15.7 Triple Integrals in Cylindrical Coordinates

Section 15.7 Triple Integrals in Cylindrical Coordinates

2. Plot the point whose cylindrical coordinates are given. Then find the rectangular coordinates of the point. (a) (2,^5π₆ , 1) (b) (8, −^2π₃, 5).

Solution:

SECTION 15.7 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES ¤ 1561

For  = 2 + 1 and  = 1:

(1) = 2^1

2( − 1)

! ^{(−1)2}

! ^= 2^2+1! ^ (2 + 1)!

0 ≤ 2 · 2²

3 · 2 ·2²(2)

5 · 4 ·2²(3)

7 · 6 · · · 2²()

(2 + 1)(2) ≤ 2 · 2

3 · 2

5 · 2

2 + 1 = 2⁴³

15(2 + 1) [for   3]

2⁴³

15(2 + 1) → 0 as  → ∞ ⇒ ^(1) → 0 as  → ∞ for  odd by the Squeeze Theorem.

Thus, lim

→∞(1) = 0.

15.7 Triple Integrals in Cylindrical Coordinates

1. (a) From Equations 1,  =  cos  = 5 cos^₂ = 5 · 0 = 0,

 =  sin  = 5 sin^₂ = 5 · 1 = 5, and  = 2, so the point is (0 5 2) in rectangular coordinates.

(b)

0

From Equations 1,  =  cos  = 6 cos

−^4

= 6 ·^√2² = 3√ 2,

 =  sin  = 6 sin

−^4

= 6

−^√2²

= −3√

2, and  = −3, so the point is

3√ 2 −3√

2 −3in rectangular coordinates.

2. (a) From Equations 1,  =  cos  = 2 cos^5₆ = 2

−^√2³

= −√ 3,

 =  sin  = 2 sin^5₆ = 2 · ¹2 = 1, and  = 1, so the point is

−√ 3 1 1 in rectangular coordinates.

(b) From Equations 1,  =  cos  = 8 cos

−^23

= 8

−¹2

= −4,

 =  sin  = 8 sin

−^23

= 8

−^√2³

= −4√

3, and  = 5, so the point is

−4 −4√

3 5in rectangular coordinates.

3. (a) (4 4 −3). From Equations 2, we have ²= ²+ ²= 4²+ 4² = 32, so  =√

32. tan  = ^_ = ⁴₄ = 1and the point (4 4)is in the first quadrant of the ­plane, so  = ^₄ + 2. Thus, one set of cylindrical coordinates is

(  ) = 4√

2^₄ −3.

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11. Sketch the solid described by the given inequalities. r²≤ z ≤ 8 − r²

Solution: SECTION 15.7 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES ¤ 587

11.  = ² ⇔  = ²+ ², a circular paraboloid opening upward with vertex the origin, and  = 8 − ² ⇔  = 8 − (²+ ²), a circular paraboloid opening downward with vertex (0 0 8). The paraboloids intersect when ²= 8 − ² ⇔ ²= 4. Thus

²≤  ≤ 8 − ²describes the solid above the paraboloid  = ²+ ²and below the paraboloid  = 8 − ²− ²for ²+ ²≤ 4.

12.  =  =

²+ ²is a cone that opens upward. Thus  ≤  ≤ 2 is the region above this cone and beneath the horizontal plane  = 2. 0 ≤  ≤ ^₂ restricts the solid to that part of this region in the first octant.

13. We can position the cylindrical shell vertically so that its axis coincides with the -axis and its base lies in the -plane. If we use centimeters as the unit of measurement, then cylindrical coordinates conveniently describe the shell as 6 ≤  ≤ 7, 0 ≤  ≤ 2, 0 ≤  ≤ 20.

14. In cylindrical coordinates, the equations are  = ²and  = 5 − ². The curve of intersection is ²= 5 − ²or  =

52. So we graph the surfaces in cylindrical coordinates, with 0 ≤  ≤

52. In Maple, we can use the coords=cylindrical option in a regular plot3d command. In Mathematica, we can use RevolutionPlot3D or

ParametricPlot3D.

15. The region of integration is given in cylindrical coordinates by

 =

(  ) | −2 ≤  ≤ 2, 0 ≤  ≤ 2, 0 ≤  ≤ ². This

represents the solid region above quadrants I and IV of the -plane enclosed by the circular cylinder  = 2, bounded above by the circular paraboloid

 = ²( = ²+ ²), and bounded below by the -plane ( = 0).

