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  # Application: Work
  
  8.1
  
  What is Work?
  
  Work, in the physics sense, is usually defined as force acting over a distance.
  Work is sometimes force times distance,1 but not always. Work is more subtle than
  that. Every time you exert a force, it is not the case that any work is done (even
  though it may feel like that to you!
  Why? Well, the work/energy equation says that work done (by the net force on
  an object) equals the objects change in kinetic energy. More simply:
  Work = Change in Kinetic Energy.
  
  ## Work will be force times distance in all
  
  of the applications we consider.
  1
  
  (8.1)
  
  This means that if an objects kinetic energy doesnt change, then no work has
  been done on the objectwhether or not a force has been exerted. In particular, a
  force will do work only if the force has a component in the direction that the object
  moves.
  Heres the distinction: if you push on a big box and it moves in the direction
  you push, then work is accomplished and it equals force times distance. If the
  box is very big and when you push it nothing happens (it does not move), then
  no work is done. If a force is applied to the box in a certain direction and the box
  moves, but not in the direction you push (maybe your friend is pushing on another
  side) but in some other direction, then work is done, but the amount depends
  on the angle the box moves relative to the direction the box moves. So again the
  answer is not simply force times distance.
  However, to keep things simple, in the context of this section we will assume the
  direction the force is applied and the direction of the motion of the object are one
  and the same. In this case, assuming that the force is constant,
  W = Work = force Distance = F x.
  The unit used to measure work vary depending on the system you are inand
  they are probably less familiar to you than the units for velocity and acceleration.
  Table 8.1 lists the units for three common systems.
  System
  
  Force
  
  Distance
  
  Work
  
  British
  cgs
  SI (international)
  
  pounds
  dynes
  Newtons
  
  feet
  centimeters
  meters
  
  foot-pounds (ft-lbs)
  ergs
  joules
  
  EXAMPLE 8.1. Calculate the work done in lifting a 147 lb object 22 feet (e.g., me walk-
  
  ## ing from the first to the third floor in Lansing)? Well,
  
  Work = force Distance = 147 22 = 3, 234 ft-lbs.
  
  ## Table 8.1: Units of work.
  
  math 131
  
  application: work
  
  ## That was easy.
  
  Now the key thing to notice is that under our assumptions, work is a product.
  One of the assumptions is that a constant force is applied. But most forces are not
  constant. Consider the following situation.
  EXAMPLE 8.2. Suppose we want to hoist a leaking bucket vertically. Because of the
  leak, the force applied to lift the bucket decreases, so that
  
  
  x2
  F ( x ) = 60 1
  ,
  0 x 50
  5000
  
  where x is measured in feet. Find the work done in lifting the bucket.
  
  So how do we calculate the work done if the force varies? Well, remember that
  Riemann sums involve products. So we will use the subdivide and conquer strategy once more. Unlike in the earlier cases there is no natural figure to draw whose
  area, volume, or arc length we are trying to calculate. This time we must apply the
  theory.
  General Situation. Assume that F ( x ) is a variable but continuous force that is a
  function of the position x and that it is applied over an interval [ a, b]. Find the
  work done. As usual, let P = { x0 , x1 , . . . , xn } be a regular partition of [ a, b] into n
  equal width subintervals of length x. Now if the intervals are short enough, since
  the force is continuous, the force will be nearly constant on each interval, though
  its value will vary from interval to interval. So let Wi denote the work done on the
  ith subinterval. Since the length of the ith subinterval is x, then
  Wi F ( xi )x.
  Consequently, adding up the pieces of work on each subinterval,
  Total Work =
  
  Wi F(xi )x.
  
  (8.2)
  
  Notice that we now have a Riemann sum involving the force function! To improve the approximation we do the standard thing: We let the number of subdivisions get large and take the limit. We find
  n
  
  Z b
  
  Wi = nlim
  F(xi )x =
  n
  
  ## Total Work = lim
  
  F ( x ) dx.
  
  We are certain that this limit of the Riemann sums exists and is, in fact, the definite
  integral because we assumed that the force F ( x ) is continuous function of the
  position.
  THEOREM 8.1 (Work Formula). If F ( x ) is a continuous force that is a function of the position
  
  x that is applied over an interval [ a, b], then the work done over the interval is
  Work =
  
  Z b
  a
  
  F ( x ) dx.
  
