applications of integration · more about areas in chapter 5 we deﬁned and calculated areas of...

6 Applications of Integration

More about Areas ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !

In Chapter 5 we defined and calculated areas of regions that lie under the graphs of functions. Here we use integrals to find areas of more general regions. First we con-sider regions that lie between the graphs of two functions. Then we look at regionsenclosed by parametric curves.

Areas between Curves

Consider the region that lies between two curves and and be-tween the vertical lines and , where and are continuous functions and

for all in . (See Figure 1.)Just as we did for areas under curves in Section 5.1, we divide S into n strips of

equal width and then we approximate the ith strip by a rectangle with base andheight . (See Figure 2. If we like, we could take all of the sample pointsto be right endpoints, in which case .) The Riemann sum

is therefore an approximation to what we intuitively think of as the area of S.

This approximation appears to become better and better as . Therefore, wedefine the area of as the limiting value of the sum of the areas of these approxi-mating rectangles.

SAn l !

FIGURE 2 (b) Approximating rectangles

x

y

b0 a

(a) Typical rectangle

x

y

b0 a

f(x i*)f(x i*)-g(x i*)

_g(x i*)x i*

Îx

!n

i!1 " f #xi*$ " t#xi*$% #x

xi* ! xi

f #xi*$ " t#xi*$#x

"a, b%xf #x$ $ t#x$tfx ! bx ! a

y ! t#x$y ! f #x$S

6.1

447

In this chapter we explore some of the applications ofthe definite integral by using it to compute areas be-tween curves, volumes of solids, lengths of curves, theaverage value of a function, the work done by a vary-ing force, the center of gravity of a plate, the force ona dam, as well as quantities of interest in biology, eco-nomics, and statistics. The common theme in most ofthese applications is the following general method,

which is similar to the one we used to find areas under curves. We break up a quantity into a largenumber of small parts. We then approximate eachsmall part by a quantity of the form and thusapproximate by a Riemann sum. Then we take thelimit and express as an integral. Finally, we evalu-ate the integral by using the Evaluation Theorem orSimpson’s Rule.

QQ

f #x i*$ #x

Q

x

y

b0 a

y=©

y=ƒ

S

FIGURE 1S=s(x,!y)!|!a¯x¯b, ©¯y¯ƒd

Guess the area of an island.Resources / Module 7

/ Areas / Start of Areas

We recognize the limit in (1) as the definite integral of . Therefore, we havethe following formula for area.

The area of the region bounded by the curves , and thelines , , where and are continuous and for all in

, is

Notice that in the special case where , is the region under the graph of and our general definition of area (1) reduces to our previous definition (Definition5.1.2).

In the case where both and are positive, you can see from Figure 3 why (2) istrue:

EXAMPLE 1 Find the area of the region bounded above by , bounded below by, and bounded on the sides by and .

SOLUTION The region is shown in Figure 4. The upper boundary curve is andthe lower boundary curve is . So we use the area formula (2) with ,

, and :

In Figure 4 we drew a typical approximating rectangle with width as a reminderof the procedure by which the area is defined in (1). In general, when we set up anintegral for an area, it’s helpful to sketch the region to identify the top curve , thebottom curve , and a typical approximating rectangle as in Figure 5. Then the areaof a typical rectangle is and the equation

summarizes the procedure of adding (in a limiting sense) the areas of all the typical rectangles.

Notice that in Figure 5 the left-hand boundary reduces to a point, whereas in Fig-ure 3 the right-hand boundary reduces to a point. In the next example both of the sideboundaries reduce to a point, so the first step is to find a and b.

A ! limn l !

!n

i!1 #yT " yB$ #x ! yb

a #yT " yB$ dx

#yT " yB$ #xyB

yT

#x

! e " 12 " 1 ! e " 1.5

A ! y1

0 #ex " x$ dx ! ex " 1

2 x 2]10

b ! 1a ! 0, t#x$ ! xf #x$ ! exy ! x

y ! ex

x ! 1x ! 0y ! xy ! ex

! yb

a f #x$ dx " yb

a t#x$ dx ! yb

a " f #x$ " t#x$% dx

A ! "area under y ! f #x$% " "area under y ! t#x$%

tf

fSt#x$ ! 0

A ! yb

a " f #x$ " t#x$% dx

"a, b%xf #x$ $ t#x$tfx ! bx ! a

y ! f #x$, y ! t#x$A2

f " t

A ! limn l !

!n

i!1 " f #xi*$ " t#xi*$% #x1

448 " CHAPTER 6 APPLICATIONS OF INTEGRATION

FIGURE 3

A=j ƒ!dx-j ©!dxa

b

a

b

0 x

y

a b

y=ƒ

y=©

S

0 x

y

1

y=´

y=x Îx

x=1

1

FIGURE 4

0 x

y

a b

yT

yB

yT-yB

Îx

FIGURE 5

EXAMPLE 2 Find the area of the region enclosed by the parabolas and.

SOLUTION We first find the points of intersection of the parabolas by solving theirequations simultaneously. This gives , or . Therefore,

, so or 1. The points of intersection are and .We see from Figure 6 that the top and bottom boundaries are

and

The area of a typical rectangle is

and the region lies between and . So the total area is

Sometimes it’s difficult, or even impossible, to find the points of intersection of twocurves exactly. As shown in the following example, we can use a graphing calculatoror computer to find approximate values for the intersection points and then proceed asbefore.

EXAMPLE 3 Find the approximate area of the region bounded by the curvesand

SOLUTION If we were to try to find the exact intersection points, we would have tosolve the equation

This looks like a very difficult equation to solve exactly (in fact, it’s impossible), soinstead we use a graphing device to draw the graphs of the two curves in Figure 7.One intersection point is the origin. We zoom in toward the other point of intersec-tion and find that . (If greater accuracy is required, we could use Newton’smethod or a rootfinder, if available on our graphing device.) Thus, an approximationto the area between the curves is

To integrate the first term we use the substitution . Then , andwhen . So

! s2.39 " 1 "#1.18$5

5%

#1.18$2

2& 0.785

! su ]1

2.39" ' x 5

5"

x 2

2 (0

1.18

A & 12 y2.39

1

dusu

" y1.18

0 #x 4 " x$ dx

x ! 1.18, we have u & 2.39du ! 2x dxu ! x 2 % 1

A & y1.18

0 ' x

sx 2 % 1" #x 4 " x$( dx

x & 1.18

xsx 2 % 1

! x 4 " x

y ! x 4 " x.y ! x)sx 2 % 1

! 2' x 2

2"

x 3

3 (0

1

! 2*12

"13+ !

13

A ! y1

0 #2x " 2x 2 $ dx ! 2 y1

0 #x " x 2 $ dx

x ! 1x ! 0

#yT " yB$ #x ! #2x " x 2 " x 2 $ #x

yB ! x 2yT ! 2x " x 2

#1, 1$#0, 0$x ! 02x#x " 1$ ! 02x 2 " 2x ! 0x 2 ! 2x " x 2

y ! 2x " x 2y ! x 2

SECTION 6.1 MORE ABOUT AREAS # 449

x

y

Îx(0, 0)

(1 , 1)

FIGURE 6

yT=2x-"

yB="

1.5

_1

_1 2

y=x$-x

y= x

œ„„„„„"+1

FIGURE 7

EXAMPLE 4 Figure 8 shows velocity curves for two cars, A and B, that start side byside and move along the same road. What does the area between the curves repre-sent? Use Simpson’s Rule to estimate it.

SOLUTION We know from Section 5.3 that the area under the velocity curve A repre-sents the distance traveled by car A during the first 16 seconds. Similarly, the areaunder curve B is the distance traveled by car B during that time period. So the areabetween these curves, which is the difference of the areas under the curves, is thedistance between the cars after 16 seconds. We read the velocities from the graphand convert them to feet per second .

Using Simpson’s Rule with intervals, so that , we estimate the dis-tance between the cars after 16 seconds:

Some regions are best treated by regarding x as a function of y. If a region isbounded by curves with equations , , , and , where and

are continuous and for (see Figure 9), then its area is

If we write for the right boundary and for the left boundary, then, as Fig-ure 10 illustrates, we have

Here a typical approximating rectangle has dimensions and .#yxR " xL

A ! yd

c #xR " xL$ dy

xLxR

x

c

d

y

0

y=dx=g(y)

x=f(y)

y=c

Îy

FIGURE 9

0 x

y

c

dxRxL

xR-xL

Îy

FIGURE 10

A ! yd

c " f #y$ " t#y$% dy

c & y & df #y$ $ t#y$tfy ! dy ! cx ! t#y$x ! f #y$

& 367 ft

& 23 "0 % 4#13$ % 2#20$ % 4#23$ % 2#25$ % 4#28$ % 2#29$ % 4#29$ % 30%

y16

0 #vA " vB$ dt

#t ! 2n ! 8

#1 mi)h ! 52803600 ft)s$


FIGURE 8

0

10

20

30

40

50

60A

B

2 4 6 8 10 12 14 16 t(seconds)

# (mi/h)

t 0 2 4 6 8 10 12 14 16

0 34 54 67 76 84 89 92 95

0 21 34 44 51 56 60 63 65

0 13 20 23 25 28 29 29 30vA " vB

vB

vA

EXAMPLE 5 Find the area enclosed by the line and the parabola.

SOLUTION By solving the two equations we find that the points of intersection areand . We solve the equation of the parabola for x and notice from

Figure 11 that the left and right boundary curves are

We must integrate between the appropriate -values, and . Thus

We could have found the area in Example 5 by integrating with respect to x insteadof y, but the calculation is much more involved. It would have meant splitting theregion in two and computing the areas labeled and in Figure 12. The method weused in Example 5 is much easier.

Areas Enclosed by Parametric Curves

We know that the area under a curve from to is , where. If the curve is given by the parametric equations and ,

, then we can calculate an area formula by using the Substitution Rule forDefinite Integrals as follows:

EXAMPLE 6 Find the area under one arch of the cycloid

(See Figure 13.)

SOLUTION One arch of the cycloid is given by . Using the SubstitutionRule with and , we have

! r 2( 32 ! 2') ! 3'r 2

! r 2[ 32 ( " 2 sin ( % 1

4 sin 2(]02'

! r 2 y2'

0 [1 " 2 cos ( % 1

2 #1 % cos 2($] d(

! r 2 y2'

0 #1 " cos ($2 d( ! r 2 y2'

0 #1 " 2 cos ( % cos2($ d(

A ! y2'r

0 y dx ! y2'

0 r#1 " cos ($r#1 " cos ($ d(

dx ! r#1 " cos ($ d(y ! r#1 " cos ($0 & ( & 2'

y ! r#1 " cos ($x ! r#( " sin ($

'or y)

* t#t$ f +#t$ dt(A ! yb

a y dx ! y*

) t#t$ f +#t$ dt

) & t & *y ! t#t$x ! f #t$F#x$ $ 0

A ! xba F#x$ dxbay ! F#x$

A2A1

! " 16 #64$ % 8 % 16 " ( 4

3 % 2 " 8) ! 18

! "12

* y 3

3 + % y 2

2% 4y(

"2

4

! y4

"2 ("1

2 y 2 % y % 4) dy

! y4

"2 [#y % 1$ " ( 1

2 y 2 " 3)] dy

A ! y4

"2 #xR " xL $ dy

y ! 4y ! "2y

xR ! y % 1xL ! 12 y 2 " 3

#5, 4$#"1, "2$

y 2 ! 2x % 6y ! x " 1

SECTION 6.1 MORE ABOUT AREAS # 451

x

y

_2

4

0

(_1, _2)

(5, 4)

xR=y+1

12xL= ¥-3

FIGURE 11

0 x

y

!3

(5,!4)

(_1, _2)

y=x-1

A¡

y=_ 2x+6

A™

y= 2x+6œ„„„„„

œ„„„„„

FIGURE 12

$ The limits of integration for are foundas usual with the Substitution Rule. When

, is either . When , is the remaining value.t

x ! b) or *tx ! a

t

0

y

x2$r

FIGURE 13

$ The result of Example 6 says that thearea under one arch of the cycloid isthree times the area of the rolling circlethat generates the cycloid (see Example7 in Section 1.7). Galileo guessed thisresult but it was first proved by theFrench mathematician Roberval and theItalian mathematician Torricelli.


19. 20." " " " " " " " " " " " "

21. Racing cars driven by Chris and Kelly are side by side atthe start of a race. The table shows the velocities of each car(in miles per hour) during the first ten seconds of the race.Use Simpson’s Rule to estimate how much farther Kellytravels than Chris does during the first ten seconds.

Two cars, and , start side by side and accelerate fromrest. The figure shows the graphs of their velocity functions.(a) Which car is ahead after one minute? Explain.(b) What is the meaning of the area of the shaded region?(c) Which car is ahead after two minutes? Explain.(d) Estimate the time at which the cars are again side by

side.

23. The widths (in meters) of a kidney-shaped swimming poolwere measured at 2-meter intervals as indicated in thefigure. Use Simpson’s Rule to estimate the area of the pool.

24. The figure shows graphs of the marginal revenue functionand the marginal cost function for a manufacturer.

[Recall from Section 4.7 that and represent therevenue and cost when units are manufactured. Assume x

C#x$R#x$C+R+

6.2

5.0

7.26.8

5.6 4.84.8

0

A

B

#

t (min)21

BA22.

y ! e x, y ! 2 " x 2y ! x 2, y ! xe"x )21–4 " Find the area of the shaded region.

5–16 " Sketch the region enclosed by the given curves. Decidewhether to integrate with respect to x or y. Draw a typicalapproximating rectangle and label its height and width. Thenfind the area of the region.

5.

6.

7. ,

8.

9. ,

10. ,

,

12. ,

13. ,

, , ,

15.

16." " " " " " " " " " " " "

; 17–20 " Use a graph to find approximate -coordinates of thepoints of intersection of the given curves. Then find (approxi-mately) the area of the region bounded by the curves.

17. , 18. , y ! 3x " x 3y ! x 4y ! 2 cos xy ! x 2

x

y ! , x ,, y ! x 2 " 2

y ! x 2, y ! 2)#x 2 % 1$

x ! ')4x ! "')4y ! sec2xy ! cos x14.

x ! y 2 " 1x ! 1 " y 2

x % y ! 0x % y 2 ! 2

x " 2y ! 3y 2 ! x11.

y ! 1 " x 2y ! x 4 " x 2

y ! x 2 % 3y ! 4x 2

y ! 1 % sx, y ! #3 % x$)3

y ! x 2y ! x

y ! sin x, y ! e x, x ! 0, x ! ')2

y ! x % 1, y ! 9 " x 2, x ! "1, x ! 2

x

x=¥-4y y

(_3,!3)

x=2y-¥

x

x=¥-2

x=ey

y=1

y=_1

y3. 4.

y=x

y=5x-"

x

y

(4,!4) x=2

y=œ„„„„x+2

y=1

x+1x

y1. 2.

Exercises ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !6.1

t t

0 0 0 6 69 801 20 22 7 75 862 32 37 8 81 933 46 52 9 86 984 54 61 10 90 1025 62 71

vKvCvKvC

; 30. Graph the parametric curve , . Findthe area enclosed between this curve and the line .

; 31. Graph the region bounded by the curve , ,, and the lines and . Then find the

area of this region.

32. Graph the astroid , and set up anintegral for the area that it encloses. Then use a computeralgebra system to evaluate the integral.

33. Find the area bounded by the loop of the curve with para-metric equations .

; 34. Estimate the area of the region enclosed by the loop of thecurve .

Find the values of such that the area of the regionbounded by the parabolas and is 576.

36. Find the area of the region bounded by the parabola ,the tangent line to this parabola at , and the -axis.

37. Find the number such that the line divides theregion bounded by the curves and into tworegions with equal area.

38. (a) Find the number such that the line bisects thearea under the curve ,

(b) Find the number such that the line bisects thearea in part (a).

39. Find a positive continuous function such that the areaunder the graph of from 0 to is for all .

40. Suppose that . For what value of is the area of the region enclosed by the curves ,

, and equal to the area of the regionenclosed by the curves , , and ?

For what values of do the line and the curveenclose a region? Find the area of the

region.y ! x)#x 2 % 1$

y ! mxm41.

y ! 0x ! 'y ! cos#x " c$x ! 0y ! cos#x " c$

y ! cos xc0 , c , ')2

t - 0A#t$ ! t 3tff

y ! bb1 & x & 4.y ! 1)x 2

x ! aa

y ! 4y ! x 2y ! bb

x#1, 1$y ! x 2

y ! c 2 " x 2y ! x 2 " c 2c35.

x ! t 3 " 12t, y ! 3t 2 % 2t % 5

x ! t 2, y ! t 3 " 3t

y ! a sin3(x ! a cos3(CAS

x ! 0y ! 10 & t & ')2y ! e tx ! cos t

y ! 2.5y ! t % 1)tx ! t " 1)tthat and are measured in thousands of dollars.] What is

the meaning of the area of the shaded region? Use the Mid-point Rule to estimate the value of this quantity.

Sketch the region that lies between the curves andand between and . Notice that the

region consists of two separate parts. Find the area of thisregion.

; 26. Graph the curves and on acommon screen and observe that the region between themconsists of two parts. Find the area of this region.

