Section 15.6 Triple Integrals

Section 15.6 Triple Integrals

11. (a) Express the triple integralRRR

Ef (x, y, z) dV as an iterated integral for the given function f and solid region E.

(b) Evaluate the iterated integral.

SECTION 15.6 Triple Integrals 1093

25. The solid enclosed by the cylinder y − x² and the planes z − 0 and y 1 z − 1

26. The solid enclosed by the cylinder x²1z²− 4 and the planes y − 21 and y 1 z − 4

27. (a) Express the volume of the wedge in the first octant that is cut from the cylinder y²1z²− 1 by the planes y − x and x − 1 as a triple integral.

(b) Use either the Table of Integrals (on Reference Pages 6 –10) or a computer algebra system to find the exact value of the triple integral in part (a).

28–30 Midpoint Rule for Triple Integrals In the Midpoint Rule for triple integrals we use a triple Riemann sum to approximate a triple integral over a box B, where fsx, y, zd is evaluated at the center sxi, yj, zkd of the box Bijk . Use the Midpoint Rule to estimate the value of the integral. Divide B into eight sub-boxes of equal size.

28. yyyBsx²1y²1z² dV, where

B −hsx, y, zd

|

0 < x < 4, 0 < y < 4, 0 < z < 4j

29. yyyB cossxyzd dV, where

B −hsx, y, zd

|

0 < x < 1, 0 < y < 1, 0 < z < 1j

30. yyyBsxe^xyz dV, where

B −hsx, y, zd

|

0 < x < 4, 0 < y < 1, 0 < z < 2j

31–32 Sketch the solid whose volume is given by the iterated integral.

31.

y

0¹

y

0^12x

y

0^222zdy dz dx

32.

y

0²

y

0^22y

y

0^42y2dx dz dy

33–36 Express the integral yyyE fsx, y, zd dV as an iterated integral in six different ways, where E is the solid bounded by the given surfaces.

33. y − 4 2 x²24z², y − 0 34. y²1z²− 9, x − 22, x − 2 35. y − x², z − 0, y 1 2z − 4

36. x − 2, y − 2, z − 0, x 1 y 2 2z − 2 11. fsx, y, zd − x 1 y 12. fsx, y, zd − 2

x=œy

E

0 x+z=2

z

y x

E z=4-x

z=4+x z=4-¥

z

y x

13–22 Evaluate the triple integral.

13. yyyE y dV, where

E −

h

^{sx, y, zd}

|

0 < x < 3, 0 < y < x, x 2 y < z < x 1 y

j

14. yyy^E e^zyy dV, where

E −

5

^{sx, y, zd}

|

0 < y < 1, y < x < 1, 0 < z < xy

6

15. yyyE s1yx³d dV, where

E −

5

^{sx, y, zd}

|

0 < y < 1, 0 < z < y², 1 < x < z 1 1

6

16. yyyE sin y dV, where E lies below the plane z − x and above the triangular region with vertices s0, 0, 0d, s, 0, 0d, and s0, , 0d

17. yyyE 6xy dV, where E lies under the plane z − 1 1 x 1 y and above the region in the xy-plane bounded by the curves y −sx, y − 0, and x − 1

18. yyyE sx 2 yd dV, where E is enclosed by the surfaces z − x²21, z − 1 2 x², y − 0, and y − 2

19. yyyT y² dV, where T is the solid tetrahedron with vertices s0, 0, 0d, s2, 0, 0d, s0, 2, 0d, and s0, 0, 2d

20. yyyT xz dV, where T is the solid tetrahedron with vertices s0, 0, 0d, s1, 0, 1d, s0, 1, 1d, and s0, 0, 1d

21. yyyE x dV, where E is bounded by the paraboloid x − 4y²14z² and the plane x − 4

22. yyyE z dV, where E is bounded by the cylinder y²1z²− 9 and the planes x − 0, y − 3x, and z − 0 in the first octant

23–26 Use a triple integral to find the volume of the given solid.

23. The tetrahedron enclosed by the coordinate planes and the plane 2x 1 y 1 z − 4

24. The solid enclosed by the paraboloids y − x²1z² and y − 8 2 x²2z²

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Solution:

SECTION 15.6 TRIPLE INTEGRALS ¤ 1543

9. (a) The solid region  can be described as  = {(  ) | −1 ≤  ≤ 1, 0 ≤  ≤ 2 − , 0 ≤  ≤ 1 − ²}.

