Physical Applications

In this section, we examine some physical applications of integration. Let’s begin with a look at calculating mass from a density function. We then turn our attention to work, and close the section with a study of hydrostatic force.

Mass and Density

We can use integration to develop a formula for calculating mass based on a density function. First we consider a thin rod or wire. Orient the rod so it aligns with the

x -axis,

([link]). Note that although we depict the rod with some thickness in the figures, for mathematical purposes we assume the rod is thin enough to be treated as a one-dimensional object.

given in terms of mass per unit length, then the mass of the rod is just the product of the density and the length of the rod:

(b - a) ρ .

If the density of the rod is not constant, however, the problem becomes a little more challenging. When the density of the rod varies from point to point, we use a linear density function,

ρ (x),

Adding the masses of all the segments gives us an approximation for the mass of the entire rod:

We now extend this concept to find the mass of a two-dimensional disk of radius

r .

As with the rod we looked at in the one-dimensional case, here we assume the disk is thin enough that, for mathematical purposes, we can treat it as a two-dimensional object. We assume the density is given in terms of mass per unit area (called area density), and further assume the density varies only along the disk’s radius (called radial density). We orient the disk in the

x y -plane,

with the center at the origin. Then, the density of the disk can be treated as a function of

x,

Now, use the partition to break up the disk into thin (two-dimensional) washers. A disk and a representative washer are depicted in the following figure.

We now approximate the density and area of the washer to calculate an approximate mass,

m_{i} .

You may recall that we had an expression similar to this when we were computing volumes by shells. As we did there, we use

x_{i}^{*} \approx (x_{i} + x_{i - 1}) / 2

to approximate the density of the washer, we approximate the mass of the washer by

We again recognize this as a Riemann sum, and take the limit as

n \to \infty .

Work Done by a Force

We now consider work. In physics, work is related to force, which is often intuitively defined as a push or pull on an object. When a force moves an object, we say the force does work on the object. In other words, work can be thought of as the amount of energy it takes to move an object. According to physics, when we have a constant force, work can be expressed as the product of force and distance.

In the English system, the unit of force is the pound and the unit of distance is the foot, so work is given in foot-pounds. In the metric system, kilograms and meters are used. One newton is the force needed to accelerate

1

m/sec². Thus, the most common unit of work is the newton-meter. This same unit is also called the joule. Both are defined as kilograms times meters squared over seconds squared

(kg \cdot m^{2} / s^{2}) .

When we have a constant force, things are pretty easy. It is rare, however, for a force to be constant. The work done to compress (or elongate) a spring, for example, varies depending on how far the spring has already been compressed (or stretched). We look at springs in more detail later in this section.

we assume the force is roughly constant over the interval, and use

F (x_{i}^{*})

Note that if F is constant, the integral evaluates to

F \cdot (b - a) = F \cdot d,

Now let’s look at the specific example of the work done to compress or elongate a spring. Consider a block attached to a horizontal spring. The block moves back and forth as the spring stretches and compresses. Although in the real world we would have to account for the force of friction between the block and the surface on which it is resting, we ignore friction here and assume the block is resting on a frictionless surface. When the spring is at its natural length (at rest), the system is said to be at equilibrium. In this state, the spring is neither elongated nor compressed, and in this equilibrium position the block does not move until some force is introduced. We orient the system such that

x = 0

According to Hooke’s law, the force required to compress or stretch a spring from an equilibrium position is given by

F (x) = k x,

is called the spring constant and is always positive. We can use this information to calculate the work done to compress or elongate a spring, as shown in the following example.

Work Done in Pumping

Consider the work done to pump water (or some other liquid) out of a tank. Pumping problems are a little more complicated than spring problems because many of the calculations depend on the shape and size of the tank. In addition, instead of being concerned about the work done to move a single mass, we are looking at the work done to move a volume of water, and it takes more work to move the water from the bottom of the tank than it does to move the water from the top of the tank.

