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Calculus Volume 2

2.5 Physical Applications

Calculus Volume 22.5 Physical Applications
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  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 2.5.1. Determine the mass of a one-dimensional object from its linear density function.
  • 2.5.2. Determine the mass of a two-dimensional circular object from its radial density function.
  • 2.5.3. Calculate the work done by a variable force acting along a line.
  • 2.5.4. Calculate the work done in pumping a liquid from one height to another.
  • 2.5.5. Find the hydrostatic force against a submerged vertical plate.

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,x-axis, with the left end of the rod at x=ax=a and the right end of the rod at x=bx=b (Figure 2.48). 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.

This figure has the x and y axes. On the x-axis is a cylinder, beginning at x=a and ending at x=b.
Figure 2.48 We can calculate the mass of a thin rod oriented along the x-axisx-axis by integrating its density function.

If the rod has constant density ρ,ρ, 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: (ba)ρ.(ba)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),ρ(x), to denote the density of the rod at any point, x.x. Let ρ(x)ρ(x) be an integrable linear density function. Now, for i=0,1,2,…,ni=0,1,2,…,n let P={xi}P={xi} be a regular partition of the interval [a,b],[a,b], and for i=1,2,…,ni=1,2,…,n choose an arbitrary point xi*[xi1,xi].xi*[xi1,xi]. Figure 2.49 shows a representative segment of the rod.

This figure has the x and y axes. On the x-axis is a cylinder, beginning at x=a and ending at x=b. The cylinder has been divided into segments. One segment in the middle begins at xsub(i-1) and ends at xsubi.
Figure 2.49 A representative segment of the rod.

The mass mimi of the segment of the rod from xi1xi1 to xixi is approximated by

miρ(xi*)(xixi1)=ρ(xi*)Δx.miρ(xi*)(xixi1)=ρ(xi*)Δx.

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

m=i=1nmii=1nρ(xi*)Δx.m=i=1nmii=1nρ(xi*)Δx.

This is a Riemann sum. Taking the limit as n,n, we get an expression for the exact mass of the rod:

m=limni=1nρ(xi*)Δx=abρ(x)dx.m=limni=1nρ(xi*)Δx=abρ(x)dx.

We state this result in the following theorem.

Theorem 2.7

Mass–Density Formula of a One-Dimensional Object

Given a thin rod oriented along the x-axisx-axis over the interval [a,b],[a,b], let ρ(x)ρ(x) denote a linear density function giving the density of the rod at a point x in the interval. Then the mass of the rod is given by

m=abρ(x)dx.m=abρ(x)dx.
2.10

We apply this theorem in the next example.

Example 2.23

Calculating Mass from Linear Density

Consider a thin rod oriented on the x-axis over the interval [π/2,π].[π/2,π]. If the density of the rod is given by ρ(x)=sinx,ρ(x)=sinx, what is the mass of the rod?

Solution

Applying Equation 2.10 directly, we have

m=abρ(x)dx=π/2πsinxdx=cosx|π/2π=1.m=abρ(x)dx=π/2πsinxdx=cosx|π/2π=1.
Checkpoint 2.23

Consider a thin rod oriented on the x-axis over the interval [1,3].[1,3]. If the density of the rod is given by ρ(x)=2x2+3,ρ(x)=2x2+3, what is the mass of the rod?

We now extend this concept to find the mass of a two-dimensional disk of radius r.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 xy-plane,xy-plane, with the center at the origin. Then, the density of the disk can be treated as a function of x,x, denoted ρ(x).ρ(x). We assume ρ(x)ρ(x) is integrable. Because density is a function of x,x, we partition the interval from [0,r][0,r] along the x-axis.x-axis. For i=0,1,2,…,n,i=0,1,2,…,n, let P={xi}P={xi} be a regular partition of the interval [0,r],[0,r], and for i=1,2,…,n,i=1,2,…,n, choose an arbitrary point xi*[xi1,xi].xi*[xi1,xi]. 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.

This figure has two images. The first is labeled “a” and is a circle with radius r. The center of the circle is labeled 0. The circle also has the positive x-axis beginning at 0, extending through the circle. The second figure is labeled “b”. It has two concentric circles with center at 0 and the x-axis extending out from 0. The concentric circles form a washer. The width of the washer is from xsub(i-1) to xsubi and is labeled delta x.
Figure 2.50 (a) A thin disk in the xy-plane. (b) A representative washer.

