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ing from x = y to x = 3. Thus it has length 3 ’ y. Then the cylinder
generated when this segment (thickened to a strip of width y) is rotated
about the x-axis has volume

V (y) = 2πy · 3 ’ y y.

The aggregate volume is then


9
V= 2πy · 3 ’ y dy
0
9
= 2π · 3y ’ y 3/2 dy
0
9
3y 2 y 5/2
= 2π · ’ dy
2 5/2 0
231
CHAPTER 8 Applications of the Integral

243 2 · 243 00
= 2π · ’ ’ ’
2 5 25
243
= 2π ·
10
243π
= .
5

You Try It: Use the method of cylindrical shells to calculate the volume enclosed
when the region 0 ¤ y ¤ sin x, 0 ¤ x ¤ π/2, is rotated about the y-axis.


8.2.3 DIFFERENT AXES
Sometimes it is convenient to rotate a curve about some line other than the
coordinate axes. We now provide a couple of examples of that type of problem.

EXAMPLE 8.11
Use the method of washers to calculate the volume of the solid enclosed

when the curve y = x, 1 ¤ x ¤ 4, is rotated about the line y = ’1. See
Fig. 8.21.


y




y = √x


x




Fig. 8.21


SOLUTION
√ The key is to notice that, at position x, the segment to be rotated has height

x ’ (’1)”the distance from the point (x, x) on the curve to the line

y = ’1. Thus the disk generated has area A(x) = π · ( x + 1)2 . The resulting
232 CHAPTER 8 Applications of the Integral

aggregate volume is

4
2
V= π· x + 1 dx
1

4
=π x + 2 x + 1 dx
1
4
x 2 2x 3/2
=π + +x
2 3/2 1
2·8 12 2 · 1
42
=π· + +4 ’π · + +1
2 3/2 2 3/2
119
= π.
6

You Try It: Calculate the volume inside the surface generated when y = x +x
3

is rotated about the line y = ’3, 1 ¤ x ¤ 4.
EXAMPLE 8.12
Calculate the volume of the solid enclosed when the area between the
curves x = (y ’ 2)2 + 1 and x = ’(y ’ 2)2 + 9 is rotated about the line
y = ’2.

y




x = (y _ 2)2 + 1 x = _(y _ 2)2 + 9




x


Fig. 8.22


SOLUTION
Solving the equations simultaneously, we ¬nd that the points of intersec-
tion are (5, 0) and (5, 4). The region between the two curves is illustrated in
Fig. 8.22.
At height y, the horizontal segment that is to be rotated stretches from
((y ’ 2)2 + 1, y) to (’(y ’ 2)2 + 9, y). Thus the cylindrical shell that is
233
CHAPTER 8 Applications of the Integral

generated has radius y ’ 2, height 8 ’ 2(y ’ 2)2 , and thickness y. It therefore
generates the element of volume given by
2π · (y ’ 2) · [8 ’ 2(y ’ 2)2 ] · y.
The aggregate volume that we seek is therefore
4
V= 2π · (y ’ 2) · [8 ’ 2(y ’ 2)2 ] dy
0
4
= 16π(y ’ 2) ’ 4π(y ’ 2)3 dy
0
4
= 8π(y ’ 2)2 ’ π(y ’ 4)4 0
= 256π.
You Try It: Calculate the volume enclosed when the curve y = cos x is rotated
about the line y = 4, π ¤ x ¤ 3π .



8.3 Work
One of the basic principles of physics is that work performed is force times distance:
If you apply force F pounds in moving an object d feet, then the work is
W =F ·d foot-pounds.
The problem becomes more interesting (than simple arithmetic) if the force is
varying from point to point. We now consider some problems of that type.
EXAMPLE 8.13
A weight is pushed in the plane from x = 0 to x = 10. Because of a
prevailing wind, the force that must be applied at point x is F (x) = 3x 2 ’
x + 10 foot-pounds. What is the total work performed?




0 10

Fig. 8.23


SOLUTION
Following the way that we usually do things in calculus, we break the problem
up into pieces. In moving the object from position x to position x + x, the
234 CHAPTER 8 Applications of the Integral

distance moved is x feet and the force applied is about F (x) = 3x 2 ’ x + 10.
See Fig. 8.23. Thus work performed in that little bit of the move is w(x) =
(3x 2 ’ x + 10) · x. The aggregate of the work is obtained by summation. In
this instance, the integral is the appropriate device:
10
10 x2
W= 3x ’ x + 10 dx = x ’ + 10x = 1050 foot-pounds.
2 3
2
0 0
EXAMPLE 8.14
A man is carrying a 100 lb sack of sand up a 20-foot ladder at the rate of
5 feet per minute. The sack has a hole in it and sand leaks out continuously
at a rate of 4 lb per minute. How much work does the man do in carrying
the sack?




