 2). (b) Only one. We know the derivative of the function and the value of the function at one value of x.

24. (a)

  corresponds to dx dy

2

C  1.

1 y4

dy here, so take dx as 12 . Then x   1y  C and, since (0, 1) lies on the curve, y

So y  11x .

(b) Only one. We know the derivative of the function and the value of the function at one value of x.

Copyright  2014 Pearson Education, Inc.

460 25.

Chapter 6 Applications of Definite Integrals x

y



dy

cos 2t dt  dx  cos 2 x  L  

0  /4

 /4

2 cos x dx  2 sin x 0

0

26.



2/3 3/2

1

1  1 x2/3 dx  

y  1 x 

 /4

2

1   cos 2 x  dx  

0

 ,

2 /4

x

 

1  1 dx  1  2/3 

 23 x 1/3

1

x

x

2/3 1/ 2

x1/3

dx  

1

2 /4



1 x  

2/3

1

        total length  8    6 2/3

 32 (1)2/3  32 42 27.

3 2

3 1 2 2

3 4

2 cos 2 x dx

2

 1 x 2/3 1/ 2   dx 1   L 2 /4 x1/3     1

1 1 dx  1 x 1/3 dx  32  x 2/3    2 /4 2 /4 x1/3 2 /4



3 4

dy

y  3  2 x, 0  x  2  dx  2  L  

2

0

2

 /4

0

 



1

1  cos 2 x dx  

 2 sin 4  2 sin(0)  1

1/2 dy 2  x  1  dx  32 1  x 2/3 4

2/3

2 /4

 /4

0

1  (2)2 dx  

2

0

2

5 dx   5 x   2 5.

0

2

d  (2  0)  (3  (1))  2 5

28. Consider the circle x 2  y 2  r 2 , we will find the length of the portion in the first quadrant, and multiply our result by 4. dy

y  r 2  x 2 , 0  x  r  dx  r

r

0

r 2  x2

 4

r

dx

0

r 2  x2

dx  4r 

x 2

r x

 L  4 2

r

0

2

2 r r   1   2 x 2  dx  4 1  2x 2 dx  4  0 0 r x  r x 

r 2 dx r  x2 2

d 9 x 2   d  y ( y  3) 2   18 x dx  2 y ( y  3)  ( y  3) 2  3( y  3)( y  1) 29. 9 x 2  y ( y  3)2  dy dy   dy  

 dx  dy 

( y 3)( y 1) ( y 3)( y 1)  dx  dy; 6x 6x

( y 3)2 ( y 1)2 4 y ( y 3)

2

2

2

2

( y 3)( y 1) ( y 3) ( y 1) ds 2  dx 2  dy 2   dy   dy 2  dy 2  dy 2 6x   36 x 2

y 2  2 y 1 4 y ( y 1) 2  ( y 1)2  dy 2  dy 2   4 y  1 dy 2  dy 2  4 y dy 2 4 y  

d  4 x 2  y 2   d 64  8 x  2 y dy  0  dy  4 x  dy  4 x dx; 30. 4 x 2  y 2  64  dx dx dx y y   dx   2 2 2  2 2 y 2 16 x 2  ds 2  dx 2  dy 2  dx 2   4yx dx   dx 2  16 x2 dx 2   1  16 x2  dx 2  dx 2  4 x 64216 x dx 2 2   y y y y   2

 20 x 264 dx 2  42 (5 x 2  16) dx 2 y

31.

y

2x

x

0

1

  dt, x  0  2  1      1  y  f ( x)   x  C where C is any real dy 2 dt

dy 2 dx

dy dx

number.

Copyright  2014 Pearson Education, Inc.

Section 6.3 Arc Length

461

32. (a) From the accompanying figure and definition of the differential (change along the tangent line) we see that dy  f ( xk 1 ) xk  length of kth tangent fin is

  xk 2  (dy)2    xk 2   f ( xk 1 ) xk 2 .

n

n

n  k 1

n k 1

  xk 2   f ( xk 1 ) xk 2

(b) Length of curve  lim  (length of kth tangent fin)  lim  n

 lim  1   f ( xk 1 )  xk   2

33.

b

a

n  k 1



1   f ( x)  dx 2

4



x 2  y 2  1  y  1  x 2 ; P  0, 14 , 12 , 43 , 1  L  

k 1

 xi  xi 1 2   yi  yi 1 2

 14  0    415  1   12  14    23  154    34  12    47  23   1  34    0  47  2

2



2

2

2

2

2

2

 1.55225 y y

dy

34. Let ( x1 , y1 ) and ( x2 , y2 ), with x2  x1 , lie on y  mx  b, where m  x2  x1 , then dx  m 2

x

x

x1

1

 L   2 1  m 2 dx  1  m2  x x2  1  m 2  x2  x1   1 

 x2  x1 2  y2  y1 2  x2  x1 2  y2  y1 2 x  x     x2  x1   2 1  x2  x1   x2  x1 2



35.

