Chapter N

Exercises N.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. 2.

    iterated integration

  3. 3.

    curve to curve, then from point to point

  4. 4.

    area

  5. 5.

    • 18⁢x2+42⁢x−117

      −108

  6. 6.

    • 2+π2⁢cos⁡y

      π2+π

  7. 7.

    • x4/2−x2+2⁢x−3/2

      23/15

  8. 8.

    • y4/2−y3+y2/2

      8/15

  9. 9.

    • sin2⁡y

      π/2

  10. 10.

    • x/(1+x2)

      12⁢ln⁡(52)

  11. 11.

    ∫14∫−21d⁡y⁢d⁡x and ∫−21∫14d⁡x⁢d⁡y.

    area of R=9⁢ units2

  12. 12.

    ∫14∫123⁢x+13d⁡y⁢d⁡x and ∫13∫32⁢y−124d⁡x⁢d⁡y.

    area of R=3⁢ units2

  13. 13.

    ∫24∫x−17−xd⁡y⁢d⁡x. The order d⁡x⁢d⁡y needs two iterated integrals as x is bounded above by two different functions. This gives:

    ∫13∫2y+1d⁡x⁢d⁡y+∫35∫27−yd⁡x⁢d⁡y.

    area of R=4⁢ units2

  14. 14.

    ∫012∫−3⁢x3⁢xd⁡y⁢d⁡x and ∫−66∫y2/312d⁡x⁢d⁡y

    area of R=96⁢ units2

  15. 15.

    ∫01∫x4xd⁡y⁢d⁡x and ∫01∫y2y4d⁡x⁢d⁡y

    area of R=7/15⁢ units2

  16. 16.

    ∫02∫x34⁢xd⁡y⁢d⁡x and ∫08∫y/4y3d⁡x⁢d⁡y

    area of R=4⁢ units2

  17. 17.

    Ry=4−x2−2224xy

    area of R=∫04∫−4−y4−yd⁡x⁢d⁡y

  18. 18.

    Ry=5−5⁢x2y=−5⁢x+50.5112345xy

    area of R=∫05∫1−y/51−y/5d⁡x⁢d⁡y

  19. 19.

    Rx2/16+y2/4=124−22xy

    area of R=∫04∫−4−x2/44−x2/4d⁡y⁢d⁡x

  20. 20.

    Rx2+y2=9−22−22xy

    area of R=∫−33∫−9−y29−y2d⁡x⁢d⁡y

  21. 21.

    Ry=x2y=x+2−1121234xy

    area of R=∫−12∫x2x+2d⁡y⁢d⁡x

  22. 22.

    R−11−11xy

    area of R=∫−10∫02⁢y+1d⁡x⁢d⁡y+∫01∫01−2⁢yd⁡x⁢d⁡y

Exercises N.2

  1. 1.

    volume

  2. 2.

    When switching the order of integration, the bounds integrals must change to reflect the bounds of the region of integration. You cannot merely change the letters x and y in a few places.

  3. 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.

  4. 4.

    No. It means that there is the same amount of signed volume under f and g over R, but the functions could be very different.

  5. 5.

    6; ∫−11∫12(xy+3)⁢d⁡y⁢d⁡x

  6. 6.

    4; ∫0π∫−π/2π/2(sin⁡x⁢cos⁡y)⁢d⁡y⁢d⁡x

  7. 7.

    112/3; ∫02∫04−2⁢y(3⁢x2−y+2)⁢d⁡x⁢d⁡y

  8. 8.

    76/15; ∫13∫1x(x2⁢y−x⁢y2)⁢d⁡y⁢d⁡x

  9. 9.

    16/5; ∫−11∫01−x2(x+y+2)⁢d⁡y⁢d⁡x

  10. 10.

    6561/40; ∫03∫x23⁢x(x⁢y2)⁢d⁡y⁢d⁡x

  11. 11.

    • Ry=xy=x211xy

      ∫01∫x2xx2⁢y⁢d⁡y⁢d⁡x=∫01∫y2yx2⁢y⁢d⁡x⁢d⁡y.

      356

  12. 12.

    • Ry=x3y=x311xy

      ∫01∫x3x3x2⁢y⁢d⁡y⁢d⁡x+∫−10∫x3x3x2⁢y⁢d⁡y⁢d⁡x =∫01∫y3y3x2⁢y⁢d⁡x⁢d⁡y+∫−10∫y3y3x2⁢y⁢d⁡x⁢d⁡y.

      0

  13. 13.

    • R−111−1xy

      ∫−11∫−11x2−y2⁢d⁡y⁢d⁡x=∫−11∫−11x2−y2⁢d⁡x⁢d⁡y.

      0

  14. 14.

    • Rx=y211xy

      ∫01∫0y2y⁢ex⁢d⁡x⁢d⁡y=∫01∫x1y⁢ex⁢d⁡y⁢d⁡x.

      e/2−1

  15. 15.

    • R3⁢x+2⁢y=612123xy

      ∫02∫03−3/2⁢x(6−3⁢x−2⁢y)⁢d⁡y⁢d⁡x=∫03∫02−2/3⁢y(6−3⁢x−2⁢y)⁢d⁡x⁢d⁡y.

      6

  16. 16.