Fig. 8.24

SOLUTION
It takes four minutes for the man to climb the ladder. At time t, the sack has
100 ’ 4t pounds of sand in it. From time t to time t + t, the man moves
5 · t feet up the ladder. He therefore performs about w(t) = (100 ’ 4t) · 5 t
foot-pounds of work. See Fig. 8.24. The total work is then the integral
4
4
W= (100 ’ 4t) 5 dt = 500t ’ 10t = 1840 foot-pounds.
2
0 0
235
CHAPTER 8 Applications of the Integral

You Try It: A man drags a 100 pound weight from x = 0 to x = 300. He resists
a wind which at position x applies a force of magnitude F (x) = x 3 + x + 40. How
much work does he perform?
EXAMPLE 8.15
According to Hooke™s Law, the amount of force exerted by a spring is propor-
tional to the distance of its displacement from the rest position.The constant
of proportionality is called the Hooke™s constant. A certain spring exerts a
force of 10 pounds when stretched 1/2 foot beyond its rest state. What is
the work performed in stretching the spring from rest to 1/3 foot beyond its
rest length?




Fig. 8.25


SOLUTION
Let the x-variable denote the position of the right end of the spring (Fig. 8.25),
with x = 0 the rest position. The left end of the spring is pinned down. Imagine
that the spring is being stretched to the right. We know that the force exerted
by the spring has the form
F (x) = kx,
with k a negative constant (since the spring will pull to the left). Also F (0.5) =
’10. It follows that k = ’20, so that
F (x) = ’20x.
Now the work done in moving the spring from position x to position x + x
will be about (20x) · x (the sign is + since we will pull the spring to the
right”against the pull of the spring). Thus the total work done in stretching
the right end of the spring from x = 0 to x = 1/3 is
1/3
1/3 10
W= (20x) dx = 10x =
2
foot-pounds.
9
0 0
236 CHAPTER 8 Applications of the Integral

EXAMPLE 8.16
Imagine that a water tank in the shape of a hemisphere of radius 10 feet
is being pumped out (Fig. 8.26). Find the work done in lowering the water
level from 1 foot from the top of the tank to 3 feet from the top of the tank.


Radius at depth x
equals √100 _ x 2




Fig. 8.26


SOLUTION
A glance at Fig. 8.27 shows that the horizontal planar slice of the tank, at the

level x feet below the top, is a disk of radius 100 ’ x 2 . This disk therefore
has area A(x) = π · (100 ’ x 2 ). Thus a slice at that level of thickness x will
have volume
V (x) = π · (100 ’ x 2 ) · x

100
x

√100 _ x 2




Fig. 8.27

and (assuming that water weights 62.4 pounds per cubic foot) weight equal to
w(x) = 62.4π · (100 ’ x 2 ) · x.
Thus the work in raising this slice to the top of the tank (where it can then
be dumped) is
W (x) = 62.4π · (100 ’ x 2 ) · x · x foot-pounds.
237
CHAPTER 8 Applications of the Integral

We calculate the total work by adding all these elements together using an
integral. The result is
3
W= 62.4π · (100 ’ x 2 ) · x dx
1
3
= 62.4π · 100x ’ x 3 dx
1
3
x4
= 62.4π 50x ’ 2
41
81 1
= 62.4π 450 ’ ’ 50 ’
4 4
= 23,712π foot-pounds.
You Try It: A spring has Hooke™s constant 5. How much work is performed in
stretching the spring half a foot from its rest position?



8.4 Averages
In ordinary conversation, when we average a collection p1 , . . . , pk of k numbers,
we add them together and divide by the number of items:
p1 + · · · + pk
σ = Average = .
k
The signi¬cance of the number σ is that if we wanted all the k numbers to be equal,
but for the total to be the same, then that common value would have to be σ .
Now suppose that we want to average a continuous function f over an interval
[a, b] of its domain. We can partition the interval,
P = {x0 , x1 , . . . , xk },
with x0 = a and xk = b as usual. We assume that this is a uniform partition, with
xj ’ xj ’1 = x = (b ’ a)/k for all j . Then an “approximate average” of f would
be given by
f (x1 ) + f (x2 ) + · · · + f (xk )
σapp = .
k
It is convenient to write this expression as
k k
b’a
1 1
= f (xj ) · = f (xj ) ·
σapp x.
b’a b’a
k
j =1 j =1
238 CHAPTER 8 Applications of the Integral

b
This last is a Riemann sum for the integral (1/[b ’ a]) · Thus, letting
a f (x) dx.
the mesh of the partition go to zero, we declare
b
1
average of f = σ = f (x) dx.
b’a a

EXAMPLE 8.17
In a tropical rain forest, the rainfall at time t is given by •(t) = 0.1 ’ 0.1t +
0.05t 2 inches per hour, 0 ¤ t ¤ 10. What is the average rainfall for times
0 ¤ t ¤ 6?
SOLUTION
We need only average the function •:
6
1
average rainfall = σ = •(t) dt
6’0 0
16
= 0.1 ’ 0.1t + 0.05t 2 dt

Y
60
FL
6
1 0.05 3
= 0.1t ’ 0.05t 2 + t
6 3 0
AM

= 0.1 ’ 0.3 + 0.6
= 0.4 inches per hour.
TE



EXAMPLE 8.18
Let f (x) = x/2 ’ sin x on the interval [’2, 5]. Compare the average value
of this function on the interval with its minimum and maximum.
SOLUTION
Observe that
1
f (x) = ’ cos x.
2
Thus the critical points occur when cos x = 1/2, or at ’π/3, π/3. We also
must consider the endpoints ’2, 5. The values at these points are
f (’2) = ’1 + sin 2 ≈ ’0.0907026

3
π
f (’π/3) = ’ + ≈ 0.3424266
6 √2
3
π
f (π/3) = ’ ≈ ’0.3424266
6 2
5
f (5) = ’ sin 5 ≈ 3.458924.
2
239
CHAPTER 8 Applications of the Integral

Plainly, the maximum value is f (5) = 5/2 ’ sin 5 ≈ 3.458924. The minimum
value is f (π/3) ≈ ’0.3424266.
The average value of our function is
5
1 x
σ= ’ sin x dx
5 ’ (’2) 2
’2
5
1 x2
= + cos x
74 ’2

1 25 4
= + cos 5 ’ + cos 2
7 4 4
1 21
= + cos 5 ’ cos 2
74
≈ 0.84997.