27. Find the area of the crescent-shaped region (called a lune)bounded by arcs of circles with radii and (see the figure).

28. Sketch the region in the -plane defined by the inequalities, and find its area.

Use the parametric equations of an ellipse, ,, , to find the area that it encloses.0 & ( & 2'y ! b sin (

x ! a cos (29.

1 " x " , y , $ 0x " 2y 2 $ 0xy

R

r

Rr

y ! x 3 " 4x 2 % 3xy ! x 2 " x

x ! ')2x ! 0y ! sin 2xy ! cos x25.

Cª(x)

y

x0 10050

1

2

3Rª(x)

CR

SECTION 6.2 VOLUMES # 453

Volumes ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !

In trying to find the volume of a solid we face the same type of problem as in findingareas. We have an intuitive idea of what volume means, but we must make this ideaprecise by using calculus to give an exact definition of volume.

We start with a simple type of solid called a cylinder (or, more precisely, a rightcylinder). As illustrated in Figure 1(a), a cylinder is bounded by a plane region ,B1

6.2

called the base, and a congruent region in a parallel plane. The cylinder consists ofall points on line segments perpendicular to the base that join to . If the area ofthe base is and the height of the cylinder (the distance from to ) is , then thevolume of the cylinder is defined as

In particular, if the base is a circle with radius , then the cylinder is a circular cylin-der with volume [see Figure 1(b)], and if the base is a rectangle with length

and width , then the cylinder is a rectangular box (also called a rectangular paral-lelepiped) with volume [see Figure 1(c)].

For a solid S that isn’t a cylinder we first “cut” S into pieces and approximate eachpiece by a cylinder. We estimate the volume of S by adding the volumes of the cylin-ders. We arrive at the exact volume of S though a limiting process in which the num-ber of pieces becomes large.

We start by intersecting S with a plane and obtaining a plane region that is called across-section of Let be the area of the cross-section of in a plane perpen-dicular to the -axis and passing through the point , where . (See Figure 2.Think of slicing with a knife through and computing the area of this slice.) Thecross-sectional area will vary as increases from to .

Let’s divide S into n “slabs” of equal width by using the planes , , . . . toslice the solid. (Think of slicing a loaf of bread.) If we choose sample points in

, we can approximate the slab (the part of that lies between the planesand ) by a cylinder with base area and “height” . (See Figure 3.)#xA#xi*$PxiPxi"1

SSiith"xi"1, xi%xi*

Px2Px1#x

FIGURE 2

y

0 xba x

S

Px

A(a)A(b)

A(x)

baxA#x$xS

a & x & bxxPxSA#x$S.

V ! lwhwl

V ! 'r 2hr

V ! Ah

VhB2B1A

B2B1

B2

FIGURE 1

h

B¡

B™

h

r

h

l

(a) CylinderV=Ah

(b) Circular cylinderV=$r@h

(c) Rectangular boxV=lwh

w


Watch an animation of Figure 2.Resources / Module 7

/ Volumes / Volumes by Cross-Section

The volume of this cylinder is , so an approximation to our intuitive con-ception of the volume of the th slab is

Adding the volumes of these slabs, we get an approximation to the total volume (thatis, what we think of intuitively as the volume):

This approximation appears to become better and better as . (Think of the slicesas becoming thinner and thinner.) Therefore, we define the volume as the limit of thesesums as . But we recognize the limit of Riemann sums as a definite integral andso we have the following definition.

Definition of Volume Let be a solid that lies between and . If thecross-sectional area of in the plane , through x and perpendicular to the x-axis, is , where is a continuous function, then the volume of is

When we use the volume formula it is important to remember that is the area of a moving cross-section obtained by slicing through perpendicu-

lar to the -axis.

EXAMPLE 1 Show that the volume of a sphere of radius is

SOLUTION If we place the sphere so that its center is at the origin (see Figure 4), thenthe plane intersects the sphere in a circle whose radius (from the PythagoreanTheorem) is . So the cross-sectional area is

A#x$ ! 'y 2 ! '#r 2 " x 2 $

y ! sr 2 " x 2

Px

V ! 43 'r 3

r

xxA#x$

V ! xba A#x$ dx

V ! limn l !

!n

i!1 A#xi*$ #x ! yb

a A#x$ dx

SAA#x$PxS

x ! bx ! aS

n l !

n l !

V & !n

i!1 A#xi*$ #x

V#Si $ & A#xi*$ #x

SiiA#xi*$ #x

y

0 xx¶=ba=x¸ ⁄ ¤ ‹ x¢ x% xß

FIGURE 3

xi-1 xi

y

0 xba x*i

Îx

S


FIGURE 4

x

y

0 x r

r

_r

y

Using the definition of volume with and , we have

(The integrand is even.)

Figure 5 illustrates the definition of volume when the solid is a sphere with radius . From the result of Example 1, we know that the volume of the sphere is

. Here the slabs are circular cylinders (disks) and the three parts of Fig-ure 5 show the geometric interpretations of the Riemann sums

when , 10, and 20 if we choose the sample points to be the midpoints .Notice that as we increase the number of approximating cylinders, the correspondingRiemann sums become closer to the true volume.

EXAMPLE 2 Find the volume of the solid obtained by rotating about the x-axis theregion under the curve from 0 to 1. Illustrate the definition of volume bysketching a typical approximating cylinder.

SOLUTION The region is shown in Figure 6(a). If we rotate about the x-axis, we get thesolid shown in Figure 6(b). When we slice through the point x, we get a disk withradius . The area of this cross-section is

and the volume of the approximating cylinder (a disk with thickness ) is

The solid lies between and , so its volume is

V ! y1

0 A#x$ dx ! y1

0 'x dx ! '

x 2

2 (0

1

!'

2

x ! 1x ! 0

A#x$ #x ! 'x #x

#x

A#x$ ! ' (sx)2 ! 'x

sx

y ! sx

(a) Using 5 disks, VÅ4.2726 (b) Using 10 disks, VÅ4.2097 (c) Using 20 disks, VÅ4.1940

xixi*n ! 5

!n

i!1 A#xi$ #x ! !

n

i!1 ' #12 " xi

2$ #x

43 ' & 4.18879r ! 1

! 43 'r 3

! 2''r 2x "x 3

3 (0

r

! 2'*r 3 "r 3

3 + ! 2' yr

0 #r 2 " x 2 $ dx

V ! yr

"r A#x$ dx ! yr

"r '#r 2 " x 2 $ dx

b ! ra ! "r


Watch an animation of Figure 5.Resources / Module 7

/ Volumes / Volumes

$ Did we get a reasonable answer inExample 2? As a check on our work,let’s replace the given region by asquare with base and height . Ifwe rotate this square, we get a cylinderwith radius , height , and volume

. We computed that thegiven solid has half this volume. Thatseems about right.

' ! 12 ! 1 ! '11

1"0, 1%

FIGURE 5Approximating the volumeof a sphere with radius 1

EXAMPLE 3 Find the volume of the solid obtained by rotating the region bounded by, , and about the -axis.

SOLUTION The region is shown in Figure 7(a) and the resulting solid is shown in Fig-ure 7(b). Because the region is rotated about the y-axis, it makes sense to slice thesolid perpendicular to the y-axis and therefore to integrate with respect to y. If weslice at height y, we get a circular disk with radius x, where . So the area of across-section through y is

and the volume of the approximating cylinder pictured in Figure 7(b) is

Since the solid lies between y ! 0 and y ! 8, its volume is

FIGURE 7

x

y

0

y=8

x=0y=˛

orx=Œ„y

x

y

0

8

y x(x,!y)Îy

(a) (b)

! '[ 35 y 5)3 ]0

8!

96'

5

V ! y8

0 A#y$ dy ! y8

0 'y 2)3 dy

A#y$ #y ! 'y 2)3 #y

A#y$ ! 'x 2 ! ' (s3 y)2 ! 'y 2)3

x ! s3 y

yx ! 0y ! 8y ! x 3

FIGURE 6 (a) (b)

x0 x

yy=œ„x

1

œ„x

Îx

0 x

y

1


See a volume of revolution being formed.Resources / Module 7

/ Volumes / Volumes of Revolution

EXAMPLE 4 The region enclosed by the curves and is rotated aboutthe -axis. Find the volume of the resulting solid.

SOLUTION The curves and intersect at the points and . Theregion between them, the solid of rotation, and a cross-section perpendicular to the -axis are shown in Figure 8. A cross-section in the plane has the shape of a

washer (an annular ring) with inner radius and outer radius , so we find thecross-sectional area by subtracting the area of the inner circle from the area of theouter circle:

Therefore, we have

EXAMPLE 5 Find the volume of the solid obtained by rotating the region in Example 4about the line .

SOLUTION The solid and a cross-section are shown in Figure 9. Again a cross-sectionis a washer, but this time the inner radius is and the outer radius is .The cross-sectional area is

and so the volume of is

! '' x 5

5" 5

x 3

3% 4

x 2

2 (0

1

!8'

15

! ' y1

0 #x 4 " 5x 2 % 4x$ dx

V ! y1

0 A#x$ dx ! ' y1

0 "#2 " x 2 $2 " #2 " x$2 % dx

S

A#x$ ! '#2 " x 2 $2 " '#2 " x$2

2 " x 22 " x

y ! 2

FIGURE 8 (b)

x

y

(0, 0)

(1,!1)

y="y=x

!

x

y

0 x

"

A(x)

(a) (c)

! '' x 3

3"

x 5

5 (0

1

!2'

15

V ! y1

0 A#x$ dx ! y1

0 '#x 2 " x 4 $ dx

A#x$ ! 'x 2 " '#x 2 $2 ! '#x 2 " x 4 $

xx 2Pxx

#1, 1$#0, 0$y ! x 2y ! x

xy ! x 2y ! x!


The solids in Examples 1–5 are all called solids of revolution because they areobtained by revolving a region about a line. In general, we calculate the volume of asolid of revolution by using the basic defining formula

and we find the cross-sectional area or in one of the following ways:

" If the cross-section is a disk (as in Examples 1–3), we find the radius of thedisk (in terms of x or y) and use

" If the cross-section is a washer (as in Examples 4 and 5), we find the innerradius and outer radius from a sketch (as in Figures 9 and 10) andcompute the area of the washer by subtracting the area of the inner diskfrom the area of the outer disk:

The next example gives a further illustration of the procedure.

FIGURE 10

rinrout

A ! ' #outer radius$2 " ' #inner radius$2

routrin

A ! '#radius$2

A#y$A#x$

V ! yb

a A#x$ dx or V ! yd

c A#y$ dy

FIGURE 9

4

2

x0 1x

y="y=x

y=2

y

xx

y=2

2-"

"

2-x

x


Module 6.2 illustrates the forma-tion and computation of volumes

using disks, washers, and shells.

EXAMPLE 6 Find the volume of the solid obtained by rotating the region inExample 4 about the line .

SOLUTION Figure 11 shows a horizontal cross-section. It is a washer with inner radiusand outer radius , so the cross-sectional area is

The volume is

We now find the volumes of two solids that are not solids of revolution.

EXAMPLE 7 Figure 12 shows a solid with a circular base of radius 1. Parallel cross-sections perpendicular to the base are equilateral triangles. Find the volume of thesolid.

SOLUTION Let’s take the circle to be . The solid, its base, and a typicalcross-section at a distance from the origin are shown in Figure 13.

FIGURE 13

y y60° 60° BA

C

œ„3y

(c) A cross-section

x

y

0

y

x

A

B(x, y)y=œ„„„„„„1-"

(b) Its base(a) The solid

0 x

y

A

B

C

1_1

xx 2 % y 2 ! 1

FIGURE 11

x0

x=œ„y

x=y

x=_1

y

y

1

1+y

1+œ„y

! ''4y 3)2

3"

y 2

2"

y 3

3 (0

1

!'

2

! ' y1

0 (2sy " y " y 2 ) dy

V ! y1

0 A#y$ dy ! ' y1

0 [(1 % sy)2 " #1 % y$2 ] dy

! ' (1 % sy )2 " '#1 % y$2

A#y$ ! '#outer radius$2 " '#inner radius$2

1 % sy1 % y

x ! "1


FIGURE 12Computer-generated picture of thesolid in Example 7

y

x

Since lies on the circle, we have and so the base of the triangleis . Since the triangle is equilateral, we see from Figure 13(c)

that its height is . The cross-sectional area is therefore

and the volume of the solid is

EXAMPLE 8 Find the volume of a pyramid whose base is a square with side andwhose height is .

SOLUTION We place the origin at the vertex of the pyramid and the -axis along itscentral axis as in Figure 14. Any plane that passes through and is perpendicularto the -axis intersects the pyramid in a square with side of length , say. We canexpress in terms of by observing from the similar triangles in Figure 15 that

and so . [Another method is to observe that the line has slope and so its equation is .] Thus, the cross-sectional area is

The pyramid lies between and , so its volume is

NOTE ! We didn’t need to place the vertex of the pyramid at the origin in Example 8.We did so merely to make the equations simple. If, instead, we had placed the centerof the base at the origin and the vertex on the positive -axis, as in Figure 16, you can y

!L2

h 2 x 3

3 (0

h

!L2h3

V ! yh

0 A#x$ dx ! yh

0 L2

h 2 x 2 dx

x ! hx ! 0

x

y

O

x h

FIGURE 14

sx

h

Lx

y

O

P

FIGURE 15

A#x$ ! s 2 !L2

h 2 x 2

y ! Lx)#2h$L)#2h$OPs ! Lx)h

xh

!s)2L)2

!sL

xssx

xPx

xO

hL

! 2 y1

0 s3 #1 " x 2 $ dx ! 2s3'x "

x 3

3 (0

1

!4s3

3

V ! y1

"1 A#x$ dx ! y1

"1 s3 #1 " x 2 $ dx

A#x$ ! 12 ! 2s1 " x 2 ! s3s1 " x 2 ! s3 #1 " x 2 $

s3 y ! s3s1 " x 2, AB , ! 2s1 " x 2ABC

y ! s1 " x 2B


Resources / Module 7/ Volumes

/ Start of Mystery of theTopless Pyramid

x

y

h

0

y

FIGURE 16

verify that we would have obtained the integral

Cylindrical Shells

Some volume problems are very difficult to handle by the slicing methods that wehave used so far. For instance, let’s consider the problem of finding the volume of the solid obtained by rotating about the -axis the region bounded by the curve

and the -axis (see Figure 17). If we slice, then we run into a severeproblem. To compute the inner radius and the outer radius of a cross-section, wewould have to solve the cubic equation for x in terms of y; that’s noteasy. Fortunately, there is a method, called the method of cylindrical shells, that iseasier to use in such a case. We illustrate it in the next example.

EXAMPLE 9 Find the volume of the solid obtained by rotating about the -axis theregion bounded by the curve and the -axis.

SOLUTION Instead of slicing, we approximate the solid using cylindrical shells. Fig-ure 18 shows a typical approximating rectangle with width . If we rotate thisrectangle about the -axis, we get a cylindrical shell whose average radius is , themidpoint of the th subinterval.

Imagine this shell to be cut and flattened, as in Figure 19. The resulting rectan-gular slab has dimensions , , and so the volume of the shell is

2'xi #2xi2 " xi

3$ #x

2xi2 " xi

3,#x2'xi

2$xiÎx

2xi@-xi#

FIGURE 18 A cylindrical shell FIGURE 19 The flattened shell

y

x

y=2"-˛

Îx

xi

2

ixiy

#x

xy ! 2x 2 " x 3y

y

x0 2

1

y=2"-˛

xL=? xR=?

FIGURE 17

y ! 2x 2 " x 3

xy ! 2x 2 " x 3y

V ! yh

0 L2

h 2 #h " y$2 dy !L2h3


If we do this for every subinterval and add the results, we get an approximation tothe volume of the solid:

This approximation improves as increases, so it seems plausible that

It can be verified that the method of shells gives the same answer as slicing.

FIGURE 20

y

x

! 2'[ 12 x 4 " 1

5 x 5 ]0

2! 2'(8 " 32

5 ) ! 165 '

! y2

0 2'x#2x 2 " x 3 $ dx ! 2' y2

0 #2x 3 " x 4 $ dx

V ! limn l !

!n

i!1 2'xi #2 xi

2 " xi3$ #x

n

V & !n

i!1 2'xi #2xi

2 " xi3$ #x


$ Notice from Figure 18 that we obtainall shells if we let increase from to .20x

$ Figure 20 shows a computer-generated picture of the solid whose volume we computed in Example 9.

Exercises ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !6.2

1–12 " Find the volume of the solid obtained by rotating theregion bounded by the given curves about the specified line.Sketch the region, the solid, and a typical disk or “washer.”

1. , , , ; about the -axis

2.


4. , ; about the -axis

, ; about the -axis


, ; about the -axis

8. , , ; about the -axis

, ; about

10. , , , ; about

11. , ; about

12. , ; about " " " " " " " " " " " " "

x ! 2y ! sxy ! x

x ! "1x ! y 2y ! x 2

y ! "1x ! 3x ! 1y ! 0y ! 1)x

y ! 1y ! sxy ! x9.

yy ! 0x ! 1y ! x 2)3

yx ! 2yy 2 ! x7.

xx ! 1x ! "1y ! 1y ! sec x

xy 2 ! xy ! x 25.

yx ! 0x ! y " y 2

yx ! 0y ! 40 & x & 2y ! x 2

y ! e x, y ! 0, x ! 0, x ! 1; about the x-axis

xy ! 0x ! 2x ! 1y ! 1)x

13. The region enclosed by the curves and inthe first quadrant is rotated about the line . Find thevolume of the resulting solid.

