9 FURTHER APPLICATIONS OF INTEGRATION

9 FURTHER APPLICATIONS OF INTEGRATION
9.1 ARC LENGTH
TRANSPARENCY AVAILABLE
#16 (Figures 2–4)
SUGGESTED TIME AND EMPHASIS
1
2
–1 class Essential material
POINTS TO STRESS
1. The derivation of the arc length integral.
2. The uses of the arc length function.
QUIZ QUESTIONS
• Text Question: So far, we have mainly used integrals to compute areas (for two-dimensional quantities)
or volumes (for three-dimensional quantities). Why do integrals come into play when computing the length
of a curve?
Answer: We are adding up infinitely many infinitesimal quantities, which constitutes an integration.
• Drill Question: Which of the following integrals gives the length of the graph y = √ x between x = a
and x = b, where 0 < a < b?
∫ b √
∫ b √
∫ √
(A) x 2 √ b
+ xdx (B) x + xdx
(C) x + 1
a
a
a 2 √ x dx
∫ √
b
(D) 1 + 1
∫ b
√
a 2 √ x dx (E) 1 + 1
a 4x dx
Answer: (E)
MATERIALS FOR LECTURE
• Derive the equation for arc length. End with a discussion about why these integrals are usually difficult to
compute exactly.
483

CHAPTER 9
FURTHER APPLICATIONS OF INTEGRATION
• Show that the area under the curve at right is exactly 2, and
comment on the simplicity of the area integral. Then set up the
integral for the arc length. This integral cannot be computed
directly, but numerical methods give
∫ π
√
0 1 + (cos x) 2 dx ≈ 3.8202. Comment on the fact that the arc
length integral is much more complicated than the area integral.
• Point out that for the arc length function s (x), ds/dx =
grows faster than x. Interpret this result geometrically.
√
1 + f ′ (x) 2 is always ≥ 1, and hence s (x)
WORKSHOP/DISCUSSION
• Use the arc length formula to compute the length along the straight line y = 3x from x = 0tox = 2.
Then compute the length using the distance formula and show that you get the same number.
• Estimate the arc length of the part of the implicitly defined function x 2/3 + y 2/3 = 1thatliesinthefirst
quadrant, then show that the length is 3 2 . y
1
0
• Derive the arc length function s (x) for f (x) = ln (cos x),0≤ x < π 2 .
1 x
GROUP WORK 1: The Bizarre Coastline
Closure is important to this exercise. Make sure the students get some idea of the wonder and mystery of the
shape that arises as k →∞. In the real world, coastline figures are notoriously inaccurate, because the more
accurately people try to measure a coast, the more crinkles they find, and the larger the result gets.
Answers:
1. L = 4 + 2π + ∫ 2π
√
0 1 + cos 2 xdx≈ 17.924, A = 4π
2. L = 4 + 2π + ∫ 2π
√
0 1 + 4cos 2 2xdx≈ 20.824, A = 4π
3. L = 4 + 2π + ∫ 2π
√
0 1 + 9cos 2 3xdx≈ 24.258, A = 4π
4. L = 4 + 2π + ∫ 2π
√
0 1 + 100 cos 2 10xdx≈ 51.120, A = 4π
∫ 2π
5. The areas remain constant, while the coastline lengths increase.
0
sin kx dx = 0 for any integer k.
Make sure to convey to the students that we can make the lengths arbitrarily large as we go down the
chain, and yet the areas remain constant.
484

SECTION 9.1
ARC LENGTH
6.
y
y
2.1
2
2
1.9
0
¹ 2¹
x
1.8
3 3.1 3.2 3.3 x
Notice that as x approaches π, the curve wiggles faster and faster, with decreasing amplitude. The limit as
x → π does exist, and is 2. We can show that L =∞either geometrically, or by comparing the resultant
∫ π
dx
integral to the improper integral 2
0 x − π .
A = 2 ∫ [ ( ) ]
π
1
0
(x − π) sin + 2 dx ≈ 17.385. (Note that this is an improper integral.)
x − π
Note that we can change the definition of f slightly, to “flip” the left half of the curve:
⎧
( ) 1
⎪⎨ − (x − π) sin + 2 if x < π
x − π
f (x) =
( ) 1
⎪⎩ (x − π) sin + 2 if x > π
x − π
y
2
0
¹ 2¹ x
The area of this new island (“Hpela”) is now the familiar 4π, and the perimeter is still infinite.
GROUP WORK 2: Cable Guy
This exercise requires the students to make several assumptions. For example, when computing the length of
cable needed for the sample segment shown on the left, they will have to decide which of the three figures on
the right most accurately represents the cross-sectional side view of the lake bottom:
10
20
10
20
10
20
10
20
Have them make an assumption and provide justification for it. Also remind students that the path of steepest
485

f (x) = 1 2 x2 − 1 4 ln x. 486
CHAPTER 9
FURTHER APPLICATIONS OF INTEGRATION
descent always runs perpendicular to the contour lines.
Notice that this exercise involves many computations and estimates. The students must divide up the work in
order to finish in a reasonable amount of time. Make sure to have several rulers on hand.
After the exercise is completed, poll the groups to see whose path used the least cable.
Answers (will vary):
1. Lake Wobegone
Bill Turner’s
House
Cable Box
12
10
11
20
8
9
10
30
40
50
7
1 23 4 5 6
60
50
40
30
20
10
Cable Source
The path above was chosen so that the line segments are approximately orthogonal to the contour lines.
1
2.
4
inch on the map represents 100 ft. In estimating the amount of
10
cable needed, we will assume that a cross-sectional view of the path
20
between two contour lines looks like the figure at right.
Segment 1 2 3 4 5 6 7 8 9 10 11 12
Horizontal 30 25 25 25 50 125 200 75 75 90 150 85
Vertical 10 10 10 10 10 10 10 10 10 10 10 10
Length 31.6 26.9 26.9 26.9 51.0 125.4 200.2 75.7 75.7 90.6 150.3 85.6
The total length is about 966.8 ft, so we should probably order at least 1000 ft to be safe.
GROUP WORK 3: Find the Function
This is a tricky problem that reduces to solving the equation
integrating. The fact that the graph of f passes through the point
√
1 + [ f ′ (x) ] 2 = x +
1
( )
1, 1 2
4x for f ′ (x) and then
is key, as it allows the students to
determine the constant of integration.
Answer: ∫ x
1
√1 + [ f ′ (t) ] √
2 dt =
1
2
x 2 + 1 4 ln x ⇔ 1 + [ f ′ (x) ] 2 1 = x + ⇔ f ′ (x) = x − 1
4x
4x
)
⇔ f (x) = 1 2 x2 − 1 4
(1, ln x + C. Because the graph of f passes through the point 1 2
, C = 0, so