You can see that the average value lies between the maximum and the minimum,
as it should. This is an instance of a general phenomenon.

You Try It: On a certain tree line, the height of trees at position x is about 100 ’
3x + sin 5x. What is the average height of trees from x = 2 to x = 200?

EXAMPLE 8.19
What is the average value of the function g(x) = sin x over the interval
[0, 2π ]?

SOLUTION
We calculate that



1 1 1
σ= sin x dx = [’ cos x] = [’1 ’ (’1)] = 0.
2π ’ 0 2π 2π
0 0

We see that this answer is consistent with our intuition: the function g(x) =
sin x takes positive values and negative values with equal weight over the
interval [0, 2π]. The average is intuitively equal to zero. And that is the actual
computed value.

You Try It: Give an example of a function on the real line whose average over
every interval of length 4 is 0.
240 CHAPTER 8 Applications of the Integral

8.5 Arc Length and Surface Area
Just as the integral may be used to calculate planar area and spatial volume, so this
tool may also be used to calculate the arc length of a curve and surface area. The
basic idea is to approximate the length of a curve by the length of its piecewise
linear approximation. A similar comment applies to the surface area. We begin by
describing the basic rubric.

8.5.1 ARC LENGTH
Suppose that f (x) is a function on the interval [a, b]. Let us see how to calculate
the length of the curve consisting of the graph of f over this interval (Fig. 8.28).
We partition the interval:
a = x0 ¤ x1 ¤ x2 ¤ · · · ¤ xk’1 ¤ xk = b.
Look at Fig. 8.29. Corresponding to each pair of points xj ’1 , xj in the partition
is a segment connecting two points on the curve; the segment has endpoints
(xj ’1 , f (xj ’1 )) and (xj , f (xj )). The length j of this segment is given by the



y = f (x)




a b


Fig. 8.28

y = f (x)




(xj , f (xj))

(xj _ 1, f (xj _ 1))




xj “1 xj

Fig. 8.29
241
CHAPTER 8 Applications of the Integral

usual planar distance formula:
1/2
= [xj ’ xj ’1 ]2 + [f (xj ) ’ f (xj ’1 )]2 .
j

We denote the quantity xj ’ xj ’1 by x and apply the de¬nition of the derivative
to obtain
f (xj ) ’ f (xj ’1 )
≈ f (xj ).
x

Now we may rewrite the formula for as
j

≈ ([ x]2 + [f (xj ) x]2 )1/2
j
= (1 + [f (xj )]2 )1/2 x.

Summing up the lengths (Fig. 8.30) gives an approximate length for the curve:
j

k k
length of curve ≈ = (1 + [f (xj )]2 )1/2 x.
j
j =1 j =1




Fig. 8.30

But this last is a Riemann sum for the integral
b
= (1 + [f (x)]2 )1/2 dx. ()
a

As the mesh of the partition becomes ¬ner, the approximating sum is ever closer to
what we think of as the length of the curve, and it also converges to this integral.
Thus the integral represents the length of the curve.

EXAMPLE 8.20
Let us calculate the arc length of the graph of f (x) = 4x 3/2 over the interval
[0, 3].
242 CHAPTER 8 Applications of the Integral

SOLUTION
The length is
3 3
(1 + [f (x)] ) dx = (1 + [6x 1/2 ]2 )1/2 dx
2 1/2
0 0
3
= (1 + 36x)1/2 dx
0
3
1
= · (1 + 36x) 3/2
54 0
1
= [1093/2 ’ 13/2 ]
54
(109)3/2 ’ 1
= .
54
EXAMPLE 8.21
Let us calculate the length of the graph of the function f (x) = (1/2)—
(ex + e’x ) over the interval [1, ln 8].

SOLUTION
We calculate that

f (x) = (1/2)(ex ’ e’x ).
Therefore the length of the curve is
ln 8
1 + [(1/2)(ex ’ e’x )]2
1/2
dx
1
1/2
1 e’2x
ln 8 e2x
= ++ dx
4 2 4
1
1 ln 8 x
e + e’x dx
=
21
1
= [ex ’ e’x ]ln 8
1
2
63 e 1
= ’+.
16 2 2e
You Try It: Set up, but do not evaluate, the integral for the arc length of the graph

of y = sin x on the interval π/4 ¤ x ¤ 3π/4.
Sometimes an arc length problem is more conveniently solved if we think of the
curve as being the graph of x = g(y). Here is an example.
243
CHAPTER 8 Applications of the Integral

EXAMPLE 8.22
Calculate the length of that portion of the graph of the curve 16x 2 = 9y 3
between the points (0, 0) and (6, 4).