14. Find the volume of the solid obtained by rotating the regionin Exercise 13 about the line .

; 15–16 " Use a graph to find approximate -coordinates of thepoints of intersection of the given curves. Then find (approxi-mately) the volume of the solid obtained by rotating about the -axis the region bounded by these curves.

15.

16." " " " " " " " " " " " "

17–18 " Each integral represents the volume of a solid.Describe the solid.

(a) (b)

18. (a) (b)

" " " " " " " " " " " " "

' y')2

0 "#1 % cos x$2 " 12 % dx' y5

2 y dy

' y1

0 #y 4 " y 8 $ dy' y')2

0 cos2x dx17.

y ! 3 sin#x 2 $, y ! e x)2 % e"2x

y ! x 2, y ! ln#x % 1$x

x

y ! 2

x ! 8y ! s3 xx ! 4y


25. A pyramid with height and rectangular base with dimen-sions and

26. A pyramid with height and base an equilateral trianglewith side (a tetrahedron)

27. A tetrahedron with three mutually perpendicular faces andthree mutually perpendicular edges with lengths 3 cm,4 cm, and 5 cm

28. The base of is a circular disk with radius . Parallel cross-sections perpendicular to the base are squares.

The base of is an elliptical region with boundary curve. Cross-sections perpendicular to the -axis

are isosceles right triangles with hypotenuse in the base.

30. The base of is the parabolic region .Cross-sections perpendicular to the -axis are equilateral triangles.

31. has the same base as in Exercise 30, but cross-sectionsperpendicular to the -axis are squares.

32. The base of is the triangular region with vertices ,, and . Cross-sections perpendicular to the -axis

are semicircles.

33. has the same base as in Exercise 32 but cross-sectionsperpendicular to the -axis are isosceles triangles withheight equal to the base.

" " " " " " " " " " " " "

34. The base of is a circular disk with radius . Parallel cross-sections perpendicular to the base are isosceles triangleswith height and unequal side in the base.(a) Set up an integral for the volume of .(b) By interpreting the integral as an area, find the volume

of .

(a) Set up an integral for the volume of a solid torus (thedonut-shaped solid shown in the figure) with radii and .

(b) By interpreting the integral as an area, find the volumeof the torus.

rR

Rr

35.

S

Sh

rS

yS

y#0, 2$#3, 0$#0, 0$S

yS

y-#x, y$ , x 2 & y & 1.S

x9x 2 % 4y 2 ! 36S29.

rS

a

aa

ah

2bbh19. A CAT scan produces equally spaced cross-sectional views

of a human organ that provide information about the organotherwise obtained only by surgery. Suppose that a CATscan of a human liver shows cross-sections spaced 1.5 cmapart. The liver is 15 cm long and the cross-sectional areas,in square centimeters, are 0, 18, 58, 79, 94, 106, 117, 128,63, 39, and 0. Use the Midpoint Rule to estimate thevolume of the liver.

20. A log 10 m long is cut at 1-meter intervals and its cross-sectional areas (at a distance from the end of the log)are listed in the table. Use the Midpoint Rule with toestimate the volume of the log.

21–33 " Find the volume of the described solid .

A right circular cone with height and base radius

22. A frustum of a right circular cone with height , lower baseradius , and top radius

A cap of a sphere with radius and height

24. A frustum of a pyramid with square base of side , squaretop of side , and height

a

b

hab

r

h

hr23.

R

h

r

rRh

rh21.

S

n ! 5xA

x (m) A ( ) x (m) A ( )

0 0.68 6 0.531 0.65 7 0.552 0.64 8 0.523 0.61 9 0.504 0.58 10 0.485 0.59

m2m2

43. Let be the solid obtained by rotating about the -axis theregion bounded by and . Explain whyit is awkward to use slicing to find the volume of . Thenfind using cylindrical shells.

44. Let be the solid obtained by rotating the region under thecurve from to about the -axis. Sketch a typical cylindrical shell and find its circumference andheight. Use shells to find the volume of . Do you think thismethod is preferable to slicing? Explain.

45. If the region shown in the figure is rotated about the -axisto form a solid, use Simpson’s Rule with to estimatethe volume of the solid.

46. Let be the volume of the solid obtained by rotating aboutthe -axis the region bounded by and . Find both by slicing and by cylindrical shells. In both cases drawa diagram to explain your method.

47. Use cylindrical shells to find the volume of the solidobtained by rotating the region bounded by and

about the line . Sketch the region and a typicalshell. Explain why this method is preferable to slicing.

48. Suppose you make napkin rings by drilling holes with dif-ferent diameters through two wooden balls (which also havedifferent diameters). You discover that both napkin ringshave the same height , as shown in the figure.(a) Guess which ring has more wood in it.(b) Check your guess: Use cylindrical shells to compute the

volume of a napkin ring created by drilling a hole withradius through the center of a sphere of radius andexpress the answer in terms of .

h

hRr

h

x ! 2y ! 0y ! x " x 2

Vy ! x 2y ! xyV

0 4

4

102 86

2

y

x

n ! 8y

S

ys'0y ! sin#x 2$S

VSV

y ! 0y ! x#x " 1$2yS36. A wedge is cut out of a circular cylinder of radius 4 by two

planes. One plane is perpendicular to the axis of the cylin-der. The other intersects the first at an angle of along adiameter of the cylinder. Find the volume of the wedge.

37. (a) Cavalieri’s Principle states that if a family of parallelplanes gives equal cross-sectional areas for two solids

and , then the volumes of and are equal. Prove this principle.

(b) Use Cavalieri’s Principle to find the volume of theoblique cylinder shown in the figure.

38. Find the volume common to two circular cylinders, eachwith radius , if the axes of the cylinders intersect at rightangles.

Find the volume common to two spheres, each with radius , if the center of each sphere lies on the surface of the

other sphere.

40. A bowl is shaped like a hemisphere with diameter 30 cm. Aball with diameter 10 cm is placed in the bowl and water ispoured into the bowl to a depth of centimeters. Find thevolume of water in the bowl.

41. A hole of radius is bored through a cylinder of radiusat right angles to the axis of the cylinder. Set up, but

do not evaluate, an integral for the volume cut out.

42. A hole of radius is bored through the center of a sphere ofradius . Find the volume of the remaining portion ofthe sphere.

R - rr

R - rr

h

r39.

r

h

r

S2S1S2S1

30.



Rotating on a Slant

We know how to find the volume of a solid of revolution obtained by rotating a regionabout a horizontal or vertical line (see Section 6.2). But what if we rotate about a slantedline, that is, a line that is neither horizontal nor vertical? In this project you are asked todiscover a formula for the volume of a solid of revolution when the axis of rotation is aslanted line.

Let be the arc of the curve between the points and andlet be the region bounded by , by the line (which lies entirely below ),and by the perpendiculars to the line from and .

1. Show that the area of is

[Hint: This formula can be verified by subtracting areas, but it will be helpful throughoutthe project to derive it by first approximating the area using rectangles perpendicular tothe line, as shown in the figure. Use part (a) of the figure to help express in terms of .]

2. Find the area of the region shown in part (b) of the figure.

3. Find a formula (similar to the one in Problem 1) for the volume of the solid obtainedby rotating about the line .

4. Find the volume of the solid obtained by rotating the region of Problem 2 about theline .y ! x " 2

y ! mx % b!

(a) (b)

y=mx+b

Îu

å

tangent to Cat { x i, f(xi)}

xi &

?

Îx

?

y

x0

(2$,!2$)

y=x+sin!x

y=x-2

#x#u

11 % m 2 yq

p " f #x$ " mx " b%"1 % mf +#x$% dx

!

FIGURE 1

P

0 x

y

qp

!

C

Q

y=ƒ

y=mx+b

Îu

QPCy ! mx % bC!

Q#q, f #q$$P#p, f #p$$y ! f #x$C

DiscoveryProject

What do we mean by the length of a curve? We might think of fitting a piece of stringto the curve in Figure 1 and then measuring the string against a ruler. But that mightbe difficult to do with much accuracy if we have a complicated curve. We need a pre-cise definition for the length of an arc of a curve, in the same spirit as the definitionswe developed for the concepts of area and volume.

If the curve is a polygon, we can easily find its length; we just add the lengths ofthe line segments that form the polygon. (We can use the distance formula to find thedistance between the endpoints of each segment.) We are going to define the length ofa general curve by first approximating it by a polygon and then taking a limit as thenumber of segments of the polygon is increased. This process is familiar for the caseof a circle, where the circumference is the limit of lengths of inscribed polygons (seeFigure 2).

Suppose that a curve is described by the parametric equations

Let’s assume that is smooth in the sense that the derivatives and are con-tinuous and not simultaneously zero for . (This ensures that has no sud-den change in direction.) We divide the parameter interval into subintervals ofequal width . If , , , . . . , are the endpoints of these subintervals, then

and are the coordinates of points that lie on and thepolygon with vertices , , . . . , approximates . (See Figure 3.) The length L of

is approximately the length of this polygon and the approximation gets better as welet n increase. (See Figure 4, where the arc of the curve between and has beenmagnified and approximations with successively smaller values of are shown.)Therefore, we define the length of to be the limit of the lengths of these inscribedpolygons:

Notice that the procedure for defining arc length is very similar to the procedurewe used for defining area and volume: We divided the curve into a large number ofsmall parts. We then found the approximate lengths of the small parts and added them.Finally, we took the limit as .

For computational purposes we need a more convenient expression for . If we letand , then the length of the th line segment of the

polygon is

, Pi"1Pi , ! s##xi$2 % ##yi$2

i#yi ! yi " yi"1#xi ! xi " xi"1

Ln l !

L ! limn l !

!n

i!1 , Pi"1Pi ,

C#t

PiPi"1

CCPnP1P0

CPi#xi, yi$yi ! t#ti$xi ! f #ti$tnt2t1t0#t

n"a, b%Ca , t , bt+#t$f +#t$C

a & t & by ! t#t$x ! f #t$

C

FIGURE 2

SECTION 6.3 ARC LENGTH # 467

Arc Length ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !6.3

FIGURE 1

0

y

x

P¸

P¡

P™ Pi _1

Pi

Pn

C

FIGURE 3

FIGURE 4

Pi

Pi-¡

Pi

Pi-¡

Pi

Pi-¡

Pi

Pi-¡

But from the definition of a derivative we know that

if is small. (We could have used any sample point in place of .) Therefore

and so

Thus

This is a Riemann sum for the function and so our argument sug-gests that

In fact, our reasoning can be made precise; this formula is correct, provided that werule out situations where a portion of the curve is traced out more than once.

Arc Length Formula If a smooth curve with parametric equations ,, , is traversed exactly once as increases from to , then

its length is

EXAMPLE 1 Find the length of the arc of the curve , that lies betweenthe points and . (See Figure 5.)

SOLUTION First we notice from the equations and that the portion of thecurve between and corresponds to the parameter interval . Sothe arc length formula (1) gives

If we substitute , then . When , ; when ,t ! 2u ! 13t ! 1du ! 18t dtu ! 4 % 9t 2

! y2

1 ts4 % 9t 2 dt

! y2

1 s4t 2 % 9t 4 dt

L ! y2

1/*dx

dt +2

% *dydt +2

dt ! y2

1 s#2t$2 % #3t2$2 dt

1 & t & 2#4, 8$#1, 1$y ! t 3x ! t 2

#4, 8$#1, 1$y ! t 3x ! t 2

L ! yb

a/*dx

dt +2

% *dydt +2

dt

bata & t & by ! t#t$x ! f #t$1

L ! yb

a s" f +#t$%2 % "t+#t$%2 dt

s" f +#t$%2 % "t+#t$%2

L & !n

i!1 s" f +#ti$%2 % "t+#ti$%2 #t

! s" f +#ti$%2 % "t+#ti$%2 #t

& s" f +#ti$ #t% 2 % "t+#ti$ #t%2

, Pi"1Pi , ! s##xi$2 % ##yi $2

#yi & t+#ti$ #t#xi & f +#ti$ #t

titi*#t

f +#ti$ &#xi

#t


$ As a check on our answer to Example1, notice from Figure 5 that it ought tobe slightly larger than the distance from

to , which is

According to our calculation inExample 1, we have

Sure enough, this is a bit greater thanthe length of the line segment.

L ! 127 (80s10 " 13s13 ) & 7.633705

s58 & 7.615773

#4, 8$#1, 1$

FIGURE 5

0 x

y

x=t@

(1,!1)

(4,!8)

y=t#

. Therefore

If we are given a curve with equation , , then we can regard as a parameter. Then parametric equations are , , and Formula 1

becomes

Similarly, if a curve has the equation , , we regard as the param-eter and the length is

Because of the presence of the root sign in Formulas 1, 2, and 3, the calculation ofan arc length often leads to an integral that is very difficult or even impossible to eval-uate explicitly. Thus, we often have to be content with finding an approximation to thelength of a curve as in the following example.

EXAMPLE 2 Estimate the length of the portion of the hyperbola from the pointto the point .

SOLUTION We have

and so, from Formula 2, the length is

It is impossible to evaluate this integral exactly, so let’s use Simpson’s Rule (seeSection 5.9) with , , , , and . Thus

& 1.1321

&#x3

" f #1$ % 4 f #1.1$ % 2 f #1.2$ % 4 f #1.3$ % / / / % 2 f #1.8$ % 4 f #1.9$ % f #2$%

L ! y2

1 /1 %

1x 4 dx

f #x$ ! s1 % 1)x 4 #x ! 0.1n ! 10b ! 2a ! 1

L ! y2

1 /1 % *dy

dx+2

dx ! y2

1 /1 %

1x 4 dx

dydx

! "1x 2y !

1x

(2, 12 )#1, 1$

xy ! 1

L ! yb

a /*dx

dy+2

% 1 dy3

ya & y & bx ! f #y$

L ! yb

a /1 % *dy

dx+2

dx2

y ! f #x$x ! xxa & x & by ! f #x$

! 127 [403)2

" 133)2 ] ! 127 (80s10 " 13s13)

L ! 118 y40

13 su du ! 1

18 ! 23 u 3)2]13

40

u ! 40


Checking the value of the definite integral with a more accurate approximation produced by a computer algebra system, we see that the approximation using Simpson’s Rule is accurate to four decimal places.

EXAMPLE 3 Find the length of the arc of the parabola from to .

SOLUTION Since , we have , and Formula 3 gives

Using either a computer algebra system or the Table of Integrals (use Formula 21after substituting ), we find that

EXAMPLE 4 Find the length of one arch of the cycloid ,.

SOLUTION From Example 7 in Section 1.7 we see that one arch is described by theparameter interval . Since

we have

This integral could be evaluated after using further trigonometric identities.Instead we use a computer algebra system:

L ! r y2'

0s2#1 " cos ($ d( ! 8r

! y2'

0 sr 2#1 " 2 cos ( % cos2( % sin2($ d( ! r y2'

0 s2#1 " cos ($ d(

L ! y2'

0 /* dx

d(+2

% * dyd(+2

d( ! y2'

0 sr 2#1 " cos ($2 % r 2 sin2( d(

dyd(

! r sin (anddxd(

! r#1 " cos ($

0 & ( & 2'

y ! r#1 " cos ($x ! r#( " sin ($

0 x

y

1

1

x=¥

FIGURE 6

L !s52

%ln(s5 % 2)

4

u ! 2y

L ! y1

0 /*dx

dy+2

% 1 dy ! y1

0 s4y 2 % 1 dy

dx)dy ! 2yx ! y 2

#1, 1$#0, 0$y 2 ! x


n

1 1.4142 1.4454 1.4648 1.472

16 1.47632 1.47864 1.479

Ln

$ Figure 6 shows the arc of theparabola whose length is computed inExample 3, together with polygonalapproximations having and line segments, respectively. For the approximate length is , thediagonal of a square. The table showsthe approximations that we get bydividing into equal subintervals.Notice that each time we double thenumber of sides of the polygon, we getcloser to the exact length, which is

L !s52

%ln(s5 % 2)

4& 1.478943

n"0, 1%Ln

L1 ! s2n ! 1

n ! 2n ! 1

$ The result of Example 4 says that thelength of one arch of a cycloid is eighttimes the radius of the generating circle(see Figure 7). This was first proved in1658 by Sir Christopher Wren, wholater became the architect of St. Paul’sCathedral in London.

FIGURE 7

0

y

x2$r

r

L=8r


intervals.) Illustrate by sketching these polygons (as inFigure 6).

(c) Set up an integral for the length of the curve.(d) If your calculator (or CAS) evaluates definite integrals,

use it to find the length of the curve to four decimalplaces. If not, use Simpson’s Rule. Compare with theapproximations in part (b).

; 16. Repeat Exercise 15 for the curve

17–20 " Use either a CAS or a table of integrals to find theexact length of the curve.

, ,

18.