SECTION 9.1
ARC LENGTH
SPECIAL PROJECT: How to Define π
This project allows the students to deduce a historic geometric result using techniques from this section.
They start by defining π as the circumference of a semi-circle of radius 1, and then show that the integral
representing this quantity is the same as the integral representing the area of a circle of radius 1.
Note that the simplification in Problem 3 requires an algebraic trick. The students may need the hint that
∫
x 2 ∫ x 2
√ dx = − 1 + 1
√ dx. They will also need to be able to integrate by parts.
1 − x 2 1 − x 2
Answers:
√
∫ 1
1. 1 +
−1
[ d
(√
1 − x 2) ] 2 ∫ 1
dx =
dx
−1
dx
√
1 − x 2
2. The term on the left is the length of a semicircle of radius 1, and the term on the right is an expression of
the area of a circle of radius 1.
3. I = ∫ √ 1 − x 2 dx.Letu = √ 1 − x 2 , dv = dx ⇒ du =−√ xdx , v = x. Then
1 − x 2
√ ∫
I = x 1 − x 2 +
x 2 ∫
dx
√
√1 = x − x 2 +
1 − x 2
dx
√
1 − x 2 − ∫ √
1 − x 2 dx
4. Using the answer to Problem 3, 2 ∫ 1
√
−1 1 − x 2 dx = x √ ∣ ∫
1 − x 2 ∣∣ 1 1
+
−1
−1
dx
√
1 − x 2
and the result follows.
HOMEWORK PROBLEMS
Core Exercises: 3, 8, 13, 17, 19, 22, 26, 31, 34, 39
Sample Assignment: 2, 3, 6, 8, 10, 13, 17, 19, 22, 25, 26, 27, 31, 34, 37, 39, 40
Exercise D A N G
2 ×
3 ×
6 ×
8 ×
10 ×
13 ×
17 ×
19 ×
22 × ×
25 ×
26 ×
27 × ×
31 × ×
34 × ×
37 × ×
39 × ×
40 × ×
487

GROUP WORK 1, SECTION 9.1
The Bizarre Coastline
In this exercise we are going to consider the areas and coastlines of some bizarre islands, and end by
analyzing a type of shape that has intrigued mathematicians for many years. Put on your shorts and
sunglasses; we are heading to the Pacific Ocean.
1. The first island on our trip is the island of Eno. The best way to model this island is to shade the region
between the curves x = 0, x = 2π, y = 0, and y = sin x + 2.
The length of the coastline of this island is merely the sum of four arc lengths. The southern border has
length 2π, and the east and west borders each have length 2. You can compute the north border’s length
by using the arc length formula to obtain a numerical estimate accurate to three decimal places.
Coastline Length of Eno:
Area of Eno:
2. Let us reluctantly leave the island of Eno, and wander over to the neighboring island of Owt. The south,
east, and west borders of Owt look the same as Eno’s do, but the northern border is given by y = sin 2x+2.
In order to be able to deduct the cost of our trip, we need to do some research here as well. Please compute
the coastline length of Owt and the area of Owt.
Coastline Length of Owt:
Area of Owt:
488

The Bizarre Coastline
3. As you probably expect, the island of Eerht is similar to the islands of Eno and Owt, except that the
northern border is given by y = sin 3x + 2. Please sketch Eerht, and compute its coastline and area.
Coastline Length of Eerht:
Area of Eerht:
4. Ah, if only we had the time to fully explore the islands of Ruof, Evif, and Xis... But there are a lot of
islands in this chain, and we don’t have all the time in the world. Let’s skip ahead a bit, and look at the
island of Net. Graph the northernmost border of Net (y = sin 10x + 2) on your calculator, and estimate
the length of the coastline of Net before computing.
Coastline Length of Net:
Area of Net:
5. You have probably noticed that there has been a pattern in the areas of this chain of islands, and the use
of this pattern will save us some in-person visits. Explain what you have observed, and show that your
conjecture holds for any one of the islands. What happens to the coastlines as we go farther along the
chain of islands?
489

The Bizarre Coastline
6. For our final stop, let’s go to the mysterious and oft-misunderstood island of Ytinifni. This island is not
in the same chain as the others; it is where the inhabitants of the other islands go for their vacations. Its
( ) 1
northern border is given by y = (x − π) sin + 2. Try your best to sketch this mysterious island.
x − π
Notice that it is very hard to draw. There is a legend of a hole in the middle of the northern border that
will claim the life of any tourist who sets foot there. (I’ve been to that area, and seen nothing unusual —
why do you think that is?) Now, using geometric reasoning or numerical approximation, complete the
following:
Coastline Length of Ytinifni:
Area of Ytinifni:
7. Is this result counterintuitive? Try to explain it, and ask your teacher for help if you cannot do so.
490

GROUP WORK 2, SECTION 9.1
Cable Guy
Billionaire businessman Bill Turner (the president of MacroHard) lives on a small island in the middle of
Lake Wobegone. What follows is a topographical map of the region.
Lake Wobegone
Bill Turner’s
House
Cable Box
10
20
30
40
50
60
50
40
30
20
10
Cable Source
1. As the Cable Guy, your job is to wire up Mr. Turner’s home for cable television. There is a cable box on
the shore, as indicated. The cable will be unwound on the bottom of the lake, from the shore to the island.
Draw a path that will minimize the amount of cable necessary to use. Don’t worry about being 100% sure,
make a guess and go with it.
2. Given that 1 4
inch on this map represents 100 feet, what is the amount of cable needed to connect Mr.
Turner’s house using your path?
491

GROUP WORK 3, SECTION 9.1
Find the Function
( )
Find a function f (x) whose arc length L (x) from 1, 1 2
to (x, f (x)), x > 1, is 1 2 x2 + 1 4
ln x.
492