SOLUTION
We express the curve as
3
x = y 3/2 , 0 ¤ y ¤ 4.
4
Then dx/dy = 9 y 1/2 . Now, reversing the roles of x and y in ( ), we ¬nd that
8
the requested length is
4 4
1 + [(9/8)y 1/2 ]2 dy = 1 + (81/64)y dy.
0 0

This integral is easily evaluated and we see that it has value [2 · (97)3/2 ’ 128]/
243.
Notice that the last example would have been considerably more dif¬cult (the
integral would have been harder to evaluate) had we expressed the curve in the
form y = f (x).

You Try It: Write the integral that represents the length of a semi-circle and
evaluate it.

8.5.2 SURFACE AREA
Let f (x) be a non-negative function on the interval [a, b]. Imagine rotating the
graph of f about the x-axis. This procedure will generate a surface of revolution,
as shown in Fig. 8.31. We will develop a procedure for determining the area of such
a surface.



y = f (x)




a b



Fig. 8.31
244 CHAPTER 8 Applications of the Integral

We partition the interval [a, b]:
a = x0 ¤ x1 ¤ x2 ¤ · · · ¤ xk’1 ¤ xk = b.
Corresponding to each pair of elements xj ’1 , xj in the partition is a portion of curve,


y = f (x)




a b




Fig. 8.32

as shown in Fig. 8.32. When that piece of curve is rotated about the x-axis, we obtain
a cylindrical surface. Now the area of a true right circular cylinder is 2π · r · h. We
do not have a true cylinder, so we proceed as follows. We may approximate the
radius by f (xj ). And the height of the cylinder can be approximated by the length
of the curve spanning the pair xj ’1 , xj . This length was determined above to be
about
1/2
1 + [f (xj )]2 xj .
Thus the area contribution of this cylindrical increment of our surface is about
1/2
2π · f (xj ) 1 + [f (xj )]2 xj .
See Fig. 8.33. If we sum up the area contribution from each subinterval of the
partition we obtain that the area of our surface of revolution is about
k
1/2
2π · f (xj ) 1 + [f (xj )]2 (—)
xj .
j =1

But this sum is also a Riemann sum for the integral
b
1/2
f (x) 1 + [f (x)]2
2π dx.
a

As the mesh of the partition gets ¬ner, the sum (—) more closely approximates
what we think of as the area of the surface, but it also converges to the integral.
245
CHAPTER 8 Applications of the Integral




,x
Fig. 8.33


We conclude that the integral
b
f (x)(1 + [f (x)]2 )1/2 dx

a
represents the area of the surface of revolution.
EXAMPLE 8.23
Let f (x) = 2x 3 . For 1 ¤ x ¤ 2 we rotate the graph of f about the x-axis.
Calculate the resulting surface area.

SOLUTION
According to our de¬nition, the area is
2
f (x)(1 + [f (x)]2 )1/2 dx

1
2
= 2π 2x 3 (1 + [6x 2 ]2 )1/2 dx
1
2 3
π
= (1 + 36x 4 )1/2 (144x 3 ) dx.
54 2
1

This integral is easily calculated using the u-substitution u = 36x 4 , du =
144x 3 dx. With this substitution the limits of integration become 36 and 576;
the area is thus equal to
576 3 576
π π
(1 + u)1/2 du = (1 + u)3/2
54 2 54 36
36
π
= [(577)3/2 ’ (37)3/2 ]
54
≈ 793.24866.
246 CHAPTER 8 Applications of the Integral

EXAMPLE 8.24
Find the surface area of a right circular cone with base of radius 4 and
height 8.

SOLUTION
It is convenient to think of such a cone as the surface obtained by rotating
the graph of f (x) = x/2, 0 ¤ x ¤ 8, about the x-axis (Fig. 8.34). According
to our de¬nition, the surface area of the cone is

8x 58
[1 + (1/2) ] dx = 2π
2 1/2
2π x dx
02 40

= 16 5π.

y




x




Fig. 8.34


You Try It: The standard formula for the surface area of a cone is

S = π r h2 + r 2 .
Derive this formula by the method of Example 8.24.
We may also consider the area of a surface obtained by rotating the graph of a
function about the y-axis. We do so by using y as the independent variable. Here
is an example:
EXAMPLE 8.25
Set up, but do not evaluate, the integral for ¬nding the area of the surface
obtained when the graph of f (x) = x 6 , 1 ¤ x ¤ 4, is rotated about the
y-axis.
247
CHAPTER 8 Applications of the Integral

SOLUTION
We think of the curve as the graph of φ(y) = y 1/6 , 1 ¤ y ¤ 4096. Then the
formula for surface area is
4096
1/2
φ(y) 1 + [φ (y)]2
2π dy.
1
Calculating φ (y) and substituting, we ¬nd that the desired surface area is the
value of the integral
4096
y 1/6 1 + (1/6)y ’5/6
2 1/2
2π dy.
1

You Try It: Write the integral that represents the surface area of a hemisphere of
radius one and evaluate it.