    • Ry=ln⁡xy=1e−1⁢(x−1)1231xy

      ∫1e∫x−1e−1ln⁡xey⁢d⁡y⁢d⁡x=∫01∫eyy⁢(e−1)+1ey⁢d⁡x⁢d⁡y.

      −12⁢e2+2⁢e−32

  17. 17.

    • R−33−33xy

      ∫−33∫09−x2(x3⁢y−x)⁢d⁡y⁢d⁡x=∫03∫−9−y29−y2(x3⁢y−x)⁢d⁡x⁢d⁡y.

      0

  18. 18.

    • R(e,1)121exy

      ∫01∫e⁢yey(4−3⁢y)⁢d⁡x⁢d⁡y=∫01∫0x/e(4−3⁢y)⁢d⁡y⁢d⁡x+∫1e∫ln⁡xx/e(4−3⁢y)⁢d⁡y⁢d⁡x.

      3⁢e−7

  19. 19.

    Integrating ex2 with respect to x is not possible in terms of elementary functions. ∫02∫02⁢xex2⁢d⁡y⁢d⁡x=e4−1.

  20. 20.

    Integrating cos⁡(y2) with respect to y is not possible in terms of elementary functions. ∫0π/2∫0ycos⁡(y2)⁢d⁡x⁢d⁡y=12.

  21. 21.

    Integrating ∫y12⁢yx2+y2⁢d⁡x gives tan−1⁡(1/y)−π/4; integrating tan−1⁡(1/y) is hard.

    ∫01∫0x2⁢yx2+y2⁢d⁡y⁢d⁡x=ln⁡2.

  22. 22.

    Integrating in the order shown is hard / impossible. By changing the order of integration, we have ∫12∫−11x⁢tan2⁡y1+ln⁡y⁢d⁡x⁢d⁡y=0, since the integrand is an odd function with respect to x. Thus the iterated integral evaluates to 0.

  23. 23.

    average value of f=6/2=3

  24. 24.

    average value of f=4/π2

  25. 25.

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

  26. 26.

    average value of f=76/152=3815≈2.53

Exercises N.3

  1. 1.

    f⁢(r⁢cos⁡θ,r⁢sin⁡θ), r⁢d⁡r⁢d⁡θ

  2. 2.

    Some regions in the x-y plane are easier to describe using polar coordinates than using rectangular coordinates. Also, some integrals are easier to evaluate one the polar substitutions have been made.

  3. 3.

    ∫02⁢π∫01(3⁢r⁢cos⁡θ−r⁢sin⁡θ+4)⁢r⁢d⁡r⁢d⁡θ=4⁢π

  4. 4.

    ∫02⁢π∫02(4⁢r⁢cos⁡θ+4⁢r⁢sin⁡θ)⁢r⁢d⁡r⁢d⁡θ=0

  5. 5.

    ∫0π∫cos⁡θ3⁢cos⁡θ(8−r⁢sin⁡θ)⁢r⁢d⁡r⁢d⁡θ=16⁢π

  6. 6.

    ∫0π/2∫0sin⁡(2⁢θ)(4)⁢r⁢d⁡r⁢d⁡θ=π/2

  7. 7.

    ∫02⁢π∫12(ln⁡(r2))⁢r⁢d⁡r⁢d⁡θ=2⁢π⁢(ln⁡16−3/2)

  8. 8.

    ∫02⁢π∫01(1−r2)⁢r⁢d⁡r⁢d⁡θ=π/2

  9. 9.

    ∫−π/2π/2∫06(r2⁢cos2⁡θ−r2⁢sin2⁡θ)⁢r⁢d⁡r⁢d⁡θ=∫−π/2π/2∫06(r2⁢cos⁡(2⁢θ))⁢r⁢d⁡r⁢d⁡θ=0

  10. 10.

    ∫0π/4∫01(cos⁡θ−sin⁡θcos⁡θ+sin⁡θ)⁢r⁢d⁡r⁢d⁡θ=ln⁡2

  11. 11.

    ∫−π/2π/2∫05(r2)⁢d⁡r⁢d⁡θ=125⁢π/3

  12. 12.

    ∫π/23⁢π/2∫04(2⁢r⁢sin⁡θ−r⁢cos⁡θ)⁢r⁢d⁡r⁢d⁡θ=128/3

  13. 13.

    ∫0π/4∫08(r⁢cos⁡θ+r⁢sin⁡θ)⁢r⁢d⁡r⁢d⁡θ=16⁢2/3

  14. 14.

    ∫0π∫12(r⁢cos⁡θ+5)⁢r⁢d⁡r⁢d⁡θ=15⁢π/2

  15. 15.

    • This is impossible to integrate with rectangular coordinates as e−(x2+y2) does not have an antiderivative in terms of elementary functions.

      ∫02⁢π∫0ar⁢er2⁢d⁡r⁢d⁡θ=π⁢(1−e−a2).

      lima→∞π⁢(1−e−a2)=π. This implies that there is a finite volume under the surface e−(x2+y2) over the entire x-y plane.

      If R=ℝ2, we can write the original integral as (∫−∞∞e−t2⁢d⁡t)2=π.

  16. 16.