We now approximate the density and area of the washer to calculate an approximate mass, mi.mi. Note that the area of the washer is given by

Ai=π(xi)2π(xi1)2=π[xi2xi12]=π(xi+xi1)(xixi1)=π(xi+xi1)Δx.Ai=π(xi)2π(xi1)2=π[xi2xi12]=π(xi+xi1)(xixi1)=π(xi+xi1)Δx.

You may recall that we had an expression similar to this when we were computing volumes by shells. As we did there, we use xi*(xi+xi1)/2xi*(xi+xi1)/2 to approximate the average radius of the washer. We obtain

Ai=π(xi+xi1)Δx2πxi*Δx.Ai=π(xi+xi1)Δx2πxi*Δx.

Using ρ(xi*)ρ(xi*) to approximate the density of the washer, we approximate the mass of the washer by

mi2πxi*ρ(xi*)Δx.mi2πxi*ρ(xi*)Δx.

Adding up the masses of the washers, we see the mass mm of the entire disk is approximated by

m=i=1nmii=1n2πxi*ρ(xi*)Δx.m=i=1nmii=1n2πxi*ρ(xi*)Δx.

We again recognize this as a Riemann sum, and take the limit as n.n. This gives us

m=limni=1n2πxi*ρ(xi*)Δx=0r2πxρ(x)dx.m=limni=1n2πxi*ρ(xi*)Δx=0r2πxρ(x)dx.

We summarize these findings in the following theorem.

Theorem 2.8

Mass–Density Formula of a Circular Object

Let ρ(x)ρ(x) be an integrable function representing the radial density of a disk of radius r.r. Then the mass of the disk is given by

m=0r2πxρ(x)dx.m=0r2πxρ(x)dx.
2.11

Example 2.24

Calculating Mass from Radial Density

Let ρ(x)=xρ(x)=x represent the radial density of a disk. Calculate the mass of a disk of radius 4.

Solution

Applying the formula, we find

m=0r2πxρ(x)dx=042πxxdx=2π04x3/2dx=2π25x5/2|04=4π5[32]=128π5.m=0r2πxρ(x)dx=042πxxdx=2π04x3/2dx=2π25x5/2|04=4π5[32]=128π5.
Checkpoint 2.24

Let ρ(x)=3x+2ρ(x)=3x+2 represent the radial density of a disk. Calculate the mass of a disk of radius 2.

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 11 kilogram of mass at the rate of 11 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·m2/s2).(kg·m2/s2).

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.

Suppose we have a variable force F(x)F(x) that moves an object in a positive direction along the x-axis from point aa to point b.b. To calculate the work done, we partition the interval [a,b][a,b] and estimate the work done over each subinterval. So, for i=0,1,2,…,n,i=0,1,2,…,n, let P={xi}P={xi} be a regular partition of the interval [a,b],[a,b], and for i=1,2,…,n,i=1,2,…,n, choose an arbitrary point xi*[xi1,xi].xi*[xi1,xi]. To calculate the work done to move an object from point xi1xi1 to point xi,xi, we assume the force is roughly constant over the interval, and use F(xi*)F(xi*) to approximate the force. The work done over the interval [xi1,xi],[xi1,xi], then, is given by

WiF(xi*)(xixi1)=F(xi*)Δx.WiF(xi*)(xixi1)=F(xi*)Δx.

Therefore, the work done over the interval [a,b][a,b] is approximately

W=i=1nWii=1nF(xi*)Δx.W=i=1nWii=1nF(xi*)Δx.

Taking the limit of this expression as nn gives us the exact value for work:

W=limni=1nF(xi*)Δx=abF(x)dx.W=limni=1nF(xi*)Δx=abF(x)dx.

Thus, we can define work as follows.

Definition

If a variable force F(x)F(x) moves an object in a positive direction along the x-axis from point a to point b, then the work done on the object is

W=abF(x)dx.W=abF(x)dx.
2.12

Note that if F is constant, the integral evaluates to F·(ba)=F·d,F·(ba)=F·d, which is the formula we stated at the beginning of this section.