  Stop and Step Back. There are a couple of things I want you to notice. First, Theorem 8.1 amounts to saying that work is the area under the force curve. Thats
  probably not how you would first think of it, but thats what the theorem says!
  Second, when I was writing these notes, I simply cut-and-pasted the earlier material on arc length and changed the a few words here and there. But it is the
  same subdivide and conquer strategy that you have seen several times now. You
  should be able to create such arguments for yourself now.
  
  math 131
  
  application: work
  
  EXAMPLE 8.3 (Return to the leaky bucket). Return to the leaking bucket. The force ap-
  
  ## plied was continuous and it was applied on the
  
  
  
  x2
  F ( x ) = 60 1
  ,
  5000
  and it was applied on the interval [0, 50] where x is measured in feet. So by Theorem 8.1 the work done in lifting the bucket is
  
  Z b
  Z 50 
  x2
  Work =
  dx
  F ( x ) dx =
  1
  5000
  a
  0
  
   50
  
  x3
  
  = 60 x
  15, 000 0
  
  
  125, 000
  = 60 50
  0
  15, 000
  
  = 2500 ft-lbs.
  Pretty straightforward!
  H
  
  8.2
  
  ## Work Done Emptying a Tank
  
  The following tank problems involve pumping liquids from one height to another
  and determining the amount of work required to do it. Heres the general question. Given a tank containing a liquid between heights a and b, how much work is
  required to pump the liquid to a height H. (Note: H may or may not be the height
  of the top of the tank.) See Figure 8.1.
  To solve the problem we do the usual thing. We subdivide and conquer. Use
  a regular partition of the interval [ a, b] on the vertical y-axis into n equal width
  subintervals, each of width y. The subintervals are used to create layers (slices)
  of liquid, each of width y. See Figure 8.1. What properties of the layer are important in determining the work done in lifting the layer out? Take a moment before
  We need
  the volume of the layer;
  the density of the layer;
  the distance the layer is moved.
  The weight of the layer is the force here (e.g., weight is measured in pounds) and
  is the density of the liquid times the volume. So the work to move the ith layer to
  height H is
  Wi = Work to move the ith layer to height H
  
  ## = Weight Distance the layer is moved
  
  = (Density Volume) Distance the layer is moved
  
  (8.3)
  
  OK, its time to use a bit of analysis and notation we saw in the section on volume.
  Lets denote the density by D (in units such as lbs/ft3 ).
  As usual, let yi denote the ith partition point of the interval [ a, b].
  
  ..........................................................................................
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  ## Figure 8.1: A tank containing a liquid
  
  between levels a and b. The liquid is to
  be moved to height H. Also shown is
  a representative layer of the liquid at
  height yi . Can we determine how much
  work is required to lift this layer to
  height H? What will this work depend
  on?
  
  math 131
  
  application: work
  
  The ith layer of liquid is approximately is approximately a cylinder (see Figure 8.1) so its volume is
  Vi = Area of the base height = Cross-sectional area height = A(yi )y,
  where A(yi ) denotes the the cross-sectional area of the tank at height yi and the
  height of the layer is y.
  Finally, since the ith layer is at height yi and has to be moved to height H, then
  the ith layer is moved a distance of H yi .
  Putting this all together, we can rewrite the work to move the ith layer to height H
  in (8.4) as
  Wi = (Density Volume) Distance the layer is moved
  
  = D Vi ( H yi )
  D [ A(yi )y]( H yi ).
  
  (8.4)
  
  But now you know how the rest of the story goes. To estimate the total work we
  add up the work to move each layer for each subinterval,
  Total Work =
  
  Wi D[ A(yi )y]( H yi ).
  
  (8.5)
  
  ## We now have a Riemann sum.2
  
  To improve the approximation we do the standard thing: We let the number of
  subdivisions get large and take the limit. We find
  Total Work = lim
  
  Z b
  
  Wi = nlim
  D[ A(yi )y]( H yi ) = D
  
  ## Take a second to compare (8.2) and
  
  (8.5). Then take another to anticipate
  the rest of this argument.
  
  A(y)[ H y] dy.
  