    ∬R f⁢(x,y)⁢d⁡A
    =∫02⁢π∫0a(h−h⁢r2⁢cos2⁡θa2+r2⁢sin2⁡θa2)⁢r⁢d⁡r⁢d⁡θ
    =∫02⁢π∫0a(h⁢r−h⁢r2a)⁢d⁡r⁢d⁡θ
    =∫02⁢π(12⁢h⁢r2−h3⁢a⁢r3)|0a⁢d⁡θ
    =∫02⁢π(16⁢a2⁢h)⁢d⁡θ
    =13⁢π⁢a2⁢h.
  17. 17.

    3⁢π/4−9⁢3/16

  18. 18.

    2⁢π

  19. 19.

    2

  20. 20.

    (1−1/2)⁢π/2

  21. 21.

    2⁢(1−a3)/3⁢(1−a2); 2/3; 1

Exercises N.4

  1. 1.

    Because they are scalar multiples of each other.

  2. 2.

    y

  3. 3.

    “little masses”

  4. 4.

    A collection of individual masses in the plane. Each mass is a point mass, i.e., mass located at a point, not across a region.

  5. 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.

  6. 6.

    If the lamina is an annulus, the center of mass will likely be in the middle, outside of the region. (See Example 14.4.9.)

  7. 7.

    x¯=5.25

  8. 8.

    x¯=1.3

  9. 9.

    (x¯,y¯)=(0,3)

  10. 10.

    (x¯,y¯)=(0,1/3)

  11. 11.

    M=150g;

  12. 12.

    M=190g

  13. 13.

    M=2lb

  14. 14.

    M=2/3lb

  15. 15.

    M=16⁢π≈50.27kg

  16. 16.

    M=325⁢π/12≈85kg

  17. 17.

    M=54⁢π≈169.65lb

  18. 18.

    M=63⁢π≈197.92lb

  19. 19.

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

  20. 20.

    M=190g; My=850; Mx=−315/2; (x¯,y¯)=(4.47,−0.83)

  21. 21.

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

  22. 22.

    M=2/3lb; My=7/30; Mx=7/30; (x¯,y¯)=(0.35,0.35)

  23. 23.

    M=16⁢π≈50.27kg; My=4⁢π; Mx=4⁢π; (x¯,y¯)=(1/4,1/4)

  24. 24.

    M=325⁢π/12≈85kg; My=2375/12; Mx=2375/12; (x¯,y¯)=(2.33,2.33)

  25. 25.

    M=54⁢π≈169.65lb; My=0; Mx=504; (x¯,y¯)=(0,2.97)

  26. 26.

    M=63⁢π≈197.92lb; My=0; Mx=1215/2; (x¯,y¯)=(0,3.07)

  27. 27.

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

  28. 28.

    Ix=16/3; Iy=256/3; IO=272/3

  29. 29.

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

  30. 30.

    Ix=16; Iy=16; IO=32

Exercises N.5

  1. 1.

    arc length

  2. 2.

    tangent

  3. 3.

    surface areas

  4. 4.

    Technology makes good approximations accessible, if not exact answers.

  5. 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.

  6. 6.

    Analytically, gx=2⁢fx and gy=2⁢fy. The double integral to compute the surface area of f over R is ∬R1+fx2+fy2⁢d⁡A; the double integral to compute the surface area of g over R is ∬R1+4⁢fx2+4⁢fy2⁢d⁡A, which is not twice the double integral used to calculate the surface area of f.

  7. 7.

    S=∫02⁢π∫02⁢π1+cos2⁡x⁢cos2⁡y+sin2⁡x⁢sin2⁡y⁢d⁡x⁢d⁡y

  8. 8.

    S=∫−33∫−9−x29−x21+4⁢x2+4⁢y2(1+x2+y2)4⁢d⁡x⁢d⁡y

    Polar offers simpler bounds:

    S=∫02⁢π∫03r⁢1+4⁢r2(1+r2)4⁢𝑑r⁢𝑑θ

  9. 9.

    S=∫−11∫−111+4⁢x2+4⁢y2⁢d⁡x⁢d⁡y

  10. 10.

    S=∫−55∫011+4⁢x2⁢e2⁢x2(1+ex2)4⁢d⁡y⁢d⁡x

  11. 11.

    S=∫03∫−111+9+49⁢d⁡x⁢d⁡y=6⁢59≈46.09

  12. 12.

    S=∫01∫01−x1+4+4⁢d⁡y⁢d⁡x=18

  13. 13.

    This is easier in polar:

    S =∫02⁢π∫04r⁢1+4⁢r2⁢cos2⁡θ+4⁢r2⁢sin2⁡θ⁢d⁡r⁢d⁡θ
    =∫02⁢π∫04r⁢1+4⁢r2⁢d⁡r⁢d⁡θ
    =π6⁢(65⁢65−1)≈273.87
  14. 14.

    S =∫01∫−yy1+4+64⁢y2⁢d⁡x⁢d⁡y
    =∫01(2⁢y⁢5+64⁢y2)⁢d⁡y
    =196⁢(69⁢69−5⁢5)≈5.85
  15. 15.

    S =∫01∫011+x+9⁢y⁢d⁡x⁢d⁡y
    =∫0123⁢((9⁢y+2)3/2−(9⁢y+1)3/2)⁢d⁡y
    =4135⁢(121⁢11−100⁢10−4⁢2+1)≈2.383
  16. 16.

