Smith R., Minton R. Calculus

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5-9 SECTION 5.1

Area Between Curves 323

y  x

y  2x

36. Give an integral equal to each area.

(a) A

+ A

(b) A

+ A

(d) A

37. Let f (t) be the area between y = sin

x and y = 1 for

0 ≤ x ≤ t. Find all critical numbers, local extrema and inﬂec-

tion points for f (t), t ≥ 0.

38. Let g be a continuous function deﬁned for x ≥ 0 with

|g(x)|≤1 for x ≥ 0. Let f (t) be the area between y = g(x)

and y = 1 for 0 ≤ x ≤ t.Ifg has a local maximum at x = a,

does f have a critical number at a? An inﬂection point at a?

What if there is a local minimum at x = a?

APPLICATIONS

39. Suppose that a country’s oil consumption for the years 1970–

1974 was approximately equal to f (t) = 16.1+1.4t million

barrels per year, where t = 0 corresponds to 1970. Following

an oil shortage in 1974, the country’s consumption changed

and was better modeled by g(t)=19.7+0.5t million barrels

per year, t ≥ 4. Show that f (4)=g(4) and explain what this

number represents. Compute the area between f (t) and g(t)

for 4 ≤ t ≤ 10.Use this number to estimate the number of bar-

rels of oil saved by the reduced oil consumption from 1974 to

1980.

y  f(t)

y  g(t)

8642

(1970) (1980)

40. Suppose that a nation’s fuelwood consumption is given by

76 + 3.2t m

/yr and new tree growth is 50 −2.4t m

/yr.

Compute and interpret the area between the curves for

0 ≤ t ≤ 10.

41. Suppose that the birthrate for a certain population is

b(t) = 2 +0.1t million people per year and the death rate for

the same population is d(t) = 2 + 0.06t million people per

year. Show that b(t) ≥ d(t) for t ≥ 0 and explain why the area

betweenthecurvesrepresentstheincrease in population. Com-

pute the increase in population for 0 ≤ t ≤ 10.

42. Suppose that the birthrate for a population is b(t) = 2 + 0.1t

million people per year and the death rate for the same

population is d(t) = 3 +0.05t million people per year. Find

the intersection T of the curves. Interpret the area between

the curves for 0 ≤ t ≤ T and the area between the curves

for T ≤ t ≤ 30. Compute the net change in population for

0 ≤ t ≤ 30.

............................................................

In exercises 43 and 44, the graph shows the rate of ﬂow of water

in gallons per hour into and out of a tank. Assuming that the

tank starts with 400 gallons, estimate the amount of water in the

tank at hours 1, 2, 3, 4 and 5 and sketch a graph of the amount

of water in the tank.

43.

1234

100

120

Out

44.

Out

1234

100

120

............................................................

45. The graph shows the supply and demand curves for a prod-

uct. The point of intersection (q

∗

, p

∗

) gives the equilibrium

quantity and equilibrium price for the product. The consumer

surplus is deﬁned to be CS =



∗

D(q)dq − p

∗

. Shade in

the area of the graph that represents the consumer surplus,

and compute this in the case where D(q) = 10 −

q and

S(q) = 2 +

120

q +

1200

250

75 100

2.5

7.5

D(q)

S(q)

46. Repeat exercise 45 for the producer surplus deﬁned by

PS = p

∗

−



q∗

S(q)dq.

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324 CHAPTER 5

Applications of the Definite Integral 5-10

47. Let C



(x) be the marginal cost of producing x thousand copies

of an item and let R



(x) be the marginal revenue from the sale

of that item, with graphs as shown.Assume that R



(x) = C



(x)

at x = 2 and x = 5. Interpret the area between the curves for

each interval: (a) 0 ≤ x ≤ 2, (b) 2 ≤ x ≤ 5, (c) 0 ≤ x ≤ 5 and

(d) 5 ≤ x ≤ 6.

C'(x)

R'(x)

1234567

48. Abasicprinciple ofeconomicsisthatproﬁtis maximizedwhen

marginalcostequalsmarginalrevenue.At whichintersectionis

proﬁtmaximizedin exercise47?Explainyouranswer.In terms

of proﬁt, what does the other intersection point represent?

EXPLORATORY EXERCISES

1. Find the area between y = x

and y = mx for any constant

m > 0. Without doing further calculations, use this area to

ﬁnd the area between y =

√

x and y = mx.