We examine the process in the context of a cylindrical tank, then look at a couple of examples using tanks of different shapes. Assume a cylindrical tank of radius

4

m is filled to a depth of 8 m. How much work does it take to pump all the water over the top edge of the tank?

represent the vertical distance below the top of the tank. That is, we orient the

x -axis

vertically, with the origin at the top of the tank and the downward direction being positive (see the following figure).

and look at the work required to lift each individual “layer” of water. So, for

i = 0, 1, 2,…, n,

In pumping problems, the force required to lift the water to the top of the tank is the force required to overcome gravity, so it is equal to the weight of the water. Given that the weight-density of water is

9800

lb/ft³, calculating the volume of each layer gives us the weight. In this case, we have

Note that this step becomes a little more difficult if we have a noncylindrical tank. We look at a noncylindrical tank in the next example.

We also need to know the distance the water must be lifted. Based on our choice of coordinate systems, we can use

x_{i}^{*}

as an approximation of the distance the layer must be lifted. Then the work to lift the

i th

Adding the work for each layer, we see the approximate work to empty the tank is given by

For pumping problems, the calculations vary depending on the shape of the tank or container. The following problem-solving strategy lays out a step-by-step process for solving pumping problems.

We now apply this problem-solving strategy in an example with a noncylindrical tank.

A Pumping Problem with a Noncylindrical Tank

Assume a tank in the shape of an inverted cone, with height $12$

ft and base radius $4$

ft. The tank is full to start with, and water is pumped over the upper edge of the tank until the height of the water remaining in the tank is $4$

ft. How much work is required to pump out that amount of water?

The tank is depicted in [link]. As we did in the example with the cylindrical tank, we orient the $x -axis$

vertically, with the origin at the top of the tank and the downward direction being positive (step 1).

This figure is an upside-down cone. The cone has an axis through the center. The top of the cone on the axis is labeled x=0.

The tank starts out full and ends with $4$

ft of water left, so, based on our chosen frame of reference, we need to partition the interval $[0, 8] .$

Then, for $i = 0, 1, 2,…, n,$

let $P = {x_{i}}$

be a regular partition of the interval $[0, 8],$

and for $i = 1, 2,…, n,$

choose an arbitrary point $x_{i}^{*} \in [x_{i - 1}, x_{i}] .$

We can approximate the volume of a layer by using a disk, then use similar triangles to find the radius of the disk (see the following figure).

From properties of similar triangles, we have

\begin{matrix} \frac{r_{i}}{12 - x_{i}^{*}} & = & \frac{4}{12} = \frac{1}{3} \\ 3 r_{i} & = & 12 - x_{i}^{*} \\ r_{i} & = & \frac{12 - x_{i}^{*}}{3} \\ = & 4 - \frac{x_{i}^{*}}{3} . \end{matrix}

Then the volume of the disk is

V_{i} = π {(4 - \frac{x_{i}^{*}}{3})}^{2} Δ x (step 2).

The weight-density of water is $62.4$

lb/ft³, so the force needed to lift each layer is approximately

F_{i} \approx 62.4 π {(4 - \frac{x_{i}^{*}}{3})}^{2} Δ x (step 3).

Based on the diagram, the distance the water must be lifted is approximately $x_{i}^{*}$

feet (step 4), so the approximate work needed to lift the layer is

W_{i} \approx 62.4 π x_{i}^{*} {(4 - \frac{x_{i}^{*}}{3})}^{2} Δ x (step 5).

Summing the work required to lift all the layers, we get an approximate value of the total work:

W = \sum_{i = 1}^{n} W_{i} \approx \sum_{i = 1}^{n} 62.4 π x_{i}^{*} {(4 - \frac{x_{i}^{*}}{3})}^{2} Δ x (step 6).