    This is easier in polar:

    S =∫02⁢π∫05r⁢1+4⁢r2⁢cos2⁡θ+4⁢r2⁢sin2⁡θr2⁢sin2⁡θ+r2⁢cos2⁡θ⁢d⁡r⁢d⁡θ
    =∫02⁢π∫05r⁢5⁢d⁡r⁢d⁡θ
    =25⁢π⁢5≈175.62
  17. 17.

    This is easier in polar:

    S =2⁢∫02⁢π∫05r⁢1+r2⁢cos2⁡θ+r2⁢sin2⁡θ25−r2⁢sin2⁡θ−r2⁢cos2⁡θ⁢d⁡r⁢d⁡θ
    =2⁢∫02⁢π∫05r⁢125−r2⁢d⁡r⁢d⁡θ
    =100⁢π≈314.16
  18. 18.

    Integrating in polar is easiest considering R:

    S =∫02⁢π∫01r⁢1+c2+d2⁢d⁡r⁢d⁡θ
    =∫02⁢π12⁢(1+c2+d2)⁢d⁡θ
    =π⁢1+c2+d2.

    The value of h does not matter as it only shifts the plane vertically (i.e., parallel to the z-axis). Different values of h do not create different ellipses in the plane.

Exercises N.6

  1. 1.

    surface to surface, curve to curve and point to point

  2. 2.

    One possible answer is “sum up lots of little volumes over D.”

  3. 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.

  4. 4.

    δ⁢V.

  5. 5.

    V=∫−11∫−11(8−x2−y2−(2⁢x+y))⁢d⁡x⁢d⁡y=88/3

  6. 6.

    V=∫02∫03(x2+y2−(−x2−y2))⁢d⁡y⁢d⁡x=52

  7. 7.

    V=∫0π∫0x(cos⁡x⁢sin⁡y+2−sin⁡x⁢cos⁡y)⁢d⁡y⁢d⁡x=π2−π≈6.728

  8. 8.

    V=∫−11∫−1−x21−x2(6−x2−y2−(2⁢x2+2⁢y2+3))⁢d⁡y⁢d⁡x. Integrating in polar is easier, giving V=∫02⁢π∫01(3−3⁢r2)⁢r⁢d⁡r⁢d⁡θ=3⁢π/2.

  9. 9.

    d⁡z⁢d⁡y⁢d⁡x: ∫03∫01−x/3∫02−2⁢x/3−2⁢yd⁡z⁢d⁡y⁢d⁡x

    d⁡z⁢d⁡x⁢d⁡y: ∫01∫03−3⁢y∫02−2⁢x/3−2⁢yd⁡z⁢d⁡x⁢d⁡y

    d⁡y⁢d⁡z⁢d⁡x: ∫03∫02−2⁢x/3∫01−x/3−z/2d⁡y⁢d⁡z⁢d⁡x

    d⁡y⁢d⁡x⁢d⁡z: ∫02∫03−3⁢z/2∫01−x/3−z/2d⁡y⁢d⁡x⁢d⁡z

    d⁡x⁢d⁡z⁢d⁡y: ∫01∫02−2⁢y∫03−3⁢y−3⁢z/2d⁡x⁢d⁡z⁢d⁡y

    d⁡x⁢d⁡y⁢d⁡z: ∫02∫01−z/2∫03−3⁢y−3⁢z/2d⁡x⁢d⁡y⁢d⁡z

    V=∫03∫01−x/3∫02−2⁢x/3−2⁢yd⁡z⁢d⁡y⁢d⁡x=1.

  10. 10.

    d⁡z⁢d⁡y⁢d⁡x: ∫13∫02∫0(3−x)/2d⁡z⁢d⁡y⁢d⁡x

    d⁡z⁢d⁡x⁢d⁡y: ∫02∫13∫0(3−x)/2d⁡z⁢d⁡x⁢d⁡y

    d⁡y⁢d⁡z⁢d⁡x: ∫13∫0(3−x)/2∫02d⁡y⁢d⁡z⁢d⁡x

    d⁡y⁢d⁡x⁢d⁡z: ∫01∫13−2⁢z∫02d⁡y⁢d⁡x⁢d⁡z

    d⁡x⁢d⁡z⁢d⁡y: ∫02∫01∫13−2⁢zd⁡x⁢d⁡z⁢d⁡y

    d⁡x⁢d⁡y⁢d⁡z: ∫01∫02∫13−2⁢zd⁡x⁢d⁡y⁢d⁡z

    V=∫01∫02∫13−2⁢zd⁡x⁢d⁡y⁢d⁡z=2.

  11. 11.

    d⁡z⁢d⁡y⁢d⁡x: ∫02∫−20∫y2/2−yd⁡z⁢d⁡y⁢d⁡x

    d⁡z⁢d⁡x⁢d⁡y: ∫−20∫02∫y2/2−yd⁡z⁢d⁡x⁢d⁡y

    d⁡y⁢d⁡z⁢d⁡x: ∫02∫02∫−2⁢z−zd⁡y⁢d⁡z⁢d⁡x

    d⁡y⁢d⁡x⁢d⁡z: ∫02∫02∫−2⁢z−zd⁡y⁢d⁡x⁢d⁡z

    d⁡x⁢d⁡z⁢d⁡y: ∫−20∫y2/2−y∫02d⁡x⁢d⁡z⁢d⁡y

    d⁡x⁢d⁡y⁢d⁡z: ∫02∫−2⁢z−z∫02d⁡x⁢d⁡y⁢d⁡z

    V=∫02∫02∫−2⁢z−zd⁡y⁢d⁡z⁢d⁡x=4/3.