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=0x=0 corresponds to the equilibrium position (see the following figure).

This figure has three images. The first is the x-axis. On the left is a vertical block. Attached to the block is a spring that ends at the y-axis and has the label x=0. The image is labeled equilibrium. The second image is the same spring that ends before the y-axis. It has x<0 and is labeled compressed. The third image is the same spring that is beyond the y-axis. It has x>0 and is labeled stretched.
Figure 2.51 A block attached to a horizontal spring at equilibrium, compressed, and elongated.

According to Hooke’s law, the force required to compress or stretch a spring from an equilibrium position is given by F(x)=kx,F(x)=kx, for some constant k.k. The value of kk depends on the physical characteristics of the spring. The constant kk 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.

Example 2.25

The Work Required to Stretch or Compress a Spring

Suppose it takes a force of 1010 N (in the negative direction) to compress a spring 0.20.2 m from the equilibrium position. How much work is done to stretch the spring 0.50.5 m from the equilibrium position?

Solution

First find the spring constant, k.k. When x=−0.2,x=−0.2, we know F(x)=−10,F(x)=−10, so

F(x)=kx10=k(−0.2)k=50F(x)=kx10=k(−0.2)k=50

and F(x)=50x.F(x)=50x. Then, to calculate work, we integrate the force function, obtaining

W=abF(x)dx=00.550xdx=25x2|00.5=6.25.W=abF(x)dx=00.550xdx=25x2|00.5=6.25.

The work done to stretch the spring is 6.256.25 J.

Checkpoint 2.25

Suppose it takes a force of 88 lb to stretch a spring 66 in. from the equilibrium position. How much work is done to stretch the spring 11 ft from the equilibrium position?

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 44 m and height 1010 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?

The first thing we need to do is define a frame of reference. We let xx represent the vertical distance below the top of the tank. That is, we orient the x-axisx-axis vertically, with the origin at the top of the tank and the downward direction being positive (see the following figure).

This figure is a right circular cylinder that is vertical. It represents a tank of water. The radius of the cylinder is 4 m, the height of the cylinder is 10 m. The height of the water inside the cylinder is 8 m. There is also a horizontal line on top of the tank representing the x=0. A line is drawn vertical beside the cylinder with a downward arrow labeled x.
Figure 2.52 How much work is needed to empty a tank partially filled with water?

Using this coordinate system, the water extends from x=2x=2 to x=10.x=10. Therefore, we partition the interval [2,10][2,10] and look at the work required to lift each individual “layer” of water. So, for i=0,1,2,…,n,i=0,1,2,…,n, let P={xi}P={xi} be a regular partition of the interval [2,10],[2,10], and for i=1,2,…,n,i=1,2,…,n, choose an arbitrary point xi*[xi1,xi].xi*[xi1,xi]. Figure 2.53 shows a representative layer.

This figure is a right circular cylinder representing a tank of water. Inside of the cylinder is a layer of water with thickness delta x. The thickness begins at xsub(i-1) and ends at xsubi.
Figure 2.53 A representative layer of water.

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 98009800 N/m3, or 62.462.4 lb/ft3, calculating the volume of each layer gives us the weight. In this case, we have

V=π(4)2Δx=16πΔx.V=π(4)2Δx=16πΔx.

Then, the force needed to lift each layer is

F=9800·16πΔx=156,800πΔx.F=9800·16πΔx=156,800πΔx.

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 xi*xi* as an approximation of the distance the layer must be lifted. Then the work to lift the ithith layer of water WiWi is approximately

Wi156,800πxi*Δx.Wi156,800πxi*Δx.

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

W=i=1nWii=1n156,800πxi*Δx.W=i=1nWii=1n156,800πxi*Δx.

This is a Riemann sum, so taking the limit as n,n, we get

W=limni=1n156,800πxi*Δx=156,800π210xdx=156,800π[x22]|210=7,526,400π23,644,883.W=limni=1n156,800πxi*Δx=156,800π210xdx=156,800π[x22]|210=7,526,400π23,644,883.

The work required to empty the tank is approximately 23,650,000 J.

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.