Thus,

  =1

−1

1−² 0

2−

0    .

(b)

 1

−1

 1−² 0

 2−

0

    =

 1

−1

 1−² 0



=2−

=0   =

 1

−1

 1−²

0 (2 − )  

=

 1

−1



2 − ² 2

=1−²

=0

 =

 1

−1

3

2 − ³− ⁵ 2





=

3² 4 −⁴

4 −⁶ 12

¹

−1

= 0

10. (a) The solid region  can be described as  = {(  ) | 0 ≤  ≤ , 0 ≤  ≤ 2, 0 ≤  ≤ 4 − ²}.

Thus,

  =2 0

 0

4−²

0    .

(b) 2 0

 0

4−²

0     =2 0



0  []^=4_=0^−2  =2 0



0 (4 − ²)   =2 0



0 (4 − ³)  

=2

0(4 − ³)

² 2

=

=0  = ¹₂2

0(4³− ⁵)  = ¹₂

⁴−^6⁶

2 0= ⁸₃ 11. (a) The solid region  can be described as  = {(  ) | 0 ≤  ≤ 2, 0 ≤  ≤ ², 0 ≤  ≤ 2 − }.

Thus,

( + )  =2 0

2−

0

²

0 ( + )   .

(b)  2 0

 2−

0

 ² 0

( + )    =

 2 0

 2−

0



 +² 2

=²

=0

  =

 2 0

 2−

0



³+⁴ 2



 

=

 2 0



³+⁴ 2

 

=2−

=0  =

 2 0



2³− ⁵ 2



 =

⁴ 2 −⁶

12

²

0

= 8 3 12. (a) The solid region  can be described as  = {(  ) |  − 4 ≤  ≤ 4 − , −2 ≤  ≤ 2, 0 ≤  ≤ 4 − ²}.

Thus,

2  =2

−2

4−² 0

4−

−4 2   .

(b) 2

−2

4−² 0

4−

−4 2    =2

−2

4−²

0 2

=4−

=−4  = 22

−2

4−²

0 (8 − 2)  

= 22

−2



8 − ²=4−²

=0  = 22

−2(16 − ⁴)  = 2

16 −^5⁵

2

−2= ⁵¹²₅ 13. 

   =3 0

 0

+

−     =3 0

 0

^=+

=−  =3 0



0 2² 

=3 0

₂

3³=

=0 =3 0

2

3³ = ¹₆⁴3

0= ⁸¹₆ = ²⁷₂ 14. 

 ^ =1 0

1





0 ^   =1 0

1





^=

=0  

=1 0

1

 (^− )   =1 0

^− =1

= =1 0

 −  − ^+ ²



=₁

2²− ¹2²− ( − 1)^+¹₃³1

0 [integrate by parts]

= ¹₂ −¹2 +¹₃− 1 = ¹2 − ⁷6

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26. Use a triple integral to find the volume of the given solid. The solid enclosed by the cylinder x²+ z²= 4 and the planes y = −1 and y + z = 4.

Solution:

572 ¤ CHAPTER 15 MULTIPLE INTEGRALS 22. Here  =

(  ) | −1 ≤  ≤ 4 −  ²+ ²≤ 4, so

 =

 2

−2

 √4−²

−√

4−²

 4−

−1

   =

 2

−2

 √4−²

−√

4−²(4 −  + 1)  

=

 2

−2

5 −¹2²^=√

4−²

=−√

4−²  =

 2

−2

10

4 − ²

= 10 2

√4 − ²+ 2 sin⁻¹ 2

²

−2

using trigonometric substitution or Formula 30 in the Table of Integrals



= 10

2 sin⁻¹(1) − 2 sin⁻¹(−1)

= 20 2 −

−^2

= 20

Alternatively, use polar coordinates to evaluate the double integral:

 2

−2

 √4−²

−√

4−²(5 − )   =

 2

0

 2

0 (5 −  sin )   

=2

0

₅

2²−¹3³sin =2

=0 

=2

0

10 −⁸3sin 



= 10 +⁸₃cos 2

0 = 20

23. (a) The wedge can be described as the region

 =

(  ) | ²+ ²≤ 1, 0 ≤  ≤ 1, 0 ≤  ≤ 

=

(  ) | 0 ≤  ≤ 1, 0 ≤  ≤ , 0 ≤  ≤

1 − ² So the integral expressing the volume of the wedge is



 =1 0

 0

 √1− ²

0   .