  12. 12.

    d⁡z⁢d⁡y⁢d⁡x: ∫03∫3⁢x9∫0y2−9⁢x2d⁡z⁢d⁡y⁢d⁡x

    d⁡z⁢d⁡x⁢d⁡y: ∫09∫0y/3∫0y2−9⁢x2d⁡z⁢d⁡x⁢d⁡y

    d⁡y⁢d⁡z⁢d⁡x: ∫03∫081−9⁢x2∫z2+9⁢x29d⁡y⁢d⁡z⁢d⁡x

    d⁡y⁢d⁡x⁢d⁡z: ∫09∫09−z2/9∫z2+9⁢x29d⁡y⁢d⁡x⁢d⁡z

    d⁡x⁢d⁡z⁢d⁡y: ∫09∫0y∫013⁢y2−z2d⁡x⁢d⁡z⁢d⁡y

    d⁡x⁢d⁡y⁢d⁡z: ∫09∫z9∫013⁢y2−z2d⁡x⁢d⁡y⁢d⁡z

  13. 13.

    d⁡z⁢d⁡y⁢d⁡x: ∫02∫1−x/21∫02⁢x+4⁢y−4d⁡z⁢d⁡y⁢d⁡x

    d⁡z⁢d⁡x⁢d⁡y: ∫01∫2−2⁢y2∫02⁢x+4⁢y−4d⁡z⁢d⁡x⁢d⁡y

    d⁡y⁢d⁡z⁢d⁡x: ∫02∫02⁢x∫z/4−x/2+11d⁡y⁢d⁡z⁢d⁡x

    d⁡y⁢d⁡x⁢d⁡z: ∫04∫z/22∫z/4−x/2+11d⁡y⁢d⁡x⁢d⁡z

    d⁡x⁢d⁡z⁢d⁡y: ∫01∫04⁢y∫z/2−2⁢y+22d⁡x⁢d⁡z⁢d⁡y

    d⁡x⁢d⁡y⁢d⁡z: ∫04∫z/41∫z/2−2⁢y+22d⁡x⁢d⁡y⁢d⁡z

    V=∫04∫z/41∫z/2−2⁢y+22d⁡x⁢d⁡y⁢d⁡z=4/3.

  14. 14.

    d⁡z⁢d⁡y⁢d⁡x: ∫−22∫04−x2∫02⁢yd⁡z⁢d⁡y⁢d⁡x

    d⁡z⁢d⁡x⁢d⁡y: ∫04∫−4−y4−y∫02⁢x+4⁢y−4d⁡z⁢d⁡x⁢d⁡y

    d⁡y⁢d⁡z⁢d⁡x: ∫−22∫08−2⁢x2∫z/24−x2d⁡y⁢d⁡z⁢d⁡x

    d⁡y⁢d⁡x⁢d⁡z: ∫08∫−4−z/24−z/2∫z/24−x2d⁡y⁢d⁡x⁢d⁡z

    d⁡x⁢d⁡z⁢d⁡y: ∫04∫02⁢y∫−4−y4−yd⁡x⁢d⁡z⁢d⁡y

    d⁡x⁢d⁡y⁢d⁡z: ∫08∫z/24∫−4−y4−yd⁡x⁢d⁡y⁢d⁡z

    V=∫−22∫04−x2∫02⁢yd⁡z⁢d⁡y⁢d⁡x=512/15.

  15. 15.

    d⁡z⁢d⁡y⁢d⁡x: ∫01∫01−x2∫01−yd⁡z⁢d⁡y⁢d⁡x

    d⁡z⁢d⁡x⁢d⁡y: ∫01∫01−y∫01−yd⁡z⁢d⁡x⁢d⁡y

    d⁡y⁢d⁡z⁢d⁡x: ∫01∫0x∫01−x2d⁡y⁢d⁡z⁢d⁡x+∫01∫x1∫01−z2d⁡y⁢d⁡z⁢d⁡x

    d⁡y⁢d⁡x⁢d⁡z: ∫01∫0z∫01−z2d⁡y⁢d⁡x⁢d⁡z+∫01∫z1∫01−x2d⁡y⁢d⁡x⁢d⁡z

    d⁡x⁢d⁡z⁢d⁡y: ∫01∫01−y∫01−yd⁡x⁢d⁡z⁢d⁡y

    d⁡x⁢d⁡y⁢d⁡z: ∫01∫01−z2∫01−yd⁡x⁢d⁡y⁢d⁡z

    Answers will vary. Neither order is particularly “hard.” The order d⁡z⁢d⁡y⁢d⁡x requires integrating a square root, so powers can be messy; the order d⁡y⁢d⁡z⁢d⁡x requires two triple integrals, but each uses only polynomials.

