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Unformatted text preview: 5E09(pp 582591) 1/17/06 6:20 PM Page 582 CHAPTER 9 Integration enables us to
calculate the force exerted
by water on a dam. Further Applications of Integration 5E09(pp 582591) 1/17/06 6:20 PM Page 583 We looked at some applications of integrals in Chapter 6:
areas, volumes, work, and average values. Here we
explore some of the many other geometric applications of
integration—the length of a curve, the area of a surface—
as well as quantities of interest in physics, engineering,
biology, economics, and statistics. For instance, we will investigate the center of
gravity of a plate, the force exerted by water pressure on a dam, the ﬂow of blood
from the human heart, and the average time spent on hold during a telephone call.  9.1 Arc Length
What do we mean by the length of a curve? We might think of ﬁtting a piece of string to
the curve in Figure 1 and then measuring the string against a ruler. But that might be
difﬁcult to do with much accuracy if we have a complicated curve. We need a precise
deﬁnition for the length of an arc of a curve, in the same spirit as the deﬁnitions we 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 of the
line segments that form the polygon. (We can use the distance formula to ﬁnd the distance
between the endpoints of each segment.) We are going to deﬁne the length of a general
curve by ﬁrst approximating it by a polygon and then taking a limit as the number of segments of the polygon is increased. This process is familiar for the case of a circle, where
the circumference is the limit of lengths of inscribed polygons (see Figure 2).
Now suppose that a curve C is deﬁned by the equation y f x , where f is continuous
and a x b. We obtain a polygonal approximation to C by dividing the interval a, b
into n subintervals with endpoints x 0 , x 1, . . . , x n and equal width x. If yi f x i , then
the point Pi x i , yi lies on C and the polygon with vertices P0 , P1 , . . . , Pn , illustrated in
Figure 3, is an approximation to C. The length L of C is approximately the length of this
polygon and the approximation gets better as we let n increase. (See Figure 4, where the
arc of the curve between Pi 1 and Pi has been magniﬁed and approximations with successively smaller values of x are shown.) Therefore, we deﬁne the length L of the curve C FIGURE 1 FIGURE 2 Pi
Pi y P™ y=ƒ P¡ Pi
Pi1 Pi Pi1 Pn P¸ Pi
Pi1
Pi1 0 F IGURE 3 a x¡ ¤ x i1 xi b x Pi1
FIGURE 4 583 5E09(pp 582591) 584 ❙❙❙❙ 1/17/06 6:21 PM Page 584 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION with equation y f x , a
(if the limit exists): b, as the limit of the lengths of these inscribed polygons x n L 1 lim nl Pi 1Pi
i1 Notice that the procedure for deﬁning arc length is very similar to the procedure we
used for deﬁning area and volume: We divided the curve into a large number of small parts.
We then found the approximate lengths of the small parts and added them. Finally, we took
the limit as n l .
The deﬁnition of arc length given by Equation 1 is not very convenient for computational purposes, but we can derive an integral formula for L in the case where f has a
continuous derivative. [Such a function f is called smooth because a small change in x
produces a small change in f x .]
If we let yi yi yi 1 , then
Pi 1Pi s xi xi 2 1 yi 2 yi x s 1 2 yi 2 By applying the Mean Value Theorem to f on the interval x i 1, x i , we ﬁnd that there is a
number xi* between x i 1 and x i such that
f xi 1 f xi* x i yi f xi f xi* that is, xi 1 x Thus, we have
s x 2 s Pi 1Pi x 2 2 yi f xi* s1 [ f xi* 2 s1 f xi* 2 x
x s 2 2 x (since x 0) f xi* 2 x Therefore, by Deﬁnition 1,
n L lim nl n Pi 1Pi lim nl i1 s1
i1 We recognize this expression as being equal to y b a s1 fx 2 dx by the deﬁnition of a deﬁnite integral. This integral exists because the function
tx
f x 2 is continuous. Thus, we have proved the following theorem:
s1
2 y The Arc Length Formula If f is continuous on a, b , then the length of the curve f x,a x b, is
L y b a s1 fx 2 dx 5E09(pp 582591) 1/17/06 6:21 PM Page 585 S ECTION 9.1 ARC LENGTH ❙❙❙❙ 585 If we use Leibniz notation for derivatives, we can write the arc length formula as
follows: y L 3 2 dy
dx b 1 a dx EXAMPLE 1 Find the length of the arc of the semicubical parabola y 2 y x 3 between the points 1, 1 and 4, 8 . (See Figure 5.)
(4, 8) SOLUTION For the top half of the curve we have
¥=˛ (1, 1) dy
dx x3 2 y 3
2 x1 2 and so the arc length formula gives
x 0 y L
F IGURE 5 1 1  As a check on our answer to Example 1,
notice from Figure 5 that it ought to be slightly
larger than the distance from 1, 1 to 4, 8 ,
which is
s58 If we substitute u
u 10. Therefore 1 9x 4, then du
L 4
9 y 10 13 4 7.615773 y dx 4 1 9
4 s1 9 d x 4. When x
4
9 su du 2
3 u3 2 x dx
13
4 1, u ; when x 4, 10
13 4 [10 3 2 ( 13 )3 2 ]
4
1
13 s13 )
27 (80 s10
8
27 According to our calculation in Example 1, we
have
1
L 27 (80 s10 13 s13 ) 7.633705
Sure enough, this is a bit greater than the length
of the line segment. 2 dy
dx 4 If a curve has the equation x t y , c y d, and t y is continuous, then by interchanging the roles of x and y in Formula 2 or Equation 3, we obtain the following formula
for its length: y L 4 d c s1 ty 2 y dy dx
dy d 1 c EXAMPLE 2 Find the length of the arc of the parabola y 2
SOLUTION Since x y 2, we have dx d y y L 1 0 2 y sec 1
2 1
2 1
4 (sec 0 [sec 1
2 sec 2 d tan
tan 1
2 y dy
1
2 We make the trigonometric substitution y
sec . When y
s1 4y 2 s1 tan 2
tan
2, so
tan 1 2
, say. Thus
L dy x from 0, 0 to 1, 1 . 2y, and Formula 4 gives
dx
dy 1 2 1 0 s1 4y 2 dy tan , which gives d y 1 sec 2 d and
2
0, tan
0, so
0; when y 1, y sec 3 d ln sec tan ln sec 0 tan 0 ) (from Example 8 in Section 8.2) 5E09(pp 582591) ❙❙❙❙ 586 1/17/06 6:22 PM Page 586 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION (We could have used Formula 21 in the Table of Integrals.) Since tan
sec 2
1 tan 2
5, so sec
s5 and
ln(s5
4 s5
2 L 2, we have 2) y
1  Figure 6 shows the arc of the parabola
whose length is computed in Example 2, together
with polygonal approximations having n 1
and n 2 line segments, respectively. For
n 1 the approximate length is L 1 s2, the
diagonal of a square. The table shows the
approximations L n that we get by dividing 0, 1
into n equal subintervals. Notice that each time
we double the number of sides of the polygon,
we get closer to the exact length, which is
L s5
2 ln(s5
4 2) n 0 1.478943 1 x Ln 1
2
4
8
16
32
64 x=¥ 1.414
1.445
1.464
1.472
1.476
1.478
1.479 FIGURE 6 Because of the presence of the square root sign in Formulas 2 and 4, the calculation of
an arc length often leads to an integral that is very difﬁcult or even impossible to evaluate
explicitly. Thus, we sometimes have to be content with ﬁnding an approximation to the
length of a curve as in the following example.
EXAMPLE 3 (a) Set up an integral for the length of the arc of the hyperbola x y
point 1, 1 to the point (2, 1 ).
2
(b) Use Simpson’s Rule with n 10 to estimate the arc length. 1 from the SOLUTION (a) We have
1
x y dy
dx 1
x2 and so the arc length is L y 2 1 1 dy
dx 2 dx y 2 1 1 (b) Using Simpson’s Rule (see Section 8.7) with a
fx
s1 1 x 4, we have
L y 2 1 1  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. x
f1
3
1.1321 1
dx
x4
1, b y 2 1 sx 4 1
dx
x2 2, n 10, x 0.1, and 1
dx
x4
4 f 1.1 2 f 1.2 4 f 1.3 2 f 1.8 4 f 1.9 f2 5E09(pp 582591) 1/17/06 6:22 PM Page 587 S ECTION 9.1 ARC LENGTH ❙❙❙❙ 587 The Arc Length Function
We will ﬁnd it useful to have a function that measures the arc length of a curve from a particular starting point to any other point on the curve. Thus, if a smooth curve C has the
equation y f x , a x b, let s x be the distance along C from the initial point
P0 a, f a to the point Q x, f x . Then s is a function, called the arc length function, and,
by Formula 2, y sx 5 x a s1 2 ft dt (We have replaced the variable of integration by t so that x does not have two meanings.)
We can use Part 1 of the Fundamental Theorem of Calculus to differentiate Equation 5
(since the integrand is continuous):
ds
dx 6 s1 fx 2 dy
dx 1 2 Equation 6 shows that the rate of change of s with respect to x is always at least 1 and is
equal to 1 when f x , the slope of the curve, is 0. The differential of arc length is
ds 7 dy
dx 1 2 dx and this equation is sometimes written in the symmetric form
y ds
Îy dx FIGURE 7 2 dx 2 2 dy The geometric interpretation of Equation 8 is shown in Figure 7. It can be used as a
mnemonic device for remembering both of the Formulas 3 and 4. If we write L x ds,
then from Equation 8 either we can solve to get (7), which gives (3), or we can solve to get dy
Îs 0 ds 8 ds x dx
dy 1 2 dy which gives (4).