,

20. ," " " " " " " " " " " " "

21. A hawk flying at at an altitude of 180 m accidentallydrops its prey. The parabolic trajectory of the falling prey isdescribed by the equation

until it hits the ground, where is its height above theground and is the horizontal distance traveled in meters.Calculate the distance traveled by the prey from the time itis dropped until the time it hits the ground. Express youranswer correct to the nearest tenth of a meter.

22. A steady wind blows a kite due west. The kite’s heightabove ground from horizontal position to is given by

Find the distance traveled by the kite.

23. A manufacturer of corrugated metal roofing wants toproduce panels that are 28 in. wide and 2 in. thick by pro-cessing flat sheets of metal as shown in the figure. The pro-file of the roofing takes the shape of a sine wave. Verify that the sine curve has equation and find

28 in2 inw

y ! sin#'x)7$

y ! 150 " 140 #x " 50$2

x ! 80 ftx ! 0

xy

y ! 180 "x 2

45

15 m)s

0 & x & 1y ! e x

0 & x & ')4y ! ln#cos x$19.

0 & y & 12x ! ln#1 " y 2$

0 & t & 1y ! t 4x ! t 317.

CAS

0 & x & 2'y ! x % sin x

1. Use the arc length formula (2) to find the length of thecurve , . Check your answer bynoting that the curve is a line segment and calculating itslength by the distance formula.

2. (a) In Example 2 in Section 1.7 we showed that theparametric equations , , ,represent the unit circle. Use these equations to showthat the length of the unit circle has the expected value.

(b) In Example 3 in Section 1.7 we showed that theequations , , , also rep-resent the unit circle. What value does the integral inFormula 1 give? How do you explain the discrepancy?

3–4 " Set up, but do not evaluate, an integral that representsthe length of the curve.

3. , ,

4. ," " " " " " " " " " " " "

; 5–10 " Graph the curve and find its exact length.

, ,

6. , ,

,

8. ,

9. , ,

10. , ,

" " " " " " " " " " " " "

11–13 " Use Simpson’s Rule with to estimate the arclength of the curve.

11. , ,

12. ,

13. ," " " " " " " " " " " " "

14. In Exercise 35 in Section 1.7 you were asked to derive theparametric equations , for thecurve called the witch of Maria Agnesi. Use Simpson’s Rulewith to estimate the length of the arc of this curvegiven by .

; 15. (a) Graph the curve , .(b) Compute the lengths of inscribed polygons with ,

, and sides. (Divide the interval into equal sub-42n ! 1

0 & x & 4y ! x s3 4 " x

')4 & ( & ')2n ! 4

y ! 2a sin2(x ! 2a cot (

0 & x & 'y ! sin x

0 & x & ')4y ! tan x

1 & t & 2y ! e"tx ! ln t

n ! 10

0 & ( & 'y ! a#sin ( " ( cos ($x ! a#cos ( % ( sin ($

"8 & t & 3y ! 4e t)2x ! e t " t

12 & x & 1y !

x 3

6%

12x

0 & y & 1x ! y 3)27.

0 & t & 3y ! 5 " 2tx ! e t % e"t

0 & t & 'y ! e t sin tx ! e t cos t5.

0 & x & 3y ! 2x

1 & t & 2y ! 43 t 3)2x ! t " t 2

0 & t & 2'y ! cos 2tx ! sin 2t

0 & t & 2'y ! sin tx ! cos t

"2 & x & 1y ! 2 " 3x

Exercises ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !6.3

Arc Length Contest

The curves shown are all examples of graphs of continuous functions that have the fol-lowing properties.

1.

2.

3. The area under the graph of from 0 to 1 is equal to 1.

The lengths of these curves, however, are different.

Try to discover formulas for two functions that satisfy the given conditions 1, 2, and 3.(Your graphs might be similar to the ones shown or could look quite different.) Then calcu-late the arc length of each graph. The winning entry will be the one with the smallest arclength.

LÅ3.249

x

y

0 1

1

LÅ2.919

x

y

0 1

1

LÅ3.152

x

y

0 1

1

LÅ3.213

x

y

0 1

1

L

f

f #x$ $ 0 for 0 & x & 1

f #0$ ! 0 and f #1$ ! 0

f

DiscoveryProject

27. (a) Graph the epitrochoid with equations

What parameter interval gives the complete curve?(b) Use your CAS to find the approximate length of this

curve.

28. A curve called Cornu’s spiral is defined by the parametricequations

where and are the Fresnel functions that were intro-duced in Section 5.4.(a) Graph this curve. What happens as and as

?(b) Find the length of Cornu’s spiral from the origin to the

point with parameter value .t

t l "!t l !

SC

y ! S#t$ ! yt

0 sin#'u 2)2$ du

x ! C#t$ ! yt

0 cos#'u 2)2$ du

CAS

y ! 11 sin t " 4 sin#11t)2$

x ! 11 cos t " 4 cos#11t)2$

CASthe width of a flat metal sheet that is needed to make a28-inch panel. (If your calculator or CAS evaluates definiteintegrals, use it. Otherwise, use Simpson’s Rule.)

24. Find the total length of the astroid , ,where

25. Show that the total length of the ellipse ,, , is

where is the eccentricity of the ellipse ( , where).

; The curves with equations , , , , . . . ,are called fat circles. Graph the curves with , , , ,and to see why. Set up an integral for the length ofthe fat circle with . Without attempting to evaluatethis integral, state the value of

limk l !

L2k

n ! 2kL2k10

864n ! 286n ! 4x n % y n ! 126.

c ! sa 2 " b 2

e ! c)ae

L ! 4a y')2

0 s1 " e 2 sin2( d(

a - b - 0y ! b cos (x ! a sin (

a - 0.y ! a sin3(x ! a cos3(

w


SECTION 6.4 AVERAGE VALUE OF A FUNCTION # 473

Average Value of a Function ! ! ! ! ! ! ! ! ! ! ! !

It is easy to calculate the average value of finitely many numbers , , . . . , :

But how do we compute the average temperature during a day if infinitely many tem-perature readings are possible? Figure 1 shows the graph of a temperature function

, where is measured in hours and in C, and a guess at the average tempera-ture, .

In general, let’s try to compute the average value of a function ,We start by dividing the interval into equal subintervals, each with length

. Then we choose points , . . . , in successive subintervals andcalculate the average of the numbers , . . . , :

(For example, if represents a temperature function and , this means that wetake temperature readings every hour and then average them.) Since ,we can write and the average value becomes

If we let increase, we would be computing the average value of a large number ofclosely spaced values. (For example, we would be averaging temperature readingstaken every minute or even every second.) The limiting value is

by the definition of a definite integral.Therefore, we define the average value of on the interval as

EXAMPLE 1 Find the average value of the function on the interval .""1, 2%

f #x$ ! 1 % x 2

fave !1

b " a yb

a f #x$ dx

"a, b%f

limn l !

1

b " a !

n

i!1 f #x i*$ #x !

1b " a

yb

a f #x$ dx

n

!1

b " a !

n

i!1 f #x i*$ #x

f #x 1*$ % / / / % f #xn*$

b " a#x

!1

b " a " f #x 1*$ #x % / / / % f #xn*$ #x%

n ! #b " a$)#x#x ! #b " a$)n

n ! 24f

f #x 1*$ % / / / % f #xn*$n

f #xn*$f #x1*$xn*x 1*#x ! #b " a$)n

n"a, b%a & x & b.y ! f #x$

Tave

.TtT#t$

yave ! y1 % y2 % / / / % yn

n

yny2y1

6.4

0 t

T

Tave

5

10

15

12

6

18 24

FIGURE 1

$ For a positive function, we can thinkof this definition as saying

areawidth

! average height

SOLUTION With and we have

If is the temperature at time , we might wonder if there is a specific time whenthe temperature is the same as the average temperature. For the temperature functiongraphed in Figure 1, we see that there are two such times––just before noon and justbefore midnight. In general, is there a number at which the value of a function isexactly equal to the average value of the function, that is, ? The followingtheorem says that this is true for continuous functions.

The Mean Value Theorem for Integrals If is continuous on , then there existsa number in such that

that is,

The Mean Value Theorem for Integrals is a consequence of the Mean Value Theo-rem for derivatives and the Fundamental Theorem of Calculus. The proof is outlinedin Exercise 17.

The geometric interpretation of the Mean Value Theorem for Integrals is that, forpositive functions , there is a number such that the rectangle with base andheight has the same area as the region under the graph of from to (see Fig-ure 2 and the more picturesque interpretation in the margin note).

EXAMPLE 2 Since is continuous on the interval , the MeanValue Theorem for Integrals says there is a number in such that

In this particular case we can find explicitly. From Example 1 we know that, so the value of c satisfies

Therefore

Thus, in this case there happen to be two numbers in the interval that work in the Mean Value Theorem for Integrals.

Examples 1 and 2 are illustrated by Figure 3.

EXAMPLE 3 Show that the average velocity of a car over a time interval is thesame as the average of its velocities during the trip.

"t1, t2 %

""1, 2%c ! 01

c 2 ! 1so1 % c 2 ! 2

f #c$ ! fave ! 2

fave ! 2c

y2

"1 #1 % x 2 $ dx ! f #c$"2 " #"1$%

""1, 2%c""1, 2%f #x$ ! 1 % x 2

baff #c$"a, b%cf

yb

a f #x$ dx ! f #c$#b " a$

f #c$ ! fave !1

b " a yb

a f #x$ dx

"a, b%c"a, b%f

f #c$ ! fave

fc

tT#t$

!13

'x %x 3

3 (2

"1

! 2

fave !1

b " a yb

a f #x$ dx !

12 " #"1$

y2

"1 #1 % x 2 $ dx

b ! 2a ! "1


$ You can always chop off the top of a (two-dimensional) mountain at a cer-tain height and use it to fill in the valleysso that the mountaintop becomes com-pletely flat.

0 x

y

1 2_1

(_1,!2)

(2,!5)

y=1+"

fave=2

FIGURE 3

FIGURE 2

0 x

y

a c b

y=ƒ

f(c)=fave

SOLUTION If is the displacement of the car at time , then, by definition, the aver-age velocity of the car over the interval is

On the other hand, the average value of the velocity function on the interval is

(by the Total Change Theorem)

!s#t2 $ " s#t1$

t2 " t1! average velocity

!1

t2 " t1 "s#t2 $ " s#t1$%

vave !1

t2 " t1 yt2

t1

v#t$ dt !1

t2 " t1 yt2

t1

s+#t$ dt

#s#t

!s#t2 $ " s#t1$

t2 " t1

ts#t$

SECTION 6.4 AVERAGE VALUE OF A FUNCTION # 475

(b) At what time was the instantaneous velocity equal to theaverage velocity?

13. Household electricity is supplied in the form of alternating current that varies from V to V with a frequency of 60 cycles per second (Hz). The voltage is thus given bythe equation

where is the time in seconds. Voltmeters read the RMS(root-mean-square) voltage, which is the square root of theaverage value of over one cycle.(a) Calculate the RMS voltage of household current.(b) Many electric stoves require an RMS voltage of 220 V.

Find the corresponding amplitude needed for the volt-age .

14. If a freely falling body starts from rest, then its displace-ment is given by . Let the velocity after a time be

. Show that if we compute the average of the velocitieswith respect to we get , but if we compute theaverage of the velocities with respect to we get .

15. Use the result of Exercise 59 in Section 5.5 to compute theaverage volume of inhaled air in the lungs in one respiratorycycle.

vave ! 23 vTs

vave ! 12 vTt

vT

Ts ! 12 tt 2

E#t$ ! A sin#120't$A

"E#t$%2

t

E#t$ ! 155 sin#120' t$

"155155

4 t (seconds)

20

0 8 12

40

60

#(km/h)

1–4 " Find the average value of the function on the given interval.

1.

2.

3.

4." " " " " " " " " " " " "

5–8 "

(a) Find the average value of on the given interval.(b) Find such that .(c) Sketch the graph of and a rectangle whose area is the

same as the area under the graph of .

5. ,

6.

; ,

; 8. ," " " " " " " " " " " " "

If is continuous and , show that takes onthe value 4 at least once on the interval .

10. Find the numbers such that the average value ofon the interval is equal to 3.

In a certain city the temperature (in F) hours after 9 A.M.was modeled by the function

Find the average temperature during the period from 9 A.M. to 9 P.M.

12. The velocity graph of an accelerating car is shown.(a) Estimate the average velocity of the car during the first

12 seconds.

T#t$ ! 50 % 14 sin 't12

t.11.

"0, b%f #x$ ! 2 % 6x " 3x 2b

"1, 3%fx3

1 f #x$ dx ! 8f9.

[0, s' ]f #x$ ! x sin#x 2 $"0, 2%f #x$ ! x 3 " x % 17.

f #x$ ! ln x, "1, 3%"0, 2%f #x$ ! 4 " x 2

ff

fave ! f #c$cf

h#r$ ! 3)#1 % r$2, "1, 6%f #t$ ! te"t 2

, "0, 5%t#x$ ! sx, "1, 4%t#x$ ! cos x, "0, ')2%

Exercises ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !6.4

Where to Sit at the Movies

A movie theater has a screen that is positioned 10 ft off the floor and is 25 ft high. The firstrow of seats is placed 9 ft from the screen and the rows are set 3 ft apart. The floor of theseating area is inclined at an angle of above the horizontal and the distance up theincline that you sit is . The theater has 21 rows of seats, so . Suppose youdecide that the best place to sit is in the row where the angle subtended by the screen at your eyes is a maximum. Let’s also suppose that your eyes are 4 ft above the floor, asshown in the figure. (In Exercise 40 in Section 4.6 we looked at a simpler version of thisproblem, where the floor is horizontal, but this project involves a more complicated situationand requires technology.)

1. Show that

where

and

2. Use a graph of as a function of to estimate the value of that maximizes . In whichrow should you sit? What is the viewing angle in this row?

3. Use your computer algebra system to differentiate and find a numerical value for theroot of the equation . Does this value confirm your result in Problem 2?

4. Use the graph of to estimate the average value of on the interval . Thenuse your CAS to compute the average value. Compare with the maximum and minimumvalues of .(

0 & x & 60((

d()dx ! 0(

((xx(

b 2 ! #9 % x cos )$2 % #x sin ) " 6$2

a 2 ! #9 % x cos )$2 % #31 " x sin )$2

( ! arccos*a 2 % b 2 " 6252ab +

(0 & x & 60x

) ! 20.

CAS

AppliedProject


25!ft

10!ft

9!ft

¨

x

å

4!ft

Among the many applications of integral calculus to physics and engineering, we con-sider three: work, force due to water pressure, and centers of mass. As with our pre-vious applications to geometry (areas, volumes, and lengths), our strategy is to breakup the physical quantity into a large number of small parts, approximate each smallpart, add the results, take the limit, and evaluate the resulting integral.

Applications to Physics and Engineering ! ! ! ! ! ! ! ! !6.5

$ As a consequence of a calculation of work, you will be able to compute the velocity needed for a rocket toescape Earth’s gravitational field. (See Exercise 18.)

16. The velocity of blood that flows in a blood vessel withradius and length at a distance from the central axis is

where is the pressure difference between the ends of thevessel and is the viscosity of the blood (see Example 7 inSection 3.3). Find the average velocity (with respect to )over the interval . Compare the average velocitywith the maximum velocity.

0 & r & Rr

1P

v#r$ !P

41l #R2 " r 2 $

rlRv Prove the Mean Value Theorem for Integrals by applying

the Mean Value Theorem for derivatives (see Section 4.3) tothe function .

18. If denotes the average value of on the intervaland , show that

fave "a, b% !c " ab " a

fave "a, c% %b " cb " a

fave "c, b%

a , c , b"a, b%ffave "a, b%

F#x$ ! xxa f #t$ dt

17.

SECTION 6.5 APPLICATIONS TO PHYSICS AND ENGINEERING # 477

Work

The term work is used in everyday language to mean the total amount of effortrequired to perform a task. In physics it has a technical meaning that depends on theidea of a force. Intuitively, you can think of a force as describing a push or pull on anobject—for example, a horizontal push of a book across a table or the downward pullof Earth’s gravity on a ball. In general, if an object moves along a straight line withposition function , then the force F on the object (in the same direction) is definedby Newton’s Second Law of Motion as the product of its mass and its acceleration:

In the SI metric system, the mass is measured in kilograms (kg), the displacement inmeters (m), the time in seconds (s), and the force in newtons ( ). Thus, aforce of 1 N acting on a mass of 1 kg produces an acceleration of 1 m)s . In the U.S.Customary system the fundamental unit is chosen to be the unit of force, which is thepound.

In the case of constant acceleration, the force is also constant and the work doneis defined to be the product of the force and the distance that the object moves:

If is measured in newtons and in meters, then the unit for is a newton-meter,which is called a joule (J). If is measured in pounds and in feet, then the unit for

is a foot-pound (ft-lb), which is about 1.36 J.For instance, suppose you lift a 1.2-kg book off the floor to put it on a desk that is

0.7 m high. The force you exert is equal and opposite to that exerted by gravity, soEquation 1 gives

and then Equation 2 gives the work done as

But if a 20-lb weight is lifted 6 ft off the ground, then the force is given as lb,so the work done is

Here we didn’t multiply by because we were given the weight (a force) and not themass.