dy

y  2 x3/2  dx  3x1/2 ; L( x)  

x

0



1  3t1/2



3

dy

x

0



x

0

1

2

19 x

1 9 x

2  u 3/2  u du  27  1

2 (1  9 x)3/2  2 ;  27 27

2(10 10 1) 27

y  x3  x 2  x  4 x1 4  dx  x 2  2 x  1  L( x)  

2

 x2  x1 2   y2  y1 2 .

1

36.

 x  x 

y2  y1 2 x2  x1

 dt  0x 1  9t dt;

[u  1  9t  du  9dt ; t  0  u  1, t  x  u  1  9 x]  19  2 (10)3 2  2  L(1)  27 27

1

1  ( x  1)2  1 2 ; 4( x 1)2 4( x 1)

2

2

x x [4(t 1)4 1]2    4(t 1)4 1    1  (t  1) 2  1 2  dt   1   dt 1 dt 2 0 0 4(t 1)  16(t 1) 4   4(t 1) 

16(t 1) 4 16(t 1)8 8(t 1)4 1 16(t 1)

4

dt  

x

0

16(t 1)8 8(t 1)4 1 16(t 1)

4

dt  

x

0

[4(t 1)4 1]2 16(t 1)

4

dt  

x 4(t 1) 4 1

0 4(t 1)2

dt

x x 1    (t  1) 2  1 2  dt ; [u  t  1  du  dt ; t  0  u  1, t  x  u  x  1]   u 2  14 u 2  du  0  1  4(t 1)  x 1

  13 u 3  14 u 1   1



 

1 ; L(1)  8  1  1  59  13 ( x  1)3  4( x11)  13  14  13 ( x  1)3  4( x11)  12 3 8 12 24

Copyright  2014 Pearson Education, Inc.

462

Chapter 6 Applications of Definite Integrals

37-42.

Example CAS commands:

Maple: with( plots ); with( Student[Calculus1] ); with( student ); f : x - sqrt(1-x^2);a : -1; b : 1; N : [2, 4, 8]; for n in N do xx : [seq( a  i*(b-a)/n, i  0..n )];

pts : [seq([x, f (x)], x  xx)]; L : simplify(add( distance(pts[i 1], pts[i]), i 1..n )); T : sprintf("#37(a) (Section 6.3)\nn  %3d L  %8.5f \n", n, L );

# (b)

P[n] : plot( [f (x), pts], x  a..b, title  T ):

# (a)

end do: display( [seq(P[n], n  N)], insequence  true, scaling  constrained ); L : ArcLength( f(x), x  a..b, output integral ): L  evalf ( L );

# (c)

Mathematica: (assigned function and values for a, b, and n may vary) Clear[x, f ] {a, b}  {1, 1}; f[x_ ]  Sqrt[1  x 2 ] p1  Plot[f[x], {x, a, b}] n  8; pts  Table[{xn, f[xn]}, {xn, a, b, (b  a)/n}]/ / N Show[p1,Graphics[{Line[pts]}]}] Sum[ Sqrt[ (pts[[i 1, 1]]  pts[[i, 1]])2  (pts[[i  1, 2]]  pts[[i, 2]])2 ], {i, 1, n}] NIntegrate[Sqrt[ 1  f '[ x]2 ], {x, a, b}]

6.4

AREAS OF SURFACES OF REVOLUTION

1. (a)

 

dy dy 2  sec2 x  dx  sec4 x dx

 S  2 

 /4

0

(c)

(b)

(tan x) 1  sec4 x dx

S  3.84

Copyright  2014 Pearson Education, Inc.

Section 6.4 Areas of Surfaces of Revolution

2. (a)

 

dy dy 2  2 x   4x2 dx dx

(b)

2

 S  2  x 2 1  4 x 2 dx 0

(c)

3. (a)

S  53.23

 

dx   1  xy  1  x  1y  dy 2

dx dy

y

2

1 y4

(b)

2  S  2  1y 1  y 4 dy 1

(c)

4. (a)

S  5.02

   cos y 2

dx  cos y  dx dy dy

(b)

2



 S  2  (sin y ) 1  cos 2 y dy 0

(c)

5. (a)

S  14.42



x1/2  y1/2  3  y  3  x1/2

(b)

2

  12 x1/2  2 dy 2   dx   1  3x 1/2  2 2 4  S  2   3  x1/2  1  1  3x 1/2  dx 1 dy





 dx  2 3  x1/2

(c) 6. (a)

S  63.37

   1  y 

dx  1  y 1/2  dx dy dy

2

(b)

1/2 2

 y  2 y  1  1  y 1/2  dx 1

 S  2  (c)

2

2

S  51.33

Copyright  2014 Pearson Education, Inc.