  We are certain that this limit of the Riemann sums exists and is, in fact, the definite
  integral as long as we assume that the cross-sectional area A(y) is continuous
  function of y.
  THEOREM 8.2 (Work Formula for Emptying a Tank). Assume the cross-sectional A(y) is a
  continuous function of the position y and that the density of the contents is a constant D.
  If the contents of the tank to be moved lie in the interval [ a, b], then the work done to move
  this material to a height H is
  
  Work = D
  
  Z b
  a
  
  A(y)[ H y] dy.
  
  Caution: The tank may not be full, the contents may be moved to a height H above the tank,
  or the entire tank may not be emptied. If the tank is being filled from a source at height H
  (either at the bottom of or below the tank), then the contents must be moved to each layer
  height y between a and b so the distance moved is y H rather than H y.
  EXAMPLE 8.4. Heres a simple first example. An above-ground backyard swimming
  
  pool has the shape of a circular cylinder with a radius of 10 ft and a depth of 8 ft.
  Assume the depth of the water in the pool is 5 ft. Find the work done in emptying the
  pool by pumping the water over the top edge of the pool. Note: The density of water
  is 62.5 lbs/ft3 . See Figure 8.2.
  
  ..................................................
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  8
  5
  
  SOLUTION. We apply Theorem 8.2. Be attentive to the different heights. The liquid
  
  lies between 0 and 5 feet but has to be moved to a height of H = 8 feet. The crosssections are circles of radius r = 10 feet. So A(y) = r2 = 100. (In most problems
  
  ## Figure 8.2: The pool is filled to a depth
  
  of 5 feet and the water is to be pumped
  out over the top edge at height 8 feet.
  
  math 131
  
  application: work
  
  ## the cross-sections will vary.) So
  
  Work = D
  
  Z b
  a
  
  A(y)[ H y] dy = 62.5
  
  Z 5
  0
  
  100 [8 y] dy
  
   5
  y2
  = 6250 8y
  2 0
  
  
  
  25
  = 6250 40
  0
  2
  
  
  ## = 17, 1875 ft-lbs.
  
  EXAMPLE 8.5. (A revision) Suppose the swimming pool in Example 8.4 above was in-
  
  ground so that the top of the pool was at ground level. How would the work integral
  change? Would the work done emptying the pool change?
  SOLUTION. Three things change in the set-up. The liquid now lies between 8 and
  
  3 feet and has to be moved to a height of H = 0 feet. The cross-sections are the
  same.
  Work = D
  
  Z b
  a
  
  A(y)[ H y] dy = 62.5
  
  Z 3
  8
  
  100 [0 y] dy
  
   3
  
  y2
  = 6250
  2 8
  
  
  9
  = 6250 + 32
  2
  
  ## = 17, 1875 ft-lbs.
  
  The work should and does remain the same.
  EXAMPLE 8.6. (A more complicated problem) An underground hemispherical tank
  with radius 10 ft is filled with oil of density 50 lbs/ft3 . Find the work done pumping
  the oil to the surface if the top of the tank is 6 feet below ground.
  SOLUTION. It will be easiest to set up the equation of the hemisphere if we think
  
  of the top of the tank at height 0 and then pump the oil to a height of 6 feet. See
  Figure 8.3.
  The cross-sections are circles. We will be able to determine the cross-sectional area
  once we determine the radius of the cross-section. The semi-circle is part of the circle
  of radius 10 centered at the origin which has equation x2 + y2 = 10. The radius of a
  cross-section is the x-coordinate of the point ( x, y) that lies on the semi-circle in fourth
  quadrant. (See Figure 8.4.) Thus,
  q
  r = x = (10)2 y2 .
  
  6
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  10
  
  ## Figure 8.3: The tank with its position
  
  re-imagined and a representative layer.
  
  ## A(y) = r2 = [102 y2 ] = (100 y2 ).
  
  10
  
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  10
  
  ## We apply Theorem 8.2. Remember the liquid is pumped to height H = 6 in our
  
  re-casting of the problem.
  
  Work = D
  
  Z b
  a
  
  A(y)[ H y] dy = 50
  
  Z 0
  10
  Z 0
  
  ((10)2 y2 )[6 y] dy
  
  
  
  600 100y 6y2 + y3 dy
  10
  !
  0
  y4
  = 50 600y 50y2 2y3 +
  
  4
  10
  
  = 50
  
  ## = 50 [0 (6000 5000 + 2000 + 2500)]
  
  = 325, 000 ft-lbs.
  
