Solutions To Selected Problems

Chapter 14

Exercises 14.1

  1. 1.

    C(y), meaning that instead of being just a constant, like the number 5, it is a function of y, which acts like a constant when taking derivatives with respect to x.

  2. 3.

    curve to curve, then from point to point

  3. 5.
    (a) 18x2+42x-117 (b) -108
  4. 7.
    (a) x4/2-x2+2x-3/2 (b) 23/15
  5. 9.
    (a) sin2y (b) π/2
  6. 11.
    14-21dydx and -2114dxdy. area of R=9 units2
  7. 13.
    24x-17-xdydx. The order dxdy needs two iterated integrals as x is bounded above by two different functions. This gives: 132y+1dxdy+3527-ydxdy. area of R=4 units2
  8. 15.
    01x4xdydx and 01y2y4dxdy area of R=7/15 units2
  9. 17.
    Ry=4-x2-2224xy

    area of R=04-4-y4-ydxdy

  10. 19.
    Rx2/16+y2/4=124-22xy

    area of R=04-4-x2/44-x2/4dydx

  11. 21.
    Ry=x2y=x+2-1121234xy

    area of R=-12x2x+2dydx

Exercises 14.2

  1. 1.

    volume

  2. 3.

    The double integral gives the signed volume under the surface. Since the surface is always positive, it is always above the x-y plane and hence produces only “positive” volume.

  3. 5.

    6; -1112(xy+3)dydx

  4. 7.

    112/3; 0204-2y(3x2-y+2)dxdy

  5. 9.

    16/5; -1101-x2(x+y+2)dydx

  6. 11.
    (a) Ry=xy=x211xy (b) 01x2xx2ydydx=01y2yx2ydxdy. (c) 356
  7. 13.
    (a) R-111-1xy (b) -11-11x2-y2dydx=-11-11x2-y2dxdy. (c) 0
  8. 15.
    (a) R3x+2y=612123xy (b) (c) 0203-3/2x(6-3x-2y)dydx=0302-2/3y(6-3x-2y)dxdy. (d) 6
  9. 17.
    (a) R-33-33xy (b) -3309-x2(x3y-x)dydx=03-9-y29-y2(x3y-x)dxdy. (c) 0
  10. 19.

    Integrating ex2 with respect to x is not possible in terms of elementary functions. 0202xex2dydx=e4-1.

  11. 21.

    Integrating y12yx2+y2dx gives tan-1(1/y)-π/4; integrating tan-1(1/y) is hard.

    010x2yx2+y2dydx=ln2.

  12. 23.

    average value of f=6/2=3

  13. 25.

    average value of f=112/34=28/3

Exercises 14.3

  1. 1.

    f(rcosθ,rsinθ), rdrdθ

  2. 3.

    02π01(3rcosθ-rsinθ+4)rdrdθ=4π

  3. 5.

    0πcosθ3cosθ(8-rsinθ)rdrdθ=16π

  4. 7.

    02π12(ln(r2))rdrdθ=2π(ln16-3/2)

  5. 9.

    -π/2π/206(r2cos2θ-r2sin2θ)rdrdθ=-π/2π/206(r2cos(2θ))rdrdθ=0

  6. 11.

    -π/2π/205(r2)drdθ=125π/3

  7. 13.

    0π/408(rcosθ+rsinθ)rdrdθ=162/3

  8. 15.
    (a) This is impossible to integrate with rectangular coordinates as e-(x2+y2) does not have an antiderivative in terms of elementary functions. (b) 02π0arer2drdθ=π(1-e-a2). (c) limaπ(1-e-a2)=π. This implies that there is a finite volume under the surface e-(x2+y2) over the entire x-y plane. (d) If R=2, we can write the original integral as (-e-t2dt)2=π.
  9. 17.

    3π/4-93/16

  10. 19.

    2

  11. 21.

    2(1-a3)/3(1-a2); 2/3; 1

Exercises 14.4

  1. 1.

    Because they are scalar multiples of each other.

  2. 3.

    “little masses”

  3. 5.

    Mx measures the moment about the x-axis, meaning we need to measure distance from the x-axis. Such measurements are measures in the y-direction.

  4. 7.

    x¯=5.25

  5. 9.

    (x¯,y¯)=(0,3)

  6. 11.

    M=150g;

  7. 13.

    M=2lb

  8. 15.

    M=16π50.27kg

  9. 17.

    M=54π169.65lb

  10. 19.

    M=150g; My=600; Mx=-75; (x¯,y¯)=(4,-0.5)

  11. 21.

    M=2lb; My=0; Mx=2/3; (x¯,y¯)=(0,1/3)

  12. 23.

    M=16π50.27kg; My=4π; Mx=4π; (x¯,y¯)=(1/4,1/4)

  13. 25.

    M=54π169.65lb; My=0; Mx=504; (x¯,y¯)=(0,2.97)

  14. 27.

    Ix=64/3; Iy=64/3; IO=128/3

  15. 29.