463

464

Chapter 6 Applications of Definite Integrals

7. (a)

dx  tan y  dx dy dy

   tan y 2

(b)

2

y

 /3  y

 2  0 tan t dt  1  tan y dy    /3 y  2    tan t dt  sec y dy 0  0  (c) S  2.08  S  2 

1

0

8. (a)

dy  dx

x2  1 

(c)

5 x

 1  1 S  8.55

0

0.2





3

y

t 2  1 dt

1

2

  dx  S   2   1  dx 

 5 4

4

y  2x  dx  12 ; S   2 y 1  dx a

x

0.8

1

dy 2

b

x

1

2

t 2  1 dt  x dx 

dy

0.6

y

0

9.

0.4

(b)

2

t 2  1 dt  1  x 2  1 dx 

 1 

1

 2 

dy 2 dx

0

0.5

   x 1

5 x

 S  2 

y

x   tan t dt

x 2

0

1 4

3

x

4

x dx  2  x2   4 5; 2 0   0  5

2

Geometry formula: base circumference  2 (2), slant height  42  22  2 5





 Lateral surface area  12 (4 ) 2 5  4 5 in agreement with the integral value

10.

  dy   2  2 y 1  2 dy  4 5  y dy  2 5  y 

d

dx dx  2; S  y  2x  x  2 y  dy  2 x 1  dy c

2

2

2

0

2 2 0

2

0

 2 5  4  8 5; Geometry formula: base circumference  2 (4), slant height  42  22  2 5





 Lateral surface area  12 (8 ) 2 5  8 5 in agreement with the integral value

11.

dx  1 ; dy 2



  dx   2 dy 2

b

S   2 y 1  dx a

 

3

1

( x 1) 2



3

  dx   2 5 1 ( x  1) dx   2 5  x2  x 1

1  12

2

3

2

  2 5  92  3  12  1    2 5 (4  2)  3 5; Geometry formula: r1  12  12  1, r2  32  12  2, slant height    (2  1)2  (3  1)2  5  Frustum surface area    r1  r2   slant height   (1  2) 5  3 5 in agreement with the integral value

12.

d

  dy   2 (2 y  1) 1  4 dy  2 5  (2 y 1) dy

dx  2; S  y  2x  12  x  2 y  1  dy  2 x 1  dx dy c

2

2

2

1

1

2

 2 5  y 2  y   2 5  (4  2)  (1  1)   4 5; Geometry formula: r1  1, r2  3,  1 slant height  (2  1)2  (3  1)2  5  Frustum surface area   (1  3) 5  4 5 in agreement with the integral value

Copyright  2014 Pearson Education, Inc.

Section 6.4 Areas of Surfaces of Revolution

13.

 

2 4 3 2 dy x 2  dy   x9  S  29x 3 dx dx 0

4



1  x9 dx;

u  1  x 4  du  4 x3 dx  1 du  x3 dx; 9 9 4 9  25/9 1/2 1 u  4 du 1

  S  2 x  0  u  1, x  2  u  25  9  25/9

 2  23 u 3/2   1 14.









  3 125  1  3 12527 27  98 27 81

 

dy dy 2  12 x 1/2  dx  41x dx

S 

15/4

3/4

2 x 1  41x dx  2 



15/4





15/4

3/4

x  14 dx







3/2 3/2  3/2     2  23 x  14  43  15  14  34  14   4   3/4 



 3   43  24  1  43 (8  1)  283   15.

(2 2 x ) dy  12  dx 2 x x2

S 

1.5

0.5

 2 

1.5

 2 

1.5

16.

2xx

2



 

2 2 x  x 2 1 

dy 2 dx

(1 x )2 2 x  x2

(1 x )2 2 x x2

dx

2 2 2 x  x 2 2 x  x 122 x  x dx

0.5 0.5

1 x

2xx

dx  2  x 0.5  2 1.5

 

dy dy 2 1    4( x11) dx dx 2 x 1 5

5

1

1

 S   2 x  1 1  4( x11) dx  2   2 

5

1





( x  1)  14 dx

5



3/2   x  54 dx  2  32 x  54   1







 



3/2 3/2  3/2 3/2     43  5  54  1  54  43  25  94   4   

 43

17.