Problem-Solving Strategy: Solving Pumping Problems
  1. Sketch a picture of the tank and select an appropriate frame of reference.
  2. Calculate the volume of a representative layer of water.
  3. Multiply the volume by the weight-density of water to get the force.
  4. Calculate the distance the layer of water must be lifted.
  5. Multiply the force and distance to get an estimate of the work needed to lift the layer of water.
  6. Sum the work required to lift all the layers. This expression is an estimate of the work required to pump out the desired amount of water, and it is in the form of a Riemann sum.
  7. Take the limit as nn and evaluate the resulting integral to get the exact work required to pump out the desired amount of water.

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

Example 2.26

A Pumping Problem with a Noncylindrical Tank

Assume a tank in the shape of an inverted cone, with height 1212 ft and base radius 44 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 44 ft. How much work is required to pump out that amount of water?

Solution

The tank is depicted in Figure 2.54. As we did in the example with the cylindrical tank, we orient the x-axisx-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.
Figure 2.54 A water tank in the shape of an inverted cone.

The tank starts out full and ends with 44 ft of water left, so, based on our chosen frame of reference, we need to partition the interval [0,8].[0,8]. Then, for i=0,1,2,…,n,i=0,1,2,…,n, let P={xi}P={xi} be a regular partition of the interval [0,8],[0,8], and for i=1,2,…,n,i=1,2,…,n, choose an arbitrary point xi*[xi1,xi].xi*[xi1,xi]. 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).

This figure has two images. The first has the x-axis. Below the axis, on a slant is a line segment extending up to the x-axis. Beside the line segment is a horizontal right circular cylinder. The second image has a triangle. The right triangle mirrors the first image with the hypotenuse the line segment in the first image. The top of the triangle is 4 units. the length of the vertical side is 12 units. The vertical side is also divided into two parts; the first is xsubi, the second is 12-xsubi. It is divided at the level where the first image has the cylinder.
Figure 2.55 Using similar triangles to express the radius of a disk of water.

From properties of similar triangles, we have

ri12xi*=412=133ri=12xi*ri=12xi*3=4xi*3.ri12xi*=412=133ri=12xi*ri=12xi*3=4xi*3.

Then the volume of the disk is

Vi=π(4xi*3)2Δx(step 2).Vi=π(4xi*3)2Δx(step 2).

The weight-density of water is 62.462.4 lb/ft3, so the force needed to lift each layer is approximately

Fi62.4π(4xi*3)2Δx(step 3).Fi62.4π(4xi*3)2Δx(step 3).

Based on the diagram, the distance the water must be lifted is approximately xi*xi* feet (step 4), so the approximate work needed to lift the layer is

Wi62.4πxi*(4xi*3)2Δx(step 5).Wi62.4πxi*(4xi*3)2Δx(step 5).

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

W=i=1nWii=1n62.4πxi*(4xi*3)2Δx(step 6).W=i=1nWii=1n62.4πxi*(4xi*3)2Δx(step 6).

Taking the limit as n,n, we obtain

W=limni=1n62.4πxi*(4xi*3)2Δx=0862.4πx(4x3)2dx=62.4π08x(168x3+x29)dx=62.4π08(16x8x23+x39)dx=62.4π[8x28x39+x436]|08=10,649.6π33,456.7.W=limni=1n62.4πxi*(4xi*3)2Δx=0862.4πx(4x3)2dx=62.4π08x(168x3+x29)dx=62.4π08(16x8x23+x39)dx=62.4π[8x28x39+x436]|08=10,649.6π33,456.7.

It takes approximately 33,45033,450 ft-lb of work to empty the tank to the desired level.

Checkpoint 2.26

A tank is in the shape of an inverted cone, with height 1010 ft and base radius 6 ft. The tank is filled to a depth of 8 ft to start with, and water is pumped over the upper edge of the tank until 3 ft of water remain in the tank. How much work is required to pump out that amount of water?

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.

Let’s begin with the simple case of a plate of area AA submerged horizontally in water at a depth s (Figure 2.56). Then, the force exerted on the plate is simply the weight of the water above it, which is given by F=ρAs,F=ρAs, where ρρ 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=ρs.p=F/A=ρs.

This image has a circular plate submerged in water. The plate is labeled A and the depth of the water is labeled s.
Figure 2.56 A plate submerged horizontally in water.