  16. 16.

    d⁡z⁢d⁡y⁢d⁡x: ∫01∫03⁢x∫01−xd⁡z⁢d⁡y⁢d⁡x+∫01∫3⁢x3∫01−y/3d⁡z⁢d⁡y⁢d⁡x d⁡z⁢d⁡x⁢d⁡y: ∫03∫0y/3∫01−y/3d⁡z⁢d⁡y⁢d⁡x+∫03∫y/31∫01−xd⁡z⁢d⁡x⁢d⁡y d⁡y⁢d⁡z⁢d⁡x: ∫01∫01−x∫03−3⁢zd⁡y⁢d⁡z⁢d⁡x d⁡y⁢d⁡x⁢d⁡z: ∫01∫01−z∫03−3⁢zd⁡y⁢d⁡x⁢d⁡z d⁡x⁢d⁡z⁢d⁡y: ∫03∫01−y/3∫01−zd⁡x⁢d⁡z⁢d⁡y d⁡x⁢d⁡y⁢d⁡z: ∫01∫03−3⁢z∫01−zd⁡x⁢d⁡y⁢d⁡z V=∫01∫03−3⁢z∫01−zd⁡x⁢d⁡y⁢d⁡z=1.

  17. 17.

    8

  18. 18.

    7/8

  19. 19.

    π

  20. 20.

    0

  21. 21.

    M=10, My⁢z=15/2, Mx⁢z=5/2, Mx⁢y=5;

    (x¯,y¯,z¯)=(3/4,1/4,1/2)

  22. 22.

    M=4, My⁢z=20/3, Mx⁢z=4, Mx⁢y=4/3;

    (x¯,y¯,z¯)=(5/3,1,1/3)

  23. 23.

    M=16/5, My⁢z=16/3, Mx⁢z=104/45, Mx⁢y=32/9;

    (x¯,y¯,z¯)=(5/3,13/18,10/9)≈(1.67,0.72,1.11)

  24. 24.

    M=65,53615≈208.05, My⁢z=0, Mx⁢z=2,097,1523465≈605.24, Mx⁢y=2,097,1523465≈605.24;

    (x¯,y¯,z¯)=(0,32/11,32/11)≈(0,2.91,2.91)

Exercises N.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. 2.

    If r=0 or ρ=0, then the point in each coordinate system lies on the z-axis regardless of the value of θ.

  3. 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.

  4. 4.

    Spheres centered at the origin; planes parallel to the z-axis that intersect the z-axis; cones centered on the z-axis with point at the origin.

  5. 5.

    • Cylindrical: (2⁢2,π/4,1) and (2,5⁢π/6,0)
      Spherical: (3,π/4,cos−1⁡(1/3)) and (2,5⁢π/6,π/2)

      Rectangular: (2,2,2) and (0,−3,−4)
      Spherical: (2⁢2,π/4,π/4) and
      (5,3⁢π/2,π−tan−1⁡(3/4))

      Rectangular: (1,1,2) and (0,0,1)
      Cylindrical: (2,π/4,2) and (0,0,1)

  6. 6.

    • Cylindrical: (1,π/2,1) and (1,π,1)
      Spherical: (2,π/2,π/4) and (2,π,π/4)

      Rectangular: (0,0,1) and (−1,−3,0)
      Spherical: (1,π,0) and (2,4⁢π/3,π/2)

      Rectangular: (3,1,0) and (0,0,−3)
      Cylindrical: (2,π/6,0) and (0,π,−3)

  7. 7.

    • Cylindrical: (4,π3,−1) and (5⁢2,3⁢π4,6)
      Spherical: (17,π3,cos−1⁡−117) and
      (86,3⁢π4,cos−1⁡686)

      Rectangular: (12,32,−2) and (−3,−1,3)
      Spherical: (5,π3,cos−1⁡(−25)) and
      (13,5⁢π6,tan−1⁡23)

      Rectangular: (2,6,2⁢2) and (1,0,0)
      Cylindrical: (2⁢2,π3,2⁢2) and (1,0,0)

  8. 8.

    • Cylindrical: (2⁢7,11⁢π6,0) and (2,π2,2)
      Spherical: (2⁢7,11⁢π6,π2) and (6,π2,cos−1⁡63)

      Rectangular: (0,12,1) and (3,−3⁢3,7)
      Spherical: (52,π2,cos⁡25) and
      (43,5⁢π3,cos−1⁡743)

      Rectangular: (0,−32,−3⁢32) and (62,22,2)
      Cylindrical: (32,3⁢π2,−3⁢32) and (2,7⁢π6,2)

  9. 9.

    • 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).

      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.

      This is upper half of the sphere of radius 3 centered at the origin (i.e., the upper hemisphere).

      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.

  10. 10.

    • A square portion of the y-z plane with corners at (0,1,0), (0,1,1), (0,2,1) and (0,2,0).

      This is a curve, a circle of radius 2, centered at (0,0,5), lying parallel to the x-y plane (i.e., in the plane z=5).

      This is a region of space, a half of a solid cone with rounded top, where the rounded top is a portion of the ball of radius 2 centered at the origin and the sides of the cone make an angle of π/4 with the positive z-axis. The bounds on θ mean only the portion “above” the x-z plane are retained.

      This is a curve, a circle of radius 1 centered at (0,0,3), lying parallel to the x-y plane.

  11. 11.

    ∫θ1θ2∫r1r2∫z1z2h⁢(r,θ,z)⁢r⁢d⁡z⁢d⁡r⁢d⁡θ

  12. 12.

    ∫φ1φ2∫θ1θ2∫ρ1ρ2h⁢(ρ,θ,φ)⁢ρ2⁢sin⁡(φ)⁢d⁡ρ⁢d⁡θ⁢d⁡φ

  13. 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∘.