    (125  27)  53 23

33 23



98  49 6 3

6

   y S

dx  y 2  dx dy dy

2

1 2 y 3 0 3

4

1  y 4 dy;

u  1  y 4  du  4 y3 dy  1 du  y3 dy; 4  2



y  0  u  1, y  1  u  2  S   2 13 u1/2 1

2

 14 du 

2

 6  u1 2 du  6  32 u 3/2   9 ( 8  1)  1 1

Copyright  2014 Pearson Education, Inc.

465

466

Chapter 6 Applications of Definite Integrals

18.

x  13 y 3/2  y1/2  0, when 1  y  3. To get positive







area, we take x   13 y 3/2  y1/2



    14  y  2  y 1  3  S    2  13 y 3/2  y1/2  1  14  y  2  y 1  dy 1 3  2   13 y3/2  y1/2  14  y  2  y 1  dy 1 

dx  dx   12 y1/2  y 1/2  dy dy

2

y y   2  dy   13 y1/2  13 y  1  y1/2  y1  dy   13  13 y  1 ( y  1) dy 1/ 2

1/ 2



3 1 3/2 y  y1/2  2 1 3

2

1/ 2

3 3 1 2 2  y3 y3  1 1  27 9        y y 1 dy  y  9 3     9  3  3  9  3  1    3 1 3 1   9 (18  1  3)  169

19.

  





dx  dy

1  dx dy 4 y

 

 4 



15/4

0

2



1  S  15/4 2  2 4 y 0





4  y 1  41 y dy  4 

15/4



15/4

 3  19  13  1

(4  y )  1 dy

0





3/2    3/2 3/2    83  5  15  53/2    83  54 5  4    

5  y dy  4  23 (5  y )3/2   0

 

 



 83 5 5  5 8 5  83 40 585 5  353 5

20.

 

dx  dy

dx 1  dy 2 y 1

2

1  S  1 2 2 y 1 5/8







ln 2 e y  e  y 2 0

e y e  y 2

ln 2





2



ln 2 e y  e  y 2



 2 

0

2

x



3/2

1 4

ln 2 e y  e  y 2 2 0

2 y1/2 dy

ln 2 2 0





 

2 y

2y

2 y

2y



 2  12 e 2 ln 2  2 ln 2  12 e2 ln 2  12  0  12   

 

15  ln 2  2 12  4  2 ln 2  12  14  2 2  18  2 ln 2   16

y  13 x 2  2

1

5/8

3/2

0

2

e y  e y 2

 2  12 e2 y  2 y  12 e 2 y   0

22.

(2 y  1)  1 dy  2 

 1    dy  2    1  (e  2  e ) dy    dy  2    dy   (e  2  e )dy

ln 2 e y  e y 2 0

 2 

1

5/8

    43 2 1  58 85   43 2  82822 25 5   12 16 2  5 5 

1

  2 2  23 y 3/2   43 2 13/2  85   5/8  21. S  2 

2 y  1 1  2 y11 dy  2 







 dy  x x 2  2 dx  ds  1  2 x 2  x 4 dx  S  2 

0

2

x 1  2 x 2  x 4 dx

 x2  1 dx  2 0 2 x  x2  1 dx  2 0 2  x3  x  dx  2  x4  x2 0  2  44  22   4 2

4

Copyright  2014 Pearson Education, Inc.

2

2

Section 6.4 Areas of Surfaces of Revolution

467

2

      23. ds  dx 2  dy 2   y3  1 3   1 dy   y 6  12  1 6   1 dy   y 6  12  1 6  dy 4y  16 y  16 y     2





2 2  2        y 3  1 3  dy   y 3  1 3  dy; S   2 y ds  2  y  y 3  1 3  dy  2π  y 4  14 y 2 dy 1 1 1 4 4 y y 4 y       2



 







 y5    2  5  14 y 1   2  32  1  15  14   2 31  1  240 (8  31  5)  253 5 8 20  5 8   1

   sin x  S  2  dy 2

dy

24.

y  cos x  dx   sin x  dx

25.

y  a 2  x 2  dx  12 a 2  x 2  S  2 

a

 /2

 /2



dy

2



1/2

  dy 2

x

( 2 x ) 

(cos x) 1  sin 2 x dx

 dx a2  x2

x2 a2  x2

a a a 2  x 2   x 2 dx  2  a dx  2 a  x  a  a a

2 a a 2  x 2 1  2x 2 dx  2 

a

a x

 2 a [a  (a)]  (2 a)(2a)  4 a 2

26.