  10 y
  
  ## Figure 8.4: The radiuspof the crosssection at height y is (10)2 y2 .
  
  math 131
  
  application: work
  
  YOU TRY IT 8.1. Set up the new integral for each modification of the example above and
  determine the work required.
  
  (a) How would the integral and work change if the tank were only 5 feet below ground?
  3 ft-lbs.)
  (b) How would the integral and work change if the top of the tank were at ground level?
  (Answer: 125, 000 ft-lbs.)
  EXAMPLE 8.7. (A tank formed by rotation) A tank is formed by rotating the region
  between y = x2 , the y-axis and the line y = 4 in the first quadrant around the y axis.
  The tank is filled with oil with density 50 lbs/ft3 .
  
  (a) Find the work done pumping the oil to the top of the tank.
  (b) Find the work done pumping the oil to the top of the tank if there is only 1 foot
  of oil in the tank.
  (c) Suppose the tank is empty and is filled from a hole in the bottom to a depth of
  3 feet. Find the work done.
  SOLUTION. Figure 8.5 shows a sketch of the tank with a representative cross-section.
  
  The circular cross-section formed by rotating the point ( x, y) about the y-axis has
  radius r = x. So the cross-sectional area is
  
  as a function of y.
  (a) We apply Theorem 8.2.
  Z b
  a
  
  A(y)[ H y] dy = 50
  
  = 50
  
  Z 4
  0
  
  Z 4
  0
  
  (y)[4 y] dy
  
  
  4y y2 dy
  
  
   4
  y3
  = 50 2y2
  3 0
  
  
  
  64
  = 50 32
  0
  3
  1600
  ft-lbs.
  =
  3
  (b) If there is only 1 foot of oil in the tank, the only change is that the upper limit of
  integration is now 1 instead of 4. (Remember the limits of integration represent
  the upper an lower levels of the content of the tank.) So
  Work = D
  
  Z b
  a
  
  A(y)[ H y] dy = 50
  
  Z 1
  0
  
  (y)[4 y] dy
  
   1
  
  y3
  = 50 2y2
  3
  
   0
  1
  = 50 2
  3
  250
  ft-lbs.
  =
  3
  (c) The key observation when a tank is filled from the bottom is that each layer
  must be moved or lifted from ground level 0 to its final height H = y. Since the
  tank is only filled to a depth of 3 feet, the limits of integration are 0 (bottom) to
  3 ft. So this time
  Work = D
  
  Z b
  a
  
  A(y)[ H y] dy = 50
  
  r = x ( x, y)
  
  A(y) = r2 = x2 = y,
  
  Work = D
  
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  Z 3
  0
  
  (y)[y 0] dy
  
   3
  y3
  3 0
  
  
  1
  = 50 2
  3
  
  
  = 50
  
  = 450 ft-lbs.
  
  ## Figure 8.5: The circular cross-section
  
  formed by rotating the point ( x, y)
  about the y-axis has radius r = x.
  
  math 131
  
  application: work
  
  YOU TRY IT 8.2. How would the integral and work change if the tank were full and the oil
  was pumped to a height 3 feet above the top of the tank? (Answer: 5200
  3 ft-lbs.)
  
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  YOU TRY IT 8.3. Suppose instead that the tank is formed by rotating the region in the first
  
  quadrant enclosed by y = ( x + 2)2 , the y-axis, and the x-axis about the y-axis (see Figure 8.6).
  R4
  (a) Find the work done pumping the oil to the top of the tank. (Answer: 50 0 ( y
  2)2 (4 y) dy = 1280
  3 ft-lbs.)
  (b) Find the work done if the oil was pumped to a height 3 feet above the top of the tank?
  3 ft-lbs.)
  (c) Find the work done pumping the oil to the top of the tank if there is only 1 foot of oil
  in the tank. (Answer: 330 ft-lbs.)
  
  ( x, y)
  
  ## Figure 8.6: The circular cross-section
  
  formed by rotating the point ( x, y)
  about the y-axis has radius r = x =
  
  y 2.
  