Equation 2 defines work as long as the force is constant, but what happens if theforce is variable? Let’s suppose that the object moves along the -axis in the positivedirection, from to , and at each point between and a force actson the object, where is a continuous function. We divide the interval into nsubintervals with endpoints and equal width . We choose a samplepoint in the th subinterval . Then the force at that point is . If islarge, then is small, and since is continuous, the values of don’t change verymuch over the interval . In other words, is almost constant on the interval f"xi"1, xi %

ff#xnf #xi*$"xi"1, xi %ixi*

#xx0, x1, . . . , xn

"a, b%ff #x$baxx ! bx ! a

x

t

W ! Fd ! 20 ! 6 ! 120 ft-lb

F ! 20

W ! Fd ! #11.76$#0.7$ & 8.2 J

F ! mt ! #1.2$#9.8$ ! 11.76 N

WdF

WdF

work ! force 2 distanceW ! Fd2

dFF

2N ! kg/m)s2

F ! m d 2sdt 21

ms#t$

and so the work that is done in moving the particle from to is approximatelygiven by Equation 2:

Thus, we can approximate the total work by

It seems that this approximation becomes better as we make larger. Therefore, wedefine the work done in moving the object from a to b as the limit of this quantityas . Since the right side of (3) is a Riemann sum, we recognize its limit as beinga definite integral and so

EXAMPLE 1 When a particle is located at a distance feet from the origin, a force of pounds acts on it. How much work is done in moving it from to ?

SOLUTION

The work done is ft-lb.

In the next example we use a law from physics: Hooke’s Law states that the forcerequired to maintain a spring stretched units beyond its natural length is proportional to :

where is a positive constant (called the spring constant). Hooke’s Law holds pro-vided that is not too large (see Figure 1).

EXAMPLE 2 A force of 40 N is required to hold a spring that has been stretched fromits natural length of 10 cm to a length of 15 cm. How much work is done in stretch-ing the spring from 15 cm to 18 cm?

SOLUTION According to Hooke’s Law, the force required to hold the spring stretched meters beyond its natural length is . When the spring is stretched

from 10 cm to 15 cm, the amount stretched is cm m. This means thatso

Thus, and the work done in stretching the spring from 15 cm to 18 cm is

! 400"#0.08$2 " #0.05$2% ! 1.56 J

W ! y0.08

0.05 800x dx ! 800

x 2

2 (0.05

0.08

f #x$ ! 800x

k ! 400.05 ! 8000.05k ! 40

f #0.05$ ! 40,! 0.055

f #x$ ! kxx

xk

f #x$ ! kx

xx

16 23

W ! y3

1 #x 2 % 2x$ dx !

x 3

3% x 2(

1

3

!503

x ! 3x ! 1x 2 % 2x

x

W ! limn l !

!n

i!1 f #xi*$ #x ! yb

a f #x$ dx

n l !

n

W & !n

i!1 f #xi*$ #x3

Wi & f #xi*$ #x

xixi"1Wi


FIGURE 1Hooke’s Law

x0frictionlesssurface

x0 x

ƒ=kx

(a) Natural position of spring

(b) Stretched position of spring

EXAMPLE 3 A tank has the shape of an inverted circular cone with height 10 m andbase radius 4 m. It is filled with water to a height of 8 m. Find the work required toempty the tank by pumping all of the water to the top of the tank. (The density ofwater is 1000 kg)m .)

SOLUTION Let’s measure depths from the top of the tank by introducing a verticalcoordinate line as in Figure 2. The water extends from a depth of 2 m to a depth of 10 m and so we divide the interval into n subintervals with endpoints

and choose in the th subinterval. This divides the water into layers. The th layer is approximated by a circular cylinder with radius and height

. We can compute from similar triangles, using Figure 3, as follows:

Thus, an approximation to the volume of the th layer of water is

and so its mass is

The force required to raise this layer must overcome the force of gravity and so

Each particle in the layer must travel a distance of approximately . The workdone to raise this layer to the top is approximately the product of the force and thedistance :

To find the total work done in emptying the entire tank, we add the contributions ofeach of the layers and then take the limit as :

Hydrostatic Pressure and Force

Deep-sea divers realize that water pressure increases as they dive deeper. This isbecause the weight of the water above them increases.

In general, suppose that a thin horizontal plate with area square meters is sub-merged in a fluid of density kilograms per cubic meter at a depth meters below thesurface of the fluid as in Figure 4. The fluid directly above the plate has volume

, so its mass is . The force exerted by the fluid on the plate ism ! 3V ! 3AdV ! Ad

d3A

! 1568'( 20483 ) & 3.4 2 106 J

! 1568' y10

2 #100x " 20x 2 % x 3 $ dx ! 1568''50x 2 "

20x 3

3%

x 4

4 (2

10

W ! limn l !

!n

i!1 1568'xi*#10 " xi*$2 #x ! y10

2 1568'x#10 " x$2 dx

n l !n

Wi & Fi xi* & 1568'xi*#10 " xi*$2 #x

xi*Fi

Wixi*

& 1568' #10 " xi*$2 #x

Fi ! mit & #9.8$160'#10 " xi*$2 #x

& 1000 !4'

25 #10 " xi*$2 #x ! 160'#10 " xi*$2 #x

mi ! density 2 volume

Vi & 'ri2 #x !

4'

25 #10 " xi*$2 #x

i

ri ! 25 #10 " xi*$

ri

10 " xi*!

410

ri#xrii

nixi*x0, x1, . . . , xn

"2, 10%

3


FIGURE 2

4

10

10-xi*

ri

FIGURE 3

0

x

2 m

4 m

10 m

x i*

Îx

ri

surface of fluid

d

A

FIGURE 4

therefore

where is the acceleration due to gravity. The pressure on the plate is defined to bethe force per unit area:

The SI unit for measuring pressure is newtons per square meter, which is called a pas-cal (abbreviation: 1 N)m Pa). Since this is a small unit, the kilopascal (kPa) isoften used. For instance, because the density of water is kg)m , the pres-sure at the bottom of a swimming pool 2 m deep is

An important principle of fluid pressure is the experimentally verified fact that atany point in a liquid the pressure is the same in all directions. (A diver feels the samepressure on nose and both ears.) Thus, the pressure in any direction at a depth in afluid with mass density is given by

This helps us determine the hydrostatic force against a vertical plate or wall or damin a fluid. This is not a straightforward problem, because the pressure is not constantbut increases as the depth increases.

EXAMPLE 4 A dam has the shape of the trapezoid shown in Figure 5. The height is20 m and the width is 50 m at the top and 30 m at the bottom. Find the force on the dam due to hydrostatic pressure if the water level is 4 m from the top of the dam.

SOLUTION We choose a vertical -axis with origin at the surface of the water as inFigure 6(a). The depth of the water is 16 m, so we divide the interval into subintervals of equal length with endpoints and we choose . The horizontal strip of the dam is approximated by a rectangle with height and width

, where, from similar triangles in Figure 6(b),

or

and so

If is the area of the strip, then

If is small, then the pressure on the strip is almost constant and we can useEquation 4 to write

Pi & 1000txi*

ithPi#x

Ai & wi #x ! #46 " xi*$ #x

ithAi

wi ! 2#15 % a$ ! 2(15 % 8 " 12 xi*) ! 46 " xi*

a !16 " xi*

2! 8 "

xi*

2a

16 " xi*!

1020

wi

#xithxi* " "xi"1, xi%xi

"0, 16%x

P ! 3td ! 4d4

3d

! 19,600 Pa ! 19.6 kPa

P ! 3td ! 1000 kg)m3 2 9.8 m)s2 2 2 m

33 ! 1000

2 ! 1

P !FA

! 3td

Pt

F ! mt ! 3tAd


$ When using U. S. Customary units,we write , where is the weight density (as opposed to , which is the mass density). For

instance, the weight density of water is.4 ! 62.5 lb)ft3

3

4 ! 3tP ! 3td ! 4d

50 m

20 m

30 m

FIGURE 5

FIGURE 6

(b)

a

10

16-xi*20

(a)x

16

0_4 15

15

15

xi* a

10

Îx

The hydrostatic force acting on the strip is the product of the pressure and thearea:

Adding these forces and taking the limit as , we obtain the total hydrostaticforce on the dam:

Moments and Centers of Mass

Our main objective here is to find the point on which a thin plate of any given shapebalances horizontally as in Figure 7. This point is called the center of mass (or cen-ter of gravity) of the plate.

We first consider the simpler situation illustrated in Figure 8, where two masses and are attached to a rod of negligible mass on opposite sides of a fulcrum and atdistances and from the fulcrum. The rod will balance if

This is an experimental fact discovered by Archimedes and called the Law of the Lever.(Think of a lighter person balancing a heavier one on a seesaw by sitting farther awayfrom the center.)

Now suppose that the rod lies along the -axis with at and at and thecenter of mass at . If we compare Figures 8 and 9, we see that and

and so Equation 5 gives

The numbers and are called the moments of the masses and (withrespect to the origin), and Equation 6 says that the center of mass is obtained byadding the moments of the masses and dividing by the total mass .

FIGURE 9

0 m¡ m™

⁄ –x ¤

x¤-–x–x-⁄

m ! m1 % m2

xm2m1m2x2m1x1

x !m1x1 % m2x2

m1 % m26

m1 x % m2 x ! m1x1 % m2 x2

m1#x " x1$ ! m2#x2 " x$

d2 ! x2 " xd1 ! x " x1x

x2m2x1m1x

m1d1 ! m2d25

d2d1

m2

m1

P

& 4.43 2 107 N

! 9800'23x 2 "x 3

3 (0

16

! 1000#9.8$ y16

0 #46x " x 2 $ dx

! y16

0 1000tx#46 " x$ dx

F ! limn l !

!n

i!1 1000txi*#46 " xi*$ #x

n l !

Fi ! Pi Ai & 1000txi*#46 " xi*$ #x

ithFi


FIGURE 7

P

FIGURE 8

m¡ m™

d¡

fulcrum

d™

In general, if we have a system of particles with masses , . . . , locatedat the points , . . . , on the -axis, it can be shown similarly that the center ofmass of the system is located at

where is the total mass of the system, and the sum of the individualmoments

is called the moment of the system with respect to the origin. Then Equation 7 couldbe rewritten as , which says that if the total mass were considered as beingconcentrated at the center of mass , then its moment would be the same as themoment of the system.

Now we consider a system of particles with masses , . . . , located at thepoints , , . . . , in the -plane as shown in Figure 10. By anal-ogy with the one-dimensional case, we define the moment of the system about they-axis to be

and the moment of the system about the x-axis as

Then measures the tendency of the system to rotate about the -axis and meas-ures the tendency to rotate about the -axis.

As in the one-dimensional case, the coordinates of the center of mass aregiven in terms of the moments by the formulas

where is the total mass. Since and , the center of mass is the point where a single particle of mass would have the same moments as

the system.

EXAMPLE 5 Find the moments and center of mass of the system of objects that havemasses 3, 4, and 8 at the points , , and .

SOLUTION We use Equations 8 and 9 to compute the moments:

Mx ! 3#1$ % 4#"1$ % 8#2$ ! 15

My ! 3#"1$ % 4#2$ % 8#3$ ! 29

#3, 2$#2, "1$#"1, 1$

m#x, y$my ! Mxmx ! Mym ! 0 mi

y !Mx

mx !

My

m10

#x, y$x

MxyMy

Mx ! !n

i!1 miyi9

My ! !n

i!1 mixi8

xy#xn, yn $#x2, y2 $#x1, y1$mnm2,m1n

xmx ! M

M ! !n

i!1 mixi

m ! 0 mi

x !!n

i!1 mixi

!n

i!1 mi

!!n

i!1 mixi

m7

xxnx2,x1

mnm2,m1n


y

0 x

m£‹

y£

m¡

m™

⁄

›

¤

fi

FIGURE 10

Since , we use Equations 10 to obtain

Thus, the center of mass is . (See Figure 11.)

Next we consider a flat plate (called a lamina) with uniform density that occu-pies a region of the plane. We wish to locate the center of mass of the plate, whichis called the centroid of . In doing so we use the following physical principles: Thesymmetry principle says that if is symmetric about a line , then the centroid of lies on . (If is reflected about , then remains the same so its centroid remainsfixed. But the only fixed points lie on .) Thus, the centroid of a rectangle is its center.Moments should be defined so that if the entire mass of a region is concentrated at thecenter of mass, then its moments remain unchanged. Also, the moment of the unionof two nonoverlapping regions should be the sum of the moments of the individualregions.

Suppose that the region is of the type shown in Figure 12(a); that is, liesbetween the lines and , above the -axis, and beneath the graph of ,where is a continuous function. We divide the interval into n subintervals withendpoints and equal width . We choose the sample point to be themidpoint of the subinterval, that is, . This determines thepolygonal approximation to shown in Figure 12(b). The centroid of the approx-imating rectangle is its center . Its area is , so its mass is

The moment of about the -axis is the product of its mass and the distance from to the -axis, which is

Adding these moments, we obtain the moment of the polygonal approximation to , and then by taking the limit as we obtain the moment of itself about the

-axis:

In a similar fashion we compute the moment of about the -axis as the productof its mass and the distance from to the -axis:

Again we add these moments and take the limit to obtain the moment of about the -axis:

Mx ! limn l !

!n

i!1 3 ! 1

2 " f #xi$%2 #x ! 3 yb

a 12 " f #x$%2 dx

x!

Mx#Ri$ ! "3 f #xi$ #x% 12 f #xi$ ! 3 ! 1

2 " f #xi$%2 #x

xCi

xRi

My ! limn l !

!n

i!1 3xi f #xi$ #x ! 3 yb

a x f #x$ dx

y!n l !!

My#Ri$ ! "3 f #xi$ #x% xi ! 3xi f #xi$ #x

xi. ThusyCiyRi

3 f #xi$ #x

f #xi$ #xCi(xi, 12 f #xi$)Ri

ith!xi ! #xi"1 % xi$)2ithxi

xi*#xx0, x1, . . . , xn

"a, b%ffxx ! bx ! a

!!

l!l!l

!l!!

!3

(11415, 1)

y !Mx

m!

1515

! 1x !My

m!

2915

m ! 3 % 4 % 8 ! 15


FIGURE 11

y

0 x

8

4

3

center of mass

FIGURE 12

y

0 xa bR¡ R™

R£xi_1 xi

xi

{xi,!f(xi)}

Ci ”xi,! f(xi)’12

(b)

y

0 xa b

y=ƒ

!

(a)

Just as for systems of particles, the center of mass of the plate is defined so thatand . But the mass of the plate is the product of its density and its

area:

and so

Notice the cancellation of the ’s. The location of the center of mass is independentof the density.

In summary, the center of mass of the plate (or the centroid of ) is located at thepoint , where

EXAMPLE 6 Find the center of mass of a semicircular plate of radius .

SOLUTION In order to use (11) we place the semicircle as in Figure 13 so thatand , . Here there is no need to use the formula to

calculate because, by the symmetry principle, the center of mass must lie on the -axis, so . The area of the semicircle is , so

The center of mass is located at the point .

x

y

r0_r

”0, ! ’4r3$

y= r@-"œ„„„„„

FIGURE 13

#0, 4r)#3'$$

!2

'r 2 2r 3

3!

4r3'

!2

'r 2 yr

0 #r 2 " x 2 $ dx !

2'r 2'r 2x "

x 3

3 (0

r

!1

'r 2)2! 1

2 yr

"r (sr 2 " x 2 )2 dx

y !1A

yr

"r 12 " f #x$%2 dx

A ! 'r 2)2x ! 0yx

b ! ra ! "rf #x$ ! sr 2 " x 2

r

y !1A

yb

a 12 " f #x$%2 dxx !

1A

yb

a xf #x$ dx11

#x, y$!

3

y !Mx

m!

3 yb

a 12 " f #x$%2 dx

3 yb

a f #x$ dx

!yb

a 12 " f #x$%2 dx

yb

a f #x$ dx

x !My

m!

3 yb

a xf #x$ dx

3 yb

a f #x$ dx

!yb

a xf #x$ dx

yb

a f #x$ dx

m ! 3A ! 3 yb

a f #x$ dx

my ! Mxmx ! My



12. A circular swimming pool has a diameter of 24 ft, the sidesare 5 ft high, and the depth of the water is 4 ft. How muchwork is required to pump all of the water out over the side?(Use the fact that water weighs .)

" " " " " " " " " " " " "

13. The tank shown is full of water.(a) Find the work required to pump the water out of the

outlet.; (b) Suppose that the pump breaks down after J of

work has been done. What is the depth of the waterremaining in the tank?

14. The tank shown is full of water. Given that water weighs62.5 lb)ft , find the work required to pump the water out ofthe tank.

When gas expands in a cylinder with radius , the pressureat any given time is a function of the volume: .The force exerted by the gas on the piston (see the figure) isthe product of the pressure and the area: . Showthat the work done by the gas when the volume expandsfrom volume to volume is

16. In a steam engine the pressure and volume of steamsatisfy the equation where is a constant. (This is true for adiabatic expansion, that is, expansion in whichthere is no heat transfer between the cylinder and its sur-roundings.) Use Exercise 15 to calculate the work done by

kPV 1.4 ! kVP

x

V

piston head

W ! yV2

V1

P dV

V2V1

F ! 'r 2P

P ! P#V $r15.