(b) A CAS gives1 0

 0

 √1− ²

0    = ^₄ −¹3. (Or use Formulas 30 and 87 from the Table of Integrals.)

24. (a) Divide  into 8 cubes of size ∆ = 8. With (  ) =

²+ ²+ ², the Midpoint Rule gives





²+ ²+ ² ≈

2

 = 1

2

 = 1

2

 = 1



 _ 

∆

= 8[ (1 1 1) +  (1 1 3) +  (1 3 1) +  (1 3 3) +  (3 1 1) +  (3 1 3) +  (3 3 1) +  (3 3 3)]

≈ 23964 (b) Using a CAS we have



²+ ²+ ² =4 0

4 0

4 0

²+ ²+ ²   ≈ 24591. This differs from the

estimate in part (a) by about 2.5%.

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1

38. The figure shows the region of integration for the integral Z 1

0

Z 1−x² 0

Z 1−x 0

f (x, y, z)dydzdx Rewrite this integral as an equivalent iterated integral in the five other orders.

1038 Chapter15 Multiple Integrals

29–32 Express the integral yyyE fsx, y, zd dV as an iterated integral in six different ways, where E is the solid bounded by the given surfaces.

29. y − 4 2 x²24z², y − 0 30. y²1z²− 9, x − 22, x − 2 31. y − x², z − 0, y 1 2z − 4

32. x − 2, y − 2, z − 0, x 1 y 2 2z − 2

33. The figure shows the region of integration for the integral

y

0¹

y

_sx¹

y

0^12y fsx, y, zd dz dy dx

Rewrite this integral as an equivalent iterated integral in the five other orders.

0 z 1

x

1 y z=1-y y=œ„x

34. The figure shows the region of integration for the integral

y

0¹

y

0^12x2

y

0^12x fsx, y, zd dy dz dx

Rewrite this integral as an equivalent iterated integral in the five other orders.

1 1

z=1-≈ 1

y=1-x 0

y x

z

35–36 Write five other iterated integrals that are equal to the given iterated integral.

35.

y

0¹

y

y¹

y

0^y fsx, y, zd dz dx dy 36.

y

0¹

y

y¹

y

0^z fsx, y, zd dx dz dy 16. yyyT xz dV, where T is the solid tetrahedron with vertices

s0, 0, 0d, s1, 0, 1d, s0, 1, 1d, and s0, 0, 1d

17. yyyE x dV, where E is bounded by the paraboloid x − 4y²14z² and the plane x − 4

18. yyyE z dV, where E is bounded by the cylinder y²1z²− 9 and the planes x − 0, y − 3x, and z − 0 in the first octant

19–22 Use a triple integral to find the volume of the given solid.

19. The tetrahedron enclosed by the coordinate planes and the plane 2x 1 y 1 z − 4

20. The solid enclosed by the paraboloids y − x²1z² and y − 8 2 x²2z²

21. The solid enclosed by the cylinder y − x² and the planes z − 0 and y 1 z − 1

22. The solid enclosed by the cylinder x²1z²− 4 and the planes y − 21 and y 1 z − 4

23. (a) Express the volume of the wedge in the first octant that is cut from the cylinder y²1z²− 1 by the planes y − x and x − 1 as a triple integral.

(b) Use either the Table of Integrals (on Reference Pages 6–10) or a computer algebra system to find the exact value of the triple integral in part (a).

24. (a) In the Midpoint Rule for triple integrals we use a triple Riemann sum to approximate a triple integral over a box B, where fsx, y, zd is evaluated at the center sxi, yj, zkd of the box Bijk. Use the Midpoint Rule to estimate yyyBsx²1y²1z² dV, where B is the cube

defined by 0 < x < 4, 0 < y < 4, 0 < z < 4. Divide B into eight cubes of equal size.

(b) Use a computer algebra system to approximate the integral in part (a) correct to the nearest integer. Com- pare with the answer to part (a).

25–26 Use the Midpoint Rule for triple integrals (Exer cise 24) to estimate the value of the integral. Divide B into eight sub- boxes of equal size.