1
8 x2 EXAMPLE 4 Find the arc length function for the curve y ln x taking P0 1, 1 as the starting point.
x2 SOLUTION If f x 1
8 ln x, then
2x fx 1 fx 2 1
8x 1 2x 4x 2 s1 fx 2 1
2 2x 1
8x 1
8x 2 1
64 x 2 4x 2 1
2x 1
8x 1
2
2 1
64 x 2 5E09(pp 582591) ❙❙❙❙ 588 1/17/06 6:23 PM Page 588 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION Thus, the arc length function is given by
x y s1 sx ft y x 2 1
8t 1 dt 2t 1
1
8 x2 ln x dt
1
8 t2 x ln t 1 1 For instance, the arc length along the curve from 1, 1 to 3, f 3 is
1
8 32 s3 ln 3 1 ln 3
8 8 8.1373 y y 1 s(x)
 Figure 8 shows the interpretation of the arc
length function in Example 4. Figure 9 shows the
graph of this arc length function. Why is s x
negative when x is less than 1? 1 P¸ 0 0 1 s(x)=≈+1 ln x1
8 x x FIGURE 8  9.1 FIGURE 9 Exercises 1. Use the arc length formula (3) to ﬁnd the length of the curve 4. y x3
6 ■ ■ 5–16  5. y
6. y ■ ,1 x ■ 6 x 3 2, 0 4x
x
6 3 4,
1
,
10 x 3 3 ln sec x , 0 x 4 ln x, cosh x, 14. y 2 1 ■ ■ ■ x
0 ■ ■ ln ■ 2, x ■ ex
ex
■ s3
x 1 y 2 x 1 1
,
1
■ a x ■ ■ 17–20 y 0 2 b, a
■ 0
■ ■ ■ ■  Set up, but do not evaluate, an integral for the length of
the curve. 17. y 1 0 e x, 0 ■ 1
x ■ x 4 x, 0 1 ■ 9 x 16. y x y ln cos x , 0 15. y 1
2 3, 1 4 13. y 3 Find the length of the curve. 5 7. y 32 1
,
2x 1
2 1 x 12. y Graph the curve and visually estimate its length. Then ﬁnd
;
its exact length.
 x2 sy y 2 11. y y s4 x 2, 0 x 2. Check your answer by noting that
the curve is a quartercircle. 2
3 1
3 9. x ln x
,
4 10. y 2. Use the arc length formula to ﬁnd the length of the curve 3. y x2
2 8. y y 2 3x, 2 x 1. Check your answer by noting that
the curve is a line segment and calculating its length by the distance formula. 3–4 x 1 y=≈ 1 ln x
8 18. y cos x, 0
x 2, 0 x
x 2
3 ■ 5E09(pp 582591) 1/17/06 6:23 PM Page 589 S ECTION 9.1 ARC LENGTH 19. x
■ y 3, y
■ 1 ■ y ■ 20. 4 ■ ■ x2
a2 ■ y2
b2 ■ ■ ■ ■ 21–24 10 to estimate the arc
 Use Simpson’s rule with n
length of the curve. Compare your answer with the value of the
integral produced by your calculator. 21. y xe x, 0 23. y sec x, ■ ■ ■ x 0 22. x x 3 ■ ■ ■ y 24. y 5 x ln x, ■ sy, ■ ■ 1
1 y
x ■ 34. A steady wind blows a kite due west. The kite’s height above ground from horizontal position x
2 y 1
40 150 3
■ 0 to x x 50 80 ft is given by 2 Find the distance traveled by the kite. ■ 3 ; 25. (a) Graph the curve y 589 until it hits the ground, where y is its height above the ground
and x is the horizontal distance traveled in meters. Calculate
the distance traveled by the prey from the time it is dropped
until the time it hits the ground. Express your answer correct to
the nearest tenth of a meter. 1 ■ ❙❙❙❙ x s4 x, 0 x 4.
(b) Compute the lengths of inscribed polygons with n 1, 2,
and 4 sides. (Divide the interval into equal subintervals.)
Illustrate by sketching these polygons (as in Figure 6).
(c) Set up an integral for the length of the curve.
(d) Use your calculator to ﬁnd the length of the curve to four
decimal places. Compare with the approximations in
part (b). 35. A manufacturer of corrugated metal rooﬁng wants to produce panels that are 28 in. wide and 2 in. thick by processing ﬂat
sheets of metal as shown in the ﬁgure. The proﬁle of the roofing takes the shape of a sine wave. Verify that the sine curve
has equation y sin x 7 and ﬁnd the width w of a ﬂat
metal sheet that is needed to make a 28inch panel. (Use your
calculator to evaluate the integral correct to four signiﬁcant
digits.) ; 26. Repeat Exercise 25 for the curve
y
C AS x sin x 0 x 2 27. Use either a computer algebra system or a table of integrals to ﬁnd the exact length of the arc of the curve x
31
lies between the points 0, 0 and (ln 4 , 2 ).
CAS ln 1 y 2 that 28. Use either a computer algebra system or a table of integrals to ﬁnd the exact length of the arc of the curve y x 4 3 that lies
between the points 0, 0 and 1, 1 . If your CAS has trouble
evaluating the integral, make a substitution that changes the
integral into one that the CAS can evaluate.
29. Sketch the curve with equation x 2 3 y2 3 1 and use symmetry to ﬁnd its length.
30. (a) Sketch the curve y 3 x 2.
(b) Use Formulas 3 and 4 to set up two integrals for the arc
length from 0, 0 to 1, 1 . Observe that one of these is
an improper integral and evaluate both of them.
(c) Find the length of the arc of this curve from 1, 1
to 8, 4 . 31. Find the arc length function for the curve y 2 in w 28 in 36. (a) The ﬁgure shows a telephone wire hanging between ; two poles at x
b and x b. It takes the shape of a
catenary with equation y c a cosh x a . Find the
length of the wire.
(b) Suppose two telephone poles are 50 ft apart and the length
of the wire between the poles is 51 ft. If the lowest point of
the wire must be 20 ft above the ground, how high up on
each pole should the wire be attached?
y 2 x 3 2 with start ing point P0 1, 2 . _b
1
3 ; 32. (a) Graph the curve y x
1 4 x , x 0.
(b) Find the arc length function for this curve with starting
7
point P0 (1, 12 ).
(c) Graph the arc length function. 33. A hawk ﬂying at 15 m s at an altitude of 180 m accidentally drops its prey. The parabolic trajectory of the falling prey is
described by the equation
y 0 bx 3 180 x2
45 37. Find the length of the curve y ; 38. The curves with equations x x1x st 3 n n 1 dt, 1 x 4. y
1, n 4, 6, 8, . . . , are
called fat circles. Graph the curves with n 2, 4, 6, 8, and 10
to see why. Set up an integral for the length L 2 k of the fat circle
with n 2 k. Without attempting to evaluate this integral, state
the value of
lim L 2 k kl 5E09(pp 582591) 590 ❙❙❙❙ 1/17/06 6:24 PM Page 590 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION DISCOVERY PROJECT
Arc Length Contest
The curves shown are all examples of graphs of continuous functions f that have the following
properties.
1. f 0 0 and f 1 2. f x 0 for 0 0
x 1 3. The area under the graph of f from 0 to 1 is equal to 1. The lengths L of these curves, however, are different.
y y y y 1 1 1 1 0 1 x 0 LÅ3.249 1 x 0 LÅ2.919 1 x 0 LÅ3.152 1 x LÅ3.213 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 calculate the arc
length of each graph. The winning entry will be the one with the smallest arc length.  9.2 Area of a Surface of Revolution cut h
r A surface of revolution is formed when a curve is rotated about a line. Such a surface is
the lateral boundary of a solid of revolution of the type discussed in Sections 6.2 and 6.3.
We want to deﬁne the area of a surface of revolution in such a way that it corresponds
to our intuition. If the surface area is A, we can imagine that painting the surface would
require the same amount of paint as does a ﬂat region with area A.
Let’s start with some simple surfaces. The lateral surface area of a circular cylinder with
radius r and height h is taken to be A 2 rh because we can imagine cutting the cylinder and unrolling it (as in Figure 1) to obtain a rectangle with dimensions 2 r and h.
Likewise, we can take a circular cone with base radius r and slant height l , cut it along
the dashed line in Figure 2, and ﬂatten it to form a sector of a circle with radius l and central angle
2 r l. We know that, in general, the area of a sector of a circle with radius
l and angle is 1 l 2 (see Exercise 35 in Section 8.3) and so in this case it is
2 h A
2πr
FIGURE 1 12
2 l 12
2 l 2r
l rl Therefore, we deﬁne the lateral surface area of a cone to be A rl. 5E09(pp 582591) 1/17/06 6:24 PM Page 591 SECTION 9.2 AREA OF A SURFACE OF REVOLUTION ❙❙❙❙ 591 2πr cut
l ¨ r l FIGURE 2 What about more complicated surfaces of revolution? If we follow the strategy we used
with arc length, we can approximate the original curve by a polygon. When this polygon
is rotated about an axis, it creates a simpler surface whose surface area approximates the
actual surface area. By taking a limit, we can determine the exact surface area.
The approximating surface, then, consists of a number of bands, each formed by rotating a line segment about an axis. To ﬁnd the surface area, each of these bands can be
considered a portion of a circular cone, as shown in Figure 3. The area of the band (or frustum of a cone) with slant height l and upper and lower radii r1 and r2 is found by subtracting the areas of two cones: l¡ r¡ 1 l A r2 l1 l r1l1 r2 r1 l1 r2 l From similar triangles we have r™ l1
r1
FIGURE 3 l1 l
r2 which gives
r1l r1l1 or A r2 l1 r1l r2 r1 l1 r1l Putting this in Equation 1, we get
y r2 l y=ƒ or
0 a b x (a) Surface of revolution
y P¸
0 Pi1 Pi a yi
Pn b x A 2 where r 1 r1 r2 is the average radius of the band.
2
Now we apply this formula to our strategy. Consider the surface shown in Figure 4,
which is obtained by rotating the curve y f x , a x b, about the xaxis, where f is
positive and has a continuous derivative. In order to deﬁne its surface area, we divide the
interval a, b into n subintervals with endpoints x0, x1, . . . , xn and equal width x, as we
did in determining arc length. If yi f x i , then the point Pi x i, yi lies on the curve. The
part of the surface between x i 1 and x i is approximated by taking the line segment Pi 1Pi
Pi 1Pi and averand rotating it about the xaxis. The result is a band with slant height l
1
age radius r 2 yi 1 yi so, by Formula 2, its surface area is (b) Approximating band
FIGURE 4 2 rl 2 yi 1 2 yi Pi 1Pi 5E09(pp 592601) 592 ❙❙❙❙ 1/17/06 6:16 PM Page 592 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION As in the proof of Theorem 9.1.2, we have
Pi 1Pi f xi* s1 2 x where xi* is some number in x i 1, x i . When x is small, we have yi
also yi 1 f x i 1
f xi* , since f is continuous. Therefore
2 yi 1 yi 2 2 f xi* s1 Pi 1Pi f xi* 2 f xi* and f xi x and so an approximation to what we think of as the area of the complete surface of revolution is
n 2 f xi* s1 3 f xi* 2 x i1 This approximation appears to become better as n l and, recognizing (3) as a Riemann
sum for the function t x
2 f x s1
f x 2, we have
n 2 f xi* s1 lim nl f xi* 2 x i1 y b a 2 f x s1 fx 2 dx Therefore, in the case where f is positive and has a continuous derivative, we deﬁne the
surface area of the surface obtained by rotating the curve y f x , a x b, about
the xaxis as 4 S y b 2 f x s1 a 2 fx dx With the Leibniz notation for derivatives, this formula becomes 5 S If the curve is described as x
becomes 6 S y b a 2y t y, c y d c 2y
dy
dx 1 y 1 2 dx d, then the formula for surface area dx
dy 2 dy and both Formulas 5 and 6 can be summarized symbolically, using the notation for arc
length given in Section 9.1, as 7 S y2 y ds 5E09(pp 592601) 1/17/06 6:16 PM Page 593 ❙❙❙❙ S ECTION 9.2 AREA OF A SURFACE OF REVOLUTION 593 For rotation about the yaxis, the surface area formula becomes S y2 dx or 8 x ds where, as before, we can use either
ds 2 dy
dx 1 ds 2 dx
dy 1 dy These formulas can be remembered by thinking of 2 y or 2 x as the circumference of a
circle traced out by the point x, y on the curve as it is rotated about the xaxis or yaxis,
respectively (see Figure 5).
y y
(x, y) y
x circumference=2πx circumference=2πy
0 FIGURE 5 (x, y) x 0 (a) Rotation about xaxis: S=j 2πy ds x (b) Rotation about yaxis: S=j 2πx ds s4 x 2, 1 x 1, is an arc of the circle x 2 y 2 4.