5 ft

hemisphere

3

2 m

3 m

8 m

3 m

4.7 2 105

62.5 lb)ft3

1. A particle is moved along the -axis by a force thatmeasures pounds at a point feet from the ori-gin. Find the work done in moving the particle from the origin to a distance of 9 ft.

2. When a particle is located at a distance meters from theorigin, a force of newtons acts on it. How muchwork is done in moving the particle from to ?Interpret your answer by considering the work done from

to and from to .

A force of 10 lb is required to hold a spring stretched 4 in.beyond its natural length. How much work is done instretching it from its natural length to 6 in. beyond its natu-ral length?

4. A spring has a natural length of 20 cm. If a 25-N force isrequired to keep it stretched to a length of 30 cm, howmuch work is required to stretch it from 20 cm to 25 cm?

Suppose that 2 J of work are needed to stretch a spring fromits natural length of 30 cm to a length of 42 cm.(a) How much work is needed to stretch it from 35 cm to

40 cm?(b) How far beyond its natural length will a force of 30 N

keep the spring stretched?

6. If 6 J of work are needed to stretch a spring from 10 cm to12 cm and another 10 J are needed to stretch it from 12 cmto 14 cm, what is the natural length of the spring?

7–12 " Show how to approximate the required work by a Rie-mann sum. Then express the work as an integral and evaluate it.

A heavy rope, 50 ft long, weighs 0.5 lb)ft and hangs overthe edge of a building 120 ft high. How much work is donein pulling the rope to the top of the building?

8. A uniform cable hanging over the edge of a tall building is40 ft long and weighs 60 lb. How much work is required topull 10 ft of the cable to the top?

9. A cable that weighs 2 lb)ft is used to lift 800 lb of coal up amineshaft 500 ft deep. Find the work done.

10. A bucket that weighs 4 lb and a rope of negligible weightare used to draw water from a well that is 80 ft deep. Thebucket starts with 40 lb of water and is pulled up at a rate of2 ft)s, but water leaks out of a hole in the bucket at a rate of0.2 lb)s. Find the work done in pulling the bucket to the topof the well.

An aquarium 2 m long, 1 m wide, and 1 m deep is full ofwater. Find the work needed to pump half of the water out of the aquarium. (Use the fact that the density of water is

.)1000 kg)m3

11.

7.

5.

3.

x ! 2x ! 1.5x ! 1.5x ! 1

x ! 2x ! 1cos#'x)3$

x

x10)#1 % x$2x

Exercises ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !6.5

24. A vertical dam has a semicircular gate as shown in the figure. Find the hydrostatic force against the gate.

25–26 " The masses are located at the points . Find themoments and and the center of mass of the system.

25. , ; ,

26. , , , ;, , ,

" " " " " " " " " " " " "

27–30 " Sketch the region bounded by the curves, and visuallyestimate the location of the centroid. Then find the exact coordi-nates of the centroid.

27. , ,

28. , ,

, , ,

30. , , ," " " " " " " " " " " " "

31–32 " Calculate the moments and and the center ofmass of a lamina with the given density and shape.

31.

32.

" " " " " " " " " " " " "

x

y

0 r

rquarter-circle

3 ! 2

x

y

0 1_1

23 ! 1

MyMx

x ! 2x ! 1y ! 0y ! 1)x

x ! 1x ! 0y ! 0y ! e x29.

x ! 9y ! 0y ! sx

x ! 2y ! 0y ! x 2

P4#6, "1$P3#"3, "7$P2#3, 4$P1#1, "2$m4 ! 4m3 ! 1m2 ! 5m1 ! 6

P2#2, 4$P1#"1, 2$m2 ! 8m1 ! 4

MyMx

Pimi

12 m

2 m

4 m

water level

the engine during a cycle when the steam starts at a pressureof 160 lb)in and a volume of 100 in and expands to a vol-ume of 800 in .

17. (a) Newton’s Law of Gravitation states that two bodies withmasses and attract each other with a force

where is the distance between the bodies and is thegravitational constant. If one of the bodies is fixed, findthe work needed to move the other from to .

(b) Compute the work required to launch a 1000-kg satellitevertically to an orbit 1000 km high. You may assumethat Earth’s mass is kg and is concentratedat its center. Take the radius of Earth to be mand .

18. (a) Use an improper integral and information from Exer-cise 17 to find the work needed to propel a 1000-kgsatellite out of Earth’s gravitational field.

(b) Find the escape velocity that is needed to propel arocket of mass out of the gravitational field of aplanet with mass and radius . (Use the fact that theinitial kinetic energy of supplies the needed work.)

19–21 " The end of a tank containing water is vertical and hasthe indicated shape. Explain how to approximate the hydrostaticforce against the end of the tank by a Riemann sum. Thenexpress the force as an integral and evaluate it.

19. 20.

" " " " " " " " " " " " "

22. A large tank is designed with ends in the shape of theregion between the curves and , measuredin feet. Find the hydrostatic force on one end of the tank ifit is filled to a depth of 8 ft with gasoline. (Assume thegasoline’s density is .)

23. A swimming pool is 20 ft wide and 40 ft long and itsbottom is an inclined plane, the shallow end having a depthof 3 ft and the deep end, 9 ft. If the pool is full of water,find the hydrostatic force on (a) the shallow end, (b) thedeep end, (c) one of the sides, and (d) the bottom of thepool.

42.0 lb)ft3

y ! 12y ! x 2)2

20 ft

8 ft

12 ft21.

10 m 10 m

5 m waterlevel

12 mv0

2RM

mv0

G ! 6.67 2 10"11 N/m2)kg26.37 2 106

5.98 2 1024

r ! br ! a

Gr

F ! G m1m2

r 2

m2m1

3

32


(b) Find the centroid of the region bounded by the lineand the parabola .

34. Let be the region that lies between the curves and, , where and are integers with

.(a) Sketch the region .(b) Find the coordinates of the centroid of .(c) Try to find values of and such that the centroid lies

outside .!nm

!

!

0 & n , mnm0 & x & 1y ! x n

y ! x m!

y ! x 2y ! x33. (a) Let be the region that lies between two curves

and , where and. By using the same sort of reasoning that led

to the formulas in (11), show that the centroid of is, where

y !1A

yb

a 12 -" f #x$% 2 " "t#x$% 2. dx

x !1A

yb

a x " f #x$ " t#x$% dx

#x, y$!

a & x & bf #x$ $ t#x$y ! t#x$y ! f #x$

!

SECTION 6.6 APPLICATIONS TO ECONOMICS AND BIOLOGY # 487

In this section we consider some applications of integration to economics (consumersurplus) and biology (blood flow, cardiac output). Others are described in the exercises.

Consumer Surplus

Recall from Section 4.7 that the demand function is the price that a company hasto charge in order to sell units of a commodity. Usually, selling larger quantitiesrequires lowering prices, so the demand function is a decreasing function. The graphof a typical demand function, called a demand curve, is shown in Figure 1. If is theamount of the commodity that is currently available, then is the current sell-ing price.

We divide the interval into subintervals, each of length , and letbe the right endpoint of the subinterval, as in Figure 2. If, after the first

units were sold, a total of only units had been available and the price per unithad been set at dollars, then the additional units could have been sold (but nomore). The consumers who would have paid dollars placed a high value on theproduct; they would have paid what it was worth to them. So, in paying only dol-lars they have saved an amount of

Considering similar groups of willing consumers for each of the subintervals andadding the savings, we get the total savings:

(This sum corresponds to the area enclosed by the rectangles in Figure 2.) If we let, this Riemann sum approaches the integral

which economists call the consumer surplus for the commodity.

yX

0 "p#x$ " P% dx1

n l !

!n

i!1 "p#xi$ " P% #x

#savings per unit$#number of units$ ! "p#xi $ " P% #x

Pp#xi$

#xp#xi$xixi"1

ithxi* ! xi

#x ! X)nn"0, X %

P ! p#X$X

xp#x$

Applications to Economics and Biology ! ! ! ! ! ! ! ! !6.6

0 x

p

P

X

(X,!P)

p=p(x)

FIGURE 1A typical demand curve

0 x

p

P

⁄ xi X

(X,!P)

FIGURE 2

The consumer surplus represents the amount of money saved by consumers in pur-chasing the commodity at price , corresponding to an amount demanded of . Fig-ure 3 shows the interpretation of the consumer surplus as the area under the demandcurve and above the line .

EXAMPLE 1 The demand for a product, in dollars, is

Find the consumer surplus when the sales level is 500.

SOLUTION Since the number of products sold is , the corresponding price is

Therefore, from Definition 1, the consumer surplus is

Blood Flow

In Example 7 in Section 3.3 we discussed the law of laminar flow:

which gives the velocity of blood that flows along a blood vessel with radius andlength at a distance from the central axis, where is the pressure differencebetween the ends of the vessel and is the viscosity of the blood. Now, in order tocompute the rate of blood flow, or flux (volume per unit time), we consider smaller,equally spaced radii . . . . The approximate area of the ring (or washer) withinner radius and outer radius is

where

(See Figure 4.) If is small, then the velocity is almost constant throughout this ringand can be approximated by . Thus, the volume of blood per unit time that flowsacross the ring is approximately

#2'ri #r$v#ri$ ! 2'riv#ri$ #r

v#ri$#r

#r ! ri " ri"1 2'ri #r

riri"1

r1, r2,

1Prl

Rv

v#r$ !P

41l #R2 " r 2 $

! $33,333.33

! #125$#500$ " #0.1$#500$2 "#0.0001$#500$3

3

! 125x " 0.1x 2 " #0.0001$* x 3

3 +(0

500

! y500

0 #125 " 0.2x " 0.0001x 2 $ dx

y500

0 "p#x$ " P% dx ! y500

0 #1200 " 0.2x " 0.0001x 2 " 1075$ dx

P ! 1200 " #0.2$#500$ " #0.0001$#500$2 ! 1075

X ! 500

p ! 1200 " 0.2x " 0.0001x 2

p ! P

XP


0 x

p

(X,!P)P

X

p=p(x)

p=P

consumersurplus

FIGURE 3

Îr

ri

FIGURE 4

and the total volume of blood that flows across a cross-section per unit time is about

This approximation is illustrated in Figure 5. Notice that the velocity (and hence thevolume per unit time) increases toward the center of the blood vessel. The approxi-mation gets better as n increases. When we take the limit we get the exact value of theflux (or discharge), which is the volume of blood that passes a cross-section per unittime:

The resulting equation

is called Poiseuille’s Law; it shows that the flux is proportional to the fourth power ofthe radius of the blood vessel.

Cardiac Output

Figure 6 shows the human cardiovascular system. Blood returns from the body throughthe veins, enters the right atrium of the heart, and is pumped to the lungs through thepulmonary arteries for oxygenation. It then flows back into the left atrium through thepulmonary veins and then out to the rest of the body through the aorta. The cardiacoutput of the heart is the volume of blood pumped by the heart per unit time, that is,the rate of flow into the aorta.

FIGURE 6

aortavein

right atrium

pulmonary arteries

left atrium

pulmonaryveins

vein

pulmonary arteries

pulmonaryveins

F !'PR4

81l2

!'P21l

'R4

2"

R4

4 ( !'PR4

81l

!'P21l

yR

0 #R2r " r 3 $ dr !

'P21l

'R2 r 2

2"

r 4

4 (r!0

r!R

! yR

0 2'r

P41l

#R2 " r 2 $ dr

F ! limn l !

!n

i!1 2'riv#ri$ #r ! yR

0 2'rv#r$ dr

!n

i!1 2'riv#ri$ #r


FIGURE 5

The dye dilution method is used to measure the cardiac output. Dye is injected intothe right atrium and flows through the heart into the aorta. A probe inserted into theaorta measures the concentration of the dye leaving the heart at equally spaced timesover a time interval until the dye has cleared. Let be the concentration ofthe dye at time . If we divide into subintervals of equal length , then theamount of dye that flows past the measuring point during the subinterval from to is approximately

where is the rate of flow that we are trying to determine. Thus, the total amount ofdye is approximately

and, letting , we find that the amount of dye is

Thus, the cardiac output is given by

where the amount of dye is known and the integral can be approximated from theconcentration readings.

EXAMPLE 2 A 5-mg bolus of dye is injected into a right atrium. The concentration ofthe dye (in milligrams per liter) is measured in the aorta at one-second intervals asshown in the chart. Estimate the cardiac output.

SOLUTION Here , , and . We use Simpson’s Rule to approximatethe integral of the concentration:

Thus, Formula 3 gives the cardiac output to be

& 0.12 L)s ! 7.2 L)min

F !A

y10

0 c#t$ dt

&5

41.87

& 41.87

& % 2#6.1$ % 4#4.0$ % 2#2.3$ % 4#1.1$ % 0%

y10

0 c#t$ dt & 1

3 "0 % 4#0.4$ % 2#2.8$ % 4#6.5$ % 2#9.8$ % 4#8.9$

T ! 10#t ! 1A ! 5

A

F !A

yT

0 c#t$ dt

3

A ! F yT

0 c#t$ dt

n l !

!n

i!1 c#ti$F #t ! F !

n

i!1 c#ti$ #t

F

#concentration$#volume$ ! c#ti$#F #t$

t ! ti

t ! ti"1

#t"0, T %tc#t$"0, T %


t t

0 0 6 6.11 0.4 7 4.02 2.8 8 2.33 6.5 9 1.14 9.8 10 05 8.9

c#t$c#t$


surplus and the producer surplus. Illustrate by sketching thesupply and demand curves and identifying the surpluses asareas.

; 9. A company modeled the demand curve for its product (indollars) by

Use a graph to estimate the sales level when the sellingprice is $16. Then find (approximately) the consumersurplus for this sales level.

A movie theater has been charging $7.50 per person andselling about 400 tickets on a typical weeknight. After sur-veying their customers, the theater estimates that for every50 cents that they lower the price, the number of movie-goers will increase by 35 per night. Find the demand func-tion and calculate the consumer surplus when the tickets are priced at $6.00.

11. If the amount of capital that a company has at time is ,then the derivative, , is called the net investment flow.Suppose that the net investment flow is million dollarsper year (where is measured in years). Find the increase incapital (the capital formation) from the fourth year to theeighth year.

12. A hot, wet summer is causing a mosquito population explo-sion in a lake resort area. The number of mosquitos isincreasing at an estimated rate of per week(where is measured in weeks). By how much does themosquito population increase between the fifth and ninthweeks of summer?

13. Use Poiseuille’s Law to calculate the rate of flow in a smallhuman artery where we can take , cm,

cm, and dynes)cm .

14. High blood pressure results from constriction of the arteries.To maintain a normal flow rate (flux), the heart has to pumpharder, thus increasing the blood pressure. Use Poiseuille’sLaw to show that if and are normal values of theradius and pressure in an artery and the constricted valuesare and , then for the flux to remain constant, and are related by the equation

Deduce that if the radius of an artery is reduced to three-fourths of its former value, then the pressure is more thantripled.

PP0

! *R0

R +4

RPPR

P0R0

2P ! 4000l ! 2R ! 0.0081 ! 0.027

t2200 % 10e0.8t

tst

f +#t$f #t$t

10.

p !800,000e"x)5000

x % 20,000

1. The marginal cost function was defined to be the derivative of the cost function. (See Sections 3.3 and 4.7.) If the marginal cost of manufacturing units of a product is

(measured in dollars per unit)and the fixed start-up cost is , use the Total Change Theorem to find the cost of producing the first2000 units.

2. The marginal revenue from selling items is .The revenue from the sale of the first 100 items is .What is the revenue from the sale of the first 200 items?

The marginal cost of producing units of a certain product is (in dollars per unit).Find the increase in cost if the production level is raisedfrom 1200 units to 1600 units.

4. The demand function for a certain commodity is. Find the consumer surplus when the sales

level is 30. Illustrate by drawing the demand curve andidentifying the consumer surplus as an area.

A demand curve is given by . Find theconsumer surplus when the selling price is .

6. The supply function for a commodity gives the rela-tion between the selling price and the number of units thatmanufacturers will produce at that price. For a higher price,manufacturers will produce more units, so is an increas-ing function of . Let be the amount of the commoditycurrently produced and let be the current price.Some producers would be willing to make and sell the com-modity for a lower selling price and are therefore receivingmore than their minimal price. The excess is called the pro-ducer surplus. An argument similar to that for consumersurplus shows that the surplus is given by the integral

Calculate the producer surplus for the supply functionat the sales level . Illustrate by

drawing the supply curve and identifying the producer sur-plus as an area.

7. A supply curve is given by . Find theproducer surplus when the selling price is .

8. For a given commodity and pure competition, the number ofunits produced and the price per unit are determined as thecoordinates of the point of intersection of the supply anddemand curves. Given the demand curve and the supply curve , find the consumer p ! 20 % x)10

p ! 50 " x)20

$10p ! 5 % 1

10 sx

X ! 10pS#x$ ! 3 % 0.01x 2

yX

0 "P " pS#x$% dx

P ! pS#X $Xx

pS

pS#x$

$10p ! 450)#x % 8$5.

p ! 5 " x)10

74 % 1.1x " 0.002x 2 % 0.00004x 3x3.