8.6 Hydrostatic Pressure
If a liquid sits in a tank, then it exerts force on the side of the tank. This force is
caused by gravity, and the greater the depth of the liquid then the greater the force.
Pascal™s principle asserts that the force exerted by a body of water depends on
depth alone, and is the same in all directions. Thus the force on a point in the side
of the tank is de¬ned to be the depth of the liquid at that point times the density of
the liquid. Naturally, if we want to design tanks which will not burst their seams, it
is important to be able to calculate this force precisely.




Fig. 8.35

Imagine a tank of liquid having density ρ pounds per cubic foot as shown in
Fig. 8.35. We want to calculate the force on one ¬‚at side wall of the tank. Thus
we will use the independent variable h to denote depth, measured down from the
surface of the water, and calculate the force on the wall of the tank between depths
h = a and h = b (Fig. 8.36). We partition the interval [a, b]:
a = h0 ¤ h1 ¤ h2 ¤ · · · ¤ hk’1 ¤ hk = b.
248 CHAPTER 8 Applications of the Integral

Assume that the width of the tank at depth h is w(h). The portion of the wall between
h = hj ’1 and h = hj is then approximated by a rectangle Rj of length w(hj ) and
width h = hj ’ hj ’1 (Fig. 8.37).




h=a
h=b




Fig. 8.36


w(hj)




Y ,h
FL
Fig. 8.37
AM

Now we have the following data:
Area of Rectangle = w(hj ) · h square feet
TE



Depth of Water ≈ hj feet
Density of Liquid = ρ pounds per cubic foot.
It follows that the force exerted on this thin portion of the wall is about
Pj = hj · ρ · w(hj ) · h.
Adding up the force on each Rj gives a total force of
k k
Pj = hj ρ w(hj ) h.
j =1 j =1

But this last expression is a Riemann sum for the integral
b
(—)
hρw(h) dh.
a

EXAMPLE 8.26
A swimming pool is rectangular in shape, with vertical sides. The bottom
of the pool has dimensions 10 feet by 20 feet and the depth of the water
249
CHAPTER 8 Applications of the Integral

is 8 feet. Refer to Fig. 8.38. The pool is full. Calculate the total force on one
of the long sides of the pool.


20
10
8




Fig. 8.38


SOLUTION
We let the independent variable h denote depth, measured vertically down
from the surface of the water. Since the pool is rectangular with vertical sides,
w(h) is constantly equal to 20 (because we are interested in the long side). We
use 62.4 pounds per cubic foot for the density of water. According to (—), the
total force on the long side is
8 8
h · 62.4 · w(h) dh = h · 62.4 · 20 dh = 39936 lbs.
0 0

You Try It: A tank full of water is in the shape of a cube of side 10 feet. How
much force is exerted against one wall of the tank between the depths of 3 feet and
6 feet?
EXAMPLE 8.27
A tank has vertical cross section in the shape of an inverted isosceles tri-
angle with horizontal base, as shown in Fig. 8.39. Notice that the base of the
tank has length 4 feet and the height is 9 feet. The tank is ¬lled with water to
a depth of 5 feet. Water has density 62.4 pounds per cubic foot. Calculate
the total force on one end of the tank.


4 ft


9 ft

5 ft


Fig. 8.39
250 CHAPTER 8 Applications of the Integral

SOLUTION
As shown in Fig. 8.40, at depth h (measured down from the surface of the
water), the tank has width corresponding to the base of an isosceles triangle
similar to the triangle describing the end of the tank. The height of this triangle
is 5 ’ h. Thus we can solve
4
w(h)
=.
5’h 9
We ¬nd that
4
w(h) = (5 ’ h).
9
According to (—), the total force on the side is then
5 4
h · 62.4 · (5 ’ h) dh ≈ 577.778 lbs.
9
0


4


9

5
5_h


Fig. 8.40


EXAMPLE 8.28
An aquarium tank is ¬lled with a mixture of water and algicide to keep the
liquid clear for viewing. The liquid has a density of 50 pounds per cubic foot.
For viewing purposes, a window is located in the side of the tank, with center
20 feet below the surface. The window is in the shape of a square of side

4 2 feet with vertical and horizontal diagonals (see Fig. 8.41). What is the
total force on this window?

SOLUTION
As usual, we measure depth downward from the surface with independent
variable h. Of course the square window has diagonal 4 feet. Then the range of
integration will be h = 20’4 = 16 to h = 20+4 = 24. Refer to Fig. 8.42. For
h between 16 and 20, we notice that the right triangle in Fig. 8.42 is isosceles
and hence has base of length h ’ 16. Therefore
w(h) = 2(h ’ 16) = 2h ’ 32.
251
CHAPTER 8 Applications of the Integral


4√2




Fig. 8.41

h = 16

4√2

h _ 16




h = 24

Fig. 8.42

According to our analysis, the total force on the upper half of the window is thus
20 44880
h · 50 · (2h ’ 32) dh = lbs.
3
16
For the lower half of the window, we examine the isosceles right triangle in
Fig. 8.43. It has base 24 ’ h. Therefore, for h ranging from 20 to 24, we have
w(h) = 2(24 ’ h) = 48 ’ 2h.
According to our analysis, the total force on the lower half of the window is
24 51200
h · 50 · (48 ’ 2h) dh = lbs.
3
20
The total force on the entire window is thus
44880 51200 96080
+ = lbs.
3 3 3
You Try It: A tank of water has ¬‚at sides. In one side, with center 4 feet below the
surface of the water, is a circular window of radius 1 foot. What is the total force
on the window?
252 CHAPTER 8 Applications of the Integral

h = 16

4√2




24 _ h


h = 24

Fig. 8.43




8.7 Numerical Methods of Integration
While there are many integrals that we can calculate explicitly, there are many
others that we cannot. For example, it is impossible to evaluate

e’x dx.
2
(—)