2. For x > 0, let f (x) be the area between y = 1 and y = sin

for 0 ≤ t ≤ x. Without calculating f (x), ﬁnd as many rela-

tionships as possible between the graphical features (zeros,

extrema, inﬂection points) of y = f (x) and the graphical fea-

tures of y = sin

5.2 VOLUME: SLICING, DISKS AND WASHERS

As we shall see throughout this chapter, the integral is an amazingly versatile tool. In this

section, we use integrals to compute the volume of a three-dimensional solid. We begin

with a simple problem.

When designing a building, architects must perform numerous detailed calculations.

For instance, in order to analyze a building’s heating and cooling systems, engineers must

calculate the volume of air being processed.

FIGURE 5.12a FIGURE 5.12b

There are probably only a few solids whose volume you know how to compute. For

instance, the building shown in Figure5.12a is essentially a rectangular box, whose volume

is given by lwh, where l is the length, w is the width and h is the height. The right circular

cylinders seen in the buildings in Figure 5.12b have volume given by πr

h, where h is the

height and r is the radius of the circular cross section. Notice in each case that the build-

ing has a familiar cross section (a rectangle in Figure 5.12a and a circle in Figure 5.12b)

that is extendedvertically. We call any such solid a cylinder (any solid whose cross sections

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5-11 SECTION 5.2

Volume: Slicing, Disks and Washers 325

perpendicular to some axis running through the solid are all the same). There is a connec-

tion between the volume formulas for these two cylinders. The volume of a right circular

cylinder is

V = (πr

)





cross-sectional area



height

while the volume of a box is

V = (length ×width)



 

cross-sectional area

× height.

In general, the volume of any cylinder is found by

V = (cross-sectional area) ×(height).

HISTORICAL

NOTES

Archimedes (ca. 287–212 B.C.)

A Greek mathematician and

scientist who was among

the first to derive formulas for

volumes and areas. Archimedes

is known for discovering the

basic laws of hydrostatics (he

reportedly leapt from his bathtub,

shouting “Eureka!’’and ran into

the streets to share his discovery)

and levers (“Give me a place to

stand on and I can move the

earth.’’). An ingenious engineer,

his catapults, grappling cranes

and reflecting mirrors terrorized

a massive Roman army that

eventually conquered his

hometown of Syracuse.

Archimedes was especially proud

of his proof that the volume of a

sphere inscribed in a cylinder is

2/3 of the volume of the cylinder

(see exercises 31–34), and

requested that this be inscribed

on his tombstone. Many of his

techniques were very similar to

those that we use in calculus

today, but many of his writings

were lost in the Middle Ages. The

amazing story of the recent

discovery of his book The Method

is told in the book The Archimedes

Codex, by Netz and Noel.

Volumes by Slicing

Even relatively simple solids, such as pyramids and domes, do not have constant cross-

sectional area, as seen in Figures 5.13a and 5.13b. To ﬁnd the volume in such a case, we

take the approach we’ve used a number of times now: ﬁrst approximate the volume and

then improve the approximation.

FIGURE 5.13a

Pyramid Entrance to the Louvre Museum in Paris

FIGURE 5.13b

U.S. Capitol Building

More generally, for a solid that extends from x = a to x = b, we start by partitioning

the interval [a, b]onthex-axis into n subintervals, each of width x =

b −a

. As usual,

we denote x

= a, x

= a + x and so on, so that

= a + ix, for i = 0, 1, 2,...,n.

We then slice the solid perpendicular to the x-axis at each of the (n − 1) points

, x

,...,x

n−1

. (See Figure 5.14a on the following page.) Notice that if n is large, then

each slice of the solid will be thin with nearly constant cross-sectional area. Suppose that

the area of the cross section corresponding to any particular value of x is given by A(x).

Observe that the slice between x = x

i−1

and x = x

is nearly a cylinder. (See Figure 5.14b.)

So, for any point c

in the interval [x

i−1

, x

], the area of the cross sections on that interval

are all approximately A(c

). The volume V

of the ith slice is then approximately the volume

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326 CHAPTER 5

Applications of the Definite Integral 5-12

i 1

x

i th slice of solid

i1

x

i th approximating cylinder

FIGURE 5.14a

Sliced solid

FIGURE 5.14b

ith slice of solid

FIGURE 5.14c

ith approximating cylinder

of the cylinder lying along the interval [x

i−1

, x

], with constant cross-sectional area A(c

)

(see Figure 5.14c), so that

≈ A(c

)





cross-sectional area

x



width

Repeating this process for each of the n slices, we ﬁnd that the total volume V of the

solid is approximately

V ≈



i=1

A(c

)x.