  14. 14.

    Bounded between the planes z=0 and z=5, between the cylinders x2+y2=9 and x2+y2=16: describes a “pipe” or “tube” of length 5, an inner radius of 3 and outer radius of 4.

  15. 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.

  16. 16.

    Bounded between y≥0, inside the cylinder x2+y2=1, above the plane z=0 and below the cone z=2−x2+y2: describes cylindrical solid of height 1 and radius 2, topped with an inverted cone of height 1 and base radius 1 with point at (0,0,2).

  17. 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.

  18. 18.

    Bounded between the plane z=0, inside the cylinder x2+y2=a2, and below the upper hemisphere z=a2−x2−y2+b, with radius a and centered at (0,0,b): describes a cylindrical solid of radius a and height b, topped with the upper hemisphere of radius a.

  19. 19.

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

  20. 20.

    Describes half of a spherical shell (i.e., y≥0) with inner radius of 1 and outer radius of 1.1 centered at the origin.

  21. 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.

  22. 22.

    It is the region is space bounded below by z=x2+y2 and above by the sphere x2+y2+z2=4, with the portion above the cone z=3⁢x2+y2 removed: 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, with the inner cone with angle π/6 removed, along with corresponding portion of the ball of radius 2.

  23. 23.

    The region in space is bounded below by the cone z=3⁢x2+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.

  24. 24.

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

  25. 25.

    In cylindrical coordinates, the density is δ⁢(r,θ,z)=r+1. Thus mass is

    ∫02⁢π∫02∫04(r+1)⁢r⁢d⁡z⁢d⁡r⁢d⁡θ=112⁢π/3.
  26. 26.

    In cylindrical coordinates, the density is δ⁢(r,θ,z)=z. Thus mass is

    ∫02⁢π∫23∫010z⁢r⁢d⁡z⁢d⁡r⁢d⁡θ=250⁢π.
  27. 27.

    In cylindrical coordinates, the density is δ⁢(r,θ,z)=1. Thus mass is

    ∫0π∫01∫04−r⁢sin⁡θr⁢d⁡z⁢d⁡r⁢d⁡θ=2⁢π−2/3≈5.617.
  28. 28.

    In cylindrical coordinates, the density is δ⁢(r,θ,z)=1. Thus mass is

    ∫02⁢π∫01∫−1−r21−r2r⁢d⁡z⁢d⁡r⁢d⁡θ=4⁢π/3.
  29. 29.

    In cylindrical coordinates, the density is δ⁢(r,θ,z)=r+1. Thus mass is

    M=∫02⁢π∫02∫04(r+1)⁢r⁢d⁡z⁢d⁡r⁢d⁡θ=112⁢π/3.

    We find My⁢z=0, Mx⁢z=0, and Mx⁢y=224⁢π/3, placing the center of mass at (0,0,2).

  30. 30.

    In cylindrical coordinates, the density is δ⁢(r,θ,z)=z. Thus mass is

    M=∫02⁢π∫23∫010z⁢r⁢d⁡z⁢d⁡r⁢d⁡θ=250⁢π.

    We find My⁢z=0, Mx⁢z=0, and Mx⁢y=5000⁢π/3, placing the center of mass at (0,0,20/3).

  31. 31.

    In cylindrical coordinates, the density is δ⁢(r,θ,z)=1. Thus mass is

    ∫0π∫01∫04−r⁢sin⁡θr⁢d⁡z⁢d⁡r⁢d⁡θ=2⁢π−2/3≈5.617.

    We find My⁢z=0, Mx⁢z=8/3−π/8, and Mx⁢y=65⁢π/16−8/3, placing the center of mass at ≈(0,0.405,1.80).

  32. 32.

    In cylindrical coordinates, the density is δ⁢(r,θ,z)=1. Thus mass is

    ∫02⁢π∫01∫−1−r21−r2r⁢d⁡z⁢d⁡r⁢d⁡θ=2⁢π/3.

    We find My⁢z=0, Mx⁢z=0, and Mx⁢y=π/4, placing the center of mass at (0,0,3/8).

  33. 33.

    In spherical coordinates, the density is δ⁢(ρ,θ,φ)=1. Thus mass is

    ∫0π/2∫02⁢π∫01ρ2⁢sin⁡(φ)⁢d⁡ρ⁢d⁡θ⁢d⁡φ=2⁢π/3.
  34. 34.

    In spherical coordinates, the density is δ⁢(ρ,θ,φ)=ρ. Thus mass is

    ∫0π∫02⁢π∫45(ρ)⁢ρ2⁢sin⁡(φ)⁢d⁡ρ⁢d⁡θ⁢d⁡φ=369⁢π.
  35. 35.

    In spherical coordinates, the density is δ⁢(ρ,θ,φ)=ρ⁢cos⁡φ. Thus mass is

    ∫0π/4∫02⁢π∫01(ρ⁢cos⁡(φ))⁢ρ2⁢sin⁡(φ)⁢d⁡ρ⁢d⁡θ⁢d⁡φ=π/8.
  36. 36.

    In spherical coordinates, the density is δ⁢(ρ,θ,φ)=ρ⁢cos⁡φ. Thus mass is

    ∫0π/4∫02⁢π∫0sec⁡(φ)(ρ⁢cos⁡(φ))⁢ρ2⁢sin⁡(φ)⁢d⁡ρ⁢d⁡θ⁢d⁡φ=π/4.
  37. 37.