That is to say, it can be proved mathematically that no closed-form antiderivative can
be written down for the function e’x . Nevertheless, (—) is one of the most important
2


integrals in all of mathematics, for it is the Gaussian probability distribution integral
that plays such an important role in statistics and probability.
Thus we need other methods for getting our hands on the value of an integral.
One method would be to return to the original de¬nition, that is to the Riemann
sums. If we need to know the value of
1
e’x dx
2

0

then we can approximate this value by a Riemann sum
1
e’x dx ≈ e’(0.25) · 0.25 + e’(0.5) · 0.25 + e’(0.75) · 0.25 + e’1 · 0.25.
2 2 2 2 2

0

A more accurate approximation could be attained with a ¬ner approximation:
10
1
’x 2
e’(j ·0.1) · 0.1
2
dx ≈ (——)
e
0 j =1
253
CHAPTER 8 Applications of the Integral

or
100
1
’x 2
e’(j ·0.01) · 0.01
2
dx ≈ ()
e
0 j =1
The trouble with these “numerical approximations” is that they are calcula-
tionally expensive: the degree of accuracy achieved compared to the number of
calculations required is not attractive.
Fortunately, there are more accurate and more rapidly converging methods for
calculating integrals with numerical techniques. We shall explore some of these in
the present section.
It should be noted, and it is nearly obvious to say so, that the techniques of
this section require the use of a computer. While the Riemann sum (——) could be
computed by hand with some considerable effort, the Riemann sum ( ) is all but
infeasible to do by hand. Many times one wishes to approximate an integral by the
sum of a thousand terms (if, perhaps, ¬ve decimal places of accuracy are needed).
In such an instance, use of a high-speed digital computer is virtually mandatory.

8.7.1 THE TRAPEZOID RULE
The method of using Riemann sums to approximate an integral is sometimes called
“the method of rectangles.” It is adequate, but it does not converge very quickly
and it begs more ef¬cient methods. In this subsection we consider the method of
approximating by trapezoids.
Let f be a continuous function on an interval [a, b] and consider a partition
P = {x0 , x1 , . . . , xk } of the interval. As usual, we take x0 = a and xk = b. We also
assume that the partition is uniform.




Fig. 8.44
In the method of rectangles we consider a sum of the areas of rectangles.
Figure 8.44 shows one rectangle, how it approximates the curve, and what error
254 CHAPTER 8 Applications of the Integral

is made in this particular approximation. The rectangle gives rise to a “triangular”
error region (the difference between the true area under the curve and the area of the
rectangle). We put quotation marks around the word “triangular” since the region
in question is not a true triangle but instead is a sort of curvilinear triangle. If we
instead approximate by trapezoids, as in Fig. 8.45 (which, again, shows just one
region), then at least visually the errors seem to be much smaller.




Fig. 8.45
In fact, letting x = xj ’ xj ’1 as usual, we see that the ¬rst trapezoid in the
¬gure has area [f (x0 )+f (x1 )]· x/2. The second has area [f (x1 )+f (x2 )]· x/2,
and so forth. In sum, the aggregate of the areas of all the trapezoids is
1 1
· {f (x0 ) + f (x1 )} · x + · {f (x1 ) + f (x2 )} · x + ···
2 2
1
+ · {f (xk’1 ) + f (xk )} · x
2
x
= · {f (x0 ) + 2f (x1 ) + 2f (x2 )
2
+ · · · + 2f (xk’1 ) + f (xk )}. (†)
It is known that, if the second derivative of f on the interval [a, b] does not exceed
M then the approximation given by the sum (†) is accurate to within
M · (b ’ a)3
.
12k 2
[By contrast, the accuracy of the method of rectangles is generally not better than
N · (b ’ a)2
,
2k
where N is an upper bound for the ¬rst derivative of f . We see that the
method of trapezoids introduces an extra power of (b ’ a) in the numerator
255
CHAPTER 8 Applications of the Integral

of the error estimate and, perhaps more importantly, an extra factor of k in the
denominator.]
EXAMPLE 8.29
Calculate the integral
1
e’x dx
2

0
to two decimal places of accuracy.
SOLUTION
We ¬rst calculate that if f (x) = e’x then f (x) = (4x 2 ’ 2)e’x and
2 2


therefore |f (x)| ¤ 2 = M for 0 ¤ x ¤ 1. In order to control the error, and to
have two decimal places of accuracy, we need to have
M · (b ’ a)3
< 0.005
12k 2
or
2 · 13
< 0.005.
12k 2
Rearranging this inequality gives
100
< k2.
3
Obviously k = 6 will do.
So we will use the partition P = {0, 1/6, 1/3, 1/2, 2/3, 5/6, 1}. The corres-
ponding trapezoidal sum is
1/6 ’02
+ 2e’(1/6) + 2e’(1/3) + 2e’(1/2)
2 2 2
S= ·e
2
+ 2e’(2/3) + 2e’(5/6) + e’1 .
2 2 2



Some tedious but feasible calculation yields then that
1
S= · {1 + 2 · .9726 + 2 · .8948 + 2 · .7880
12
+ 2 · .6412 + 2 · .4994 + .3679}
8.9599
= = .7451.
12
We may use a computer algebra utility like Mathematica or Maple to
calculate the integral exactly (to six decimal places) to equal 0.746824. We thus
see that the answer we obtained with the Trapezoid Rule is certainly accurate
to two decimal places. It is not accurate to three decimal places.
256 CHAPTER 8 Applications of the Integral

It should be noted that Maple and Mathematica both use numerical tech-
niques, like the ones being developed in this section, to calculate integrals. So our
calculations merely emulate what these computer algebra utilities do so swiftly and
so well.