Taking the limit as $n \to \infty,$

we obtain

\begin{matrix} W & = lim_{n \to \infty} \sum_{i = 1}^{n} 62.4 π x_{i}^{*} {(4 - \frac{x_{i}^{*}}{3})}^{2} Δ x \\ = \int_{0}^{8} 62.4 π x {(4 - \frac{x}{3})}^{2} d x \\ = 62.4 π \int_{0}^{8} x (16 - \frac{8 x}{3} + \frac{x^{2}}{9}) d x = 62.4 π \int_{0}^{8} (16 x - \frac{8 x^{2}}{3} + \frac{x^{3}}{9}) d x \\ = 62.4 π {[8 x^{2} - \frac{8 x^{3}}{9} + \frac{x^{4}}{36}] \|}_{0}^{8} = 10,649.6 π \approx 33,456.7. \end{matrix}

It takes approximately $33,450$

ft-lb of work to empty the tank to the desired level.

Hydrostatic Force and Pressure

In this last section, we look at the force and pressure exerted on an object submerged in a liquid. In the English system, force is measured in pounds. In the metric system, it is measured in newtons. Pressure is force per unit area, so in the English system we have pounds per square foot (or, perhaps more commonly, pounds per square inch, denoted psi). In the metric system we have newtons per square meter, also called pascals.

submerged horizontally in water at a depth s ([link]). Then, the force exerted on the plate is simply the weight of the water above it, which is given by

F = ρ A s,

is the weight density of water (weight per unit volume). To find the hydrostatic pressure—that is, the pressure exerted by water on a submerged object—we divide the force by the area. So the pressure is

p = F / A = ρ s .

By Pascal’s principle, the pressure at a given depth is the same in all directions, so it does not matter if the plate is submerged horizontally or vertically. So, as long as we know the depth, we know the pressure. We can apply Pascal’s principle to find the force exerted on surfaces, such as dams, that are oriented vertically. We cannot apply the formula

F = ρ A s

directly, because the depth varies from point to point on a vertically oriented surface. So, as we have done many times before, we form a partition, a Riemann sum, and, ultimately, a definite integral to calculate the force.

Suppose a thin plate is submerged in water. We choose our frame of reference such that the x-axis is oriented vertically, with the downward direction being positive, and point

x = 0

correspond to the surface of the water. In this case, depth at any point is simply given by

s (x) = x .

However, in some cases we may want to select a different reference point for

x = 0,

The partition divides the plate into several thin, rectangular strips (see the following figure).

Let’s now estimate the force on a representative strip. If the strip is thin enough, we can treat it as if it is at a constant depth,

s (x_{i}^{*}) .

Evaluating this integral gives us the force on the plate. We summarize this in the following problem-solving strategy.

Chapter Opener: Finding Hydrostatic Force

We now return our attention to the Hoover Dam, mentioned at the beginning of this chapter. The actual dam is arched, rather than flat, but we are going to make some simplifying assumptions to help us with the calculations. Assume the face of the Hoover Dam is shaped like an isosceles trapezoid with lower base $750$

ft, upper base $1250$

ft, and height $750$

ft (see the following figure).

This figure has two images. The first is a picture of a dam. The second image beside the dam is a trapezoidal figure representing the dimensions of the dam. The top is 1250 feet, the bottom is 750 feet. The height is 750 feet. When the reservoir is full, Lake Mead’s maximum depth is about 530 ft, and the surface of the lake is about 10 ft below the top of the dam (see the following figure).

This figure is a trapezoid with the longer side on top. There is a smaller trapezoid inside the first with height labeled 530 feet. It is also 10 feet below the top of the larger trapezoid.

Find the force on the face of the dam when the reservoir is full.
The southwest United States has been experiencing a drought, and the surface of Lake Mead is about 125 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?

We begin by establishing a frame of reference. As usual, we choose to orient the $x -axis$
vertically, with the downward direction being positive. This time, however, we are going to let
$x = 0$
represent the top of the dam, rather than the surface of the water. When the reservoir is full, the surface of the water is
$10$
ft below the top of the dam, so
$s (x) = x - 10$
(see the following figure).