  EXAMPLE 8.8. (A more complicated problem) A large gasoline storage tank lies 3 feet
  
  ## underground. It is in the shape of a half-cylinder (flat side up) 4 ft in radius and 8 ft
  
  long. If the tank is full of gas with density 60 lbs/ft3 , find the work done pumping the
  gas to a level 2 ft above ground.
  SOLUTION. It will be easiest to set up the equation of the hemisphere if we think of
  
  the top of the tank at height 0. Since the original tank was 3 feet underground and the
  contents were to be pumped to 2 feet above ground, in our model we need to pump
  the gas to a total of 5 feet above ground. See Figure 8.7.
  2
  2
  The cross-sections
  p are rectangles of width 2x and length 8. Since x + y = 4 on the
  semi-circle, x = 16 y2 . So the cross-sectional area is
  q
  A(y) = 2x (8) = 16 16 y2 .
  We apply Theorem 8.2. Remember the liquid is pumped to height H = 5 in our
  re-casting of the problem.
  5
  
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  x2 + y2 = 42
  
  Work = D
  
  Z b
  a
  
  ## Figure 8.7: Left: Think of the tank top
  
  at ground level and the liquid being
  pumped to height H = 5 ft. Right: The
  width of
  p the cross-section at height y is
  2x = 2 42 y2 .
  
  3
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  x2 + y2 = 42
  
  Z 0
  
  q
  (16 16 y2 [5 y] dy
  4
  
  Z 0  q
  q
  = 800
  5 16 y2 y 16 y2 dy
  
  A(y)[ H y] dy = 50
  
  = 4000
  
  Z 0 q
  4
  
  16 y2 dy 800
  
  Z 0
  4
  
  16 y2 dy.
  
  (8.6)
  
  R0 p
  The first integral 4 16 y2 dy represents the area of a quarter circle of radius
  2
  r = 4 and so its value3 (area) is 4 4 = 4.
  The second integral in (8.6) can be done using the substitution u = 16 y2 with
  1
  2 du = y dy. Be careful to change the limits!
  Z 0
  4
  
  16 y2 dy =
  
  16
  Z 16
  
  1 1/2
  1
  64
  u
  du = u3/2 =
  .
  2
  3
  3
  0
  0
  
  ## Remember that we have p
  
  not yet developed an antiderivative of 16 y2 , as
  we noted in the section on arc length,
  so we have to use the geometric interpretation of the definite integral
  here.
  3
  
  math 131
  
  application: work
  
  ## Consequently, (8.6) becomes
  
  Z 0 q
  
  Z 0
  
  16 y2 dy 800
  y
  4
  4
   
  64
  = 4000(4 ) 800
  3
  
  Work = 4000
  
  16 y2 dy
  
  33172.14913 ft-lbs.
  YOU TRY IT 8.4. How would the integral and work change if the tank in the previous problem had the flat side facing down? Hint: Put the center of the semi-circle at the origin. The
  tank now lies above the x-axis. What height is the gas pumped to this time? Now work out
  ft-lbs.)
  3
  YOU TRY IT 8.5. A tank in the form of a truncated cone is formed by rotating the segment
  between (2, 0) and (4, 4) around the y-axis. It is filled with sludge (density 80 lbs/ft3 ). If
  the sludge is pumped 3 feet upwards into a tank truck, how much work was required?
  ft-lbs.)
  3
  YOU TRY IT 8.6 (Rotation about a non-axis line). This time a tank in the form of a truncated
  
  cone is formed by rotating the segment between (0, 4) and (2, 0) around the line x = 4. It
  is filled with sludge (density 80 lbs/ft3 ). If the sludge is pumped 3 feet upwards into a tank
  ft-lbs.)
  truck, how much work was required? (Answer: 40960
  3
  YOU TRY IT 8.7 (Non-rotation). A triangular trough for cattle is 8 ft long. The ends are trian-
  
  gles with with a base of 3 ft and height of 2 ft (but the vertex points down). See Figure 8.8.
  Find the work done by the cattle in emptying just the top foot of water (density 62.5 lbs/ft3 )
  over the edge. (Answer: 500 ft-lbs.)
  
  YOU TRY IT 8.8. A cup shaped tank is obtained by rotating the curve y = x3 about the y-axis
  
  where 0 x 2.
  
  lbs/ft3 ?).
  