    Ix=16/3; Iy=64/3; IO=80/3

Exercises 14.5

  1. 1.

    arc length

  2. 3.

    surface areas

  3. 5.

    Intuitively, adding h to f only shifts f up (i.e., parallel to the z-axis) and does not change its shape. Therefore it will not change the surface area over R.

    Analytically, fx=gx and fy=gy; therefore, the surface area of each is computed with identical double integrals.

  4. 7.

    S=02π02π1+cos2xcos2y+sin2xsin2ydxdy

  5. 9.

    S=-11-111+4x2+4y2dxdy

  6. 11.

    S=03-111+9+49dxdy=65946.09

  7. 13.
    This is easier in polar: S =02π04r1+4r2cos2θ+4r2sin2θdrdθ =02π04r1+4r2drdθ =π6(6565-1)273.87
  8. 15.
    S =01011+x+9ydxdy =0123((9y+2)3/2-(9y+1)3/2)dy =4135(12111-10010-42+1)2.383
  9. 17.
    This is easier in polar: S =202π05r1+r2cos2θ+r2sin2θ25-r2sin2θ-r2cos2θdrdθ =202π05r125-r2drdθ =100π314.16

Exercises 14.6

  1. 1.

    surface to surface, curve to curve and point to point

  2. 3.

    Answers can vary. From this section we used triple integration to find the volume of a solid region, the mass of a solid, and the center of mass of a solid.

  3. 5.

    V=-11-11(8-x2-y2-(2x+y))dxdy=88/3

  4. 7.

    V=0π0x(cosxsiny+2-sinxcosy)dydx=π2-π6.728

  5. 9.
    dzdydx: 0301-x/302-2x/3-2ydzdydx dzdxdy: 0103-3y02-2x/3-2ydzdxdy dydzdx: 0302-2x/301-x/3-z/2dydzdx dydxdz: 0203-3z/201-x/3-z/2dydxdz dxdzdy: 0102-2y03-3y-3z/2dxdzdy dxdydz: 0201-z/203-3y-3z/2dxdydz V=0301-x/302-2x/3-2ydzdydx=1.
  6. 11.
    dzdydx: 02-20y2/2-ydzdydx dzdxdy: -2002y2/2-ydzdxdy dydzdx: 0202-2z-zdydzdx dydxdz: 0202-2z-zdydxdz dxdzdy: -20y2/2-y02dxdzdy dxdydz: 02-2z-z02dxdydz V=0202-2z-zdydzdx=4/3.
  7. 13.
    dzdydx: 021-x/2102x+4y-4dzdydx dzdxdy: 012-2y202x+4y-4dzdxdy dydzdx: 0202xz/4-x/2+11dydzdx dydxdz: 04z/22z/4-x/2+11dydxdz dxdzdy: 0104yz/2-2y+22dxdzdy dxdydz: 04z/41z/2-2y+22dxdydz V=04z/41z/2-2y+22dxdydz=4/3.
  8. 15.
    dzdydx: 0101-x201-ydzdydx dzdxdy: 0101-y01-ydzdxdy dydzdx: 010x01-x2dydzdx+01x101-z2dydzdx dydxdz: 010z01-z2dydxdz+01z101-x2dydxdz dxdzdy: 0101-y01-ydxdzdy dxdydz: 0101-z201-ydxdydz Answers will vary. Neither order is particularly “hard.” The order dzdydx requires integrating a square root, so powers can be messy; the order dydzdx requires two triple integrals, but each uses only polynomials.
  9. 17.

    8

  10. 19.

    π

  11. 21.
    M=10, Myz=15/2, Mxz=5/2, Mxy=5; (x¯,y¯,z¯)=(3/4,1/4,1/2)
  12. 23.
    M=16/5, Myz=16/3, Mxz=104/45, Mxy=32/9; (x¯,y¯,z¯)=(5/3,13/18,10/9)(1.67,0.72,1.11)

Exercises 14.7

  1. 1.

    In cylindrical, r determines how far from the origin one goes in the x-y plane before considering the z-component. Equivalently, if on projects a point in cylindrical coordinates onto the x-y plane, r will be the distance of this projection from the origin.

    In spherical, ρ is the distance from the origin to the point.

  2. 3.

    Cylinders (tubes) centered at the origin, parallel to the z-axis; planes parallel to the z-axis that intersect the z-axis; planes parallel to the x-y plane.