SPECIAL PROJECT FOR SECTION 9.1
How to Define π
(Idea due to H. Wu, The Joy of Lecturing, pages 5–6)
In this age of calculators, people sometimes get confused about how the number π is defined. The constant
π is not the number 22 7
, nor 3.1415926, nor even
3.14159265358979323846264338327950288419716939937510582097
All of those are just approximations.
One can find many expressions that equal π. Forexample,
( )
2 × 2 × 4 × 4 × 6 × 6 × 8 ×···
π = 2
1 × 3 × 3 × 5 × 5 × 7 × 7 ×···
π = 4
(1 − 1 3 + 1 5 − 1 7 + 1 9 − 1 +···)
11
π =
√
6
1 2 + 6 2 2 + 6 3 2 + 6 4 2 +···
π = 3
(1 + 12
4 × 6 + 1 2 × 3 2
4 × 6 × 8 × 10 + 1 2 × 3 2 × 5 2
+···)
4 × 6 × 8 × 10 × 12 × 14
Usually, however, people define π as the ratio of the circumference of a circle to its diameter. Then we say
that the circumference of a circle is 2πr, and its area is πr 2 . But it is easy to forget how amazing an idea this
originally was; that the circumference, the area, and the radius of a general circle are so tightly related. In
this exercise, we go back to basics. We define π as the length of a semicircle of radius 1 (that is, one-half the
circumference) and then prove that it is equal to the area of a circle of radius 1.
1. Use the arc length formula to show that the length of the semicircle of radius 1, x 2 + y 2 = 1, y ≥ 0, can
be written as
∫ 1
−1
dx
√ .
1 − x 2
2. Show that the statement “The length of a semicircle of radius 1 equals the area of a circle of radius 1” can
be written as
∫ 1
−1
∫
dx
1 √
√ = 2 1 − x 2 dx
1 − x 2
−1
493

How to Define π
3. Using integration by parts with u = √ 1 − x 2 , show that
∫ √ ∫
∫
1 − x 2 dx = x√
1 − x 2 dx √
+ √ − 1 − x 2 dx
1 − x 2
4. Show that the formula of Problem 2 is true.
494

DISCOVERY PROJECT
Arc Length Contest
This interesting project requires students to discover formulas for two positive functions with area 1 over
[0, 1] satisfying f (0) = f (1) = 0. The basic idea is to find the graph with minimal arc length. You may
want to add the condition that the functions be differentiable. Since one example is given with L < 3, this is
a challenging project.
495

9.2 AREA OF A SURFACE OF REVOLUTION
TRANSPARENCY AVAILABLE
#17 (Figure 4)
SUGGESTED TIME AND EMPHASIS
1
2
–1 class Optional material
POINTS TO STRESS
1. The derivation of the formulas for the surface area of a solid of revolution about the x-andy-axes.
2. The approximation of surface areas by using simpler surfaces.
QUIZ QUESTIONS
• Text Question: The arc of the parabola y = x 2 from (1, 1) to (2, 4) is rotated around the y-axis. Which
is larger, the area of the resultant surface, or the surface area of the frustum of a cone with the same top
and bottom edges?
3
_3 3
_3
Answer: The first
• Drill Question: Set up, but do not evaluate, an integral for the area of the surface obtained by rotating
the curve y = sec x,0≤ x ≤ π 4
about the y-axis.
Answer: S = ∫ π/4
0
2π sec x √ 1 + tan 2 x sec 2 xdx
MATERIALS FOR LECTURE
• Derive one of the formulas for the surface area of rotation about the x-axis, using the surface area of a
frustum.
• Derive the surface area of the sphere of radius R by rotating the top half of a circle about the x-axis.
• Compute the surface area formed by rotating f (x) = sin x, 0≤ x ≤ π about the x-axis. (The integral is
∫ π
0 2π sin x√ 1 + cos 2 xdx= 4π ∫ 1
√
0 1 + u 2 du.)
• Show that the surface formed by rotating f (x) = 1/x, x ≥ 1 about the x-axis has infinite area.
WORKSHOP/DISCUSSION
• We want to consider the area of the surface formed when the curve x 2/3 + y 2/3 = 1, x ≥ 0, y ≥ 0is
rotated about the x-axis. Estimate this area, first using the cone generated by the line segment connecting
496

SECTION 9.2
AREA OF A SURFACE OF REVOLUTION
the points (0, 1) and (1, 0), and then the cone generated by the line segment of slope m =−1 through the
( ) 1
point
2 3/2 , 1
2 3/2 .
• Compute the surface area when the function f (x) = 2 3 x3/2 ,1≤ x ≤ 2 is rotated about the y-axis.
• Show that the area of the surface of revolution formed when f (x) = x 4 is rotated about the x-axis cannot
be directly computed. Show that this is also true when the function is rotated about the y-axis.
GROUP WORK 1: Gabriel’s Horn
There are two forms of this exercise. Some groups should be given the first, some the second. The students
should not be told that there are two different forms. After they have made their conclusions, poll the class.
Then have the students who think that the horn can be painted try to defend their case, and the ones who say
it cannot defend theirs. Close by having the students try to understand how a surface with infinite area can
contain a finite volume (see Exercise 25).
Answers:
∫ ∞
( ) 1 2
Form One π dx = π ft 3
1 x
∫ ∞
√
2π
Form Two
1 + 1 ∫ ∞
x x 4 dx diverges, by comparison to 2π
x dx
1
GROUP WORK 2: Mind Your p’s and q’s
This activity involves some subtle comparisons and proves an important mathematical point.
Answers:
1. 2π
√
( p
) 2
∫
4π ∞
x p 1 + <
x p+1 x p and 4π
dx converges.
1 x p
2. 2π
√
( q ) 2
∫
2π ∞
x q 1 + >
x q+1 x q and 2π
dx diverges.
xq HOMEWORK PROBLEMS
Core Exercises: 1, 6, 15, 17, 28
Sample Assignment: 1, 6, 9, 12, 14, 15, 17, 19, 22, 25, 28, 33
1
1
Exercise D A N G
1 ×
6 ×
9 ×
12 ×
14 ×
15 ×
Exercise D A N G
17 × ×
19 × ×
22 ×
25 ×
28 × ×
33 ×
497