Find the area of the surface obtained by rotating this arc about the xaxis. (The surface is
a portion of a sphere of radius 2. See Figure 6.)
EXAMPLE 1 The curve y SOLUTION We have y dy
dx 1
2 x2 4 12 x 2x x2 s4 and so, by Formula 5, the surface area is
1 x S y 1
1 2y 2 y 2 y 4 y 1
1 1
1 FIGURE 6
 Figure 6 shows the portion of the sphere
whose surface area is computed in Example 1. 1
1 dy
dx 1 s4 x2 s4 x2 1 dx 4 1 2 dx
x2
x2 4 2
s4 x2 2 8 dx dx 5E09(pp 592601) 594 ❙❙❙❙ 1/17/06 6:17 PM Page 594 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION  Figure 7 shows the surface of revolution
whose area is computed in Example 2.
y x 2 from 1, 1 to 2, 4 is rotated about the
yaxis. Find the area of the resulting surface. EXAMPLE 2 The arc of the parabola y
SOLUTION 1 Using (2, 4) y
y=≈ 0 1 2 x2 S y2 x y 2 1 x ds 2x y 2 S  As a check on our answer to Example 2,
notice from Figure 7 that the surface area
should be close to that of a circular cylinder with
the same height and radius halfway between
the upper and lower radius of the surface:
2 1.5 3
28.27. We computed that
the surface area was
5 s5 ) 30.85 y 4 2 1 Substituting u 1 4 x 2, we have du
integration, we have (17 s17 2x we have, from Formula 8, FIGURE 7 6 dy
dx and 17 5 dx 4x 2 dx x s1 8 x d x. Remembering to change the limits of su du (17 s17 6 2 dy
dx 1 4 [ 2 u 3 2 ]17
5
3 5 s5 ) SOLUTION 2 Using x dx
dy and sy 1
2 sy we have which seems reasonable. Alternatively, the surface area should be slightly larger than the area
of a frustum of a cone with the same top and
bottom edges. From Equation 2, this is
2 1.5 (s10 ) 29.80. S y2
2 4
6 x ds y 4 1 y 17 5 y sy 4 1 2x 1 1
dy
4y 1 su du (17 s17 dx
dy y 4 1 (where u 5 s5 ) 2 dy
s4y
1 1 dy
4y) (as in Solution 1) EXAMPLE 3 Find the area of the surface generated by rotating the curve y 0
 Another method: Use Formula 6 with
x ln y. x 1, about the xaxis. SOLUTION Using Formula 5 with y ex and dy
dx ex e x, 5E09(pp 592601) 1/17/06 6:17 PM Page 595 ❙❙❙❙ S ECTION 9.2 AREA OF A SURFACE OF REVOLUTION 595 we have y S 1 0 2y 2 y 2 e y 1 1
2 2  Or use Formula 21 in the Table of Integrals. [sec tan tan 1. y ln x, 1 2. y sin 2x, 0 x 2; 3. y sec x, 0 x 4; 4. y e x, 1
■ x 3; y ■ xaxis 2; ■ ■ ■ ■ ■ ■ ■ ■ ■ x 3,
y 0 2 x 2 18, s x, 4 x 8. y cos 2 x, 9. y cosh x, 12. x
■ y 2 6
1 1
2 x 2y , 1 y y , 2 1
■ ■ 2 13. y sx, 1 14. y 1 x 2, ■ sa 2 y
0 ■ x 0 ■ y, 0 ■ ■ ■ sx, s1
■ 1 x x e, 0
■ ■ ■ ■ ■ ■ ■ a ■ 1 x, 1
■ ■ a
■ ■ ■ ■ ■ x 1
■ x
■ 22. y 2
■ ■ ■ sx 2
■ 1,
■ 0
■ ■ x 3 ■ ■ 23–24  Use a CAS to ﬁnd the exact area of the surface obtained
by rotating the curve about the yaxis. If your CAS has trouble
evaluating the integral, express the surface area as an integral in the
other variable. x 3,
■ 0
■ y
■ 24. y 1
■ ■ ■
ln x
■ 1,
■ 0
■ x 1 ■ x, y x 1, 0 y 1 x is rotated
about the xaxis, the volume of the resulting solid is ﬁnite (see
Exercise 63 in Section 8.8). Show that the surface area is inﬁnite. (The surface is shown in the ﬁgure and is known as
Gabriel’s horn.)
y 0 a2
y 2
■ 1
y ■ 20. y 3
■ x 1
y= x x 2 18. y 3
x ■ 2 a cosh y a ,
■ 1)] 25. If the region  The given curve is rotated about the yaxis. Find the area
of the resulting surface.
3 ■ ■ 23. y 1 13–16 16. x ln(s2 s2  Use either a CAS or a table of integrals to ﬁnd the exact
area of the surface obtained by rotating the given curve about the
xaxis. ■ 2 ■ CAS x 32 2 6 x 1
,
2x 1
■ 15. x e2) 21–22 ■ 9 0 x3
6 x 1 sec x,
■ 21. y 2 0 11. x 1) e 2 and 1 s1 ln(s2 ■ CAS 1
3 (by Example 8 in Section 8.2) 4 s2 tan 2 ln x, 19. y yaxis Find the area of the surface obtained by rotating the curve
about the xaxis. 10. y tan ln(e 17. y  7. y tan  Use Simpson’s Rule with n
10 to approximate the area
of the surface obtained by rotating the curve about the xaxis. Compare your answer with the value of the integral produced by your
calculator. yaxis 6. 9 x tan 1e) 17–20 xaxis 5. y and Exercises  Set up, but do not evaluate, an integral for the area of the
surface obtained by rotating the curve about the given axis. 5–12 e 2x dx tan 1 e2 e x s1 ex) ln sec 1–4 ■ 0 ln sec [e s1 S 1 (where u e, we have sec 2 Since tan y 2 (where u sec 3 d [sec  9.2 dx u 2 du s1 4 2 dy
dx 1 1 x ■ 5E09(pp 592601) 596 ❙❙❙❙ 1/17/06 6:18 PM Page 596 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION e x, x 0, is rotated about the xaxis,
ﬁnd the area of the resulting surface. 26. If the inﬁnite curve y CAS area of the surface generated by rotating the curve y sx,
0 x 4, about the line y 4. Then use a CAS to evaluate
the integral. 27. (a) If a 0, ﬁnd the area of the surface generated by rotating
the loop of the curve 3ay 2 x a x 2 about the xaxis.
(b) Find the surface area if the loop is rotated about the
yaxis. 33. Find the area of the surface obtained by rotating the circle x2 28. A group of engineers is building a parabolic satellite dish r 2 about the line y r. between two parallel planes is S
d h, where d is the diameter of the sphere and h is the distance between the planes.
(Notice that S depends only on the distance between the planes
and not on their location, provided that both planes intersect
the sphere.) 29. The ellipse y2
b2 y2 34. Show that the surface area of a zone of a sphere that lies whose shape will be formed by rotating the curve y ax 2
about the yaxis. If the dish is to have a 10ft diameter and a
maximum depth of 2 ft, ﬁnd the value of a and the surface area
of the dish. x2
a2 32. Use the result of Exercise 31 to set up an integral to ﬁnd the 35. Formula 4 is valid only when f x 1 a 0. Show that when f x
is not necessarily positive, the formula for surface area becomes b S is rotated about the xaxis to form a surface called an ellipsoid.
Find the surface area of this ellipsoid. y b a 2 f x s1 fx 2 dx 36. Let L be the length of the curve y f x , a x b, where
f is positive and has a continuous derivative. Let S f be the surface area generated by rotating the curve about the xaxis. If c
is a positive constant, deﬁne t x
fx
c and let St be the
corresponding surface area generated by the curve y t x ,
a x b. Express St in terms of S f and L. 30. Find the surface area of the torus in Exercise 61 in Section 6.2.
31. If the curve y f x , a x b, is rotated about the horizontal line y c, where f x
c, ﬁnd a formula for the area of
the resulting surface. DISCOVERY PROJECT
Rotating on a Slant
We know how to ﬁnd the volume of a solid of revolution obtained by rotating a region about a
horizontal or vertical line (see Section 6.2). We also know how to ﬁnd the surface area of a
surface of revolution if we rotate a curve about a horizontal or vertical line (see Section 9.2). But
what if we rotate about a slanted line, that is, a line that is neither horizontal nor vertical? In this
project you are asked to discover formulas for the volume of a solid of revolution and for the
area of a surface of revolution when the axis of rotation is a slanted line.
Let C be the arc of the curve y f x between the points P p, f p and Q q, f q and let
be the region bounded by C, by the line y m x b (which lies entirely below C ), and by the
perpendiculars to the line from P and Q.
y Q y=ƒ P y=m x+b C Îu
0 p q x 5E09(pp 592601) 1/17/06 6:18 PM Page 597 S ECTION 9.3 APPLICATIONS TO PHYSICS AND ENGINEERING 1. Show that the area of ❙❙❙❙ 597 is
1
m2 1 y q p fx mx b1 m f x dx [Hint: This formula can be veriﬁed by subtracting areas, but it will be helpful throughout the
project to derive it by ﬁrst approximating the area using rectangles perpendicular to the line,
as shown in the ﬁgure. Use the ﬁgure to help express u in terms of x.] ? tangent to C
at { x i , f(x i )} ? y=m x+b
Îu xi å y ∫ Îx (2π, 2π) 2. Find the area of the region shown in the ﬁgure at the left.
y=x+sin x 3. Find a formula similar to the one in Problem 1 for the volume of the solid obtained by rotating y=x2 about the line y mx b. 4. Find the volume of the solid obtained by rotating the region of Problem 2 about the line y
0 x  9.3 2. 5. Find a formula for the area of the surface obtained by rotating C about the line y
CAS FIGURE FOR PROBLEM 2 x mx b. 6. Use a computer algebra system to ﬁnd the exact area of the surface obtained by rotating the curve y sx, 0
mal places. x 4, about the line y 1
2 x. Then approximate your result to three deci Applications to Physics and Engineering
Among the many applications of integral calculus to physics and engineering, we consider
two here: force due to water pressure and centers of mass. As with our previous applications to geometry (areas, volumes, and lengths) and to work, our strategy is to break up the
physical quantity into a large number of small parts, approximate each small part, add the
results, take the limit, and then evaluate the resulting integral. Hydrostatic Pressure and Force
surface of ﬂuid d
A FIGURE 1 Deepsea divers realize that water pressure increases as they dive deeper. This is because
the weight of the water above them increases.