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=ρAsF=ρAs 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=0x=0 corresponding to a logical reference point. Let s(x)s(x) denote the depth at point x. Note we often let x=0x=0 correspond to the surface of the water. In this case, depth at any point is simply given by s(x)=x.s(x)=x. However, in some cases we may want to select a different reference point for x=0,x=0, so we proceed with the development in the more general case. Last, let w(x)w(x) denote the width of the plate at the point x.x.

Assume the top edge of the plate is at point x=ax=a and the bottom edge of the plate is at point x=b.x=b. Then, for i=0,1,2,…,n,i=0,1,2,…,n, let P={xi}P={xi} be a regular partition of the interval [a,b],[a,b], and for i=1,2,…,n,i=1,2,…,n, choose an arbitrary point xi*[xi1,xi].xi*[xi1,xi]. The partition divides the plate into several thin, rectangular strips (see the following figure).

This image is the overhead view of a submerged circular plate. The x-axis is to the side of the plate. The plate’s diameter goes from x=a to x=b. There is a strip in the middle of the plate with thickness of delta x. On the axis this thickness begins at x=xsub(i-1) and ends at x=xsubi. The length of the strip in the plate is labeled w(csubi).
Figure 2.57 A thin plate submerged vertically in water.

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(xi*).s(xi*). We then have

Fi=ρAs=ρ[w(xi*)Δx]s(xi*).Fi=ρAs=ρ[w(xi*)Δx]s(xi*).

Adding the forces, we get an estimate for the force on the plate:

Fi=1nFi=i=1nρ[w(xi*)Δx]s(xi*).Fi=1nFi=i=1nρ[w(xi*)Δx]s(xi*).

This is a Riemann sum, so taking the limit gives us the exact force. We obtain

F=limni=1nρ[w(xi*)Δx]s(xi*)=abρw(x)s(x)dx.F=limni=1nρ[w(xi*)Δx]s(xi*)=abρw(x)s(x)dx.
2.13

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

Problem-Solving Strategy: Finding Hydrostatic Force
  1. Sketch a picture and select an appropriate frame of reference. (Note that if we select a frame of reference other than the one used earlier, we may have to adjust Equation 2.13 accordingly.)
  2. Determine the depth and width functions, s(x)s(x) and w(x).w(x).
  3. Determine the weight-density of whatever liquid with which you are working. The weight-density of water is 62.462.4 lb/ft3, or 9800 N/m3.
  4. Use the equation to calculate the total force.

Example 2.27

Finding Hydrostatic Force

A water trough 15 ft long has ends shaped like inverted isosceles triangles, with base 8 ft and height 3 ft. Find the force on one end of the trough if the trough is full of water.

Solution

Figure 2.58 shows the trough and a more detailed view of one end.

This figure has two images. The first is a water trough with rectangular sides. The length of the trough is 15 feet, the depth is 3 feet, and the width is 8 feet. The second image is a cross section of the trough. It is a triangle. The top has length of 8 feet and the sides have length 5 feet. The altitude is labeled with 3 feet.
Figure 2.58 (a) A water trough with a triangular cross-section. (b) Dimensions of one end of the water trough.

Select a frame of reference with the x-axisx-axis oriented vertically and the downward direction being positive. Select the top of the trough as the point corresponding to x=0x=0 (step 1). The depth function, then, is s(x)=x.s(x)=x. Using similar triangles, we see that w(x)=8(8/3)xw(x)=8(8/3)x (step 2). Now, the weight density of water is 62.462.4 lb/ft3 (step 3), so applying Equation 2.13, we obtain

F=abρw(x)s(x)dx=0362.4(883x)xdx=62.403(8x83x2)dx=62.4[4x289x3]|03=748.8.F=abρw(x)s(x)dx=0362.4(883x)xdx=62.403(8x83x2)dx=62.4[4x289x3]|03=748.8.

The water exerts a force of 748.8 lb on the end of the trough (step 4).

Checkpoint 2.27

A water trough 12 m long has ends shaped like inverted isosceles triangles, with base 6 m and height 4 m. Find the force on one end of the trough if the trough is full of water.

Example 2.28

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 750750 ft, upper base 12501250 ft, and height 750750 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.
Figure 2.59 A simplified model of the Hoover Dam with assumed dimensions.
  1. Find the force on the face of the dam when the reservoir is full.
  2. 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?