You Try It: How ¬ne a partition would we have needed to use if we wanted four
decimal places of accuracy in the last example? If you have some facility with a
computer, use the Trapezoid Rule with that partition and con¬rm that your answer
agrees with Mathematica™s answer to four decimal places.
EXAMPLE 8.30
Use the Trapezoid Rule with k = 4 to estimate
1
1
dx.
1 + x2
0

SOLUTION
Of course we could calculate this integral precisely by hand, but the point
here is to get some practice with the Trapezoid Rule. We calculate
± 
 
1/4  1 1
1 1 1
S= · +2· +2· +2· + .
2  1+02 1+12 
2 2 2
 
1 2 3
1+ 4 1+ 4 1+ 4

A bit of calculation reveals that
1 5323
S= · ≈ 0.782794 . . . .
8 850
Now if we take f (x) = 1/(1 + x 2 ) then f (x) = (6x 2 ’ 2)/(1 + x 2 )3 .
Thus, on the interval [0, 1], we have that |f (x)| ¤ 4 = M. Thus the error
estimate for the Trapezoid Rule predicts accuracy of
M · (b ’ a)3 4 · 13
= ≈ 0.020833 . . . .
12 · 42
12k 2
This suggests accuracy of one decimal place.
Now we know that the true and exact value of the integral is arctan 1 ≈
0.78539816 . . .. Thus our Trapezoid Rule approximation is good to one, and
nearly to two, decimal places”better than predicted.

8.7.2 SIMPSON™S RULE
Simpson™s Rule takes our philosophy another step: If rectangles are good, and
trapezoids better, then why not approximate by curves? In Simpson™s Rule, we
approximate by parabolas.
257
CHAPTER 8 Applications of the Integral

We have a continuous function f on the interval [a, b] and we have a partition
P = {x0 , x1 , . . . , xk } of our partition as usual. It is convenient in this technique to
assume that we have an even number of intervals in the partition.




Fig. 8.46

Now each rectangle, over each segment of the partition, is capped off by an
arc of a parabola. Figure 8.46 shows just one such rectangle. In fact, for each pair
of intervals [x2j ’2 , x2j ’1 ], [x2j ’1 , x2j ], we consider the unique parabola passing
through the endpoints

(—)
(x2j ’2 , f (x2j ’2 )), (x2j ’1 , f (x2j ’1 )), (x2j , f (x2j )).

Note that a parabola y = Ax 2 + Bx + C has three undetermined coef¬cients, so
three points as in (—) will determine A, B, C and pin down the parabola.
In fact (pictorially) the difference between the parabola and the graph of f is
so small that the error is almost indiscernible. This should therefore give rise to a
startling accurate approximation, and it does.
Summing up the areas under all the approximating parabolas (we shall not
perform the calculations) gives the following approximation to the integral:
b x
f (x) dx ≈ {f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 )
3
a
+ 2f (x4 ) + · · · + 2f (xk’2 ) + 4f (xk’1 ) + f (xk )}.

If it is known that the fourth derivative f (iv) (x) satis¬es |f (iv) (x)| ¤ M on [a, b],
then the error resulting from Simpson™s method does not exceed

M · (b ’ a)5
.
180 · k 4
258 CHAPTER 8 Applications of the Integral

EXAMPLE 8.31
e’x dx to two decimal places of
2
1
Use Simpson™s Rule to calculate 0
accuracy.
SOLUTION
If we set f (x) = e’x then it is easy to calculate that
2



f (iv) (x) = e’x · [12 ’ 72x 2 + 32x 4 ].
2



Thus |f (x)| ¤ 12 = M.
In order to achieve the desired degree of accuracy, we require that
M · (b ’ a)5
< 0.005
180 · k 4
or
12 · 15
< 0.005.
180 · k 4


Y
Simple manipulation yields
FL
200
< k4.
15
AM

This condition is satis¬ed when k = 2.
Thus our job is easy. We take the partition P = {0, 1/2, 1}. The sum arising
from Simpson™s Rule is then
TE



1/2
S= {f (0) + 4f (1/2) + f (1)}
3
1
= {e’0 + 4 · e’(1/2) + e’1 }
2 2 2

6
1
= {1 + 3.1152 + 0.3679}
6
1
≈ · 4.4831
6
≈ 0.7472
Comparing with the “exact value” 0.746824 for the integral that we noted in
Example 8.29, we ¬nd that we have achieved two decimal places of accuracy.
It is interesting to note that if we had chosen a partition with k = 6, as we did
in Example 8.29, then Simpson™s Rule would have guaranteed an accuracy of
M · (b ’ a)5 12 · 15
= ≈ 0.00005144,
180 · k 4 180 · 64
or nearly four decimal places of accuracy.
259
CHAPTER 8 Applications of the Integral

EXAMPLE 8.32
Estimate the integral
1
1
dx
1 + x2
0

using Simpson™s Rule with a partition having four intervals. What degree of
accuracy does this represent?