    S  2  x 1  dx  2  x h r    h r    r h  r   dy 2

dy

 22r

2

h

h x2 2 0

2

hr 0 h

r2 h2

y  hr x  dx  hr  dx

2 r h2

2 h2 2

2

hr 0 h

r2 h2

2

  dy. Now, x  y  16  x  16  y

d

c

 2 

7

16

y 2

16  y

2

 

 dx dy

2

y2 2

16  y

2

; S

7

16



2

27. The area of the surface of one work is S   2 x 1  dx dy  dx  dy

h 2  r 2 h x dx 0 h2

h 2  r 2 dx  2 r h h2

2

2

2 162  y 2 1 

y2 2

16  y

2

dy  

2

2

7

16

2

2

2

162  y 2   y 2 dy

16 dy  32  9  288  904.78 cm 2 . The enamel needed to cover one surface of one wok is

V  S  0.5 mm  S  0.05 cm  (904.78)(0.05) cm3  45.24 cm3 . For 5000 woks, we need 5000  V  5000  45.24 cm3  (5)(45.24) L  226.2 L  226.2 liters of each color are needed.

28.

dy

y  r 2  x 2  dx   12  2 

ah

a

29.

dy

 2 

a

2

r x

 2

x 2

r x

 

dx  dy 2

2

x2 ; r 2  x2

S  2 

a h

2

r 2  x 2 1  2x 2 dx r x

a

 r 2  x2   x2 dx  2 r aah dx  2 rh, which is independent of a.

y  R 2  x 2  dx   12 ah

2x

2x 2

R x

2



x 2

R x

2

 

dx  dy

2

x2 ; R  x2 2

S  2  

 R2  x2   x2 dx  2 R a dx  2 Rh a h

Copyright  2014 Pearson Education, Inc.

a h

a

R2  x2 1 

x2 dx R  x2 2

468

Chapter 6 Applications of Definite Integrals

30. (a)

x 2  y 2  452  x  452  y 2  dx  dy S

45

22.5

2 452  y 2 1 

y2 452  y 2

y 2

45  y

dy  2 

2

  2

 dx dy

y2 2

45  y 2

;

 452  y 2   y 2 dy  2  454522.5 dy

45

22.5

 (2 )(45)(67.5)  6075 square feet (b) 19,085 square feet 31. (a) An equation of the tangent line segment is (see figure) y  f (mk )  f (mk )  x  mk  . When x  xk 1 we have r1  f (mk )  f (mk )( xk 1  mk )

  x

x

 f (mk )  f (mk )  2k  f (mk )  f (mk ) 2k ; when x  xk we have

r2  f (mk )  f (mk )  xk  mk  x

 f (mk )  f (mk ) 2k ;





2

x x 2 2 2 2 2 (b) L2k   xk    r2  r1    xk    f (mk ) 2k   f (mk ) 2k    xk    f (mk )xk   

 xk 2   f (mk )xk 2 , as claimed

 Lk 

(c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent line segment about the x-axis is given by Sk    r1  r2  Lk    2 f (mk )

 xk 2   f (mk )xk 2 using

parts (a) and (b) above. Thus, Sk  2 f (mk ) 1   f (mk ) xk .

2

n

n

b

n k 1

n  k 1

a





2 2 (d) S  lim  Sk  lim  2 f (mk ) 1   f (mk )  xk   2 f ( x) 1   f ( x)  dx

32.



y  1 x

2/3 3/2



1

 S  2 2 1  x 0

 

1/2 dy  dx  32 1  x 2/3



2/3 3/2

1



 23 x 1/3



1 x   2/3

1/ 2

1/3

x

 dx  4  1  x  1

1 1 x 2/3

2/3 3/2

0

  dy 2

 dx

1 x 2/3  1  1 x 2/3 x 2/3 1



x 2/3 dx  4  1  x 2/3 0

 x1/3 dx; 3/2

u 1  x 2/3  du   23 x 1/3dx   32 du  x1/3 dx; x  0  u  1, x  1  u  0 0







0



 S  4  u 3/2  32 du  6  25 u 5/2   6 0  25  125  1 1

6.5

WORK AND FLUID FORCES

1. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F ( x )  kx.

3

3

3

The work done by F is W   F ( x) dx  k  x dx  k2  x 2   92k . This work is equal to 1800 J  0 0 0  92 k  1800  k  400 N/m

Copyright  2014 Pearson Education, Inc.

Section 6.5 Work and Fluid Forces

469

 200 lb/in.

2. (a) We find the force constant from Hooke’s Law: F  kx  k  Fx  k  800 4 2

2

0

0

(b) The work done to stretch the spring 2 inches beyond its natural length is W   kx dx  200 x dx 2

 200  x2   200(2  0)  400 in-lb  33.3 ft-lb  0 (c) We substitute F  1600 into the equation F  200 x to find 1600  200 x  x  8 in.