  3. 5.
    (a) Cylindrical: (22,π/4,1) and (2,5π/6,0) Spherical: (3,π/4,cos-1(1/3)) and (2,5π/6,π/2) (b) Rectangular: (2,2,2) and (0,-3,-4) Spherical: (22,π/4,π/4) and (5,3π/2,π-tan-1(3/4)) (c) Rectangular: (1,1,2) and (0,0,1) Cylindrical: (2,π/4,2) and (0,0,1)
  4. 7.
    (a) Cylindrical: (4,π3,-1) and (52,3π4,6) Spherical: (17,π3,cos-1-117) and (86,3π4,cos-1686) (b) Rectangular: (12,32,-2) and (-3,-1,3) Spherical: (5,π3,cos-1(-25)) and (13,5π6,tan-123) (c) Rectangular: (2,6,22) and (1,0,0) Cylindrical: (22,π3,22) and (1,0,0)
  5. 9.
    (a) A cylindrical surface or tube, centered along the z-axis of radius 1, extending from the x-y plane up to the plane z=1 (i.e., the tube has a length of 1). (b) This is a region of space, being half of a tube with “thick” walls of inner radius 1 and outer radius 2, centered along the z-axis with a length of 1, where the half “below” the x-z plane is removed. (c) This is upper half of the sphere of radius 3 centered at the origin (i.e., the upper hemisphere). (d) This is a region of space, where the ball of radius 2, centered at the origin, is removed from the ball of radius 3, centered at the origin.
  6. 11.

    θ1θ2r1r2z1z2h(r,θ,z)rdzdrdθ

  7. 13.

    The region in space is bounded between the planes z=0 and z=2, inside of the cylinder x2+y2=4, and the planes θ=0 and θ=π/2: describes a “wedge” of a cylinder of height 2 and radius 2; the angle of the wedge is π/2, or 90.

  8. 15.

    Bounded between the plane z=0 and the cone z=1-x2+y2: describes an inverted cone, with height of 1, point at (0,0,1) and base radius of 1.

  9. 17.

    Describes a quarter of a ball of radius 3, centered at the origin; the quarter resides above the x-y plane and above the x-z plane.

  10. 19.

    Describes the portion of the unit ball that resides in the first octant.

  11. 21.

    Bounded above the cone z=x2+y2 and below the sphere x2+y2+z2=4: describes a shape that is somewhat “diamond”-like; some think of it as looking like an ice cream cone (see Figure 14.7.8). It describes a cone, where the side makes an angle of π/4 with the positive z-axis, topped by the portion of the ball of radius 2, centered at the origin.

  12. 23.

    The region in space is bounded below by the cone z=3x2+y2 and above by the plane z=1: it describes a cone, with point at the origin, centered along the positive z-axis, with height of 1 and base radius of tan(π/6)=1/3.

  13. 25.
    In cylindrical coordinates, the density is δ(r,θ,z)=r+1. Thus mass is 02π0204(r+1)rdzdrdθ=112π/3.
  14. 27.
    In cylindrical coordinates, the density is δ(r,θ,z)=1. Thus mass is 0π0104-rsinθrdzdrdθ=2π-2/35.617.
  15. 29.
    In cylindrical coordinates, the density is δ(r,θ,z)=r+1. Thus mass is M=02π0204(r+1)rdzdrdθ=112π/3.

    We find Myz=0, Mxz=0, and Mxy=224π/3, placing the center of mass at (0,0,2).

  16. 31.
    In cylindrical coordinates, the density is δ(r,θ,z)=1. Thus mass is 0π0104-rsinθrdzdrdθ=2π-2/35.617.

    We find Myz=0, Mxz=8/3-π/8, and Mxy=65π/16-8/3, placing the center of mass at (0,0.405,1.80).

  17. 33.
    In spherical coordinates, the density is δ(ρ,θ,φ)=1. Thus mass is 0π/202π01ρ2sin(φ)dρdθdφ=2π/3.
  18. 35.
    In spherical coordinates, the density is δ(ρ,θ,φ)=ρcosφ. Thus mass is 0π/402π01(ρcos(φ))ρ2sin(φ)dρdθdφ=π/8.
  19. 37.
    In spherical coordinates, the density is δ(ρ,θ,φ)=1. Thus mass is 0π/202π01ρ2sin(φ)dρdθdφ=2π/3.

    We find Myz=0, Mxz=0, and Mxy=π/4, placing the center of mass at (0,0,3/8).

  20. 39.
    In spherical coordinates, the density is δ(ρ,θ,φ)=ρcosφ. Thus mass is 0π/402π01(ρcos(φ))ρ2sin(φ)dρdθdφ=π/8.

    We find Myz=0, Mxz=0, and Mxy=(4-2)π/30, placing the center of mass at (0,0,4(4-2)/15).

  21. 41.

    Rectangular: -11-1-x21-x2-1-x2-y21-x2-y2dzdydx

    Cylindrical: 02π01-1-r21-r2rdzdrdθ

    Spherical: 0π02π01ρ2sin(φ)dρdθdφ

    Spherical appears simplest, avoiding the integration of square-roots and using techniques such as Substitution; all bounds are constants.

  22. 43.

    Rectangular: -11-1-x21-x2x2+y21dzdydx

    Cylindrical: 02π01r1rdzdrdθ

    Spherical: 0π/402π0secφρ2sin(φ)dρdθdφ

    Cylindrical appears simplest, avoiding the integration of square-roots that rectangular uses. Spherical is not difficult, though it requires Substitution, an extra step.

  23. 45.

    center: (2,3,-1), radius: 14

  24. 47.

    (a,θ,acotϕ)

  25. 49.

    Hint: Use the distance formula for Cartesian coordinates.

  26. 51.
  27. 53.

    1-sin2/2

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