  ## (a) Assume the tank is full of water (density 62.5
  
  How much work is done in
  emptying the tank by removing the water over the top edge of the tank? (Answer:
  3600 ft-lbs?)
  (b) How much work would be done in raising the water 2 feet above the tanks top?
  (c) Suppose the depth of the liquid in the tank is 1 foot. Find the work required to pump
  the liquid to the top edge of the tank.
  YOU TRY IT 8.9. (a) A cone-shaped reservoir has a 10 foot radius across the top and a 15 foot
  
  depth. If the reservoir has 9 feet of oil (density 54 lbs/ft3 ) in it, how much work is
  required to empty it by bringing the water to the top of the reservoir? (First figure out
  the equation of the line that determines the cone.)
  (b) Same question with the reservoir being completely full. (Answer: 101250 ft-lbs?)
  YOU TRY IT 8.10 (Expanding Gas). Boyles law says that the pressure (force) exerted by a
  
  gas is inversely proportional to the volume. A quantity of gas with an initial volume of 2
  cubic feet and a pressure of 1000 pounds per square foot expands to a volume of 5 cubic
  feet. Find the work done. Hint: p = k/V, so first solve for k. Next, the work done by an
  expanding gas is
  W=
  
  Z V1
  V0
  
  p dv.
  
  ## Answer: 2000(ln 5 ln 2) ft. lbs.)
  
  YOU TRY IT 8.11 (Extra Credit). A heavy rope is 60 feet long and has a density of 1.5 lbs/ft. It
  
  is hanging over the edge of a building 100 ft tall. Find the work done in pulling the rope to
  the top of the building. Hint: Model your answer like a tank problem. Think of the rope in
  sections. What force must be applied to move each section of the rope? What distance must
  each section be used? OK, set up the integral and do it!
  
  8.3
  
  Problems
  
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  .
  
  ## Figure 8.8: Find the work done in
  
  emptying the top foot of the trough
  over the top edge.
  
  math 131
  1. Assume 25 ft-lb of work is required to stretch a spring 3 feet beyond its natural length.
  Find the work done in stretching the spring from 2 to 4 ft beyond its natural length.
  Hint: First find the spring constant. (Answer: k = 50/9 lbs/ft and 100/3 ft-lbs.)
  2. Let R be the region in the first quadrant bounded by the x-axis, the y-axis, and the curve
  y = 4 x2 . Rotate the R around the y-axis to form a silo tank. If the tank is filled with
  wheat (100 lbs per cu. ft) how much work is done in raising the wheat to the top of the
  silo. (Ans: 6400
  3 ft lbs.)
  3. Let R be the region in the first quadrant bounded by the y-axis and the curves y = x2
  and y = 6 x. Rotate the R around the y-axis to form a tank. If the tank is filled with
  honey (90 lbs per cu. ft) how much work is done in raising the honey to the top of the
  tank. Note: you will have to divide the work into two pieces! Be careful of the limits.
  (Ans: 2760 ft lbs.)
  4. A beer ball with radius 1 foot is located with its center at the origin. It has a tap 0.5 ft
  long sticking up out the top. If beer has a density of 60 pounds per cubic foot, find the
  work done in emptying the beer ball. Hint: Locate the center of the ball at the origin.
  (Ans: 120 ft-lbs)
  5. Ants excavate a chamber underground that is described as follows: Let S be the region in
  
  the fourth quadrant enclosed by y = x, y = 1, and the y-axis; revolve S around the
  y-axis.
  (a) Set up and simplify the integral for the volume of the chamber using the shell
  method. (For future reference: Ans: /5)
  (b) [From an exam] Suppose that the chamber contained soil which weighed 50 lbs per
  cubic foot. How much work did the ants do in raising the soil to ground level? (Ans:
  50/6 ft. lbs.)
  If you have time, set up each of the following integrals which we will do with Maple.
  6. A small farm elevated water tank is in the shape obtained from rotating the region in the
  first quadrant enclosed by the curves y = 10 21 x2 , y = 8, and the y-axis about the y-axis.
  (a) Find the work lost" if the water (62.5 lbs/ft3 ) leaks onto the ground from a hole in
  the bottom of the tank. (Answer: 6500/3 ft-lbs.)
  (b) Find the work lost" if the water leaks onto the ground from a hole in the side of the
  tank at height 9 feet. (Answer: 1750/3 ft-lbs.)
  7. The line segment between the points (1, 3) and (3, 1) is rotated around the y-axis to
  form a truncated conical gas storage tank. If the tank has only 1 foot of gasoline (density
  60 lbs/ft3 ) in it, set up the integral for the work done to pump all of the gas to ground
  level.
  
  application: work
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