GROUP WORK 1, SECTION 9.2
Gabriel’s Horn
Sketch the graph of f (x) = 1/x from x = 1tox =∞. (Okay, you don’t actually have to graph it all the
way out to infinity.)
Now think of the object formed by rotating this curve about the x-axis. This object is called Gabriel’s Horn.
Your job is to paint Gabriel’s Horn. You are given a brush and as many cans of paint as you want. But before
you start, it occurs to you that you might be wasting your time. After all, the horn is unbounded. You don’t
want to spend hours and hours on a job that might never be finished!
Here’s an idea: calculate the volume of Gabriel’s Horn. That is, find the volume for the solid obtained by
rotating the region bounded by f (x) = 1/x, x = 1, and y = 0 about the x-axis. If this volume is finite, you
could fill the horn with paint, shake it around a little, pour out the extra and be done with the job. (You are
allowed to waste paint in Heaven.)
So, can you paint the horn?
498

GROUP WORK 1, SECTION 9.2
Gabriel’s Horn
Sketch the graph of f (x) = 1/x from x = 1tox =∞. (Okay, you don’t actually have to graph it all the
way out to infinity.)
Now think of the object formed by rotating this curve about the x-axis. This object is called Gabriel’s Horn.
Your job is to paint Gabriel’s Horn. You are given a brush and as many cans of paint as you want. But before
you start, it occurs to you that you might be wasting your time. After all, the horn is unbounded. You don’t
want to spend hours and hours on a job that might never be finished!
Here’s an idea: examine the surface area of the inside of the horn. That is, look at the surface area of the solid
obtained by rotating the region bounded by f (x) = 1/x, x = 1, and y = 0 about the x-axis. If this surface
area is finite, then it can be covered with a coat of paint. Use the Comparison Theorem to help you determine
if it is finite. (The thickness of the paint is irrelevant; this is, after all, Heaven.)
So, can you paint the horn?
499

GROUP WORK 2, SECTION 9.2
Mind Your p’s and q’s
1. Show that if p > 1, then the surface formed by rotating y = 1/x p , x ≥ 1 about the x-axis has finite area.
2. Show that if 0 < q < 1, then the surface formed by rotating y = 1/x q , x ≥ 1 about the y-axis does not
have finite area.
500

DISCOVERY PROJECT
Rotating on a Slant
This project tests the students’ understanding of the mathematics behind the volume and surface area formulas.
As an extended project, it would make a good challenge for an advanced group of students. Problems 1 and 2
could be assigned as an in-class exercise, while the later problems will require a good deal of time and effort.
Be prepared to give assistance where needed.
501

9.3 APPLICATIONS TO PHYSICS AND ENGINEERING
SUGGESTED TIME AND EMPHASIS
1
2
–1 class Optional material
POINTS TO STRESS
1. Two applications (hydrostatic pressure; moments and centers of mass) are discussed in the text. It is
recommended that the instructor stress one, and perhaps touch on the other.
2. Centroids and the symmetry principle, if moments and centers of mass are covered.
QUIZ QUESTIONS
• Text Question: Both applications in the text start by a summation of discrete or sampled data, and
eventually replace the summation by an integral. We have seen this technique earlier in the course. Name
two previous occasions where we’ve used the technique of changing the sum of sample values into an
integral of a continuous function.
Answer: Answers include finding the area under a curve, finding average values of a function, and finding
volumes.
• Drill Question: Set up, but do not evaluate, an integral or integrals that will give the y-coordinate of the
centroid of the region sketched below.
y
4
3
y=4-x@
2
1
_2 _1 0 1 2 x
_1
Answer: y =
∫ 2
−2 1 2
( 4 − x
2 ) 2 dx
∫ 2 (
−2 4 − x 2 ) dx
MATERIALS FOR LECTURE AND WORKSHOP/DISCUSSION
Lecture coverage will, of course, be determined by which applications are to be emphasized. I recommend
an in-depth treatment of one of the two applications in lecture, and touching on the other in discussion.
Throughout, emphasize problem solving skills such as drawing a picture and setting up the problem carefully.
• Hydrostatic Pressure and Force: Work through Exercise 14. This problem requires resolving a force on a
dam with inclined sides.
502

SECTION 9.3
APPLICATIONS TO PHYSICS AND ENGINEERING
• Center of Mass: Model the state of Minnesota as shown below, and compute the center of mass of the
state. (Many other states can be modeled similarly.)
• If centers of mass are covered, state the Theorem of Pappus and use it to compute the volume of the solid
generated when a equilateral triangular plate of constant density and side length r with vertices at (0, 0),
)
(r, 0),and(
12
r, is rotated about the y-axis. Note that by the symmetry principle, the centroid lies on
the line x = 1 2 r.
√
3
2
r
• Find the centroid (x, y) of the region R between y = x 4 and y = x 6 for 0 ≤ x ≤ 1.
• Show how the center of mass of a “one-dimensional” object, such as a yardstick with various weights
taped to it, can be found experimentally: Start with one finger at each end, and slowly draw them together.
See how close the experimental result is to the mathematical prediction.
GROUP WORK 1 (Center of Mass): The Floating Center
A CAS can be used in lieu of a graphing calculator. You may need to give the students the hint that ∫ sec xdx=
ln |sec x + tan x| + C.
Answers:
1. ȳ = 1 ∫ ( )
1/2
2 −1/2 sec2 xdx= tan 12
≈ 0. 546
3. If the region were mounted on a light sheet of plastic, it
would balance at that point.
4. The limit is infinity. Physically, this means that the region
(as ε → 0) has mass approaching ∞ above any horizontal
line y = k.
2.
y
100 (0, 100)
4
(0, 3.9)
y=sec x
2
¹ _2 _
· 0
· ¹ _2
x
503

CHAPTER 9
FURTHER APPLICATIONS OF INTEGRATION
GROUP WORK 2 (Hydrostatic Pressure): Under Pressure
HOMEWORK PROBLEMS
Core Exercises: 1, 4, 13, 18, 19, 21, 26, 31, 35, 41
Sample Assignment: 1, 4, 10, 13, 14, 17, 18, 19, 21, 24, 26, 27, 31, 32, 35, 36, 38, 41, 44
Exercise D A N G
1 × ×
4 × ×
10 × ×
13 ×
14 ×
17 ×
18 ×
19 × ×
21 × ×
24 ×
26 × ×
27 × ×
31 ×
32 ×
35 × ×
36 ×
38 × ×
41 × ×
44 ×
504