To find the width function, we again turn to similar triangles as shown in the figure below.

From the figure, we see that
$w (x) = 750 + 2 r .$
Using properties of similar triangles, we get
$r = 250 - (1 / 3) x .$
Thus,

$w (x) = 1250 - \frac{2}{3} x (step 2).$

Using a weight-density of
$62.4$
lb/ft³ (step 3) and applying [link], we get

$\begin{matrix} F & = \int_{a}^{b} ρ w (x) s (x) d x \\ = \int_{10}^{540} 62.4 (1250 - \frac{2}{3} x) (x - 10) d x = 62.4 \int_{10}^{540} - \frac{2}{3} [x^{2} - 1885 x + 18750] d x \\ = −62.4 (\frac{2}{3}) {[\frac{x^{3}}{3} - \frac{1885 x^{2}}{2} + 18750 x] \|}_{10}^{540} \approx 8,832,245,000 lb = 4,416,122.5 t . \end{matrix}$

Note the change from pounds to tons $(2000$

lb = $1$

ton) (step 4). This changes our depth function, $s (x),$

and our limits of integration. We have $s (x) = x - 135 .$

The lower limit of integration is $135 .$

The upper limit remains $540 .$

Evaluating the integral, we get

\begin{matrix} F & = \int_{a}^{b} ρ w (x) s (x) d x \\ = \int_{135}^{540} 62.4 (1250 - \frac{2}{3} x) (x - 135) d x \\ = −62.4 (\frac{2}{3}) \int_{135}^{540} (x - 1875) (x - 135) d x = −62.4 (\frac{2}{3}) \int_{135}^{540} (x^{2} - 2010 x + 253125) d x \\ = −62.4 (\frac{2}{3}) {[\frac{x^{3}}{3} - 1005 x^{2} + 253125 x] \|}_{135}^{540} \approx 5,015,230,000 lb = 2,507,615 t . \end{matrix}

Key Concepts

Key Equations

For the following exercises, find the work done.

Find the work done when a constant force $F = 12$

lb moves a chair from $x = 0.9$

to $x = 1.1$

ft.

How much work is done when a person lifts a $50$

lb box of comics onto a truck that is $3$

ft off the ground?

150

ft-lb

What is the work done lifting a $20$

kg child from the floor to a height of $2$

m? (Note that $1$

kg equates to $9.8$

Find the work done when you push a box along the floor $2$

m, when you apply a constant force of $F = 100 N .$

200 J

Compute the work done for a force $F = 12 / x^{2}$

N from $x = 1$

to $x = 2$

What is the work done moving a particle from $x = 0$

to $x = 1$

m if the force acting on it is $F = 3 x^{2}$

1

For the following exercises, find the mass of the one-dimensional object.

A wire that is $2$

ft long (starting at $x = 0)$

and has a density function of $ρ (x) = x^{2} + 2 x$

lb/ft

A car antenna that is $3$

ft long (starting at $x = 0)$

and has a density function of $ρ (x) = 3 x + 2$

lb/ft

\frac{39}{2}

A metal rod that is $8$

in. long (starting at $x = 0)$

and has a density function of $ρ (x) = e^{1 / 2 x}$

lb/in.

A pencil that is $4$

in. long (starting at $x = 2)$

and has a density function of $ρ (x) = 5 / x$

oz/in.

ln (243)

A ruler that is $12$

in. long (starting at $x = 5)$

and has a density function of $ρ (x) = ln (x) + (1 / 2) x^{2}$

oz/in.

For the following exercises, find the mass of the two-dimensional object that is centered at the origin.