25. yyyB cossxyzd dV, where

B −hsx, y, zd

|

0 < x < 1, 0 < y < 1, 0 < z < 1j 26. yyyBsxe^xyz dV, where

B −hsx, y, zd

|

0 < x < 4, 0 < y < 1, 0 < z < 2j

27–28 Sketch the solid whose volume is given by the iterated integral.

27.

y

0¹

y

0^12x

y

0^222zdy dz dx 28.

y

0²

y

0^22y

y

0^42y2dx dz dy

CAS

CAS

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Solution:

SECTION 15.6 TRIPLE INTEGRALS ¤ 577 Then 

 (  )  =2 0

2 2−

(+−2)2

0  (  )    =2 0

2 2−

(+−2)2

0  (  )   

=2 0

2 0

2

2−+2 (  )    =1 0

2 2

2

2−+2 (  )   

=2 0

2 0

2

2−+2 (  )    =1 0

2 2

2

2−+2 (  )   

33.

The diagrams show the projections of  onto the -, -, and -planes.

Therefore

1 0

1

√

1− 

0  (  )    =1 0

² 0

1−

0  (  )    =1 0

1−

0

²

0  (  )   

=1 0

1−

0

²

0  (  )    =1 0

1−√ 0

√1−

  (  )   

=1 0

(1−)² 0

1−

√  (  )   

34.

The projections of  onto the

- and -planes are as in the first two diagrams and so

1 0

1−² 0

1− 

0  (  )    =1 0

^√1−

0

1−

0  (  )   

=1 0

1−

0

1−²

0  (  )    =1 0

1−

0

1−²

0  (  )   

Now the surface  = 1 − ²intersects the plane  = 1 −  in a curve whose projection in the -plane is  = 1 − (1 − )² or  = 2 − ². So we must split up the projection of  on the -plane into two regions as in the third diagram. For ( ) in 1, 0 ≤  ≤ 1 −  and for ( ) in ², 0 ≤  ≤√

1 − , and so the given integral is also equal to

1 0

1−√1−

0

^√1−

0  (  )    +1 0

1 1−√1−

1−

0  (  )   

=1 0

2−² 0

1−

0  (  )    +1 0

1 2−²

^√1−

0  (  )   

35.

1 0

1





0  (  )    =

 (  )  where  = {(  ) | 0 ≤  ≤ ,  ≤  ≤ 1, 0 ≤  ≤ 1}.

[continued]

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

42. Evaluate the triple integral using only geometric interpretation and symmetry.

Z Z Z

B

(z³+ sin y + 3)dV , where B is the unit ball x²+ y²+ z²≤ 1.

Solution:

SECTION 15.6 TRIPLE INTEGRALS ¤ 579 37. The region  is the solid bounded by a circular cylinder of radius 2 with axis the -axis for −2 ≤  ≤ 2. We can write



(4 + 5²²)  =

4  +

5²², but (  ) = 5²²is an odd function with respect to . Since  is symmetrical about the -plane, we have

5²² = 0. Thus



(4 + 5²²)  =

4  = 4 ·  () = 4 · (2)²(4) = 64.

38. We can write

(³+ sin  + 3)  =

³ +

sin   +

3 . But ³is an odd function with respect to  and the region  is symmetric about the -plane, so

³ = 0. Similarly, sin  is an odd function with respect to  and  is symmetric about the -plane, so

sin   = 0. Thus



(³+ sin  + 3)  =

3  = 3 ·  () = 3 ·⁴3(1)³= 4.

39. The projection of  onto the -plane is the disk  =

( ) | ²+ ²≤ 1.

 =

 (  )  =



1−²−²

0 3 

 =

3(1 − ²− ²) 

= 31 0

2

0 (1 − ²)    = 32

0  1

0( − ³) 

= 3

2

0

1

2²−¹4⁴1

0= 3 (2)1 2−¹4

= ³₂

 =

 (  )  =



1−²−²

0 3 

 =

3(1 − ²− ²) 

= 31 0

2

0 ( cos )(1 − ²)    = 32

0 cos   1

0(²− ⁴) 

= 3

sin 2

0

₁

3³−¹5⁵1

0= 3 (0)₁

3−¹5

= 0

 =

 (  )  =



1−²−²

0 3 

 =

3(1 − ²− ²) 

= 31 0

2

0 ( sin )(1 − ²)    = 32

0 sin   1

0(²− ⁴) 