2

−2

2 0

²

0     =2

−2

2 0

=²

=0   =2

−2

2

0 ³ 

=2

−2  2

0 ³ =

2

−2

1 4⁴2

0

=  (4 − 0) = 4

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16. (a) Express the triple integral RRR

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

(b) Evaluate the iterated integral.

1100 CHAPTER 15 Multiple Integrals

15.7 Exercises

1–2 Plot the point whose cylindrical coordinates are given.

Then find the rectangular coordinates of the point.

1. (a) s5, y2, 2d (b) s6, 2y4, 23d 2. (a) s2, 5y6, 1d (b) s8, 22y3, 5d

3–4 Change from rectangular to cylindrical coordinates.

3. (a) s4, 4, 23d (b)

(

5s3, 25, s3

)

4. (a)

(

0, 22, 9

)

(b)

(

21, s3 , 6

)

5–6 Describe in words the surface whose equation is given.

5. r − 2 6.  − y6

7–8 Identify the surface whose equation is given.

7. r²1z²− 4 8. r − 2 sin 

9–10 Write the equations in cylindrical coordinates.

9. (a) x²2x 1 y²1z²− 1 (b) z − x²2y²

10. (a) 2x²12y²2z²− 4 (b) 2x 2 y 1 z − 1

11–12 Sketch the solid described by the given inequalities.

11. r²<z <8 2 r²

12. 0 <  < y2, r < z < 2

13. A cylindrical shell is 20 cm long, with inner radius 6 cm and outer radius 7 cm. Write inequalities that describe the shell in an appropriate coordinate system. Explain how you have positioned the coordinate system with respect to the shell.

14. Use graphing software to draw the solid enclosed by the paraboloids z − x²1y² and z − 5 2 x²2y².

15–16

(a) Express the triple integral yyyE fsx, y, zd dV as an iterated integral in cylindrical coordinates for the given function f and solid region E.

(b) Evaluate the iterated integral.

;

15. fsx, y, zd − x²1y² 16. fsx, y, zd − xy

E

z=2-≈-¥

≈+¥=1

E

z=6-≈-¥

0

z=œ„„„„„≈+¥

z z

x y

x y

17–18 Sketch the solid whose volume is given by the integral and evaluate the integral.

17.

y

y2^3y2

y

0³

y

_r⁹2 r dz dr d

18.

y

0²

y

0²

y

0^r r dz d dr

19–30 Use cylindrical coordinates.

19. Evaluate yyyEsx²1y² dV, where E is the region that lies inside the cylinder x²1y²− 16 and between the planes z − 25 and z − 4.

20. Evaluate yyyE z dV, where E is enclosed by the paraboloid z − x²1y² and the plane z − 4.

21. Evaluate yyyE sx 1 y 1 zd dV, where E is the solid in the first octant that lies under the paraboloid z − 4 2 x²2y². 22. Evaluate yyyE sx 2 yd dV, where E is the solid that lies

between the cylinders x²1y²− 1 and x²1y²− 16, above the xy-plane, and below the plane z − y 1 4.

23. Evaluate yyyE x² dV, where E is the solid that lies within the cylinder x²1y²− 1, above the plane z − 0, and below the cone z²− 4x²14y².

24. Find the volume of the solid that lies within both the cylin- der x²1y²− 1 and the sphere x²1y²1z²− 4.

25. Find the volume of the solid that is enclosed by the cone z −sx²1y² and the sphere x²1y²1z²− 2.

26. Find the volume of the solid that lies between the parabo- loid z − x²1y² and the sphere x²1y²1z²− 2.

27. (a) Find the volume of the region E that lies between the paraboloid z − 24 2 x²2y² and the cone

z − 2sx²1y² .

(b) Find the centroid of E (the center of mass in the case where the density is constant).

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Solution:

1

SECTION 15.7 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES ¤ 1563

12.  =  =

²+ ²is a cone that opens upward. Thus  ≤  ≤ 2 is the region above this cone and beneath the horizontal plane  = 2. 0 ≤  ≤ ^2 restricts the solid to that part of this region in the first octant.

13. We can position the cylindrical shell vertically so that its axis coincides with the ­axis and its base lies in the ­plane. If we use centimeters as the unit of measurement, then cylindrical coordinates conveniently describe the shell as 6 ≤  ≤ 7, 0 ≤  ≤ 2, 0 ≤  ≤ 20.

14. In cylindrical coordinates, the equations are  = ²and  = 5 − ². The curve of intersection is ²= 5 − ²or  =52. So we graph the surfaces in cylindrical coordinates, with 0 ≤  ≤

52. In Maple, we can use the coords=cylindrical option in a regular plot3d command. In Mathematica, we can use RevolutionPlot3D or

ParametricPlot3D.