2

3. We find the force constant from Hooke’s law: F  kx. A force of 2 N stretches the spring to 0.02 m N . The force of 4 N will stretch the rubber band y m, where F  ky  y  F  2  k  (0.02)  k  100 m k 0.04 4N  y  0.04 m  4 cm. The work done to stretch the rubber band 0.04 m is W  kx dx N 0 100 m



y  100 

0.04

0.04

x dx  100  x2    0 2

0



(100)(0.04)2  0.08 J 2

N . The work done to  k  90 m 4. We find the force constant from Hooke’s law: F  kx  k  Fx  k  90 1 5

5

5

 

stretch the spring 5 m beyond its natural length is W   kx dx  90 x dx  90  x2   (90) 25  1125 J 2   0 0 0 2

21,714

21,714 lb  k  7238 in 3 0.5

5. (a) We find the spring’s constant from Hooke’s law: F  kx  k  Fx  85  (b) The work done to compress the assembly the first half inch is W  

0.5

0

0.5

(0.5)2

 7238  x2   (7238) 2   0 second half inch is: 2

W 

1.0

0.5



0

x dx

(7238)(0.25)  905 in-lb. The work done to compress the assembly the 2

1.0

1.0

1  (0.5)2   x dx  7238  x2   7238 2   0.5  0.5

kx dx  7238

kx dx  7238

2

(7238)(0.75)  2714 in-lb 2

lb . If someone  16 150  2, 400 in 6. First, we find the force constant from Hooke’s law: F  kx  k  Fx  150 1



 16 

compresses the scale x  18 in, he/she must weigh F  kx  2, 400 18  300 lb. The work done to compress the scale this far is W  

1/8

0

2 1/8

 18.75 lb  in.  2.5 kx dx  2400  x2   2400 ft-lb 264 16  0

7. The force required to haul up the rope is equal to the rope’s weight, which varies steadily and is proportional to x, the length of the rope still hanging: F ( x )  0.624 x. The work done is: W  

50

0

F ( x) dx  

50

0

50

 0.624  x2   780 J   0 2

8. The weight of sand decreases steadily by 72 lb over the 18 ft, at 4 lb/ft. So the weight of sand when the b

18

a

0

bag is x ft off the ground is F ( x)  144  4 x. The work done is: W   F ( x)dx   (144  4 x) dx 18

 144 x  2 x 2   1944 ft-lb  0

Copyright  2014 Pearson Education, Inc.

0.624x dx

470

Chapter 6 Applications of Definite Integrals

9. The force required to lift the cable is equal to the weight of the cable paid out: F ( x )  (4.5)(180  x ) where x is the position of the car off the first floor. The work done is: W  

180

0

180

2  4.5 180 x  x2   0



 4.5 1802  180 2

2



F ( x) dx  4.5

180

0

(180  x ) dx

4.5180  2  72,900 ft-lb 2

10. Since the force is acting toward the origin, it acts opposite to the positive x-direction. Thus F ( x)   k2 . x

b

b

b

The work done is W    k2 dx  k   12 dx  k  1x   k a

a

k

a

x





1  1  k ( a b ) b a ab

11. Let r  the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to (20  x), the distance the bucket is being raised. The leakage rate of the water is 0.8 lb/ft raised and the weight of the water in the bucket is 20

20

2 0.8 (20  x )dx  0.8  20 x  x2   160 ft-lb.  0 0 

F  0.8(20  x). So: W  

12. Let r  the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to (20  x), the distance the bucket is being raised. The leakage rate of the water is 2 lb/ft raised and the weight of the water in the bucket is F  2(20  x ).

20

20

2(20  x ) dx  2  20 x  x2   400 ft-lb.

0  0

So: W  

2

Note that since the force in Exercise 12 is 2.5 times the force in Exercise 11 at each elevation, the total work is also 2.5 times as great.

13. We will use the coordinate system given. (a) The typical slab between the planes at y and y  y has a volume of V  (10)(12)y  120y ft3. The force F required to lift the slab is equal to its weight: F  62.4 V  62.4 120y lb. The distance through which F must act is about y ft, so the work done lifting the slab is about W  force  distance  62.4  120  y  y ft-lb The work it takes to lift all the water is approximately 20

20

0

0

W   W   62.4  120 y  y ft-lb.