Solution

  1. We begin by establishing a frame of reference. As usual, we choose to orient the x-axisx-axis vertically, with the downward direction being positive. This time, however, we are going to let x=0x=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 1010 ft below the top of the dam, so s(x)=x10s(x)=x10 (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 s(x)=x-10. It represents the depth of the water. It is also 10 feet below the top of the larger trapezoid. The top of the larger trapezoid is at x=0.
    Figure 2.60 We first choose a frame of reference.

    To find the width function, we again turn to similar triangles as shown in the figure below.
    This figure has two images. The first is a trapezoid with larger side on the top. The length of the top is divided into 3 measures. The first measure is 250 feet, the second is 750 feet, and the third is 250 feet. The height of the trapezoid is 750 feet. The length of the bottom is 750 feet. Inside of the trapezoid the width is labeled w(x). Inside if one of the triangular sides is the width r. The second image is the same trapezoid. It has the height labeled as 750 feet. Inside the trapezoid it has the height divided into two segments. The first is labeled x, and the second is labeled 750-x. On the side of the trapezoid a triangle has been formed by a vertical line from the bottom side to the top. Inside of the triangle is a horizontal line segment labeled r.
    Figure 2.61 We use similar triangles to determine a function for the width of the dam. (a) Assumed dimensions of the dam; (b) highlighting the similar triangles.

    From the figure, we see that w(x)=750+2r.w(x)=750+2r. Using properties of similar triangles, we get r=250(1/3)x.r=250(1/3)x. Thus,
    w(x)=125023x(step 2).w(x)=125023x(step 2).

    Using a weight-density of 62.462.4 lb/ft3 (step 3) and applying Equation 2.13, we get
    F=abρw(x)s(x)dx=1054062.4(125023x)(x10)dx=62.41054023[x21885x+18750]dx=−62.4(23)[x331885x22+18750x]|105408,832,245,000lb=4,416,122.5t.F=abρw(x)s(x)dx=1054062.4(125023x)(x10)dx=62.41054023[x21885x+18750]dx=−62.4(23)[x331885x22+18750x]|105408,832,245,000lb=4,416,122.5t.

Note the change from pounds to tons (2000(2000 lb = 11 ton) (step 4). This changes our depth function, s(x),s(x), and our limits of integration. We have s(x)=x135.s(x)=x135. The lower limit of integration is 135.135. The upper limit remains 540.540. Evaluating the integral, we get

F=abρw(x)s(x)dx=13554062.4(125023x)(x135)dx=−62.4(23)135540(x1875)(x135)dx=−62.4(23)135540(x22010x+253125)dx=−62.4(23)[x331005x2+253125x]|1355405,015,230,000lb=2,507,615t.F=abρw(x)s(x)dx=13554062.4(125023x)(x135)dx=−62.4(23)135540(x1875)(x135)dx=−62.4(23)135540(x22010x+253125)dx=−62.4(23)[x331005x2+253125x]|1355405,015,230,000lb=2,507,615t.
Checkpoint 2.28

When the reservoir is at its average level, the surface of the water is about 50 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?

Media

To learn more about Hoover Dam, see this article published by the History Channel.

Section 2.5 Exercises

For the following exercises, find the work done.

218.

Find the work done when a constant force F=12F=12 lb moves a chair from x=0.9x=0.9 to x=1.1x=1.1 ft.

219.

How much work is done when a person lifts a 5050 lb box of comics onto a truck that is 33 ft off the ground?

220.

What is the work done lifting a 2020 kg child from the floor to a height of 22 m? (Note that 11 kg equates to 9.89.8 N)

221.

Find the work done when you push a box along the floor 22 m, when you apply a constant force of F=100N.F=100N.

222.

Compute the work done for a force F=12/x2F=12/x2 N from x=1x=1 to x=2x=2 m.

223.

What is the work done moving a particle from x=0x=0 to x=1x=1 m if the force acting on it is F=3x2F=3x2 N?

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

224.

A wire that is 22 ft long (starting at x=0)x=0) and has a density function of ρ(x)=x2+2xρ(x)=x2+2x lb/ft

225.