15. (a) In cylindrical coordinates, the region can be described as  = {(  ) | 0 ≤  ≤ 1, 0 ≤  ≤ , 0 ≤  ≤ 2 − ²}.

Thus,

(²+ ²)  = 0

1 0

2−²

0 ²·    .

(b)   0

 1 0

 2−² 0

³   =

  0

 1 0

³

=2−²

=0   =

  0

 1 0

(2³− ⁵)  

=

  0



 1 0

(2³− ⁵)  =

=

=0 ·

⁴ 2 −⁶

6

^=1

=0

=  3

16. (a) In cylindrical coordinates, the region  is bounded above by the paraboloid  = 6 − ²and below by the cone  = .

The paraboloid and cone intersect when 6 − ² =  ⇒ ²+  − 6 = 0 ⇒  = 2 (  0), so the region can be described as  = {(  ) | 0 ≤  ≤ 2, 0 ≤  ≤ 2,  ≤  ≤ 6 − ²}. Then



()  =2

0

2 0

6−²

  cos  ·  sin  ·    .

(b) 2

0

2 0

6−²

 ³cos  sin     =2

0

2

0 ³cos  sin 

=6−²

=  

=2

0

2

0 ³cos  sin  (6³− ⁴− ⁵)  

=2

0 cos  sin  2

0(6³− ⁴− ⁵)  = 0 ·2

0(6³− ⁴− ⁵)  = 0 17. The region of integration represents the solid enclosed by the paraboloid  = ²,

( = ²+ ²), below the plane  = 9 in the second and third quadrants.

32

2

3 0

9

²    =32

2

3

0 []^=9_=2   =32

2

3 0

9 − ³

 

=32

2 3 0

9 − ³

 = 

9 2²−^4⁴

3 0= ^81₄

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22. EvaluateRRR

E(x − y)dV , where E is the solid that lies between the cylinders x²+ y²= 1 and x²+ y²= 16, above the xy-plane, and below the plane z = y + 4.

Solution:

588 ¤ CHAPTER 15 MULTIPLE INTEGRALS

16. The region of integration is given in cylindrical coordinates by

 = {(  ) | 0 ≤  ≤ 2, 0 ≤  ≤ 2, 0 ≤  ≤ }. This represents the solid region enclosed by the circular cylinder  = 2, bounded above by the cone  = , and bounded below by the -plane.

2 0

2

0



0     =2 0

2

0

=

=0   =2 0

2

0 ² 

=2

0 ²2

0  =1 3³2

0

2

0 = ⁸₃· 2 = ¹⁶3 17. In cylindrical coordinates,  is given by {(  ) | 0 ≤  ≤ 2 0 ≤  ≤ 4 −5 ≤  ≤ 4}. So





²+ ² =2

0

4 0

4

−5

√²    =2

0 4

0 ² 4

−5

=

2

0

₁

3³4 0

4

−5= (2)₆₄

3

(9) = 384

18. The paraboloid  = ²+ ²= ²intersects the plane  = 4 in the circle ²+ ²= 4or ²= 4 ⇒  = 2, so in cylindrical coordinates,  is given by

(  )

 0 ≤  ≤ 2 0 ≤  ≤ 2 ²≤  ≤ 4. Thus



   =2

0

2 0

4

²()     =2

0

2 0

1

2²=4

=² 

=2

0

2 0

8 −¹2⁵

  =2

0  2 0

8 − ¹2⁵

 = 2

4²− 12¹⁶2 0

= 2 16 −¹⁶3

= ⁶⁴₃

19. The paraboloid  = 4 − ²− ²= 4 − ²intersects the -plane in the circle ²+ ²= 4or ²= 4 ⇒  = 2, so in cylindrical coordinates,  is given by

(  )

 0 ≤  ≤ 2 0 ≤  ≤ 2 0 ≤  ≤ 4 − ² . Thus



( +  + )  =2 0

2 0

4−²

0 ( cos  +  sin  + )    

=2 0

2 0

²(cos  + sin ) +¹₂²=4−²

=0  

=2 0

2 0

(4²− ⁴)(cos  + sin ) +¹₂(4 − ²)²

 