In general, suppose that a thin horizontal plate with area A square meters is submerged
in a ﬂuid of density kilograms per cubic meter at a depth d meters below the surface of
the ﬂuid as in Figure 1. The ﬂuid directly above the plate has volume V Ad, so its mass
is m
V
Ad. The force exerted by the ﬂuid on the plate is therefore
F mt tAd 5E09(pp 592601) 598 ❙❙❙❙ 1/17/06 6:18 PM Page 598 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION where t is the acceleration due to gravity. The pressure P on the plate is deﬁned to be the
force per unit area:
F
A P
 When using U.S. Customary units, we write
P
td
d, where
t is the weight
density (as opposed to , which is the mass
density). For instance, the weight density of
62.5 lb ft 3.
water is td The SI unit for measuring pressure is newtons per square meter, which is called a pascal
(abbreviation: 1 N m2 1 Pa). Since this is a small unit, the kilopascal (kPa) is often
used. For instance, because the density of water is
1000 kg m3 , the pressure at the
bottom of a swimming pool 2 m deep is
P 1000 kg m 3 td 19,600 Pa 9.8 m s 2 2m 19.6 kPa An important principle of ﬂuid pressure is the experimentally veriﬁed fact that at any
point in a liquid the pressure is the same in all directions. (A diver feels the same pressure
on nose and both ears.) Thus, the pressure in any direction at a depth d in a ﬂuid with mass
density is given by
P 1 td d This helps us determine the hydrostatic force against a vertical plate or wall or dam in a
ﬂuid. This is not a straightforward problem because the pressure is not constant but
increases as the depth increases.
50 m 20 m SOLUTION We choose a vertical xaxis with origin at the surface of the water as in
Figure 3(a). The depth of the water is 16 m, so we divide the interval 0, 16 into subintervals of equal length with endpoints x i and we choose xi*
x i 1, x i . The i th horizontal strip of the dam is approximated by a rectangle with height x and width wi ,
where, from similar triangles in Figure 3(b), 30 m
FIGURE 2 _4
0 15 x i* Îx EXAMPLE 1 A dam has the shape of the trapezoid shown in Figure 2. The height is 20 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. 15 10 a
16 a and so wi 10
20 xi*
2 15 or
2(15 a xi* 16 a 8 2
1
2 8 xi*) 46 xi*
2
xi* 16 15 If Ai is the area of the i th strip, then x Ai (a)
10 wi 46 xi* x If x is small, then the pressure Pi on the i th strip is almost constant and we can use
Equation 1 to write a 20 x Pi 1000 txi* 16x i*
(b)
FIGURE 3 The hydrostatic force Fi acting on the i th strip is the product of the pressure and the
area:
Fi Pi Ai 1000 txi* 46 xi* x 5E09(pp 592601) 1/17/06 6:18 PM Page 599 S ECTION 9.3 APPLICATIONS TO PHYSICS AND ENGINEERING ❙❙❙❙ 599 Adding these forces and taking the limit as n l , we obtain the total hydrostatic force
on the dam:
n F 1000 txi* 46 lim nl y 16 0 xi* x i1 1000 tx 46 1000 9.8 y 9800 23x 2 4.43 16 x dx
x 2 dx 7 46 x 0 16 x3
3 0 10 N EXAMPLE 2 Find the hydrostatic force on one end of a cylindrical drum with radius 3 ft if
the drum is submerged in water 10 ft deep.
y di SOLUTION In this example it is convenient to choose the axes as in Figure 4 so that the
origin is placed at the center of the drum. Then the circle has a simple equation,
x 2 y 2 9. As in Example 1 we divide the circular region into horizontal strips of
equal width. From the equation of the circle, we see that the length of the i th strip is
2 s9
yi* 2 and so its area is œ„„„„„„„
œ9(yi*)@ 7
10 Îy Ai y i*
0 x 2 y The pressure on this strip is approximately ≈+¥=9
F IGURE 4 yi* 2 s9 di yi* 62.5 7 and so the force on the strip is approximately
di Ai yi* 2 s9 62.5 7 yi* 2 y The total force is obtained by adding the forces on all the strips and taking the limit:
n F lim nl yi* 2 s9 62.5 7 yi* 2 y i1 125 y 3
3 7 y s9 3 125 7 y s9 y 2 dy y 2 dy 3 3 125 y y s9 y 2 dy 3 The second integral is 0 because the integrand is an odd function (see Theorem 5.5.6).
The ﬁrst integral can be evaluated using the trigonometric substitution y 3 sin , but
it’s simpler to observe that it is the area of a semicircular disk with radius 3. Thus
F 3 875 y s9
3 7875
2 y 2 dy 12,370 lb 875 1
2 3 2 5E09(pp 592601) 600 ❙❙❙❙ 1/17/06 6:19 PM Page 600 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION Moments and Centers of Mass
Our main objective here is to ﬁnd the point P on which a thin plate of any given shape balances horizontally as in Figure 5. This point is called the center of mass (or center of gravity) of the plate.
We ﬁrst consider the simpler situation illustrated in Figure 6, where two masses m1 and
m2 are attached to a rod of negligible mass on opposite sides of a fulcrum and at distances
d1 and d2 from the fulcrum. The rod will balance if P FIGURE 5 d¡ d™ m¡ m1d1 2 m™ fulcrum m2 d2 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 away from
the center.)
Now suppose that the rod lies along the xaxis with m1 at x 1 and m2 at x 2 and the center
of mass at x. If we compare Figures 6 and 7, we see that d1 x x 1 and d2 x 2 x and
so Equation 2 gives
m1 x x1 m2 x 2 x m1 x m2 x m1 x 1 m2 x 2 x FIGURE 6 m1 x 1
m1 m2 x 2
m2 3 The numbers m1 x 1 and m2 x 2 are called the moments of the masses m1 and m2 (with respect
to the origin), and Equation 3 says that the center of mass x is obtained by adding the
moments of the masses and dividing by the total mass m m1 m2 .
–
x ⁄
0 m¡ ¤ –
x⁄ m™ –
¤x x FIGURE 7 In general, if we have a system of n particles with masses m1, m2, . . . , mn located at the
points x 1, x 2, . . . , x n on the xaxis, it can be shown similarly that the center of mass of the
system is located at
n n mi xi
4 x i1
n mi xi
i1 m mi
i1 where m mi is the total mass of the system, and the sum of the individual moments
n M mi xi
i1 is called the moment of the system about the origin. Then Equation 4 could be rewritten as mx M , which says that if the total mass were considered as being concentrated at
the center of mass x, then its moment would be the same as the moment of the system. 5E09(pp 592601) 1/17/06 6:19 PM Page 601 S ECTION 9.3 APPLICATIONS TO PHYSICS AND ENGINEERING y
m£ ⁄ ‹ ›
0 x ﬁ
¤ 601 Now we consider a system of n particles with masses m1, m2, . . . , mn located at the
points x 1, y1 , x 2 , y2 , . . . , x n , yn in the xyplane as shown in Figure 8. By analogy with
the onedimensional case, we deﬁne the moment of the system about the yaxis to be m¡ y£ ❙❙❙❙ n My 5 m™ mi x i
i1 and the moment of the system about the xaxis as FIGURE 8 n Mx 6 mi yi
i1 Then My measures the tendency of the system to rotate about the yaxis and Mx measures
the tendency to rotate about the xaxis.
As in the onedimensional case, the coordinates x, y of the center of mass are given
in terms of the moments by the formulas
My
m x 7 y Mx
m where m
mi is the total mass. Since mx My and my Mx , the center of mass
x, y is the point where a single particle of mass m would have the same moments as the
system.
EXAMPLE 3 Find the moments and center of mass of the system of objects that have
masses 3, 4, and 8 at the points 1, 1 , 2, 1 , and 3, 2 .
SOLUTION We use Equations 5 and 6 to compute the moments: My
y Mx 3 Since m
4 3 4 8 31 42 83 29 4 1 82 15 15, we use Equations 7 to obtain x x
FIGURE 9 1 center of mass
8 0 3 My
m 29
15 y Mx
m 15
15 1 Thus, the center of mass is (1 14 , 1). (See Figure 9.)
15
Next we consider a ﬂat plate (called a lamina) with uniform density that occupies a
region of the plane. We wish to locate the center of mass of the plate, which is called
the centroid of . In doing so we use the following physical principles: The symmetry
principle says that if is symmetric about a line l , then the centroid of lies on l . (If
is reﬂected about l , then remains the same so its centroid remains ﬁxed. But the only
ﬁxed points lie on l .) 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 the center of mass, then its
moments remain unchanged. Also, the moment of the union of two nonoverlapping
regions should be the sum of the moments of the individual regions. 5E09(pp 602611) 602 ❙❙❙❙ 1/17/06 Page 602 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION y 0 6:08 PM y=ƒ a b x Suppose that the region is of the type shown in Figure 10(a); that is, lies between
the lines x a and x b, above the xaxis, and beneath the graph of f , where f is a
continuous function. We divide the interval a, b into n subintervals with endpoints
x 0 , x 1, . . . , x n and equal width x. We choose the sample point xi* to be the midpoint xi of
xi 1 xi 2. This determines the polygonal approximathe i th subinterval, that is, xi
tion to shown in Figure 10(b). The centroid of the i th approximating rectangle Ri is its
1
center Ci (xi , 2 f xi ). Its area is f xi x, so its mass is
f xi (a)
y x The moment of Ri about the yaxis is the product of its mass and the distance from Ci to
the yaxis, which is xi. Thus { xi , f(xi )}
1
Ci ”xi , 2 f(xi )’ My Ri f xi x xi xi f xi x Adding these moments, we obtain the moment of the polygonal approximation to , and
then by taking the limit as n l we obtain the moment of itself about the yaxis:
0 a R¡ R™ R£ xi _1 xi xi b x
n My lim xi f xi nl (b) y x b x f x dx a i1 F IGURE 10 In a similar fashion we compute the moment of Ri about the xaxis as the product of its
mass and the distance from Ci to the xaxis:
Mx Ri f xi x 1
2 1
2 f xi f xi 2 x Again we add these moments and take the limit to obtain the moment of
xaxis:
n Mx mx 1
2 lim nl f xi 2 b1
2
a y x i1 2 fx about the dx Just as for systems of particles, the center of mass of the plate is deﬁned so that
My and my Mx . But the mass of the plate is the product of its density and its area:
m y A b a f x dx and so
x My
m y b a y b y Mx
m b1
2 a y b a fx 2 b a y f x dx a y y x f x dx b a dx f x dx x f x dx
f x dx y b1
2 a y b a fx 2 dx f x dx Notice the cancellation of the ’s. The location of the center of mass is independent of the
density. 5E09(pp 602611) 1/17/06 6:08 PM Page 603 S ECTION 9.3 APPLICATIONS TO PHYSICS AND ENGINEERING In summary, the center of mass of the plate (or the centroid of
x, y , where
1
A x 8 y b a x f x dx 1
A y b1
2
a y fx ❙❙❙❙ 603 ) is located at the point 2 dx EXAMPLE 4 Find the center of mass of a semicircular plate of radius r.
y SOLUTION In order to use (8) we place the semicircle as in Figure 11 so that
y=œ„„„„„
r@≈ fx
r, b r. Here there is no need to use the formula to calcusr 2 x 2 and a
late x because, by the symmetry principle, the center of mass must lie on the yaxis, so
x 0. The area of the semicircle is A
r 2 2, so 4r ” 0, 3π ’
0 _r x r 1
A y FIGURE 11 y r1
2
r 1
r2 2
2
r2 y 2 fx
r y (sr 1
2 r r 0 2 2r
r2 3 dx
x 2 )2 d x 2 r 2 x dx 3 x3
3 2
r 2x
r2 2 . EXAMPLE 5 Find the centroid of the region bounded by the curves y 0, and x 0 4r
3 The center of mass is located at the point 0, 4r 3
x r cos x, y 2. SOLUTION The area of the region is y A 2 2 cos x d x 0 sin x 0 1 so Formulas 8 give
1
A x 2 y 0 x sin x 1
A y y=Ł x
π π
2 0 2 y 0 x cos x d x sin x dx (by integration by parts) y 0 21
2 fx 2 dx 1
2 2 y 0 cos 2x dx π ” 2 1, 8 ’
0 2 0 2 1 2
y y x f x dx 1
4 y 2 0 1 cos 2 x dx 1
4 [x 1
2 sin 2 x]0 x 8
FIGURE 12 The centroid is 2 1, 8 and is shown in Figure 12. 2 0, 5E09(pp 602611) 604 ❙❙❙❙ 1/17/06 6:08 PM Page 604 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION y 1
C i ” xi , 2 f(xi )+g(xi ) ’ If the region lies between two curves y f x and y t x , where f x
t x , as
illustrated in Figure 13, then the same sort of argument that led to Formulas 8 can be used
to show that the centroid of is x, y , where y=ƒ 9 y=©
0 a b xi x 1
A y b a xfx t x dx 1
A y b1
2
a y 2 fx dx (See Exercise 43.)