Notice that as the number of slices increases, the volume approximation should improve

and we get the exact volume by computing the limit

V = lim

n→∞



i=1

A(c

)x,

assuming the limit exists. You should recognize this limit as the deﬁnite integral

V =



A(x)dx. (2.1)

Volume of a solid with

cross-sectional area A(x)

REMARK 2.1

We use the same process followed here to derive many important formulas. In each

case, we divide an object into n smaller pieces, approximate the quantity of interest

for each of the small pieces, sum the approximations and then take a limit, ultimately

recognizing that we have derived a deﬁnite integral. For this reason, it is essential that

you understand the concept behind formula (2.1). Memorization will not do this for

you. However, if you understand how the various pieces of this puzzle ﬁt together,

then the rest of this chapter should fall into place for you nicely.

EXAMPLE 2.1 Computing Volume from Cross-Sectional Areas

The Pyramid Arena in Memphis has a square base of side approximately 600 feet and

a height of approximately 320 feet. Find the volume of the pyramid with these

measurements.

Solution Since the pyramid has square horizontal cross sections, we need only ﬁnd a

formula for the size of the square at each height. Let x represent the height above the

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5-13 SECTION 5.2

Volume: Slicing, Disks and Washers 327

ground. At x = 0, the cross section is a square of side 600 feet. At x = 320, the cross

section can be thought of as a square of side 0 feet. If f (x) represents the side length of

the square cross section at height x, we know that f (0) = 600, f (320) = 0 and f (x)

must be a linear function. (Think about this; the sides of the pyramid do not curve.) The

slope of the line is m =

600 − 0

0 − 320

=−

and we use the y-intercept of 600 to get

f (x) =−

x +600.

The cross-sectional area is simply the square of f (x), so that from (2.1), we have

V =



320

A(x)dx =



320



−

x +600



dx.

Observe that we can evaluate this integral by substitution, by taking u =−

x +600,

so that du =−

dx. This gives us

V =



320



−

x +600



dx =−



600



600

du =



600

= 38,400,000ft



In many important applications,the cross-sectional area is not known exactly, but must

be approximated using measurements. In such cases, we can approximate the volume using

numerical integration.

A Pyramid

EXAMPLE 2.2 Estimating Volume from Cross-Sectional Data

In medical imaging, such as CT (computerized tomography) and MRI (magnetic

resonance imaging) processes, numerous measurements are taken and processed by a

computer to construct a three-dimensional image of the tissue the physician wishes to

study. The process is similar to the slicing process we have used to ﬁnd the volume of a

solid. In this case, however, mathematical representations of various slices of the tissue

are combined to produce a three-dimensional image that physicians view to determine

the health of the tissue. Suppose that an MRI scan indicates that the cross-sectional

areas of adjacent slices of a tumor are given by the values in the table.

x (cm) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

A(x) (cm

) 0.0 0.1 0.4 0.3 0.6 0.9 1.2 0.8 0.6 0.2 0.1

Estimate the volume of the tumor.

Solution To ﬁnd the volume of the tumor, we would compute [following (2.1)]

V =



A(x)dx,

except that we only know A(x) at a ﬁnite number of points. Although we can’t compute

this exactly, we can use Simpson’s Rule with x = 0.1 to estimate the value of this

integral:

V =



A(x) dx

≈

b −a



A(0) +4A(0.1) +2A(0.2) +4A(0.3) + 2A(0.4) + 4A(0.5)

+2A(0.6) +4A(0.7) +2A(0.8) +4A(0.9) + A(1)



0.1

(0 + 0.4 + 0.8 + 1.2 + 1.2 + 3.6 + 2.4 + 3.2 +1.2 +0.8 +0.1)

≈ 0.49667 cm



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328 CHAPTER 5

Applications of the Definite Integral 5-14

We now turn to the problem of ﬁnding the volume of the dome in Figure 5.13b. Since

the horizontal cross sections are circles, we need only to determine the radius of each circle.

4545

FIGURE 5.15

y =−

+ 90.