=2 0

₄

3³−¹5⁵

(cos  + sin ) −12¹(4 − ²)³=2

=0

=2 0

64

15(cos  + sin ) + ¹⁶₃ 

 =64

15(sin  − cos ) + ¹⁶32 0

= ⁶⁴₁₅(1 − 0) +¹⁶3 · ^2 −⁶⁴15(0 − 1) − 0 = ⁸3 +¹²⁸₁₅

20. In cylindrical coordinates  is bounded by the planes  = 0,  =  sin  + 4 and the cylinders  = 1 and  = 4, so  is given by {(  ) | 0 ≤  ≤ 2 1 ≤  ≤ 4 0 ≤  ≤  sin  + 4}. Thus



( − )  =2

0

4 1

 sin +4

0 ( cos  −  sin )     =2

0

4

1 (²cos  − ²sin )[  ]= sin +4

=0  

=2

0

4

1 (²cos  − ²sin )( sin  + 4)  

=2

0

4 1

³(sin  cos  − sin²) + 4²(cos  − sin )

 

=2

0

1

4⁴(sin  cos  − sin²) +⁴₃³(cos  − sin )^=4

=1 

=2

0

64 − ¹4

(sin  cos  − sin²) +256 3 −⁴3

(cos  − sin )



=2

0

255

4 (sin  cos  − sin²) + 84(cos  − sin )



=255 4

1

2sin² −1

2 − ¹4sin 2

+ 84 (sin  + cos )2

0 = ²⁵⁵₄ (−) + 84(1) − 0 − 84(1) = −²⁵⁵4 

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30. Find the mass of a ball B given by x²+ y²+ z²≤ a²if the density at any point is proportional to its distance from the z-axis.

Solution:

SECTION 15.7 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES ¤ 591

Since the region is homogeneous and symmetric, = = 0and

=

 2

0

 ^√2 0

  4²

    = 

 2

0

 ^√2 0

1

2² − 8⁵

 

= 

 2

0

1

4²²−⁴3⁶^{=√2}

=0  = 

 2

0 1

24³ = ₁₂¹³

Hence (  ) =

0 0²₃.

28. Since density is proportional to the distance from the -axis, we can say (  ) = 

²+ ². Then

 = 22

0

 0

 √²−²

0 ²   = 22

0

 0 ²√

²− ² 

= 22

0

1

8(2²− ²)√

²− ²+¹₈⁴sin⁻¹()^=

=0 = 22

0

1 8⁴

2

 = ¹₄⁴²

29. The region of integration is the region above the cone  =

²+ ², or  = , and below the plane  = 2. Also, we have

−2 ≤  ≤ 2 with −

4 − ²≤  ≤

4 − ²which describes a circle of radius 2 in the -plane centered at (0 0). Thus,

 2

−2

 √4−²

−√

4−²

 2

√²+²

    =

 2

0

 2 0

 2



( cos )      =

 2

0

 2 0

 2



²(cos )    

=2

0

2

0 ²(cos )₁

2²=2

=   = ¹₂2

0

2

0 ²(cos ) 4 − ²

 

= ¹₂2

0 cos   2 0

4²− ⁴

 = ¹₂[sin ]^2₀ ₄

3³−¹5⁵2 0= 0 30. The region of integration is the region above the plane  = 0 and below the paraboloid  = 9 − ²− ². Also, we have

−3 ≤  ≤ 3 with 0 ≤  ≤√

9 − ²which describes the upper half of a circle of radius 3 in the -plane centered at (0 0).

Thus,

 3

−3

 √9−² 0

 9−²−² 0

²+ ²   =

  0

 3 0

 9−² 0

√²    =

  0

 3 0

 9−² 0

²  

= 0

3 0 ²

9 − ²

  = 0 3

0

9²− ⁴



=

 0

3³−¹5⁵3 0= 

81 − ²⁴³5

= ¹⁶²₅ 

31. (a) The mountain comprises a solid conical region . The work done in lifting a small volume of material ∆ with density

( )to a height ( ) above sea level is ( )( ) ∆ . Summing over the whole mountain we get

 =

( )( ) .

(b) Here  is a solid right circular cone with radius  = 19 000 m, height  = 3800 m, and density ( ) = 3200 kgm³at all points  in . We use cylindrical coordinates:

 =2

0

 0

(1−)

0  · 3200    = 2

0 3200₁

2²=(1−)

=0 

= 6400

  0

² 2

1 − 



2

 = 3200²

  0



 −2²

 + ³

²





= 3200²

² 2 −2³

3 + ⁴ 4²

^

0

= 3200²

²

2 − 2² 3 +²

4



= 3200²₁

12²

= ⁸⁰⁰₃ ²²= ⁸⁰⁰₃ (19 000)²(3800)²

≈ 44 × 10¹⁸J



 =  − 

 = 1 − 



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