This is a Riemann sum for the function 62.4 120 y over the interval 0  y  20. The work of pumping the tank empty is the limit of these sums: 20

 

y2  (62.4)(120)(200)  1,497,600 ft-lb 62.4  120 y dy  (62.4)(120)  2   (62.4)(120) 400 2 0  0

W 

20

 

5  hp motor is t  W (b) The time t it takes to empty the full tank with 11  250 ft-lb sec

1,497,600 ftlb  5990.4 sec 250 ft-lb sec

 1.664 hr  t  1 hr and 40 min (c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 ft is 10

 

10 y2  374,400 ft-lb and the time is t  Wft-lb W   62.4  120 y dy  (62.4)(120)  2   (62.4)(120) 100 2 250 sec 0  0  1497.6 sec  0.416 hr  25 min

Copyright  2014 Pearson Education, Inc.

Section 6.5 Work and Fluid Forces

471

(d) In a location where water weighs 62.26 lb3 : ft

a) W  (62.26)(24,000)  1,494,240 ft-lb .

1,494,240  5976.96 sec  1.660 hr  t  1 hr and 40 min 250 In a location where water weighs 62.59 lb3 ft

b) t 

a) W  (62.59)(24,000)  1,502,160  ft-lb b) t 

1,502,160  6008.64 sec  1.669 hr  t  1 hr and 40.1 min 250

14. We will use the coordinate system given. (a) The typical slab between the planes at y and y  y has a volume of V  (20)(12) y  240y ft 3 . The force F required to lift the slab is equal to its weight: F  62.4V  62.4  240y lb. The distance through which F must act is about y ft, so the work done lifting the slab is about W  force  distance 20

 62.4  240  y  y ft-lb. The work it takes to lift all the water is approximately W   W 10

20

  62.4  240 y  y ft-lb. This is a Riemann sum for the function 62.4  240 y over the interval 10

10  y  20. The work it takes to empty the cistern is the limit of these sums: 20

 y2  62.4  240 y dy  (62.4)(240)  2   (62.4)(240)(200  50)  (62.4)(240)(150)  2,246,400 ft-lb 10  10

W 

(b) t 

20

2,246,400 ft-lb W   8168.73 sec  2.27 hours  2 hr and 16.1 min 275 275 ft-lb sec

(c) Following all the steps of part (a), we find that the work it takes to empty the tank halfway is 15





 

15  y2  W   62.4  240 y dy  (62.4)(240)  2   (62.4)(240) 225  100  (62.4)(240) 125  936, 000 ft.

2 2 2 10  10

Then the time is t 

W  936,000  3403.64 sec  56.7 min 275 275 ft-lb sec

(d) In a location where water weighs 62.26 lb3 : ft

a) W  (62.26)(240)(150)  2, 241,360 ft-lb.

2,241,360  8150.40 sec  2.264 hours  2 hr and 15.8 min 275 933,900  933,900 ft-lb; t  275  3396 sec  0.94 hours  56.6 min c) W  (62.26)(240) 125 2 In a location where water weighs 62.59 lb3 : ft

b) t 

 

a) W  (62.59)(240)(150)  2,253, 240 ft-lb.

2,253,240  8193.60 sec  2.276 hours  2 hr and 16.56 min 275 938,850 W  (62.59)(240) 125  938,850 ft-lb; t  275  3414 sec  0.95 hours  56.9 min 2

b) t  c)

 

Copyright  2014 Pearson Education, Inc.

472

Chapter 6 Applications of Definite Integrals

  , thickness y, and height below the top of the tank (10  y). So the work to pump the oil in this slab, W , is 57 (10  y ) π   . The work to pump all the oil to top of the tank is y 2

15. The slab is a disk of area  x 2   2

y 2 2

10 57 0 4

W 

10

10 y 2  y3  dy  57π4 103y  y4 0  11,875 ft  lb  37,306 ft-lb 3

4

  and since y 2

16. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is (14  y )( ) 2

 

 ft 3 , half the the tank is half full and the volume of the original cone is V  13  r 2 h  13  52 (10)  250 3 y2

  1 ft 3 , and with half the volume the cone is filled to a height y, 250 y  y  3 500 ft. volume  250π 6 3 4 6 3

500 57 4 0

So W  



2

14 y  y

3



 14 y dy  57π 4  3

3

y4  4  

3

500

 60,042 ft-lb.