EXAMPLE 6 Find the centroid of the region bounded by the line y y
y x and the parabola x 2. SOLUTION The region is sketched in Figure 14. We take f x b (1, 1) x, t x
1 in Formulas 9. First we note that the area of the region is 1
” 2 , 2 ’
5 y A 1 x 0 x2
2 2 x dx 1 x3
3 x x 2, a 0, and 1
6 0 y=≈
0 2 x FIGURE 13 y=x tx Therefore
1
A x
FIGURE 14 y 1 0 xfx 1 6 y x2 x 3 dx 0 1
A y 3 x3
3 y The centroid is ( , 12
25 11
2 fx 1 6 2 x5
5 0 1 t x dx tx 1
6 y x3
3
2 1 0 1 x4
4
dx x 2 dx xx 1
2 0 1
1
6 y 11
2 0 x2 x 4 dx 2
5 0 ). We end this section by showing a surprising connection between centroids and volumes
of revolution.  This theorem is named after the Greek
mathematician Pappus of Alexandria, who lived
in the fourth century A.D. Theorem of Pappus Let be a plane region that lies entirely on one side of a line l
in the plane. If is rotated about l , then the volume of the resulting solid is the
product of the area A of and the distance d traveled by the centroid of . Proof We give the proof for the special case in which the region lies between y fx
and y t x as in Figure 13 and the line l is the yaxis. Using the method of cylindrical
shells (see Section 6.3), we have
V y b a 2 xfx
b 2 y 2 xA a t x dx xfx 2 xA t x dx
(by Formulas 9) Ad 5E09(pp 602611) 1/17/06 6:08 PM Page 605 S ECTION 9.3 APPLICATIONS TO PHYSICS AND ENGINEERING where d
yaxis. ❙❙❙❙ 605 2 x is the distance traveled by the centroid during one rotation about the EXAMPLE 7 A torus is formed by rotating a circle of radius r about a line in the plane of
the circle that is a distance R
r from the center of the circle. Find the volume of the
torus. r 2. By the symmetry principle, its centroid is its
center and so the distance traveled by the centroid during a rotation is d 2 R. Therefore, by the Theorem of Pappus, the volume of the torus is
SOLUTION The circle has area A V Ad r2 2R 22 2 rR The method of Example 7 should be compared with the method of Exercise 61 in
Section 6.2.  9.3 Exercises
9. 1. An aquarium 5 ft long, 2 ft wide, and 3 ft deep is full of water. Find (a) the hydrostatic pressure on the bottom of the
aquarium, (b) the hydrostatic force on the bottom, and (c) the
hydrostatic force on one end of the aquarium. r 2. A swimming pool 5 m wide, 10 m long, and 3 m deep is ﬁlled with seawater of density 1030 kg m3 to a depth of 2.5 m. Find
(a) the hydrostatic pressure at the bottom of the pool, (b) the
hydrostatic force on the bottom, and (c) the hydrostatic force
on one end of the pool.
3–9 A vertical plate is submerged in water and has the indicated
shape. Explain how to approximate the hydrostatic force against
the end of the tank by a Riemann sum. Then express the force as
an integral and evaluate it.
 3. 4.
2 ft ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 10. A large tank is designed with ends in the shape of the region between the curves y x 2 2 and y 12, measured in feet.
Find the hydrostatic force on one end of the tank if it is ﬁlled
to a depth of 8 ft with gasoline. (Assume the gasoline’s density
is 42.0 lb ft3.)
of the trough are equilateral triangles with sides 8 m long and
vertex at the bottom. Find the hydrostatic force on one end of
the trough. 4 ft 12. A vertical dam has a semicircular gate as shown in the ﬁgure. Find the hydrostatic force against the gate. 6 ft
5. ■ 11. A trough is ﬁlled with a liquid of density 840 kg m3 . The ends
1 ft 3 ft 4 ft ■ 2m 6. 20 m 5m water level 12 m 4m
4m
7. 8. 12 ft a 13. A cube with 20cmlong sides is sitting on the bottom of an 8 ft
h
20 ft
b aquarium in which the water is one meter deep. Find the hydrostatic force on (a) the top of the cube and (b) one of the sides
of the cube.
14. A dam is inclined at an angle of 30 from the vertical and has the shape of an isosceles trapezoid 100 ft wide at the top and 5E09(pp 602611) 606 ❙❙❙❙ 1/17/06 6:09 PM Page 606 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION 23–26 50 ft wide at the bottom and with a slant height of 70 ft. Find
the hydrostatic force on the dam when it is full of water.
15. A swimming pool is 20 ft wide and 40 ft long and its bottom is  Sketch the region bounded by the curves, and visually
estimate the location of the centroid. Then ﬁnd the exact coordinates of the centroid. an inclined plane, the shallow end having a depth of 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) the deep end, (c) one of
the sides, and (d) the bottom of the pool. 23. y 16. Suppose that a plate is immersed vertically in a ﬂuid with density and the width of the plate is w x at a depth of x meters 26. y beneath the surface of the ﬂuid. If the top of the plate is at
depth a and the bottom is at depth b, show that the hydrostatic
force on one side of the plate is y F x 2, 4 24. 3 x 2y ■ 2.0 2.5 3.0 3.5 4.0 4.5 x 29. y sin x, 30. y x, 31. x 5.0 Plate width (m) 0 0.8 1.7 2.4 2.9 3.3 3.6 ■ ■ ■ ■ y y cos x, 0, 5 y, y x ■ x 1 x, x 0,
x 4 2 0 ■ ■ ■ ■ ■ ■ ■ ■ ■ 33. 5 1
y semicircle 2 tx A 5 m¡=25
_2 0 _1 1 x
0 _1 1 x _2 34. 2
y quartercircle
r 0 m™=20
3 7 ■ ■ m£=10 0
■ x ■ ■ ■ ■ ■ ■ ■ ■ 1 6, m2 5, m3 1, m4 4;
P1 1, 2 , P2 3, 4 , P3 3, 7 , P4 6,
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 1
■ ■ ■ ■ ■ 2x
and y x 2, 0 x 2, to three decimal places. Sketch the
region and plot the centroid to see if your answer is reasonable. 37. Prove that the centroid of any triangle is located at the point 22. m1 ■ ■ intersection of the curves y x ln x and y x 3 x. Then
ﬁnd (approximately) the centroid of the region bounded by
these curves. 21. m1 ■ ■ ; 36. Use a graph to ﬁnd approximate xcoordinates of the points of  The masses mi are located at the points Pi. Find the
moments Mx and My and the center of mass of the system. 6, m2 5, m3 10;
P1 1, 5 , P2 3, 2 , P3 2, ■ x r ; 35. Find the centroid of the region bounded by the curves y x 21–22 ■ ■ m™=30 0 ■ 2
■ x2 y y
1  ■ x ■  Calculate the moments Mx and My and the center of mass
of a lamina with the given density and shape. Pointmasses m i are located on the xaxis as shown. Find
the moment M of the system about the origin and the center of
mass x. ■ 1, 1 32–34 where x is the xcoordinate of the centroid of the plate and
A is its area. This equation shows that the hydrostatic force
against a vertical plane region is the same as if the region
were horizontal at the depth of the centroid of the region.
(b) Use the result of part (a) to give another solution to
Exercise 9. 20. 0
x x 2 18. (a) Use the formula of Exercise 16 to show that 2 x
■ y
2, ■ 32. F 0, x
0, Find the centroid of the region bounded by the given
sx, ■ Depth (m) 0,
x ■ 28. y txw x d x a m¡=40 y 27. y b y
0, ■  0 curves. table shows measurements of its width, taken at the indicated
depths. Use Simpson’s rule to estimate the force of the water
against the plate. 19. y 1 x, ■ 27–31 17. A vertical, irregularly shaped plate is submerged in water. The 19–20 6, e x, 25. y y of intersection of the medians. [Hints: Place the axes so that
the vertices are a, 0 , 0, b , and c, 0 . Recall that a median
is a line segment from a vertex to the midpoint of the oppo 5E09(pp 602611) 1/17/06 6:09 PM Page 607 S ECTION 9.4 APPLICATIONS TO ECONOMICS AND BIOLOGY  Use the Theorem of Pappus to ﬁnd the volume of the
given solid. 40. A sphere of radius r 38–39  Find the centroid of the region shown, not by integration,
but by locating the centroids of the rectangles and triangles (from
Exercise 37) and using additivity of moments. 2
1 1
_2 3x 1 ■ ■ ■  9.4 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 0 _1 ■ ■ ■ be the region that lies between the curves y x m
and y x n, 0 x 1, where m and n are integers with
0 n m.
(a) Sketch the region .
(b) Find the coordinates of the centroid of .
(c) Try to ﬁnd values of m and n such that the centroid lies
outside . 44. Let 1 2 x _1
■ 2, 5 , and 5, 4 about the xaxis
43. Prove Formulas 9. 2 0 42. The solid obtained by rotating the triangle with vertices 2, 3 , y 3 _2 (Use Example 4.) 41. A cone with height h and base radius r ■ 39. y 607 40–42 site side. Recall also that the medians intersect at a point twothirds of the way from each vertex (along the median) to the
opposite side.] 38. ❙❙❙❙ ■ ■ ■ Applications to Economics and Biology
In this section we consider some applications of integration to economics (consumer surplus) and biology (blood ﬂow, cardiac output). Others are described in the exercises. Consumer Surplus
p p=p(x) (X, P ) P 0 X x FIGURE 1 A typical demand curve
p Recall from Section 4.8 that the demand function p x is the price that a company has to
charge in order to sell x units of a commodity. Usually, selling larger quantities requires
lowering prices, so the demand function is a decreasing function. The graph of a typical
demand function, called a demand curve, is shown in Figure 1. If X is the amount of the
commodity that is currently available, then P p X is the current selling price.