$880090 " 0.02xx

C#0$ ! $1,500,000C+#x$ ! 0.006x 2 " 1.5x % 8

x

C+#x$

Exercises ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !6.6

Calculus plays a role in the analysis of random behavior. Suppose we consider thecholesterol level of a person chosen at random from a certain age group, or the heightof an adult female chosen at random, or the lifetime of a randomly chosen battery ofa certain type. Such quantities are called continuous random variables because theirvalues actually range over an interval of real numbers, although they might be meas-ured or recorded only to the nearest integer. We might want to know the probabilitythat a blood cholesterol level is greater than 250, or the probability that the height ofan adult female is between 60 and 70 inches, or the probability that the battery we arebuying lasts between 100 and 200 hours. If X represents the lifetime of that type ofbattery, we denote this last probability as follows:

According to the frequency interpretation of probability, this number is the long-runproportion of all batteries of the specified type whose lifetimes are between 100 and200 hours. Since it represents a proportion, the probability naturally falls between 0and 1.

Every continuous random variable X has a probability density function . Thismeans that the probability that X lies between a and b is found by integrating froma to b:

For example, Figure 1 shows the graph of a model of the probability density functionfor a random variable X defined to be the height in inches of an adult female in the

United States (according to data from the National Health Survey). The probabilitythat the height of a woman chosen at random from this population is between 60 and70 inches is equal to the area under the graph of from 60 to 70.

In general, the probability density function of a random variable X satisfies thecondition for all x. Because probabilities are measured on a scale from 0 f #x$ $ 0

f

FIGURE 1Probability density function

for the height of an adult femalex

y

0 6560 70

y=ƒ

area=probability that theheight of a womanis between 60 and70 inches

f

f

P#a & X & b$ ! yb

a f #x$ dx1

ff

P#100 & X & 200$

Probability ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !6.7

The dye dilution method is used to measure cardiac outputwith 8 mg of dye. The dye concentrations, in mg)L, aremodeled by , , where is meas-ured in seconds. Find the cardiac output.

16. After an 8-mg injection of dye, the readings of dye concen-tration at two-second intervals are as shown in the table atthe right. Use Simpson’s Rule to estimate the cardiacoutput.

t0 & t & 12c#t$ ! 14 t#12 " t$

15.


t t

0 0 12 3.92 2.4 14 2.34 5.1 16 1.66 7.8 18 0.78 7.6 20 0

10 5.4

c#t$c#t$

SECTION 6.7 PROBABILITY # 493

to 1, it follows that

EXAMPLE 1 Let for and for all othervalues of .(a) Verify that is a probability density function.(b) Find .

SOLUTION(a) For we have , so for all . We alsoneed to check that Equation 2 is satisfied:

Therefore, is a probability density function.

(b) The probability that lies between 4 and 8 is

EXAMPLE 2 Phenomena such as waiting times and equipment failure times are com-monly modeled by exponentially decreasing probability density functions. Find theexact form of such a function.

SOLUTION Think of the random variable as being the time you wait on hold before anagent of a company you’re telephoning answers your call. So instead of x, let’s use tto represent time, in minutes. If is the probability density function and you call attime , then, from Definition 1, represents the probability that an agentanswers within the first two minutes and is the probability that your call isanswered during the fifth minute.

It’s clear that for (the agent can’t answer before you place thecall). For we are told to use an exponentially decreasing function, that is, afunction of the form , where A and c are positive constants. Thus

We use condition 2 to determine the value of A:

!Ac

! limx l !

'"Ac

e"ct(0

x

! limx l !

Ac

#1 " e"cx$

! y!

0 Ae"ct dt ! lim

x l ! y x

0 Ae"ct dt

1 ! y!

"! f #t$ dt ! y0

"! f #t$ dt % y!

0 f #t$ dt

f #t$ ! 10Ae"ct

if t , 0if t $ 0

f #t$ ! Ae"ctt - 0

t , 0f #t$ ! 0

x54 f #t$ dt

x20 f #t$ dtt ! 0

f

! 0.006[5x 2 " 13 x 3]4

8! 0.544

P#4 & X & 8$ ! y8

4 f #x$ dx ! 0.006 y8

4 #10x " x 2$ dx

X

f

! 0.006[5x 2 " 13 x 3]0

10! 0.006(500 " 1000

3 ) ! 1

y!

"! f #x$ dx ! y10

0 0.006x#10 " x$ dx ! 0.006 y10

0 #10x " x 2$ dx

xf #x$ $ 00.006x#10 " x$ $ 00 & x & 10

P#4 & X & 8$f

xf #x$ ! 00 & x & 10f #x$ ! 0.006x#10 " x$

y!

"! f #x$ dx ! 12

Therefore, and so . Thus, every exponential density function has theform

A typical graph is shown in Figure 2.

Average Values

Suppose you’re waiting for a company to answer your phone call and you wonder howlong, on the average, you could expect to wait. Let be the corresponding densityfunction, where t is measured in minutes, and think of a sample of N people who havecalled this company. Most likely, none of them had to wait more than an hour, so let’srestrict our attention to the interval . Let’s divide that interval into n inter-vals of length and endpoints . . . . (Think of as lasting a minute, or halfa minute, or 10 seconds, or even a second.) The probability that somebody’s call getsanswered during the time period from to is the area under the curve from to , which is approximately equal to . (This is the area of the approx-imating rectangle in Figure 3, where is the midpoint of the interval.)

Since the long-run proportion of calls that get answered in the time period from to is , we expect that, out of our sample of N callers, the number whose callwas answered in that time period is approximately and the time that eachwaited is about . Therefore, the total time they waited is the product of these num-bers: approximately . Adding over all such intervals, we get the approxi-mate total of everybody’s waiting times:

If we now divide by the number of callers N, we get the approximate average waitingtime:

We recognize this as a Riemann sum for the function . As the time interval shrinks(that is, and ), this Riemann sum approaches the integral

This integral is called the mean waiting time.In general, the mean of any probability density function is defined to be

The mean can be interpreted as the long-run average value of the random variable X.It can also be interpreted as a measure of centrality of the probability density function.

5 ! y!

"! xf #x$ dx

f

y60

0 t f #t$ dt

n l !#t l 0t f #t$

!n

i!1 ti f # ti$ #t

!n

i!1 N ti f # ti $ #t

ti"Nf # ti $ #t%ti

Nf # ti$ #tf # ti $ #tti

ti"1

ti

f # ti $ #ttiti"1

y ! f #t$titi"1

#t0, t1, t2,#t0 & t & 60

f #t$

f #t$ ! 10ce"ct

if t , 0if t $ 0

A ! cA)c ! 1


FIGURE 2An exponential density function

t

y

0

cf(t)=

0ce _ct

if t<0if t˘0

t

y

0

Ît

t i-1 t i

t i

y=f(t )

FIGURE 3

$ It is traditional to denote the mean bythe Greek letter (mu).5

The expression for the mean resembles an integral we have seen before. If is the region that lies under the graph of , we know from Formula 6.5.11 that the x-coordinate of the centroid of is

because of Equation 2. So a thin plate in the shape of balances at a point on the ver-tical line . (See Figure 4.)

EXAMPLE 3 Find the mean of the exponential distribution of Example 2:

SOLUTION According to the definition of a mean, we have

To evaluate this integral we use integration by parts, with and :

The mean is , so we can rewrite the probability density function as

EXAMPLE 4 Suppose the average waiting time for a customer’s call to be answered bya company representative is five minutes.(a) Find the probability that a call is answered during the first minute.(b) Find the probability that a customer waits more than five minutes to beanswered.

SOLUTION(a) We are given that the mean of the exponential distribution is min and so,from the result of Example 3, we know that the probability density function is

f #t$ ! 100.2e"t)5

if t , 0if t $ 0

5 ! 5

f #t$ ! 105"1e"t)5

if t , 0if t $ 0

5 ! 1)c

!1c

! limx l !

*"xe"cx %1c

"e"cx

c +y!

0 tce"ct dt ! lim

x l ! y x

0 tce"ct dt ! lim

x l ! *"te"ct]x

0 % yx

0 e"ct dt+

dv ! ce"ct dtu ! t

5 ! y!

"! t f #t$ dt ! y!

0 tce"ct dt

f #t$ ! 10ce"ct

if t , 0if t $ 0

x ! 5!

x !y!

"! xf #x$ dx

y!

"! f #x$ dx

! y!

"! xf #x$ dx ! 5

!f

!


FIGURE 4T balances at a point on the line x=m

t

y

0 m

y=ƒ

x=mT

$ The limit of the first term is by l’Hospital’s Rule.

0

Thus, the probability that a call is answered during the first minute is

So about 18% of customers’ calls are answered during the first minute.

(b) The probability that a customer waits more than five minutes is

About 37% of customers wait more than five minutes before their calls areanswered.

Notice the result of Example 4(b): Even though the mean waiting time is 5 min-utes, only 37% of callers wait more than 5 minutes. The reason is that some callershave to wait much longer (maybe 10 or 15 minutes), and this brings up the average.

Another measure of centrality of a probability density function is the median. Thatis a number m such that half the callers have a waiting time less than m and the othercallers have a waiting time longer than m. In general, the median of a probability den-sity function is the number m such that

This means that half the area under the graph of lies to the right of m. In Exercise 5you are asked to show that the median waiting time for the company described inExample 4 is approximately 3.5 minutes.

Normal Distributions

Many important random phenomena—such as test scores on aptitude tests, heightsand weights of individuals from a homogeneous population, annual rainfall in a givenlocation—are modeled by a normal distribution. That means that the probabilitydensity function of the random variable X is a member of the family of functions

You can verify that the mean for this function is . The positive constant is calledthe standard deviation; it measures how spread out the values of X are. From the bell-shaped graphs of members of the family in Figure 5, we see that for small values of

65

f #x$ !1

6s2' e"#x"5$2)#26 2$3

f

y!

m f #x$ dx !

12

!1e

& 0.368

! limx l !

y x

5 0.2e"t)5 dt ! lim

x l ! #e"1 " e"x)5 $

P#T - 5$ ! y!

5 f #t$ dt ! y!

5 0.2e"t)5 dt

! 1 " e"1)5 & 0.1813

! 0.2#"5$e"t)5]0

1

! y1

0 0.2e"t)5 dt

P#0 & T & 1$ ! y1

0 f #t$ dt


$ The standard deviation is denoted bythe lowercase Greek letter (sigma).6

the values of X are clustered about the mean, whereas for larger values of the values of X are more spread out. Statisticians have methods for using sets of data toestimate and .

The factor is needed to make a probability density function. In fact,it can be verified using the methods of multivariable calculus that

EXAMPLE 5 Intelligence Quotient (IQ) scores are distributed normally with mean 100 and standard deviation 15. (Figure 6 shows the corresponding probability den-sity function.)(a) What percentage of the population has an IQ score between 85 and 115?(b) What percentage of the population has an IQ above 140?

SOLUTION(a) Since IQ scores are normally distributed, we use the probability density functiongiven by Equation 3 with and :

Recall from Section 5.8 that the function doesn’t have an elementary anti-derivative, so we can’t evaluate the integral exactly. But we can use the numericalintegration capability of a calculator or computer (or the Midpoint Rule or Simp-son’s Rule) to estimate the integral. Doing so, we find that

So about 68% of the population has an IQ between 85 and 115, that is, within onestandard deviation of the mean.

(b) The probability that the IQ score of a person chosen at random is more than 140 is

To avoid the improper integral we could approximate it by the integral from 140 to200. (It’s quite safe to say that people with an IQ over 200 are extremely rare.) Then

Therefore, about 0.4% of the population has an IQ over 140.

P#X - 140$ & y200

140

115s2'

e"#x"100$2)450 dx & 0.0038

P#X - 140$ ! y!

140

115s2'

e"#x"100$2)450 dx

P#85 & X & 115$ & 0.68

y ! e"x 2

P#85 & X & 115$ ! y115

85

115s2'

e"#x"100$2)#2!152$

dx

6 ! 155 ! 100

y!

"!

16s2'

e"#x"5$2)#26 2$

dx ! 1

f1)(6s2' )

FIGURE 5Normal distributions

x

y

0 m

12

=1

=2

=

s

s

s

65

66


FIGURE 6Distribution of IQ scores

x

y

0 60

0.01

80 100 120 140

0.02


7. The manager of a fast-food restaurant determines that the average time that her customers wait for service is2.5 minutes.(a) Find the probability that a customer has to wait for more

than 4 minutes.(b) Find the probability that a customer is served within the

first 2 minutes.(c) The manager wants to advertise that anybody who isn’t

served within a certain number of minutes gets a freehamburger. But she doesn’t want to give away free ham-burgers to more than 2% of her customers. What shouldthe advertisement say?

8. According to the National Health Survey, the heights ofadult males in the United States are normally distributedwith mean 69.0 inches and standard deviation 2.8 inches.(a) What is the probability that an adult male chosen at ran-

dom is between 65 inches and 73 inches tall?(b) What percentage of the adult male population is more

than 6 feet tall?

The “Garbage Project” at the University of Arizona reports that the amount of paper discarded by households per weekis normally distributed with mean 9.4 lb and standard devia-tion 4.2 lb. What percentage of households throw out atleast 10 lb of paper a week?

10. Boxes are labeled as containing 500 g of cereal. Themachine filling the boxes produces weights that are nor-mally distributed with standard deviation 12 g.(a) If the target weight is 500 g, what is the probability that

the machine produces a box with less than 480 g ofcereal?

(b) Suppose a law states that no more than 5% of a manu-facturer’s cereal boxes can contain less than the statedweight of 500 g. At what target weight should the manu-facturer set its filling machine?

11. For any normal distribution, find the probability that therandom variable lies within two standard deviations of themean.

12. The standard deviation for a random variable with probabil-ity density function and mean is defined by

Find the standard deviation for an exponential density func-tion with mean .

13. The hydrogen atom is composed of one proton in thenucleus and one electron, which moves about the nucleus.

5

6 ! 'y!

"! #x " 5$2 f #x$ dx(1)2

5f

9.

Let be the probability density function for the lifetimeof a manufacturer’s highest quality car tire, where ismeasured in miles. Explain the meaning of each integral.

(a) (b)

2. Let be the probability density function for the time ittakes you to drive to school in the morning, where ismeasured in minutes. Express the following probabilities asintegrals.(a) The probability that you drive to school in less than

15 minutes(b) The probability that it takes you more than half an hour

to get to school

A spinner from a board game randomly indicates a realnumber between 0 and 10. The spinner is fair in the sensethat it indicates a number in a given interval with the sameprobability as it indicates a number in any other interval ofthe same length.(a) Explain why the function

is a probability density function for the spinner’s values.(b) What does your intuition tell you about the value of the

mean? Check your guess by evaluating an integral.

(a) Explain why the function whose graph is shown is aprobability density function.

(b) Use the graph to find the following probabilities:(i) (ii)

(c) Calculate the mean.

5. Show that the median waiting time for a phone call to thecompany described in Example 4 is about 3.5 minutes.

6. (a) A type of lightbulb is labeled as having an average life-time of 1000 hours. It’s reasonable to model the proba-bility of failure of these bulbs by an exponential densityfunction with mean . Use this model to findthe probability that a bulb(i) fails within the first 200 hours,

(ii) burns for more than 800 hours.(b) What is the median lifetime of these lightbulbs?

5 ! 1000

y=ƒ

4 6 8 10 x

y

0 2

0.1

0.2

P#3 & X & 8$P#X , 3$

4.

f #x$ ! 10.10

if 0 & x & 10if x , 0 or x - 10

3.

tf #t$

y!

25,000 f #x$ dxy40,000

30,000 f #x$ dx

xf #x$1.

Exercises ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !6.7

The integral

gives the probability that the electron will be found withinthe sphere of radius meters centered at the nucleus.(a) Verify that is a probability density function.(b) Find . For what value of does have its

maximum value?; (c) Graph the density function.

(d) Find the probability that the electron will be within thesphere of radius centered at the nucleus.

(e) Calculate the mean distance of the electron from thenucleus in the ground state of the hydrogen atom.

4a0

p#r$rlimr l ! p#r$p#r$

r

P#r$ ! yr

0

4a 3

0 s 2e"2s)a0 ds

In the quantum theory of atomic structure, it is assumed thatthe electron does not move in a well-defined orbit. Instead,it occupies a state known as an orbital, which may bethought of as a “cloud” of negative charge surrounding thenucleus. At the state of lowest energy, called the groundstate, or 1s-orbital, the shape of this cloud is assumed to bea sphere centered at the nucleus. This sphere is described interms of the probability density function

where is the Bohr radius . #a0 & 5.59 2 10"11 m$a0

r $ 0p#r$ !4a 3

0 r 2e"2r)a0

CHAPTER 6 REVIEW # 499

6. Suppose that you push a book across a 6-meter-long tableby exerting a force at each point from to .What does represent? If is measured in new-tons, what are the units for the integral?

7. Describe how we can find the hydrostatic force against avertical wall submersed in a fluid.

8. (a) What is the physical significance of the center of massof a thin plate?

(b) If the plate lies between and , where, write expressions for the coordinates of the

center of mass.