0

17. The typical slab between the planes at y and y  y has a volume of V   (radius)2 (thickness)

 

2   20 y   100 y ft 3 . The force F required to lift the slab is equal to its weight: 2

F  51.2V  51.2 100 y lb  F  5120 y lb The distance through which F must act is about 30

30

0

0

(30  y ) ft. The work it takes to lift all the kerosene is approximately W   W   5120 (30  y )y ft-lb which is a Riemann sum. The work to pump the tank dry is the limit of these sums: 30

 

30 y2  (5120)(450 )  7,238,229.48 ft-lb W   5120 (30  y ) dy  5120 30 y  2   5120 900 2 0  0

18. (a) Follow all the steps of Example 5 but make the substitution of 64.5 lb3 for 57 lb3 . Then, ft

3 4 8 64.5   10 y  y  (10  y ) y 2 dy  64.5   4 4 3 4 0 

W 

8 0



3

4

 

 108  8  64.5  64.5 4 3 4 4

ft

 83   103  2  64.53 8

3

3

 21.5  8  34,582.65 ft-lb (b) Exactly as done in Example 5 but change the distance through which F acts to distance  (13  y ) ft.

8

   8    2  



3 4 8  13 y3 y 4  Then W   574 (13  y ) y 2 dy  574  3  4   574 1338  84  574 0  0

3

13 3

57 83 7 34

 (19 )(82 )(7)(2)  53,482.5 ft-lb 19. The typical slab between the planes at y and y  y has a volume of about V   (radius) 2 (thickness)

 y  y ft3. The force F ( y) required to lift this slab is equal to its weight: F ( y )  73  V 2  73  y  y  73 y y lb. The distance through which F ( y ) must act to lift the slab to the top of the 

2

reservoir is about (4  y ) ft, so the work done is approximately W  73 y (4  y ) y ft-lb. The work done n

lifting all the slabs from y  0 ft to y  4 ft is approximately W   73 yk  4  yk  y ft-lb. Taking the limit k 0

Copyright  2014 Pearson Education, Inc.

Section 6.5 Work and Fluid Forces 4

4

0

0

of these Riemann sums as n  , we get W   73 y (4  y ) dy  73 



4

473

 4 y  y2  dy



 ft-lb  2446.25 ft-lb.  73  2 y 2  13 y 3   73 32  64  2336 3 3  0

20. The typical slab between the planes at y and y  y has volume of about V  (length)(width)(thickness)   2 25  y 2  (10)y ft 3 . The force F ( y ) required to lift this slab is equal to its weight:   F ( y )  53  V  53  2 25  y 2  (10) y  1060 25  y 2 y lb. The distance through which F ( y ) must act to   lift the slab to the level of 15 m above the top of the reservoir is about (20  y ) ft, so the work done is

approximately W  1060 25  y 2 (20  y ) y ft-lb. The work done lifting all the slabs from y  5 ft to n

y  5 ft is approximately W   1060 25  yk2  20  yk  y ft-lb. Taking the limit of these Riemann sums as k 0

5

5

5

5

n  , we get W   1060 25  y 2 (20  y )dy  1060 (20  y ) 25  y 2 dy 5 5  1060   20 25  y 2 dy   y 25  y 2 dy  . To evaluate the first integral, we use we can interpret 5  5  5

5

2

5

2

2

5 25  y dy as the area of the semicircle whose radius is 5, thus 5 20 25  y dy  205 25  y dy  20  12  (5)2   250 . To evaluate the second integral let u  25  y 2  du  2 y dy; y  5  u  0,   5 0 5 y  5  u  0, thus  y 25  y 2 dy   12  u du  0. Thus, 1060   20 25  y 2 dy  0 5  5 5 5  1060   20 25  y 2 dy   y 25  y 2 dy   1060(250  0)  265000  832522 ft-lb  5 5 

21. The typical slab between the planes at y and y  y has a volume of about V   (radius) 2 (thickness) 2

   25  y 2  y m3 . The force F ( y ) required to lift this slab is equal to its weight:   2





F ( y )  9800  V  9800  25  y 2  y  9800 25  y 2 y N. The distance through which F ( y ) must   act to lift the slab to the level of 4 m above the top of the reservoir is about (4  y ) m, so the work done is





approximately W  9800 25  y 2 (4  y )y N  m. The work done lifting all the slabs from y  5 m to 0





y  0 m is approximately W   9800 25  y 2 (4  y )y N  m. Taking the limit of these Riemann sums,



0



5

we get W   9800 25  y 2 (4  y ) dy  9800  5

0

5

0

100  25 y  4 y2  y3  dy





y4   y 2  43 y 3  4   9800 500  25225  34 125  625  9800 100 y  25  15, 073, 099.75 J  2 4   5

22. The typical slab between the planes at y and y  y has a volume of about V   (radius) 2 (thickness) 2









56 lb    100  y 2  y   100  y 2 y ft 3 . The force is F ( y )  3  V  56 100  y 2 y lb. The ft  

Copyright  2014 Pearson Education, Inc.