EXAMPLE 2.3 Computing the Volume of a Dome

Suppose that a dome has circular cross sections, with outline y =−

+ 90, for

−45 ≤ x ≤ 45. (In units of feet, this gives dimensions similar to the Capitol dome in

Figure 5.13b. A graph is shown in Figure 5.15.) Find the volume of the dome.

Solution As seen in Figure 5.15, the circular cross sections occur at each value of y,

with 0 ≤ y ≤ 90. For a given y, the radius extends from x = 0tox =



(90 − y). The

radius for this value of y is given by r(y) =



(90 − y), so that the cross-sectional

areas are given by

A(y) = π





(90 − y)



for 0 ≤ y ≤ 90. The volume is then given by

V =



A(y) dy =







(90 − y)



dy =





2025 −



= π



2025y −



= 91,125π ≈ 286,278ft



Observe that an alternative way of stating the problem in example 2.3 is to say: Find

the volume formed by revolving the region bounded by the curve x =



(90 − y) and the

y-axis, for 0 ≤ y ≤ 90, about the y-axis.

Example 2.3 can be generalized to the method of disks used to compute the volume of

a solid formed by revolving a two-dimensional region about a vertical or horizontal line.

We consider this general method next.

The Method of Disks

Supposethat f (x) ≥ 0and f iscontinuousontheinterval[a, b].Taketheregionbounded by

thecurve y = f (x)and the x-axis, for a ≤ x ≤ b,and revolveit about the x-axis, generating

a solid. (See Figures 5.16a and 5.16b.) We can ﬁnd the volume of this solid by slicing it

perpendiculartothex-axisandrecognizingthateachcrosssectionisa circular disk of radius

r = f (x). (See Figure 5.16b.) From (2.1), we then have that the volume of the solid is

Volume of a solid of revolution

(Method of disks)

V =



π[ f (x)]

  

cross-sectional area = πr

dx. (2.2)

y  f(x)

a b

FIGURE 5.16a

y = f (x) ≥ 0

FIGURE 5.16b

Solid of revolution

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5-15 SECTION 5.2

Volume: Slicing, Disks and Washers 329

Since the cross sections of such a solid of revolution are all disks, we refer to this method

of ﬁnding volume as the method of disks.

EXAMPLE 2.4 Using the Method of Disks to Compute Volume

Revolve the region under the curve y =

√

x on the interval [0, 4] about the x-axis and

ﬁnd the volume of the resulting solid of revolution.

Solution It’s critical to draw a picture of the region and the solid of revolution, so

that you get a clear idea of the radii of the circular cross sections. You can see from

Figures 5.17a and 5.17b that the radius of each cross section is given by r =

√

From (2.2), we then get the volume:

V =





√



  

cross-sectional area = πr

dx = π



xdx= π



= 8π.

321

r  x

FIGURE 5.17a

y =

√

FIGURE 5.17b

Solid of revolution



In a similar way, suppose that g(y) ≥ 0 and g is continuous on the interval [c, d].

Then, revolving the region bounded by the curve x = g(y) and the y-axis, for c ≤ y ≤ d,

about the y-axis generates a solid. (See Figures 5.18a and 5.18b.) Once again, notice from

Figure 5.18b that the cross sections of the resulting solid of revolution are circular disks

of radius r = g(y). All that has changed here is that we have interchanged the roles of the

variables x and y. The volume of the solid is then given by

V =



π[g(y)]

  

cross-sectional area = πr

dy. (2.3)

Volume of a solid of revolution

(Method of disks)

x  g(y)

FIGURE 5.18a

Revolve about the y-axis

FIGURE 5.18b

Solid of revolution

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330 CHAPTER 5

Applications of the Definite Integral 5-16

REMARK 2.2

When using the method of disks, the variable of integration depends solely on the axis

about which you revolve the two-dimensional region: revolving about the x-axis

requires integration with respect to x, while revolving about the y-axis requires

integration with respect to y. This is easily determined by looking at a sketch of the

solid. Don’t make the mistake of simply looking for what you can plug in where. This

is a recipe for disaster, for the rest of this chapter will require you to make similar

choices, each based on distinctive requirements of the problem at hand.

EXAMPLE 2.5 Using the Method of Disks with y as the

Independent Variable

Find the volume of the solid resulting from revolving the region bounded by the curves

y = 4 − x

and y = 1 from x = 0tox =

√

3 about the y-axis.

Solution You will ﬁnd a graph of the curve in Figure 5.19a and of the solid in

Figure 5.19b.