GROUP WORK 1, SECTION 9.3
TheFloatingCenter
1. Consider the region bounded by y = sec x, x =− 1 2 , x = 1 2 ,andthex-axis.
y
2
y=sec x
_1
0
1
x
Computethecenterofmass,anddrawitonthefigure.(Hint: x = 0 by symmetry. y can be computed by
evaluating a relatively simple integral.)
2. Now consider the region defined by y = sec x, x =− π 2 + ε, x = π 2
− ε, and the x-axis.
y
4
y=sec x
2
¹ _2 _
· 0
· ¹ _2
x
Draw in the centers of mass for ε = 0.25 and for ε = 0.01.
3. Some define the center of mass to be the point where a region will balance. What does your second result
mean physically in light of the above definition?
4. Make a conjecture about lim
ε→0
y. What does this mean, physically?
505

GROUP WORK 2, SECTION 9.3
Under Pressure
An irregular solid with a circular top plate of diameter 4 feet and four elliptical side plates of height 6 feet and
width 4 feet is placed on the bottom of a pond that is 12 feet deep.
2
3
2
1. Find the hydrostatic pressure on the top plate.
2. Find the hydrostatic pressure on one of the elliptical plates.
506

DISCOVERY PROJECT
Complementary Coffee Cups
Students might miss the point that k is a parameter, and conclude that cup A (the fat one) holds more liquid.
It might be best to ask the students to describe cup B for a very large value of k, so they can see that it is
possible for cup B to have the largest capacity.
It should be emphasized that it does not suffice to answer Problem 1 by choosing a specific function f (y),
although that may be a first step in understanding the problem. The idea is to get a general solution.
If cups can be found that physically resemble cups A and B, students can test their answer to Problem 4
directly.
507

9.4 APPLICATIONS TO ECONOMICS AND BIOLOGY
SUGGESTED TIME AND EMPHASIS
Sections 9.2–9.5 are not covered directly on the Calculus AB exam. However, a major part of the study of
calculus is applications. Something would be lost if students left a calculus course without a good idea of the
variety of ways in which their knowledge could be applied. It is therefore recommended that some time be
spent on a subset of these sections, either immediately after covering Section 9.1, or during the interval of
time between the AP exam and the end of the course.
POINTS TO STRESS
Three applications (consumer surplus, blood flow, and cardiac output) are discussed in the text. It is recommended
that the instructor choose either consumer surplus or blood flow to stress in lecture, and perhaps touch
on the other two examples in workshop.
QUIZ QUESTIONS
• Text Question: What is the meaning of the term “flux”?
Answer: In the context of blood flow, it refers to the volume per unit time of blood that passes through a
cross-section of a blood vessel.
• Drill Question: The marginal cost of producing x cases of olive oil is 74+1.1x−0.002x 2 +0.00004x 3 (in
dollars per case). Find the increase in cost if the production level is raised from 1200 cases to 1600 cases.
Answer: $43,866,933.33
MATERIALS FOR LECTURE AND WORKSHOP/DISCUSSION
• Provide a complete description of one of the three examples in the text. Touch on the other two examples.
• Surplus: Bring a quantity of dice, stickers, or any other inexpensive product to class, one for every student.
Secretly determine your selling price P. Have the students write down the most that they would pay for
one of the objects. (In a smaller class, give a definite range of prices.) For example, if Brandon’s price
is lower than your secret price P, no transaction will take place. If his price is higher than P, hewillbuy
the object. Daniel’s individual savings is the difference between the price he was willing to pay and P. In
the example below, his savings would be 50 − 35 = 15 cents. The consumer surplus for the class will be
the sum of these savings. Construct a demand curve p (x) by tallying their bids in decreasing order and
computing the consumer surplus numerically using a step function. Discuss why we would model p (x)
as a continuous function in some cases.
Example: Suppose you bring in your collection of McGovern/Eagleton buttons, and you are willing to part
with them for 35 cents each. Assume your students wrote down the most they would pay for one:
Michael 0.15 Daniel 0.50 Christopher 1.50 Jessica 0.25 Ashley 0.35
Joshua 0.10 Samantha 0.80 Amanda 0.40 Brittany 0.60 Matthew 1.25
508

SECTION 9.4
APPLICATIONS TO ECONOMICS AND BIOLOGY
Then p (1) = 1.50, p (2) = 1.25, and the demand curve would look like this:
p(x)
1.60
1.40
1.20
1.00
p(x)
1.60
1.40
1.20
1.00
Consumer Surplus=$2.95
0.80
0.60
0.40
0.20
Demand Curve
0.80
0.60
0.40
0.20
0 2 4 6 8 10 x
0 2 4 6 8 10
x
• Blood Flow: Bring a turkey baster and several sizes of rubber tubing. Demonstrate, with the aid of
volunteers (and a bucket of colored water), how pressure and flow change with different sizes of tubing.
GROUP WORK: Homer’s Blood
This is an elaboration of Exercise 14 from the text.
Answers:
1. As the heart pumps, the total amount of blood moving through the body isn’t going to change significantly,
and the rate at which it moves isn’t going to change significantly.
2. P = P 0 R0
4
R 4
3. (a) 1.22P 0 (b) 1.52P 0 (c) 2.44P 0 (d) 7.72P 0
4. After about 15.910 years
HOMEWORK PROBLEMS
Core Exercises: 1, 4, 7, 9, 14, 18
Sample Assignment: 1, 4, 7, 9, 10, 12, 14, 15, 18, 19
Exercise D A N G
1 × ×
4 × ×
7 ×
9 × ×
10 × ×
12 ×
14 × ×
15 ×
18 × ×
19 × ×
509