9. Given a demand function , explain what is meant by the consumer surplus when the amount of a commodity cur-rently available is and the current selling price is . Illus-trate with a sketch.

10. (a) What is the cardiac output of the heart?(b) Explain how the cardiac output can be measured by the

dye dilution method.

11. What is a probability density function? What propertiesdoes such a function have?

12. Suppose is the probability density function for theweight of a female college student, where is measured inpounds.(a) What is the meaning of the integral ?(b) Write an expression for the mean of this density

function.(c) How can we find the median of this density function?

x1000 f #x$ dx

xf #x$

PX

p#x$

a & x & by ! 0y ! f #x$

f #x$x60 f #x$ dx

x ! 6x ! 0f #x$1. (a) Draw two typical curves and , where

for . Show how to approximatethe area between these curves by a Riemann sum andsketch the corresponding approximating rectangles.Then write an expression for the exact area.

(b) Explain how the situation changes if the curves have equations and , where for

.

2. Suppose that Sue runs faster than Kathy throughout a 1500-meter race. What is the physical meaning of the areabetween their velocity curves for the first minute of therace?

3. (a) Suppose is a solid with known cross-sectional areas.Explain how to approximate the volume of by a Riemann sum. Then write an expression for the exactvolume.

(b) If is a solid of revolution, how do you find the cross-sectional areas?

4. (a) How is the length of a curve defined?(b) Write an expression for the length of a smooth curve

with parametric equations , , .(c) How does the expression in part (b) simplify if the curve

is described by giving terms of , that is, ,? What if is given as a function of ?

5. (a) What is the average value of a function on an interval?

(b) What does the Mean Value Theorem for Integrals say?What is its geometric interpretation?

"a, b%f

yxa & x & by ! f #x$xy

a & t & by ! t#t$x ! f #t$

S

SS

c & y & df #y$ $ t#y$x ! t#y$x ! f #y$

a & x & bf #x$ $ t#x$y ! t#x$y ! f #x$

6 ReviewC O N C E P T C H E C K


widest point to be 53 cm. The circumference 7 cm fromeach end is 45 cm. Use Simpson’s Rule to make your estimate.

11. The base of a solid is a circular disk with radius 3. Find thevolume of the solid if parallel cross-sections perpendicularto the base are isosceles right triangles with hypotenuselying along the base.

12. The base of a solid is the region bounded by the parabolasand . Find the volume of the solid if the

cross-sections perpendicular to the -axis are squares withone side lying along the base.

13. The height of a monument is 20 m. A horizontal cross-section at a distance meters from the top is an equilateraltriangle with side meters. Find the volume of the monument.

14. (a) The base of a solid is a square with vertices located at, and . Each cross-section

perpendicular to the -axis is a semicircle. Find the vol-ume of the solid.

(b) Show that by cutting the solid of part (a), we canrearrange it to form a cone. Thus compute its volumemore simply.

15. Find the length of the curve with parametric equations, , .

16. Use Simpson’s Rule with to estimate the length ofthe arc of the curve from to .

17. A force of 30 N is required to maintain a spring stretchedfrom its natural length of 12 cm to a length of 15 cm. Howmuch work is done in stretching the spring from 12 cm to20 cm?

18. A 1600-lb elevator is suspended by a 200-ft cable thatweighs 10 lb)ft. How much work is required to raise theelevator from the basement to the third floor, a distance of30 ft?

19. A tank full of water has the shape of a paraboloid of revolu-tion as shown in the figure; that is, its shape is obtained byrotating a parabola about a vertical axis.

(2, 14)#1, 1$y ! 1)x 2n ! 10

0 & t & 2y ! 2t 3x ! 3t 2

x#0, "1$#1, 0$, #0, 1$, #"1, 0$

x)4x

xy ! 2 " x 2y ! x 2

28 cm

1–2 " Find the area of the region bounded by the given curves.

1.

2. ," " " " " " " " " " " " "

3. The curve traced out by a point at a distance 1 m from thecenter of a circle of radius 2 m as the circle rolls along the -axis is called a trochoid and has parametric equations

One arch of the trochoid is given by the parameter interval. Find the area under one arch of this trochoid.

4. Find the volume of the solid obtained by rotating about the -axis the region bounded by the curves

, and .

5. Let be the region bounded by the curves , and . Use the Midpoint Rule with to

estimate the following.(a) The area of (b) The volume obtained by rotating about the -axis

6. Let be the region in the first quadrant bounded by the curves and . Calculate the following quantities:(a) The area of (b) The volume obtained by rotating about the -axis(c) The volume obtained by rotating about the -axis

7. Find the volumes of the solids obtained by rotating theregion bounded by the curves and about thefollowing lines:(a) The -axis (b) The -axis (c)

; 8. Let be the region bounded by the curves and. Estimate the following quantities:

(a) The -coordinates of the points of intersection of thecurves

(b) The area of (c) The volume generated when is rotated about the

-axis(d) The volume generated when is rotated about the

-axis

9. Describe the solid whose volume is given by the integral.

(a)

(b)

10. Suppose you are asked to estimate the volume of a football.You measure and find that a football is 28 cm long. You use a piece of string and measure the circumference at its

y1

0 ' [#2 " x 2 $2 " (2 " sx )2 ] dx

y')2

0 2' cos2x dx

y!

x!

!

xy ! x 6 " x % 1

y ! 1 " x 2!

y ! 2yx

y ! x 2y ! x

y!x!

!

y ! 2x " x 2y ! x 3!

x!!

n ! 4y ! 0x ! 1y ! tan#x 2 $,!

x ! 1y ! 1 % xy ! e"2x,x

0 & ( & 2'

x ! 2( " sin ( y ! 2 " cos (

x

y 2 " 6y " x ! 0x " 2y % 7 ! 0

y ! e x " 1, y ! x 2 " x, x ! 1

E X E R C I S E S

24. Find the average value of the function on the interval .

25. If is a continuous function, what is the limit as ofthe average value of on the interval ?

26. After a 6-mg injection of dye into a heart, the readings of dye concentration at two-second intervals are as shown in the table. Use Simpson’s Rule to estimate the cardiacoutput.

27. (a) Explain why the function

is a probability density function.(b) Find .(c) Calculate the mean. Is the value what you would expect?

28. Lengths of human pregnancies are normally distributed with mean 268 days and standard deviation 15 days. Whatpercentage of pregnancies last between 250 days and280 days?

29. The length of time spent waiting in line at a certain bank is modeled by an exponential density function with mean 8 minutes.(a) What is the probability that a customer is served in the

first 3 minutes?(b) What is the probability that a customer has to wait more

than 10 minutes?(c) What is the median waiting time?

P#X , 4$

f #x$ ! 10

'

20 sin*'x

10 + if

if

0 & x & 10

x , 0 or x - 10

"x, x % h%fh l 0f

"0, 2%f #x$ ! x 2 s1 % x 3(a) If its height is 4 ft and the radius at the top is 4 ft, find

the work required to pump the water out of the tank.; (b) After 4000 ft-lb of work has been done, what is the

depth of the water remaining in the tank?

20. A trough is filled with water and its vertical ends have theshape of the parabolic region in the figure. Find the hydro-static force on one end of the trough.

21. A gate in an irrigation canal is constructed in the form of atrapezoid 3 ft wide at the bottom, 5 ft wide at the top, and2 ft high. It is placed vertically in the canal, with the waterextending to its top. Find the hydrostatic force on one sideof the gate.

22. Find the centroid of the region shown.

23. The demand function for a commodity is given by. Find the consumer surplus

when the sales level is 100.p ! 2000 " 0.1x " 0.01x 2

(3,!2)

x

y

0

4 ft

8 ft

4 ft

4 ft

CHAPTER 6 REVIEW # 501

t t

0 0 14 4.72 1.9 16 3.34 3.3 18 2.16 5.1 20 1.18 7.6 22 0.5

10 7.1 24 012 5.8

c#t$c#t$

502

1. Find the area of the region .

2. There is a line through the origin that divides the region bounded by the parabolaand the -axis into two regions with equal area. What is the slope of that

line?

3. A clepsydra, or water clock, is a glass container with a small hole in the bottom throughwhich water can flow. The “clock” is calibrated for measuring time by placing markingson the container corresponding to water levels at equally spaced times. Let becontinuous on the interval and assume that the container is formed by rotating thegraph of about the -axis. Let denote the volume of water and the height of thewater level at time .(a) Determine as a function of .(b) Show that

(c) Suppose that is the area of the hole in the bottom of the container. It follows from Torricelli’s Law that the rate of change of the volume of the water is given by

where is a negative constant. Determine a formula for the function such thatis a constant . What is the advantage in having ?

4. The figure shows a horizontal line intersecting the curve . Find thenumber such that the areas of the shaded regions are equal.

5. A solid is generated by rotating about the -axis the region under the curve ,where is a positive function and . The volume generated by the part of the curvefrom to is for all . Find the function .

6. A cylindrical glass of radius and height is filled with water and then tilted until thewater remaining in the glass exactly covers its base. (a) Determine a way to “slice” the water into parallel rectangular cross-sections and

then set up a definite integral for the volume of the water in the glass.(b) Determine a way to “slice” the water into parallel cross-sections that are trapezoids

and then set up a definite integral for the volume of the water.(c) Find the volume of water in the glass by evaluating one of the integrals in part (a)

or part (b).(d) Find the volume of the water in the glass from purely geometric considerations.(e) Suppose the glass is tilted until the water exactly covers half the base. In what

direction can you “slice” the water into triangular cross-sections? Rectangularcross-sections? Cross-sections that are segments of circles? Find the volume ofwater in the glass.

7. (a) Show that the volume of a segment of height of a sphere of radius is

V ! 13 'h 2#3r " h$

rh

r

L L

r

Lr

fb - 0b 2x ! bx ! 0x $ 0f

y ! f #x$x

cy ! 8x " 27x 3y ! c

dh)dt ! CCdh)dtfk

dVdt

! kAsh

A

dVdt

! ' " f #h$%2 dhdt

hVt

hVyf"0, b%

x ! f #y$

xy ! x " x 2

S ! -#x, y$ , x $ 0, y & 1, x 2 % y 2 & 4y.

FIGURE FOR PROBLEM 3

x

y

x=f(y)

h

b

0 x

yy=8x-27˛

y=c


Focus onProblemSolving

(b) Show that if a sphere of radius 1 is sliced by a plane at a distance from the centerin such a way that the volume of one segment is twice the volume of the other, then

is a solution of the equation

where . Use Newton’s method to find accurate to four decimal places.(c) Using the formula for the volume of a segment of a sphere, it can be shown that the

depth to which a floating sphere of radius sinks in water is a root of the equation

where is the specific gravity of the sphere. Suppose a wooden sphere of radius0.5 m has specific gravity 0.75. Calculate, to four-decimal-place accuracy, the depthto which the sphere will sink.

(d) A hemispherical bowl has radius 5 inches and water is running into the bowl at therate of .(i) How fast is the water level in the bowl rising at the instant the water is 3 inches

deep?(ii) At a certain instant, the water is 4 inches deep. How long will it take to fill the

bowl?

8. Archimedes’ Principle states that the buoyant force on an object partially or fully sub-merged in a fluid is equal to the weight of the fluid that the object displaces. Thus, foran object of density floating partly submerged in a fluid of density , the buoyant force is given by , where is the acceleration due to gravity and

is the area of a typical cross-section of the object. The weight of the object isgiven by

(a) Show that the percentage of the volume of the object above the surface of the liquid is

(b) The density of ice is and the density of seawater is . Whatpercentage of the volume of an iceberg is above water?

(c) An ice cube floats in a glass filled to the brim with water. Does the water overflowwhen the ice melts?

(d) A sphere of radius 0.4 m and having negligible weight is floating in a large fresh-water lake. How much work is required to completely submerge the sphere? Thedensity of the water is .

9. Water in an open bowl evaporates at a rate proportional to the area of the surface of thewater. (This means that the rate of decrease of the volume is proportional to the area ofthe surface.) Show that the depth of the water decreases at a constant rate, regardless ofthe shape of the bowl.

10. A sphere of radius 1 overlaps a smaller sphere of radius in such a way that their inter-section is a circle of radius . (In other words, they intersect in a great circle of thesmall sphere.) Find so that the volume inside the small sphere and outside the largesphere is as large as possible.

11. Suppose that the density of seawater, , varies with the depth below the surface.(a) Show that the hydrostatic pressure is governed by the differential equation

dPdz

! 3#z$t

z3 ! 3#z$

rr

r

1000 kg)m3

1030 kg)m3917 kg)m3

100 3 f " 30

3 f

W ! 30 t yL"h

"h A#y$ dy

A#y$tF ! 3 f t x0

"h A#y$ dy3 f30

0.2 in3)s

s

x 3 " 3rx 2 % 4r 3s ! 0

rx

x0 , x , 1

3x 3 " 9x % 2 ! 0

x

x

503


h

r


y=0

h

y=_h

y=L-h

L

where is the acceleration due to gravity. Let and be the pressure and densityat . Express the pressure at depth as an integral.

(b) Suppose the density of seawater at depth is given by where is apositive constant. Find the total force, expressed as an integral, exerted on a verticalcircular porthole of radius whose center is at a distance below the surface.

12. The figure shows a semicircle with radius 1, horizontal diameter , and tangent linesat and . At what height above the diameter should the horizontal line be placed so asto minimize the shaded area?

13. Let be a pyramid with a square base of side and suppose that is a sphere with itscenter on the base of and is tangent to all eight edges of . Find the height of . Thenfind the volume of the intersection of and .

14. A paper drinking cup filled with water has the shape of a cone with height and semi-vertical angle (see the figure). A ball is placed carefully in the cup, thereby displacingsome of the water and making it overflow. What is the radius of the ball that causes thegreatest volume of water to spill out of the cup?

15. A curve is defined by the parametric equations

Find the length of the arc of the curve from the origin to the nearest point where there isa vertical tangent line.

16. Suppose we are planning to make a taco from a round tortilla with diameter 8 inches bybending the tortilla so that it is shaped as if it is partially wrapped around a circularcylinder. We will fill the tortilla to the edge (but no more) with meat, cheese, and otheringredients. Our problem is to decide how to curve the tortilla in order to maximize thevolume of food it can hold.(a) We start by placing a circular cylinder of radius along a diameter of the tortilla

and folding the tortilla around the cylinder. Let represent the distance from thecenter of the tortilla to a point on the diameter (see the figure). Show that thecross-sectional area of the filled taco in the plane through perpendicular to theaxis of the cylinder is

and write an expression for the volume of the filled taco.(b) Determine (approximately) the value of that maximizes the volume of the taco.

(Use a graphical approach with your CAS.)

17. A string is wound around a circle and then unwound while being held taut. The curvetraced by the point at the end of the string is called the involute of the circle. If theP

r

A#x$ ! rs16 " x 2 " 12 r 2 sin*2

r s16 " x 2+

PP

xr

CAS

x ! yt

1 cos u

u du y ! yt

1 sin u

u du

h

¨

(h

PSPPP

S2bP

QPPQ

L - rr

H3 ! 30e z)Hzzz ! 0

30P0t

504

P Q


P

x


circle has radius and center and the initial position of is , and if the parame-ter is chosen as in the figure, show that parametric equations of the involute are

18. A cow is tied to a silo with radius by a rope just long enough to reach the oppositeside of the silo. Find the area available for grazing by the cow.

19. In a famous 18th-century problem, known as Buffon’s needle problem, a needle oflength is dropped onto a flat surface (for example, a table) on which parallel lines

units apart, , have been drawn. The problem is to determine the probability thatthe needle will come to rest intersecting one of the lines. Assume that the lines run east-west, parallel to the -axis in a rectangular coordinate system (as in the figure). Let bethe distance from the “southern” end of the needle to the nearest line to the north. (Ifthe needle’s southern end lies on a line, let . If the needle happens to lie east-west,let the “western” end be the “southern” end.) Let be the angle that the needle makeswith a ray extending eastward from the “southern” end. Then and

. Note that the needle intersects one of the lines only when .Now, the total set of possibilities for the needle can be identified with the rectangularregion , , and the proportion of times that the needle intersects aline is the ratio

This ratio is the probability that the needle intersects a line. Find the probability that the needle will intersect a line if . What if ?

20. If the needle in Problem 19 has length , it’s possible for the needle to intersectmore than one line.(a) If , find the probability that a needle of length 7 will intersect at least one

line. [Hint: We can proceed as in Problem 19. Define as before; then the totalset of possibilities for the needle can be identified with the same rectangular region

, . What portion of the rectangle corresponds to the needleintersecting a line?]

(b) If , find the probability that a needle of length 7 will intersect two lines.(c) If , find a general formula for the probability that the needle intersects

three lines.2L , h & 3LL ! 4

0 & ( & '0 & y & L

yL ! 4

h - L

h ! L)2h ! L

area under y ! h sin (area of rectangle

0 & ( & '0 & y & L

y , h sin (0 & ( & '0 & y & L

(y ! 0

yx

L $ hLh

r

x ! r#cos ( % ( sin ($ y ! r#sin ( " ( cos ($

(#r, 0$POr

505

xO

y

r¨ P

T


y h sin ¨¨

h

L


$2

$

y

¨

Lh

applications of integration · more about areas in chapter 5 we deﬁned and calculated areas of...

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