A car antenna that is 33 ft long (starting at x=0)x=0) and has a density function of ρ(x)=3x+2ρ(x)=3x+2 lb/ft

226.

A metal rod that is 88 in. long (starting at x=0)x=0) and has a density function of ρ(x)=e1/2xρ(x)=e1/2x lb/in.

227.

A pencil that is 44 in. long (starting at x=2)x=2) and has a density function of ρ(x)=5/xρ(x)=5/x oz/in.

228.

A ruler that is 1212 in. long (starting at x=5)x=5) and has a density function of ρ(x)=ln(x)+(1/2)x2ρ(x)=ln(x)+(1/2)x2 oz/in.

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

229.

An oversized hockey puck of radius 22 in. with density function ρ(x)=x32x+5ρ(x)=x32x+5

230.

A frisbee of radius 66 in. with density function ρ(x)=exρ(x)=ex

231.

A plate of radius 1010 in. with density function ρ(x)=1+cos(πx)ρ(x)=1+cos(πx)

232.

A jar lid of radius 33 in. with density function ρ(x)=ln(x+1)ρ(x)=ln(x+1)

233.

A disk of radius 55 cm with density function ρ(x)=3xρ(x)=3x

234.

A 1212-in. spring is stretched to 1515 in. by a force of 7575 lb. What is the spring constant?

235.

A spring has a natural length of 1010 cm. It takes 22 J to stretch the spring to 1515 cm. How much work would it take to stretch the spring from 1515 cm to 2020 cm?

236.

A 11-m spring requires 1010 J to stretch the spring to 1.11.1 m. How much work would it take to stretch the spring from 11 m to 1.21.2 m?

237.

A spring requires 55 J to stretch the spring from 88 cm to 1212 cm, and an additional 44 J to stretch the spring from 1212 cm to 1414 cm. What is the natural length of the spring?

238.

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

239.

A force of F=20xx3F=20xx3 N stretches a nonlinear spring by xx meters. What work is required to stretch the spring from x=0x=0 to x=2x=2 m?

240.

Find the work done by winding up a hanging cable of length 100100 ft and weight-density 55 lb/ft.

241.

For the cable in the preceding exercise, how much work is done to lift the cable 5050 ft?

242.

For the cable in the preceding exercise, how much additional work is done by hanging a 200200 lb weight at the end of the cable?

243.

[T] A pyramid of height 500500 ft has a square base 800800 ft by 800800 ft. Find the area AA at height h.h. If the rock used to build the pyramid weighs approximately w=100lb/ft3,w=100lb/ft3, how much work did it take to lift all the rock?

244.

[T] For the pyramid in the preceding exercise, assume there were 10001000 workers each working 1010 hours a day, 55 days a week, 5050 weeks a year. If the workers, on average, lifted 10 100 lb rocks 22 ft/hr, how long did it take to build the pyramid?

245.

[T] The force of gravity on a mass mm is F=((GMm)/x2)F=((GMm)/x2) newtons. For a rocket of mass m=1000kg,m=1000kg, compute the work to lift the rocket from x=6400x=6400 to x=6500x=6500 km. (Note: G=6×10−17N m2/kg2G=6×10−17N m2/kg2 and M=6×1024kg.)M=6×1024kg.)

246.

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

247.

[T] A rectangular dam is 4040 ft high and 6060 ft wide. Compute the total force FF on the dam when

  1. the surface of the water is at the top of the dam and
  2. the surface of the water is halfway down the dam.
248.

[T] Find the work required to pump all the water out of a cylinder that has a circular base of radius 55 ft and height 200200 ft. Use the fact that the density of water is 6262 lb/ft3.

249.

[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.

250.

[T] How much work is required to pump out a swimming pool if the area of the base is 800800 ft2, the water is 44 ft deep, and the top is 11 ft above the water level? Assume that the density of water is 6262 lb/ft3.

251.

A cylinder of depth HH and cross-sectional area AA stands full of water at density ρ.ρ. Compute the work to pump all the water to the top.

252.

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.

253.

A cone-shaped tank has a cross-sectional area that increases with its depth: A=(πr2h2)/H3.A=(πr2h2)/H3. Show that the work to empty it is half the work for a cylinder with the same height and base.

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