GROUP WORK, SECTION 9.4
Homer’s Blood
The volume of blood that passes through a cross-section of an artery (of radius R, with pressure difference
P) per unit time is given by the formula
F = πPR4
8ηl
(l is the length of the artery and η is viscosity of the blood — both are assumed to be constant).
F is often called or defined as the flux. High blood pressure results from constriction of the arteries. As the
radius of the arteries shrinks, the blood pressure needs to increase to maintain a constant flux.
1. Why should we expect the flux to remain constant?
2. Suppose P 0 and R 0 are the initial radius and the initial pressure. Then the flux is πP 0R0
4 . If the pressure
8ηl
and radius at some other time are P and R, then the flux is πPR4 . Using the fact that flux must remain
8ηl
constant, determine an expression for P in terms of R, R 0 ,andP 0 .
(
3. As the years go by, the diameter of Homer’s arteries is given by R (t) = R 0 1 − t )
,wheret is the
100
number of years from the present. What is Homer’s blood pressure in terms of his original blood pressure
(a) 5 years from now?
(b) 10 years from now?
(c) 20 years from now?
(d) 40 years from now?
4. Homer’s doctor would like to perform angioplasty to enlarge the arteries when his blood pressure has
doubled. When will this happen?
510

9.5 PROBABILITY
TRANSPARENCY AVAILABLE
#18 (Figure 5)
SUGGESTED TIME AND EMPHASIS
1 class Optional material
POINTS TO STRESS
1. The definition of a probability density function, and how it can be used to determine P (a ≤ x ≤ b).
2. The definition of a mean, algebraically and geometrically.
3. The uses of the normal distribution.
QUIZ QUESTIONS
• Text Question: Estimate the probability that a North American male is precisely six feet tall.
Answer: 0
• Drill Question: Assume that the probability density function of a continuous random variable X is given
1
by f (X) =
π ( 1 + X 2). What is the probability that 0 < X < √ 3?
Answer: 1 π
∫ √ 3
0
dx
1 + x 2 = 1 3
MATERIALS FOR LECTURE
• Describe Buffon’s needle problem as follows (see Problem 11 in Problems Plus after this chapter):
In a famous 18th-century problem, known as Buffon’s needle problem, a needle of length h is dropped
onto a flat surface (for example, a table) on which parallel lines L units apart, L ≥ h, have been drawn.
The problem is to determine the probability that the needle will come to rest intersecting one of the lines.
Assume that the lines run east-west, parallel to the x-axis in a rectangular coordinate system (as in the
figure below).
L
y
h
¬
h sin ¬
y
L
h
y=h sin ¬
¹ _2 ¹
¬
Let y be the distance from the “southern” end of the needle to the nearest line to the north. (If the needle’s
southern end lies on a line, let y = 0. If the needle happens to lie precisely east-west, let the “western”
end be the “southern” end.) Then 0 ≤ y ≤ L and 0 ≤ θ ≤ π. Note that the needle intersects one of the
lines only when y < h sin θ. Now, the total set of possibilities for the needle can be identified with the
rectangular region 0 ≤ y ≤ L, 0≤ θ ≤ π, and the proportion of times that the needle intersects a line is
511

CHAPTER 9
FURTHER APPLICATIONS OF INTEGRATION
area under y = h sin θ
the ratio . This ratio is the probability that the needle intersects a line.
area of rectangle
Demonstrate one case, such as L = 1, h = 1 2
. If there is time, also consider the case L = h.
1
• Prove that this version of the Witch of Agnesi, f (x) =
π [ (x − 1) 2 ], is a probability density function.
+ 1
Calculate the mean and median (m = μ = 1).
Answer: 1 ∫ ∞
dx
lim arctan (x − 1) − lim arctan (x − 1)
π (x − 1) 2 + 1 = m→∞ n→−∞
= 1
π
−∞
WORKSHOP/DISCUSSION
• Describe a probability density function in detail, such as the one in Example 3, including its mean μ.
• Describe the median m geometrically (half the area is to the left, half is to the right) and algebraically
[ ∫ ∞
m
f (x) dx = ∫ m
−∞ f (x) dx = 1 2].
GROUP WORK 1: Foretelling the Future
This is an exercise that will enable students to discover the link between abstract probability density functions
and concrete random events.
First have the students graph the function f (x) = √ 1
( )
(x − 7)2
exp − on their calculators, using the
12π 12
range 0 ≤ x ≤ 13. Once they have this graph, abruptly change gears, merely promising that their graphs will
prove useful later in the exercise. Give a quick lecture on random events. Note that rolling a pair of dice will
give them a random number between 2 and 12. After this discussion, group the students into threes and fours,
and hand out (six-sided) dice, one pair per student. Hand out the density worksheet. While they are rolling
along, draw a sketch of f (x) on an obscure blackboard. Make sure that the students plot their own data first,
and then create an aggregate plot with their group’s data.
If one group finishes before the others, have them draw their histogram on the board. If there is still more
time, they can add data to their totals.
When everyone is done, put a class histogram on the board. Finally, have them numerically integrate f (x)
between 6.5 and 7.5, and notice that the answer ( 1 6
) is close to what they got as a class for the frequency of
7. Point out that each of their frequency values is similarly close to an area under f (x). Close by discussing
the strengths and weaknesses of the model. (One strength is that the probabilities can be calculated quickly,
without rolling thousands of dice. One weakness is that this continuous model isn’t completely accurate: the
model gives non-zero probabilities for rolling a 13, a 14, or even a 0!)
512

SECTION 9.5
PROBABILITY
GROUP WORK 2: The Wheel of Fate
There are two ways to look at Problems 1(b) and 2(b). The students can make a decision based on the chance
of having to pay over $1, or they can think about the expected value of a sundae (or a grade). Discuss both
approaches at the end.
Answers:
1. (a) 54% (b) 94.6 cents (c) See instructions; the answer is controversial.
2. (a) θ ≥ 150
(b) If θ = 100, there is a 39% chance of getting less than 80% on the test. The expected value is 80.97%.
The students can discuss whether or not the risk is worth it.
POSSIBLE PROJECT
Following the discussion of Buffon’s needle problem in workshop/discussion, Problem 11 of Problems Plus
would make a nice group project.
HOMEWORK PROBLEMS
Core Exercises: 2, 5, 8, 10, 13
Sample Assignment: 1, 2, 4, 5, 7, 8, 10, 11, 13, 15, 17, 19
Exercise D A N G
1 ×
2 ×
4 ×
5 ×
7 × ×
8 × × ×
10 × ×
11 × ×
13 × ×
15 × ×
17 ×
19 × × ×
513