SOLUTION
Of course this example is parallel to Example 8.30, and you should compare
the two examples. Our function is f (x) = 1/(1 + x 2 ) and our partition is
P = {0, 1/4, 2/4, 3/4, 1}. The sum from Simpson™s Rule is

1/4
S= · {f (0) + 4f (1/4) + 2f (1/2) + 4f (3/4) + f (1)}
3
1 1 1
= · +4·
1 + 02 1 + (1/4)2
12
1 1 1
+ 2· +4· +
1 + (1/2)2 1 + (3/4)2 1 + 12
1
≈ · {1 + 3.7647 + 1.6 + 2.56 + 0.5}
12
≈ 0.785392.

Comparing with Example 8.30, we see that this answer is accurate to four
decimal places. We invite the reader to do the necessary calculation with the
Simpson™s Rule error to term to con¬rm that we could have predicted this
degree of accuracy.

You Try It: Estimate the integral

e2 1
dx
ln x
e

using both the Trapezoid Rule and Simpson™s Rule with a partition having six
points. Use the error term estimate to state what the accuracy prediction of each of
your calculations is. If the software Mathematica or Maple is available to you,
check the answers you have obtained against those provided by these computer
algebra systems.
260 CHAPTER 8 Applications of the Integral

Exercises
1. A solid has base the unit circle and vertical slices, parallel to the y-axis,
which are half-disks. Calculate the volume of this solid.
2. A solid has base a unit square with center at the origin and vertices on the
x- and y-axes. The vertical cross-section of this solid, parallel to the y-axis,
is a disk. What is the volume of this solid?
3. Set up the integral to calculate the volume enclosed when the indicated
curve over the indicated interval is rotated about the indicated line. Do not
evaluate the integral.
y = x 2 2 ¤ x ¤ 5 x-axis
(a)

y = x 1 ¤ x ¤ 9 y-axis
(b)
y = x 3/2 0 ¤ x ¤ 2 y = ’1
(c)
y = x + 3 ’1 ¤ x ¤ 2 y = 5
(d)
y = x 1/2 4 ¤ x ¤ 6 x = ’2
(e)
y = sin x 0 ¤ x ¤ π/2 y = 0
(f )
4. Set up the integral to evaluate the indicated surface area. Do not evaluate.
The area of the surface obtained when y = x 2/3 , 0 ¤ x ¤ 4, is
(a)
rotated about the x-axis.
The area of the surface obtained when y = x 1/2 , 0 ¤ x ¤ 3, is
(b)
rotated about the y-axis.
The area of the surface obtained when y = x 2 , 0 ¤ x ¤ 3, is rotated
(c)
about the line y = ’2.
The area of the surface obtained when y = sin x, 0 ¤ x ¤ π, is
(d)
rotated about the x-axis.
The area of the surface obtained when y = x 1/2 , 1 ¤ x ¤ 4, is
(e)
rotated about the line x = ’2.
The area of the surface obtained when y = x 3 , 0 ¤ x ¤ 1, is rotated
(f )
about the x-axis.
5. A water tank has a submerged window that is in the shape of a circle of
radius 2 feet. The center of this circular window is 8 feet below the surface.
Set up, but do not calculate, the integral for the pressure on the lower half
of this window”assuming that water weighs 62.4 pounds per cubic foot.
6. Aswimming pool is V-shaped. Each end of the pool is an inverted equilateral
triangle of side 10 feet. The pool is 25 feet long. The pool is full. Set up,
but do not calculate, the integral for the pressure on one end of the pool.
7. A man climbs a ladder with a 100 pound sack of sand that is leaking one
pound per minute. If he climbs steadily at the rate of 5 feet per minute, and
261
CHAPTER 8 Applications of the Integral

if the ladder is 40 feet high, then how much work does he do in climbing
the ladder?
8. Because of a prevailing wind, the force that opposes a certain runner is
3x 2 + 4x + 6 pounds at position x. How much work does this runner
perform as he runs from x = 3 to x = 100 (with distance measured in
feet)?
9. Set up, but do not evaluate, the integrals for each of the following arc length
problems.
The length of the curve y = sin x, 0 ¤ x ¤ π
(a)
The length of the curve x 2 = y 3 , 1 ¤ x ¤ 8
(b)
The length of the curve cos y = x, 0 ¤ y ¤ π/2
(c)
The length of the curve y = x 2 , 1 ¤ x ¤ 4
(d)
10. Set up the integral for, but do not calculate, the average value of the given
function on the given interval.
f (x) = sin2 x [2, 5]
(a)
g(x) = tan x [0, π/4]
(b)
x
(c) h(x) = , [’2, 2]
x+1
sin x
(d) f (x) = [’π, 2π ]
2 + cos x
11. Write down the sum that will estimate the given integral using the method
of rectangles with mesh of size k. You need not actually evaluate the sum.
4

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