2√3

x 



4  y

y  4  x

FIGURE 5.19a

y = 4 − x

FIGURE 5.19b

Solid of revolution

Notice from Figures 5.19a and 5.19b that the radius of any of the circular cross

sectionsisgivenbyx.So,wemustsolvetheequation y = 4 − x

forx,togetx =

√

4 − y.

Since the surface extends from y = 1toy = 4, the volume is given by (2.3) to be

V =







4 − y



  

πr

dy =



π(4 − y) dy

= π



4y −



= π



(16 − 8) −



4 −



9π



The Method of Washers

One complication that occurs in computing volumes is that the solid may have a cavity or

“hole” in it. Another occurs when a region is revolved about a line other than the x-axis

or the y-axis. Neither case will present you with any signiﬁcant difﬁculties, if you look

carefully at the ﬁgures. We illustrate these ideas in examples 2.6 and 2.7.

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5-17 SECTION 5.2

Volume: Slicing, Disks and Washers 331

EXAMPLE 2.6 Computing Volumes of Solids with and Without Cavities

Let R be the region bounded by the graphs of y =

, x = 0 and y = 1. Compute the

volume of the solid formed by revolving R about (a) the y-axis, (b) the x-axis and (c) the

line y = 2.

Solution (a) The region R is shown in Figure 5.20a and the solid formed by revolving

it about the y-axis is shown in Figure 5.20b. Notice that this part of the problem is

similar to example 2.5.

x  4y

FIGURE 5.20a

x =

√

FIGURE 5.20b

Solid of revolution

Outer radius:

y  1

Inner radius:

y 

FIGURE 5.21a

y =

1

Washer-shaped

cross sections

FIGURE 5.21b

Solid with cavity

From (2.3), the volume is given by

V =









  

πr

dy = π



= 2π.

(b) Revolving the region R about the x-axis produces a cavity in the middle of the solid.

See Figure 5.21a for a graph of the region R and Figure 5.21b for a sketch of the solid.

Our strategy is to compute the volume of the outside of the object (as if it were solid)

and then subtract the volume of the cavity. Before diving into a computation, be sure to

visualize the geometry behind this. Here, the outside surface of the solid is formed by

revolving the line y = 1 about the x-axis. The cavity is formed by revolving the curve

y =

about the x-axis. Look carefully at Figures 5.21a and 5.21b and make certain

that you see this. The outer radius, r

, is the distance from the x-axis to the line y = 1,

or r

= 1. The inner radius, r

, is the distance from the x-axis to the curve y =

,or

. Applying (2.2) twice, we see that the volume is given by

V =



π(1)



π (outer radius)

dx −







  

π (inner radius)

= π





1 −



dx = π



x −





= π



2 −



π.

cylindrical hole in the middle. The region R is shown in Figure 5.22a and the solid is

shown in Figure 5.22b (on the following page).

The volume is computed in the same way as in part (b), by subtracting the volume

of the cavity from the volume of the outside solid. From Figures 5.22a and 5.22b, notice

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332 CHAPTER 5

Applications of the Definite Integral 5-18

Outer radius:

2  y  2 

Inner radius:

y  2  1  1

y  2

FIGURE 5.22a

Revolve about y = 2

FIGURE 5.22b

Solid of revolution

that the radius of the outer surface is the distance from the line y = 2 to the curve

y =

or r

= 2 −

. The radius of the inner hole is the distance from the line

y = 2 to the line y = 1orr

= 2 − 1 = 1. From (2.2), the volume is given by

V =





2 −



  

π(outer radius)

dx −



π(2 − 1)

  

π(inner radius)

= π





4 − x



− 1



dx = π



3x −



= π



6 −



π.



In parts (b) and (c) of example 2.6, the volume was computed by subtracting an inner

volume from an outer volume in order to compensate for a cavity inside the solid. This

technique is a slight generalization of the method of disks and is referred to as the method

of washers, since the cross sections of the solids look like washers.

EXAMPLE 2.7 Revolving a Region About Different Lines

Let R be the region bounded by y = 4 − x

and y = 0. Find the volume of the solids

obtained by revolving R about each of the following: (a) the y-axis, (b) the line y =−3,

Solution For part (a), we draw the region R in Figure 5.23a and the solid of revolution

in Figure 5.23b.

From Figure 5.23b, notice that each cross section of the solid is a circular disk,

whose radius is simply x. Solving for x,wegetx =

√

4 − y, where we have selected x