    In spherical coordinates, the density is δ⁢(ρ,θ,φ)=1. Thus mass is

    ∫0π/2∫02⁢π∫01ρ2⁢sin⁡(φ)⁢d⁡ρ⁢d⁡θ⁢d⁡φ=2⁢π/3.

    We find My⁢z=0, Mx⁢z=0, and Mx⁢y=π/4, placing the center of mass at (0,0,3/8).

  38. 38.

    In spherical coordinates, the density is δ⁢(ρ,θ,φ)=ρ. Thus mass is

    ∫0π∫02⁢π∫45(ρ)⁢ρ2⁢sin⁡(φ)⁢d⁡ρ⁢d⁡θ⁢d⁡φ=369⁢π.

    We find My⁢z=0, Mx⁢z=0, and Mx⁢y=0, placing the center of mass at (0,0,0).

  39. 39.

    In spherical coordinates, the density is δ⁢(ρ,θ,φ)=ρ⁢cos⁡φ. Thus mass is

    ∫0π/4∫02⁢π∫01(ρ⁢cos⁡(φ))⁢ρ2⁢sin⁡(φ)⁢d⁡ρ⁢d⁡θ⁢d⁡φ=π/8.

    We find My⁢z=0, Mx⁢z=0, and Mx⁢y=(4−2)⁢π/30, placing the center of mass at (0,0,4⁢(4−2)/15).

  40. 40.

    In spherical coordinates, the density is δ⁢(ρ,θ,φ)=ρ⁢cos⁡φ. Thus mass is

    ∫0π/4∫02⁢π∫0sec⁡(φ)(ρ⁢cos⁡(φ))⁢ρ2⁢sin⁡(φ)⁢d⁡ρ⁢d⁡θ⁢d⁡φ=π/4.

    We find My⁢z=0, Mx⁢z=0, and Mx⁢y=π/5, placing the center of mass at (0,0,4/5).

  41. 41.

    Rectangular: ∫−11∫−1−x21−x2∫−1−x2−y21−x2−y2d⁡z⁢d⁡y⁢d⁡x

    Cylindrical: ∫02⁢π∫01∫−1−r21−r2r⁢d⁡z⁢d⁡r⁢d⁡θ

    Spherical: ∫0π∫02⁢π∫01ρ2⁢sin⁡(φ)⁢d⁡ρ⁢d⁡θ⁢d⁡φ

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

  42. 42.

    Rectangular: ∫−11∫−1−x21−x2∫01d⁡z⁢d⁡y⁢d⁡x

    Cylindrical: ∫02⁢π∫01∫01r⁢d⁡z⁢d⁡r⁢d⁡θ

    Spherical: ∫0π/4∫02⁢π∫0sec⁡φρ2⁢sin⁡(φ)⁢d⁡ρ⁢d⁡θ⁢d⁡φ+∫π/4π/2∫02⁢π∫0csc⁡φρ2⁢sin⁡(φ)⁢d⁡ρ⁢d⁡θ⁢d⁡φ

    Cylindrical appears simplest, avoiding the integration of square-roots and two triple integrals; all bounds are constants.

  43. 43.

    Rectangular: ∫−11∫−1−x21−x2∫x2+y21d⁡z⁢d⁡y⁢d⁡x

    Cylindrical: ∫02⁢π∫01∫r1r⁢d⁡z⁢d⁡r⁢d⁡θ

    Spherical: ∫0π/4∫02⁢π∫0sec⁡φρ2⁢sin⁡(φ)⁢d⁡ρ⁢d⁡θ⁢d⁡φ

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

  44. 44.

    Rectangular: ∫01∫01∫01d⁡z⁢d⁡y⁢d⁡x

    Cylindrical: ∫0π/4∫0sec⁡θ∫01r⁢d⁡z⁢d⁡r⁢d⁡θ+∫π/4π/2∫0csc⁡θ∫01r⁢d⁡z⁢d⁡r⁢d⁡θ

    Spherical: ∫0π/4∫0tan−1⁡(sec⁡θ)∫0sec⁡φρ2⁢sin⁡φ⁢d⁡ρ⁢d⁡φ⁢d⁡θ+∫0π/4∫tan−1⁡(sec⁡θ)π/2∫0sec⁡θ⁢csc⁡φρ2⁢sin⁡φ⁢d⁡ρ⁢d⁡φ⁢d⁡θ+∫π/4π/2∫0tan−1⁡(csc⁡θ)∫0sec⁡φρ2⁢sin⁡φ⁢d⁡ρ⁢d⁡φ⁢d⁡θ+∫π/4π/2∫tan−1⁡(csc⁡θ)π/2∫0csc⁡θ⁢csc⁡φρ2⁢sin⁡φ⁢d⁡ρ⁢d⁡φ⁢d⁡θ.

    Rectangular is clearly the simplest.

  45. 45.

    center: (2,3,−1), radius: 14

  46. 46.
  47. 47.

    (a,θ,a⁢cot⁡ϕ)

  48. 48.
  49. 49.

    Hint: Use the distance formula for Cartesian coordinates.

  50. 50.
  51. 51.
  52. 52.
  53. 53.

    1−sin⁡2/2

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