GROUP WORK 1, SECTION 9.5
Foretelling the Future
In this exercise we will be rolling the laughing bones. For Parts 1 and 2, each group member should use his
or her own data. Then consolidate everyone’s data and do Step 3 as a group.
1. Roll a pair of dice ten times and write down the results, then fill out the table.
Roll 1:
Roll 2:
Roll 3:
Roll 4:
Roll 5:
Roll 6:
Roll 7:
Roll 8:
Roll 9:
Roll 10:
Number
2
3
4
5
6
7
8
9
10
11
Number of
Times Rolled
Frequency (decimal)
12
2. Now draw a density graph of your results. This will be a bar graph where the x-axis is the number shown
on the dice, and the y-axis is the frequency of that number.
514

Foretelling the Future
3. Now fill out the table and density graph again, this time for the combined data from everyone in your
group.
Number
Number of
Times Rolled
Frequency (decimal)
2
3
4
5
6
7
8
9
10
11
12
515

GROUP WORK 2, SECTION 9.5
The Wheel of Fate
1. In Westbury, New York, there used to be an ice-cream store that offered its young customers a special
deal. They could buy a hot-fudge sundae for $1.00, or spin the Sundae Wheel and pay the price upon
which the spinner stopped. The Sundae Wheel (and the store that housed it) no longer exist, but the author
remembers it as looking something like this:
(a) What is the probability of a sundae costing over $1.00?
(b) What is the expected value for the cost of a sundae?
(c) In light of parts (a) and (b), would you rather pay a dollar for the sundae, or choose to spin the wheel?
Why?
516

The Wheel of Fate
2. One young person who loved eating sundaes at the store grew up and became a math professor. He had
the idea of allowing students to take a certain quiz, or (instead) opt to spin...
the Wheel of Fate!
(While this idea was still in its planning stages, it was vetoed by the department head.)
Now, assume that you think you can score 80% on the quiz.
(a) What angle would θ have to be in order to give you at least a 75% chance of spinning a grade of at
least 80%?
(b) If θ was 100 ◦ , would you take the quiz or spin? Why? Which would give you the best score, on
average?
517

9 SAMPLE EXAM
Problems marked with an asterisk (*) are particularly challenging and should be given careful consideration.
1. Let f be a differentiable function defined for x ≥ 0 satisfying f (0) = 0.
(a) Write an equation for the arc length function s (x) of the graph of f from (0, 0) to (x, f (x)).
(b) Find the function f (x) satisfying f (0) = 0 whose arc length function is s (x) = (x + 1) 3/2 .
2. Consider the solids of revolution formed by rotating f (x) = cos x, − π 2 ≤ x ≤ π 2
0 ≤ x ≤ π, about the x-axis.
and g (x) = sin x,
(a) Why do these two solids have equal surface area?
(b) Set up an integral to compute the surface area of one of these solids.
[
w √ 1 + w 2 + ln
(c) Given the formula ∫ √ 1 + w 2 dw = 1 2
setupinpart(b).
3. Consider the following region R.
(w + √ 1 + w 2 )]
, compute the integral you
Find an equation for the perimeter P of the region R. Do not evaluate your equation.
4. A well-known formula states that the surface area of a sphere is 4 times the area enclosed by an equatorial
circle.
(a) What is the area of an equatorial circle of a sphere of radius R?
(b) Express the above claim in terms of your answer to part (a).
(c) Using an integral to compute the surface area of a solid of revolution, show that the equation you
wrote in part (b) is true..
*5. Let f and g be differentiable functions such that 2 ≤ f ′ (x) ≤ 4 and 2 ≤ g (x) ≤ 4 for all x.
(a) Find good upper and lower bounds on the arc length of the graph of f (x) from x = 0tox = 4.
(b) Can we find a good lower bound on the length of g (x) from x = 0tox = 4? Why or why not? If
so, give a bound.
(c) Now suppose we know that 1 ≤ f (x) ≤ 3and−1 ≤ f ′ (x) ≤ 1ontheinterval[0, 4]. Find good
upper and lower bounds for the area of the surface generated between x = 0andx = 4whenthe
graph of f (x) is rotated about the x-axis.
518

CHAPTER 9
SAMPLE EXAM
6. Let f (x) = 3√ x. We wish to put upper and lower bounds on the length of f (x) from x = 0tox = 1.
(a) Write an integral that, when evaluated, will give us the length that we want.
(b) Explain why the integral you wrote in part (a) is improper.
(c) Give a geometrical reason why the integral in part (a) converges.
7. An airline notes that if the price of an 8 A.M. round trip ticket from Boston to New York is set at $250,
they can sell 1000 tickets. For every $50 increase in price, they can sell 100 fewer tickets. For every $50
decrease in price, they can sell 100 more, up to their capacity of 2000 seats.
(a) Find the demand function.
(b) If they go ahead and charge $250, what is the total consumer surplus?
(c) What does the quantity you found in part (b) represent?
(d) How many people are willing to pay $500?
* (e) It annoys the airline that there are people willing to pay $500, but only paying $250. If they set the
price at $500, sell all the tickets they can at that price, and then have a “sale”, changing the price of
tickets to $250, would the consumer surplus increase or decrease? Why?
8. Which of the following can be probability density functions? Why or why not?
(a) f (x) =
1
π ( 1 + x 2)
(b) g (x) = 2
1 + e x
{ 12
e x if x < 0
(c) h (x) =
1
2 e−x if x ≥ 0
9. Let R be the region shaded below.
(a) Find the x-coordinate of the centroid of R without integrating. Explain why your answer is correct.
(b) Find the y-coordinate of the centroid of R by evaluating the appropriate integral.
519