An oversized hockey puck of radius $2$

in. with density function $ρ (x) = x^{3} - 2 x + 5$

\frac{332 π}{15}

A frisbee of radius $6$

in. with density function $ρ (x) = e^{− x}$

A plate of radius $10$

in. with density function $ρ (x) = 1 + cos (π x)$

100 π

A jar lid of radius $3$

in. with density function $ρ (x) = ln (x + 1)$

A disk of radius $5$

cm with density function $ρ (x) = \sqrt{3 x}$

20 π \sqrt{15}

A $12$

-in. spring is stretched to $15$

in. by a force of $75$

lb. What is the spring constant?

A spring has a natural length of $10$

cm. It takes $2$

J to stretch the spring to $15$

cm. How much work would it take to stretch the spring from $15$

cm to $20$

cm?

6

A $1$

-m spring requires $10$

J to stretch the spring to $1.1$

m. How much work would it take to stretch the spring from $1$

m to $1.2$

A spring requires $5$

J to stretch the spring from $8$

cm to $12$

cm, and an additional $4$

J to stretch the spring from $12$

cm to $14$

cm. What is the natural length of the spring?

5

A shock absorber is compressed 1 in. by a weight of 1 t. What is the spring constant?

A force of $F = 20 x - x^{3}$

N stretches a nonlinear spring by $x$

meters. What work is required to stretch the spring from $x = 0$

to $x = 2$

36

Find the work done by winding up a hanging cable of length $100$

ft and weight-density $5$

lb/ft.

For the cable in the preceding exercise, how much work is done to lift the cable $50$

ft?

18,750

ft-lb

For the cable in the preceding exercise, how much additional work is done by hanging a $200$

lb weight at the end of the cable?

[T] A pyramid of height $500$

ft has a square base $800$

ft by $800$

ft. Find the area $A$

at height $h .$

If the rock used to build the pyramid weighs approximately $w = 100 {lb/ft}^{3},$

how much work did it take to lift all the rock?

\frac{32}{3} \times 10^{9} ft-lb

[T] For the pyramid in the preceding exercise, assume there were $1000$

workers each working $10$

hours a day, $5$

days a week, $50$

weeks a year. If the workers, on average, lifted 10 100 lb rocks $2$

ft/hr, how long did it take to build the pyramid?

[T] The force of gravity on a mass $m$

is $F = − ((G M m) / x^{2})$

newtons. For a rocket of mass $m = 1000 kg,$

compute the work to lift the rocket from $x = 6400$

to $x = 6500$

km. (Note: $G = 6 \times 10^{−17} {N m}^{2} / {kg}^{2}$

and $M = 6 \times 10^{24} kg .)$

8.65 \times 10^{5} J

[T] For the rocket in the preceding exercise, find the work to lift the rocket from $x = 6400$

to $x = \infty .$

[T] A rectangular dam is $40$

ft high and $60$

ft wide. Compute the total force $F$

on the dam when

the surface of the water is at the top of the dam and
the surface of the water is halfway down the dam.

a. $3,000,000$

lb, b. $749,000$

[T] Find the work required to pump all the water out of a cylinder that has a circular base of radius $5$

ft and height $200$

ft. Use the fact that the density of water is $62$

lb/ft³.

[T] Find the work required to pump all the water out of the cylinder in the preceding exercise if the cylinder is only half full.

23.25 π

million ft-lb

[T] How much work is required to pump out a swimming pool if the area of the base is $800$

ft², the water is $4$

ft deep, and the top is $1$

ft above the water level? Assume that the density of water is $62$

lb/ft³.

A cylinder of depth $H$

and cross-sectional area $A$

stands full of water at density $ρ .$

Compute the work to pump all the water to the top.

\frac{A ρ H^{2}}{2}

For the cylinder in the preceding exercise, compute the work to pump all the water to the top if the cylinder is only half full.

A cone-shaped tank has a cross-sectional area that increases with its depth: $A = (π r^{2} h^{2}) / H^{3} .$

Show that the work to empty it is half the work for a cylinder with the same height and base.

Answers may vary