We divide the interval 0, X into n subintervals, each of length x X n, and let
xi* x i be the right endpoint of the i th subinterval, as in Figure 2. If, after the ﬁrst x i 1
units were sold, a total of only x i units had been available and the price per unit had been
set at p x i dollars, then the additional x units could have been sold (but no more). The
consumers who would have paid p x i dollars placed a high value on the product; they
would have paid what it was worth to them. So, in paying only P dollars they have saved
an amount of
savings per unit number of units
p xi
Px
Considering similar groups of willing consumers for each of the subintervals and adding
the savings, we get the total savings:
n p xi P x i1 0 (This sum corresponds to the area enclosed by the rectangles in Figure 2.) If we let n l ,
this Riemann sum approaches the integral (X, P ) P ⁄ FIGURE 2 xi X x 1 y X 0 px P dx which economists call the consumer surplus for the commodity. 5E09(pp 602611) 608 ❙❙❙❙ 1/17/06 6:09 PM Page 608 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION p The consumer surplus represents the amount of money saved by consumers in purchasing the commodity at price P, corresponding to an amount demanded of X . Figure 3
shows the interpretation of the consumer surplus as the area under the demand curve and
above the line p P. p=p(x)
consumer
surplus EXAMPLE 1 The demand for a product, in dollars, is
P (X, P ) p p=P 0 X x 1200 0.0001x 2 0.2 x Find the consumer surplus when the sales level is 500.
SOLUTION Since the number of products sold is X 500, the corresponding price is FIGURE 3 P 1200 0.2 500 0.0001 500 2 1075 Therefore, from Deﬁnition 1, the consumer surplus is y 500 0 px P dx y 500 1200 0 y 500 125 0 125x 0.0001x 2 0.2 x 0.0001x 2 d x 0.2 x 0.1x 2 125 500 1075 d x 0.1 500 500 x3
3 0.0001 0 0.0001 500
3 2 3 $33,333.33 Blood Flow
In Example 7 in Section 3.4 we discussed the law of laminar ﬂow:
P
R2
4l vr r2 which gives the velocity v of blood that ﬂows along a blood vessel with radius R and length
l at a distance r from the central axis, where P is the pressure difference between the ends
of the vessel and is the viscosity of the blood. Now, in order to compute the rate of blood
ﬂow, or ﬂux (volume per unit time), we consider smaller, equally spaced radii r1, r2 , . . . .
The approximate area of the ring (or washer) with inner radius ri 1 and outer radius ri is
2 ri r Îr
ri where r ri ri 1 (See Figure 4.) If r is small, then the velocity is almost constant throughout this ring and
can be approximated by v ri . Thus, the volume of blood per unit time that ﬂows across the
ring is approximately
2 ri r v ri
2 ri v ri r
and the total volume of blood that ﬂows across a crosssection per unit time is approximately
n FIGURE 4 2 ri v ri
i1 r 5E09(pp 602611) 1/17/06 6:09 PM Page 609 SECTION 9.4 APPLICATIONS TO ECONOMICS AND BIOLOGY ❙❙❙❙ 609 This approximation is illustrated in Figure 5. Notice that the velocity (and hence the volume per unit time) increases toward the center of the blood vessel. The approximation gets
better as n increases. When we take the limit we get the exact value of the ﬂux (or discharge), which is the volume of blood that passes a crosssection per unit time:
n F
FIGURE 5 lim nl y R 0 y r P
R2
4l r 2 dr 0 i1 2r P
2l
P
2l R 2 ri v ri y R 0 R 2r P
2l r 3 dr 4 4 R
2 2 r v r dr R
4 PR
8l R2 r2
2 r4
4 rR r0 4 The resulting equation
PR 4
8l F 2 is called Poiseuille’s Law; it shows that the ﬂux is proportional to the fourth power of the
radius of the blood vessel. Cardiac Output
vein
pulmonary
arteries right
atrium
pulmonary
veins vein aorta
pulmonary
arteries
pulmonary
veins left
atrium Figure 6 shows the human cardiovascular system. Blood returns from the body through the
veins, enters the right atrium of the heart, and is pumped to the lungs through the pulmonary arteries for oxygenation. It then ﬂows back into the left atrium through the pulmonary veins and then out to the rest of the body through the aorta. The cardiac output of
the heart is the volume of blood pumped by the heart per unit time, that is, the rate of ﬂow
into the aorta.
The dye dilution method is used to measure the cardiac output. Dye is injected into the
right atrium and ﬂows through the heart into the aorta. A probe inserted into the aorta measures the concentration of the dye leaving the heart at equally spaced times over a time
interval 0, T until the dye has cleared. Let c t be the concentration of the dye at time t.
If we divide 0, T into subintervals of equal length t, then the amount of dye that ﬂows
past the measuring point during the subinterval from t ti 1 to t ti is approximately
concentration volume c ti F t FIGURE 6 where F is the rate of ﬂow that we are trying to determine. Thus, the total amount of dye
is approximately
n n c ti F t F i1 c ti
i1 and, letting n l , we ﬁnd that the amount of dye is
A T F y c t dt
0 t 5E09(pp 602611) ❙❙❙❙ 610 1/17/06 6:10 PM Page 610 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION Thus, the cardiac output is given by
A F 3 y T 0 c t dt where the amount of dye A is known and the integral can be approximated from the concentration readings.
t ct t ct 0
1
2
3
4
5 0
0.4
2.8
6.5
9.8
8.9 6
7
8
9
10 6.1
4.0
2.3
1.1
0 EXAMPLE 2 A 5mg bolus of dye is injected into a right atrium. The concentration of the
dye (in milligrams per liter) is measured in the aorta at onesecond intervals as shown in
the chart. Estimate the cardiac output.
SOLUTION Here A
5, t 1, and T
integral of the concentration: y 10 0 c t dt 1
3 0 10. We use Simpson’s Rule to approximate the 4 0.4 2 2.8 2 6.1 4 6.5 4 4.0 2 9.8 2 2.3 4 1.1 4 8.9
0 41.87
Thus, Formula 3 gives the cardiac output to be
A F y 10 0 c t dt 0.12 L s  9.4 5
41.87
7.2 L min Exercises 1. The marginal cost function C x was deﬁned to be the derivative of the cost function. (See Sections 3.4 and 4.8.)
If the marginal cost of maufacturing x meters of a fabric is
Cx
5 0.008 x 0.000009 x 2 (measured in dollars per
meter) and the ﬁxed startup cost is C 0
$20,000, use the
Net Change Theorem to ﬁnd the cost of producing the ﬁrst
2000 units.
2. The marginal revenue from the sale of x units of a product is 12 0.0004 x. If the revenue from the sale of the ﬁrst
1000 units is $12,400, ﬁnd the revenue from the sale of the
ﬁrst 5000 units.
3. The marginal cost of producing x units of a certain product is 74 1.1x 0.002 x 2 0.00004 x 3 (in dollars per unit).
Find the increase in cost if the production level is raised from
1200 units to 1600 units.
4. The demand function for a certain commodity is p 5 x 10.
Find the consumer surplus when the sales level is 30. Illustrate
by drawing the demand curve and identifying the consumer
surplus as an area. 5. A demand curve is given by p 450 x 8 . Find the consumer surplus when the selling price is $10. 6. The supply function pS x for a commodity gives the rela tion between the selling price and the number of units that
manufacturers will produce at that price. For a higher price,
manufacturers will produce more units, so pS is an increasing
function of x. Let X be the amount of the commodity currently
produced and let P pS X be the current price. Some producers would be willing to make and sell the commodity for a
lower selling price and are therefore receiving more than their
minimal price. The excess is called the producer surplus. An
argument similar to that for consumer surplus shows that the
surplus is given by the integral y X 0 P pS x d x Calculate the producer surplus for the supply function
pS x
3 0.01x 2 at the sales level X 10. Illustrate by
drawing the supply curve and identifying the producer surplus
as an area.
7. If a supply curve is modeled by the equation p 200 0.2 x 3 / 2, ﬁnd the producer surplus when the selling
price is $400. 5E09(pp 602611) 1/17/06 6:10 PM Page 611 S ECTION 9.5 PROBABILITY 8. For a given commodity and pure competition, the number of units produced and the price per unit are determined as the
coordinates of the point of intersection of the supply and
demand curves. Given the demand curve p 50 x 20 and
the supply curve p 20 x 10, ﬁnd the consumer surplus
and the producer surplus. Illustrate by sketching the supply and
demand curves and identifying the surpluses as areas. ; 9. A company modeled the demand curve for its product
(in dollars) by
p 800,000e x 5000
x 20,000 ❙❙❙❙ 611 13. Use Poiseuille’s Law to calculate the rate of ﬂow in a small human artery where we can take
0.027, R
l 2 cm, and P 4000 dynes cm2. 0.008 cm, 14. High blood pressure results from constriction of the arteries. To maintain a normal ﬂow rate (ﬂux), the heart has to pump
harder, thus increasing the blood pressure. Use Poiseuille’s
Law to show that if R0 and P0 are normal values of the radius
and pressure in an artery and the constricted values are R and
P, then for the ﬂux to remain constant, P and R are related by
the equation
P
P0 R0
R 4 Use a graph to estimate the sales level when the selling price is
$16. Then ﬁnd (approximately) the consumer surplus for this
sales level. Deduce that if the radius of an artery is reduced to threefourths
of its former value, then the pressure is more than tripled. 10. A movie theater has been charging $7.50 per person and selling 15. The dye dilution method is used to measure cardiac output with about 400 tickets on a typical weeknight. After surveying their
customers, the theater estimates that for every 50 cents that
they lower the price, the number of moviegoers will increase
by 35 per night. Find the demand function 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 t is f t , then the derivative, f t , is called the net investment ﬂow. Suppose that the net investment ﬂow is st million dollars per year
(where t is measured in years). Find the increase in capital (the
capital formation) from the fourth year to the eighth year. 8 mg of dye. The dye concentrations, in mg L, are modeled by
1
ct
t , 0 t 12, where t is measured in seconds.
4 t 12
Find the cardiac output.
16. After an 8mg injection of dye, the readings of dye concentra tion at twosecond intervals are as shown in the table. Use
Simpson’s Rule to estimate the cardiac output.
t in a lake resort area. The number of mosquitos is increasing at
an estimated rate of 2200 10e 0.8t per week (where t is measured in weeks). By how much does the mosquito population
increase between the ﬁfth and ninth weeks of summer?  9.5 t ct 0
2
4
6
8
10 12. A hot, wet summer is causing a mosquito population explosion ct
0
2.4
5.1
7.8
7.6
5.4 12
14
16
18
20 3.9
2.3
1.6
0.7
0 Probability
Calculus plays a role in the analysis of random behavior. Suppose we consider the cholesterol level of a person chosen at random from a certain age group, or the height of an adult
female chosen at random, or the lifetime of a randomly chosen battery of a certain type.