= 3

− cos 2

0

1

3³−¹5⁵1

0= 3 (0)1 3− ¹5

= 0

 =

 (  )  =



1−²−²

0 3 

 =



₃

2²=1−²−²

=0 

= ³₂

(1 − ²− ²)² = ³₂1 0

2

0 (1 − ²)²  

= ³₂2

0  1

0( − 2³+ ⁵)  = ³₂

2

0

1

2²−¹2⁴+ ¹₆⁶1 0

= ³₂(2)1

2 −¹2+¹₆

= ¹₂

Thus the mass is³₂and the center of mass is (  ) =



 

 





=

 0 01

3

 .

40.  =1

−1

1−² 0

1−

0 4    = 41

−1

1−²

0 (1 − )   = 41

−1

 −¹2²=1−²

=0  = 21

−1(1 − ⁴)  =¹⁶₅,

 =1

−1

1−² 0

1−

0 4    = 21

−1

1−²

0 (1 − )²  = 21

−1

−¹3(1 − )³=1−²

=0 

= ²₃1

−1

1 − ⁶

 =4 3

6 7

= ²⁴₂₁

[continued]

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

58. Average Value The average value of a function f (x, y, z) over a solid region E is defined to be favg= 1

V (E) Z Z Z

E

f (x, y, z)dV

2

where V (E) is the volume of E. For instance, if ρ is a density function, then ρaveis the average density of E.

Find the average height of the points in the solid hemisphere x²+ y²+ z²≤ 1, z ≥ 0.

Solution:

SECTION 15.6 TRIPLE INTEGRALS ¤ 583 (c)  ( ≤ 1  ≤ 1  ≤ 1) =1

−∞

1

−∞

1

−∞ (  )    =1 0

1 0

1 0

1

100−(05+02+01)  

= ₁₀₀¹ 1

0 ^−051

0 ^−021

0 ^−01

= ₁₀₀¹ 

−2^−051 0

−5^−021 0

−10^−011 0

= (1 − ^−05)(1 − ^−02)(1 − ^−01) ≈ 0006787

53.  () = ³ ⇒ ^ave= 1

³

  0

  0

  0

    = 1

³

  0

 

  0

 

  0

 

= 1

³

² 2

^

0

² 2

^

0

² 2

^

0

= 1

³

² 2

² 2

² 2 = ³

8

54. The height of each point is given by its -coordinate, so the average height of the points in

 =

(  ) | ²+ ²+ ²≤ 1  ≥ 0 is

1

 ()





 

Here  () = ¹₂· ⁴3(1)³= ²₃ [half the volume of a sphere], so

1

 ()



   = _23¹ 1

−1

 √1−²

−√

1−²

 √1−²−²

0     = _2³ 1

−1

 √1−²

−√

1−²

1

2²=√

1−²−²

=0  

= _2³ ·¹2

1

−1

 √1−²

−√

1−²(1 − ²− ²)   =_4³ 2

0

1

0 (1 − ²)   

= _4³ 2

0 1

0( − ³)  = _4³ (2)1

2²−¹4⁴1 0= ³₂1

4

= ³₈

55. (a) The triple integral will attain its maximum when the integrand 1 − ²− 2²− 3²is positive in the region  and negative everywhere else. For if  contains some region  where the integrand is negative, the integral could be increased by excluding  from , and if  fails to contain some part  of the region where the integrand is positive, the integral could be increased by including  in . So we require that ²+ 2²+ 3²≤ 1. This describes the region bounded by the ellipsoid ²+ 2²+ 3²= 1.

(b) The maximum value of

(1 − ²− 2²− 3²)  occurs when  is the solid region bounded by the ellipsoid

²+ 2²+ 3²= 1. The projection of  on the -plane is the planar region bounded by the ellipse ²+ 2²= 1, so

 =

(  ) | −1 ≤  ≤ 1 −

1

2(1 − ²) ≤  ≤

1

2(1 − ²) −

1

3(1 − ²− 2²) ≤  ≤

1

3(1 − ²− 2²) and



(1 − ²− 2²− 3²)  =

 1

−1

 ^1₂(¹−²)

− 1 2(¹−²)

 ^1₃(¹−²−2²)

− 1

3(¹−²−2²)(1 − ²− 2²− 3²)    = 4√ 6 45  using a CAS.

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

3