10. Let R be a thin plate whose cross-section is shown below.
Assume that R has uniform density. Find the coordinates of the center of mass of R.
11. Let R and S be the regions sketched below:
y
y
f(x)
g(x)=f(x+4)
R
S
a
b
x
a+4 b+4
x
(a) Write integrals for the moments M x about the x-axis of R and S. (Assume that both regions have the
same uniform density.)
(b) Give a physical reason why the two moments are the same.
(c) Use substitution to show that the two moments are the same.
(d) Which of the moments M y about the y-axis will be larger? Explain your answer.
1. (a) s (x) = ∫ x
0
9 SAMPLE EXAM SOLUTIONS
√
1 + [ f ′ (t) ] 2 dt
(b) Setting ∫ √
x
0
1 + [ f ′ (t) ] √
2 dt = (x + 1) 3/2 and differentiating, we solve for f ′ 9
(x) =
4 x + 5 4
and so
( ) 3/2 ( 3/2
f (x) =
27
8 94
x + 5 4 +C. Since f (0) = 0, the correct answer is f (x) =
8 94
27
x + 4) 5 5 3/2
−
27 .
2. (a) The surface areas are equal because the graph of g (x) = sin x is just the graph of f (x) = cos x
shifted to the right by π 2 .
(b) S = ∫ π
0 2π sin x√ 1 + cos 2 xdx
520

CHAPTER 9
SAMPLE EXAM SOLUTIONS
3.
(c) Substitute u = cos x. Thendu =−sin xdx,so
S = − ∫ √
−1
1
2π 1 + u 2 du = 2π ∫ √
1
−1
1 + u 2 du
[ √
= 2π u 1 + u 2 + ln
y
1
0
y=Ïx
R
LÁ
y=x@
Lª
1 x
( √ )] 1
u 1 + u 2 = 2√ 2π + π ln
−1
P = L 1 + L 2 =
∫ 1
0
√
1 +
( 1
2 √ x
( √ )
√ 2+1
2−1
) 2 ∫ 1 √
dx + 1 + (2x) 2 dx
4. (a) The area enclosed by an equatorial circle is πR 2 . A sphere of radius R is formed by rotating the curve
y = √ R 2 − x 2 about the x-axis.
(b) The surface area is S =
(c)
∫ R
−R
∫ R
−R
∫ R
−R
2π √ ( )
R 2 − x
√1 2 −x 2
+ √ dx, so the claim means that
R 2 − x 2
2π √ ( )
R 2 − x
√1 2 −x 2
+ √ dx = 4πR 2 .
R 2 − x 2
2π √ ( )
R 2 − x
√1 2 −x 2
+ √ dx =
R 2 − x 2
∫ R
−R
2π √ R 2 − x 2 √
R 2
R 2 − x 2 dx
= ∫ R
−R2πRdx = 4πR2
5. (a) The arc length L = ∫ √
4
0
1 + [ f ′ (x) ] 2 dx,sosince2≤ f ′ (x) ≤ 4, we can say that 4 √ 5 ≤ L ≤
4 √ 17.
(b) Here L = ∫ 4
0
√1 + [ g ′ (x) ] √
2 dx ≥ 4, since 1 + [ g ′ (x) ] 2 ≥ 1forallx.
(c) The surface area S = ∫ 4
0
√1 2π f (x) + [ f ′ (x) ] 2 dx,sosince1≤ f (x) ≤ 3and−1 ≤ f ′ (x) ≤ 1
on the interval [0, 4], we can say 8π ≤ S ≤ 24 √ 2π.
6. (a) L = ∫ √
( )
1
2
0
1 + 13
x −2/3 dx
(b) The integral is improper because lim
interval of integration.
x→0 + √
1 +
(
13
x −2/3 ) 2
=∞. The integrand is unbounded on the
(c) Wecanseefromagraphof f (x) = 3√ x that the curve has finite length; therefore, the integral must
converge.
521
0

CHAPTER 9
FURTHER APPLICATIONS OF INTEGRATION
7. (a) P (x) = 1000 − x
2
(b) Consumer surplus = ∫ 1000
0
+ 250 = 750 − x 2
(750 − 1 2 x − 250 )
dx = $250,000
(c) The consumer surplus represents the total savings for customers who would have paid more than $250
for a ticket.
(d) Set P (x) = 750 − 1 2
x = 500 and solve for x: x = 500, so 500 people are willing to pay $500.
(e) The sale would cause the consumer surplus to decrease, since although the same number of tickets are
sold, many people are paying the higher price.
8. (a) f (x) can be a probability density function, since
∫ ∞
∫
1
∞
−∞ π ( 1 + x 2) dx = 2 1
0 π ( 2
1 + x2) dx = lim
t→∞ π arctan t = 1
∫ ∞
2
2
(b) g (x) cannot, since
dx diverges (for one thing, lim
−∞ 1 + ex x→−∞ 1 + e x = 2.)
(c) h (x) can, since ∫ ∞
−∞ h (x) dx = 2 ∫ ∞ 1
0 2
e −x (
dx = lim 1 − e
−t ) = 1.
t→∞
9. Let R 1 be the region above the x-axis and R 2 the region below.
(a) x = π 2
by symmetry.
(b) y = (area of R ∫
1 π
1)(y of R 1 ) + (area of R 2 )(y of R 2 ) 2 0
=
(sin x)2 dx − 1 ∫ π
2 0 (−2sinx)2 dx
∫
area of R
π
0 sin xdx+ ∫ π
0
[ ] 2sinxdx
π
− 3 12
2
x − 1 2 sin x cos x 0
=
[−3cosx] π =− π
0
8
10. Let R 1 be the rectangular region, R 2 be the semicircle, and R 3 be the triangular region.
(a) x = 2bysymmetry.
(b) y = (area of R 1)(y of R 1 ) + (area of R 2 )(y of R 2 ) + (area of R 3 )(y of R 3 )
area of R
5
2 + 1 ∫ 6
(√ ) 2 ∫
2 2 4 − (x − 4)
2 dx − 2 · 1 4
2 2 (4 − 2x)2 dx 5
2
=
=
+ 8 − 32
3
4 + 2π + 8
12 + 2π =− 1
72 + 12π
∫ b
f (x) 2
∫ b+4
f (x − 4) 2
11. (a) For R, M x = dx.ForS, M x =
dx.
a 2
a+4 2
(b) The moments about the x-axis should be the same because region S is just region R shifted 4 units to
the right. The distribution of the mass relative to the x-axis is unchanged.
∫ b
f (u) 2
(c) Let u = x − 4. Then du = dx and M x for S becomes du which is the same as M x for R.
a 2
(d) M y for S is larger than M y for R because region S is further from the y-axis than region R.
522