Such quantities are called continuous random variables because their values actually
range over an interval of real numbers, although they might be measured or recorded only
to the nearest integer. We might want to know the probability that a blood cholesterol level
is greater than 250, or the probability that the height of an adult female is between 60 and
70 inches, or the probability that the battery we are buying lasts between 100 and 200
hours. If X represents the lifetime of that type of battery, we denote this last probability as
follows:
P 100 X 200 According to the frequency interpretation of probability, this number is the longrun proportion of all batteries of the speciﬁed type whose lifetimes are between 100 and 200
hours. Since it represents a proportion, the probability naturally falls between 0 and 1. 5E09(pp 612619) 612 ❙❙❙❙ 1/17/06 6:13 PM Page 612 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION Every continuous random variable X has a probability density function f . This means
that the probability that X lies between a and b is found by integrating f from a to b :
Pa 1 X y b b a f x dx For example, Figure 1 shows the graph of a model of the probability density function f
for 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 probability that the height
of a woman chosen at random from this population is between 60 and 70 inches is equal
to the area under the graph of f from 60 to 70.
y area=probability that the
height of a woman
is between 60 and
70 inches y=ƒ FIGURE 1 Probability density function
for the height of an adult female 0 60 65 x 70 In general, the probability density function f of a random variable X satisﬁes the condition f x
0 for all x. Because probabilities are measured on a scale from 0 to 1, it follows that y 2 f x dx 1 EXAMPLE 1 Let f x
0.006 x 10 x for 0 x
of x.
(a) Verify that f is a probability density function.
(b) Find P 4 X 8 . 10 and f x 0 for all other values SOLUTION (a) For 0 x 10 we have 0.006 x 10
check that Equation 2 is satisﬁed: y y f x dx 10 0 x 0.006 x 10 [ 0.006 5x 2 0, so f x 0.006 y x dx 1
3 x3 10
0 0 for all x. We also need to
10 0 0.006(500 10 x
1000
3 ) x 2 dx
1 Therefore, f is a probability density function.
(b) The probability that X lies between 4 and 8 is
P4 X 8 y 8 4 f x dx [ 0.006 5x 2 8 0.006 y 10 x
4 1
3 x3 8
4 x 2 dx 0.544 E XAMPLE 2 Phenomena such as waiting times and equipment failure times are commonly
modeled by exponentially decreasing probability density functions. Find the exact form
of such a function. 5E09(pp 612619) 1/17/06 6:13 PM Page 613 S ECTION 9.5 PROBABILITY ❙❙❙❙ 613 SOLUTION Think of the random variable as being the time you wait on hold before an
agent of a company you’re telephoning answers your call. So instead of x, let’s use t to
represent time, in minutes. If f is the probability density function and you call at time
t 0, then, from Deﬁnition 1, x02 f t dt represents the probability that an agent answers
within the ﬁrst two minutes and x45 f t dt is the probability that your call is answered
during the ﬁfth minute.
It’s clear that f t
0 for t 0 (the agent can’t answer before you place the call).
For t 0 we are told to use an exponentially decreasing function, that is, a function of
the form f t
Ae ct, where A and c are positive constants. Thus 0
Ae ft if t
if t 0
0 f t dt y ct We use condition 2 to determine the value of A:
1 y
y 0 y
c Ae ct dt lim lim y x 0 Ae x
ct lim 0 xl 0 ct f t dt dt A
1
c e cx A
c
Therefore, A c 1 and so A c. Thus, every exponential density function has the form t 0 0 xl A
e
c xl if t<0
0
f(t)=
ce _ct if t˘0 y f t dt 0
ce ft ct if t
if t 0
0 FIGURE 2 A typical graph is shown in Figure 2. An exponential density function Average Values
y y=f(t)
Ît 0 t i1 ti
ti FIGURE 3 t Suppose you’re waiting for a company to answer your phone call and you wonder how
long, on average, you can expect to wait. Let f t be the corresponding density function,
where t is measured in minutes, and think of a sample of N people who have called this
company. Most likely, none of them had to wait more than an hour, so let’s restrict our
attention to the interval 0 t 60. Let’s divide that interval into n intervals of length t
and endpoints 0, t1, t2, . . . . (Think of t as lasting a minute, or half a minute, or 10 seconds, or even a second.) The probability that somebody’s call gets answered during the
time period from ti 1 to ti is the area under the curve y f t from ti 1 to ti , which is
approximately equal to f ti t. (This is the area of the approximating rectangle in Figure 3, where ti is the midpoint of the interval.)
Since the longrun proportion of calls that get answered in the time period from ti 1 to
ti is f ti t, we expect that, out of our sample of N callers, the number whose call was
answered in that time period is approximately Nf ti t and the time that each waited is
about ti . Therefore, the total time they waited is the product of these numbers: approximately ti Nf ti t . Adding over all such intervals, we get the approximate total of everybody’s waiting times:
n N ti f ti
i1 t 5E09(pp 612619) 614 ❙❙❙❙ 1/17/06 6:13 PM Page 614 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION If we now divide by the number of callers N, we get the approximate average waiting time:
n ti f ti t i1 We recognize this as a Riemann sum for the function tf t . As the time interval shrinks
(that is, t l 0 and n l ), this Riemann sum approaches the integral y 60 0 t f t dt This integral is called the mean waiting time.
In general, the mean of any probability density function f is deﬁned to be  It is traditional to denote the mean by the
Greek letter (mu). y The mean can be interpreted as the longrun average value of the random variable X. It can
also be interpreted as a measure of centrality of the probability density function.
The expression for the mean resembles an integral we have seen before. If
is the
region that lies under the graph of f , we know from Formula 9.3.8 that the xcoordinate
of the centroid of is y y= ƒ y x f x dx y x=m
T
0 x f x dx f x dx y x
m t F IGURE 4 T balances at a point on the line x=m x f x dx because of Equation 2. So a thin plate in the shape of
line x
. (See Figure 4.) balances at a point on the vertical EXAMPLE 3 Find the mean of the exponential distribution of Example 2: 0
ce ft ct if t
if t 0
0 SOLUTION According to the deﬁnition of a mean, we have y y t f t dt 0 tce ct dt To evaluate this integral we use integration by parts, with u y 0 tce ct dt lim xl y lim  The limit of the ﬁrst term is 0 by
l’Hospital’s Rule. xl x 0 tce xe ct dt lim 1
c cx te xl e t and dv ct x
0 y x 0 e ct ce
dt cx c 1
c
The mean is 1 c, so we can rewrite the probability density function as
ft 0
1 e t if t
if t 0
0 ct dt: 5E09(pp 612619) 1/17/06 6:13 PM Page 615 S ECTION 9.5 PROBABILITY ❙❙❙❙ 615 EXAMPLE 4 Suppose the average waiting time for a customer’s call to be answered by a
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 be answered.
SOLUTION (a) We are given that the mean of the exponential distribution is
5 min and so,
from the result of Example 3, we know that the probability density function is
0
0.2e ft if t
if t t5 0
0 Thus, the probability that a call is answered during the ﬁrst minute is
P0 T y 1 1 f t dt 0 y 1 t5 0.2e 0 0.2 5e 1 e dt
t5 15 1 0 0.1813 So about 18% of customers’ calls are answered during the ﬁrst minute.
(b) The probability that a customer waits more than ﬁve minutes is
PT 5 y 5 f t dt y x t5 lim xl 1
e y 5 0.2e 0.2e 5 dt t5 dt lim e xl 1 e x5 0.368 About 37% of customers wait more than ﬁve minutes before their calls are answered. Notice the result of Example 4(b): Even though the mean waiting time is 5 minutes,
only 37% of callers wait more than 5 minutes. The reason is that some callers have 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. That is a
number m such that half the callers have a waiting time less than m and the other callers
have a waiting time longer than m. In general, the median of a probability density function is the number m such that y m f x dx 1
2 This means that half the area under the graph of f lies to the right of m. In Exercise 7 you
are asked to show that the median waiting time for the company described in Example 4
is approximately 3.5 minutes. 5E09(pp 612619) 616 ❙❙❙❙ 1/17/06 6:14 PM Page 616 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION Normal Distributions
Many important random phenomena—such as test scores on aptitude tests, heights and
weights of individuals from a homogeneous population, annual rainfall in a given location—are modeled by a normal distribution. This means that the probability density
function of the random variable X is a member of the family of functions  The standard deviation is denoted by the
lowercase Greek letter (sigma). 1
s2 fx 3 2 x e 2 2 You can verify that the mean for this function is . The positive constant is called the
standard deviation; it measures how spread out the values of X are. From the bellshaped
graphs of members of the family in Figure 5, we see that for small values of 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 to estimate and .
y 1 s= 2 s=1
s=2
FIGURE 5 0 x m Normal distributions The factor 1 ( s2 ) is needed to make f a probability density function. In fact, it can
be veriﬁed using the methods of multivariable calculus that y
y e x 2 2 2 dx 1 EXAMPLE 5 Intelligence Quotient (IQ) scores are distributed normally with mean
100 and standard deviation 15. (Figure 6 shows the corresponding probability density
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? 0.02
0.01 0 1
s2 60 80 100 120 140 FIGURE 6 x SOLUTION (a) Since IQ scores are normally distributed, we use the probability density function
100 and
15:
given by Equation 3 with Distribution of IQ scores P 85 X 115 y 115 85 1
15 s2 e x 100 2 2 15 2 dx 2 Recall from Section 8.5 that the function y e x doesn’t have an elementary antiderivative, so we can’t evaluate the integral exactly. But we can use the numerical integration
capability of a calculator or computer (or the Midpoint Rule or Simpson’s Rule) to estimate the integral. Doing so, we ﬁnd that
P 85 X 115 0.68 So about 68% of the population has an IQ between 85 and 115, that is, within one standard deviation of the mean. 5E09(pp 612619) 1/17/06 6:14 PM Page 617 ❙❙❙❙ S ECTION 9.5 PROBABILITY 617 (b) The probability that the IQ score of a person chosen at random is more than 140 is
1
2
P X 140
y140 15 s2 e x 100 450 d x
To avoid the improper integral we could approximate it by the integral from 140 to 200.
(It’s quite safe to say that people with an IQ over 200 are extremely rare.) Then
PX 140 y 200 140 1
15 s2 e x 100 2 450 dx 0.0038 Therefore, about 0.4% of the population has an IQ over 140.  9.5 Exercises 1. Let f x be the probability density function for the lifetime of a manufacturer’s highest quality car tire, where x is measured in
miles. Explain the meaning of each integral.
(a) y 40,000 30,000 f x dx (b) y 25,000 (b) Use the graph to ﬁnd the following probabilities:
(i) P X 3
(ii) P 3 X 8
(c) Calculate the mean.
y f x dx 0.2 2. Let f t be the probability density function for the time it takes you to drive to school in the morning, where t is measured in
minutes. Express the following probabilities as integrals.
(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
3. Let f x 3
64 4. Let f x kx 2 1 x s16 x 2 for 0 x 4 and f x
other values of x.
(a) Verify that f is a probability density function.
(b) Find P ( X 2 . 0 for all x if 0 x 1 and f x
0 if x 0
or x 1.
(a) For what value of k is f a probability density function?
(b) For that value of k, ﬁnd P ( X 1 ).
2
(c) Find the mean. 5. A spinner from a board game randomly indicates a real number between 0 and 10. The spinner is fair in the sense that it indicates a number in a given interval with the same probability as
it indicates a number in any other interval of the same length.
(a) Explain why the function
fx 0.1
0 if 0
if x x 10
0 or x 10 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.
6. (a) Explain why the function whose graph is shown is a proba bility density function. y=ƒ 0.1
0 2 4 6 8 10 x 7. Show that the median waiting time for a phone call to the com pany described in Example 4 is about 3.5 minutes.
8. (a) A type of lightbulb is labeled as having an average lifetime of 1000 hours. It’s reasonable to model the probability of
failure of these bulbs by an exponential density function
with mean
1000. Use this model to ﬁnd the 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?
9. The manager of a fastfood restaurant determines that the average time that her customers wait for service is 2.5 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 free hamburger. But she doesn’t want to give away free hamburgers
to more than 2% of her customers. What should the advertisement say?
10. According to the National Health Survey, the heights of adult males in the United States are normally distributed with mean
69.0 inches and standard deviation 2.8 inches.
(a) What is the probability that an adult male chosen at random
is between 65 inches and 73 inches tall? 5E09(pp 612619) ❙❙❙❙ 618 1/17/06 6:14 PM Page 618 CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION 15. The hydrogen atom is composed of one proton in the nucleus (b) What percentage of the adult male population is more than
6 feet tall? and one electron, which moves about the nucleus. In the quantum theory of atomic structure, it is assumed that the electron
does not move in a welldeﬁned orbit. Instead, it occupies a
state known as an orbital, which may be thought of as a
“cloud” of negative charge surrounding the nucleus. At the
state of lowest energy, called the ground state, or 1sorbital,
the shape of this cloud is assumed to be a sphere centered at
the nucleus. This sphere is described in terms of the probability
density function 11. The “Garbage Project” at the University of Arizona reports that the amount of paper discarded by households per week is
normally distributed with mean 9.4 lb and standard deviation
4.2 lb. What percentage of households throw out at least 10 lb
of paper a week?
12. Boxes are labeled as containing 500 g of cereal. The machine ﬁlling the boxes produces weights that are normally 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 of cereal?
(b) Suppose a law states that no more than 5% of a manufacturer’s cereal boxes can contain less than the stated weight
of 500 g. At what target weight should the manufacturer set
its ﬁlling machine? pr where a0 is the Bohr radius a 0
integral
Pr 13. For any normal distribution, ﬁnd the probability that the random variable lies within two standard deviations of the
mean.
is deﬁned by
12 y x 2 ; f x dx Find the standard deviation for an exponential density function
with mean .  9 Review ■ CONCEPT CHECK 1. (a) How is the length of a curve deﬁned? 2. (a) Write an expression for the surface area of the surface b, about 3. Describe how we can ﬁnd the hydrostatic force against a verti cal wall submersed in a ﬂuid.
4. (a) What is the physical signiﬁcance of the center of mass of a thin plate?
(b) If the plate lies between y f x and y 0, where
a x b, write expressions for the coordinates of the
center of mass.
5. What does the Theorem of Pappus say? r 0 r 5.59 42
se
a3
0 2s a 0 0
10 11 m . The ds ■ 6. Given a demand function p x , explain what is meant by the (b) Write an expression for the length of a smooth curve given
by y f x , a x b.
(c) What if x is given as a function of y?
obtained by rotating the curve y f x , a x
the xaxis.
(b) What if x is given as a function of y?
(c) What if the curve is rotated about the yaxis? y 2r a 0 gives the probability that the electron will be found within the
sphere of radius r meters centered at the nucleus.
(a) Verify that p r is a probability density function.
(b) Find lim r l p r . For what value of r does p r have its
maximum value?
(c) Graph the density function.
(d) Find the probability that the electron will be within the
sphere of radius 4a0 centered at the nucleus.
(e) Calculate the mean distance of the electron from the
nucleus in the ground state of the hydrogen atom. 14. The standard deviation for a random variable with probability density function f and mean 42
re
a3
0 consumer surplus when the amount of a commodity currently
available is X and the current selling price is P. Illustrate with a
sketch.
7. (a) What is the cardiac output of the heart? (b) Explain how the cardiac output can be measured by the dye
dilution method.
8. What is a probability density function? What properties does such a function have?
9. Suppose f x is the probability density function for the weight of a female college student, where x is measured in pounds.
(a) What is the meaning of the integral x0100 f x dx ?
(b) Write an expression for the mean of this density function.
(c) How can we ﬁnd the median of this density function?
10. What is a normal distribution? What is the signiﬁcance of the standard deviation? 5E09(pp 612619) 1/17/06 6:14 PM Page 619 CHAPTER 9 R EVIEW ■ 1–2 1. y 1
6 2. y 2 ln sin( x) , x 2 4 32 , 0 x 1
2 ■ ■ 3 ■ Find the centroid of the region shown.
14. y x ■  13. 3 ■ ■ ■ ■ ■ ■ 619 ■ 13–14 Find the length of the curve.  ■ EXERCISES ❙❙❙❙ y
3 (3, 2) 2 ■ 1 3. (a) Find the length of the curve x4
16 y 1
2x 2 0 1 x 2 ■ e ,0 ■ x sst ■ ■ ■ ■ 18. After a 6mg injection of dye into a heart, the readings of dye concentration at twosecond intervals are as shown in
the table. Use Simpson’s Rule to estimate the cardiac output.
t ct t ct 0
2
4
6
8
10
12 0
1.9
3.3
5.1
7.6
7.1
5.8 14
16
18
20
22
24 4.7
3.3
2.1
1.1
0.5
0 7. Find the length of the curve 1 ■ p 2000 0.1x 0.01x 2. Find the consumer surplus
when the sales level is 100. 6 to estimate the length of the 6 to estimate the area of the
surface obtained by rotating the curve in Exercise 5 about the
xaxis. y ■ 17. The demand function for a commodity is given by 6. Use Simpson’s Rule with n y ■ sphere of radius r is 4 r 3 to ﬁnd the centroid of the semi3
circular region bounded by the curve y sr 2 x 2 and
the xaxis. 3. x ■ x 16. Use the Theorem of Pappus and the fact that the volume of a x 2, 0 x 1, is rotated about the yaxis.
Find the area of the resulting surface.
(b) Find the area of the surface obtained by rotating the curve
in part (a) about the xaxis.
curve y ■ 3 ter 1, 0 is rotated about the yaxis. 4. (a) The curve y x2 ■ 0 _2 15. Find the volume obtained when the circle of radius 1 with cen (b) Find the area of the surface obtained by rotating the curve
in part (a) about the yaxis. 5. Use Simpson’s Rule with n x 1 dt 1 x 16 8. Find the area of the surface obtained by rotating the curve in Exercise 7 about the yaxis.
9. A gate in an irrigation canal is constructed in the form of a trapezoid 3 ft wide at the bottom, 5 ft wide at the top, and 2 ft
high. It is placed vertically in the canal, with the water extending to its top. Find the hydrostatic force on one side of the gate.
10. A trough is ﬁlled with water and its vertical ends have the shape of the parabolic region in the ﬁgure. Find the hydrostatic
force on one end of the trough. 19. (a) Explain why the function fx 20
0 sin x
10 if 0 x 10 if x 0 or x 10 is a probability density function.
(b) Find P X 4 .
(c) Calculate the mean. Is the value what you would expect? 8 ft 20. Lengths of human pregnancies are normally distributed 4 ft with mean 268 days and standard deviation 15 days. What percentage of pregnancies last between 250 days and 280 days?
21. The length of time spent waiting in line at a certain bank 11–12  Find the centroid of the region bounded by the given curves.
11. y
12. y
■ x 2, 4 sin x,
■ ■ y
y x
0, ■ 2
x ■ 4,
■ x
■ 3
■ 4
■ ■ ■ ■ 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? 5E09(pp 620621) 1/17/06 6:11 PM P ROBLEMS
PLUS Page 620 1. Find the area of the region S x, y x 0, y 1, x 2 y2 4y . 2. Find the centroid of the region enclosed by the loop of the curve y 2 x3 x 4. 3. If a sphere of radius r is sliced by a plane whose distance from the center of the sphere is d, then the sphere is divided into two pieces called segments of one base. The corresponding
surfaces are called spherical zones of one base.
(a) Determine the surface areas of the two spherical zones indicated in the ﬁgure.
(b) Determine the approximate area of the Arctic Ocean by assuming that it is approximately
circular in shape, with center at the North Pole and “circumference” at 75 north latitude.
Use r 3960 mi for the radius of Earth.
(c) A sphere of radius r is inscribed in a right circular cylinder of radius r . Two planes perpendicular to the central axis of the cylinder and a distance h apart cut off a spherical zone
of two bases on the sphere. Show that the surface area of the spherical zone equals the
surface area of the region that the two planes cut off on the cylinder.
(d) The Torrid Zone is the region on the surface of Earth that is between the Tropic of Cancer
(23.45 north latitude) and the Tropic of Capricorn (23.45 south latitude). What is the
area of the Torrid Zone? d
h r 4. (a) Show that an observer at height H above the north pole of a sphere of radius r can see a part of the sphere that has area
2 r 2H
rH
(b) Two spheres with radii r and R are placed so that the distance between their centers is d,
where d r R. Where should a light be placed on the line joining the centers of the
spheres in order to illuminate the largest total surface?
5. Suppose that the density of seawater, z , varies with the depth z below the surface.
(a) Show that the hydrostatic pressure is governed by the differential equation
dP
dz zt where t is the acceleration due to gravity. Let P0 and 0 be the pressure and density at
z 0. Express the pressure at depth z as an integral.
zH
(b) Suppose the density of seawater at depth z is given by
, where H is a positive
0e
constant. Find the total force, expressed as an integral, exerted on a vertical circular porthole of radius r whose center is located at a distance L r below the surface.
6. The ﬁgure shows a semicircle with radius 1, horizontal diameter PQ, and tangent lines at P and Q. At what height above the diameter should the horizontal line be placed so as to minimize the shaded area?
P
FIGURE FOR PROBLEM 6 620 Q 7. Let P be a pyramid with a square base of side 2b and suppose that S is a sphere with its center on the base of P and is tangent to all eight edges of P. Find the height of P. Then ﬁnd the
volume of the intersection of S and P. 5E09(pp 620621) 1/17/06 6:11 PM Page 621 8. Consider a ﬂat metal plate to be placed vertically under water with its top 2 m below the surface of the water. Determine a shape for the plate so that if the plate is divided into any
number of horizontal strips of equal height, the hydrostatic force on each strip is the same.
9. A uniform disk with radius 1 m is to be cut by a line so that the center of mass of the smaller piece lies halfway along a radius. How close to the center of the disk should the cut be made?
(Express your answer correct to two decimal places.)
10. A triangle with area 30 cm 2 is cut from a corner of a square with side 10 cm, as shown in the ﬁgure. If the centroid of the remaining region is 4 cm from the right side of the square, how
far is it from the bottom of the square? 10 cm 11. In a famous 18thcentury problem, known as Buffon’s needle problem, a needle of length h is FIGURE FOR PROBLEM 10 h
¨ y dropped onto a ﬂat surface (for example, a table) on which parallel lines L units apart, L h,
have been drawn. The problem is to determine the probability that the needle will come to rest
intersecting one of the lines. Assume that the lines run eastwest, parallel to the xaxis in a
rectangular coordinate system (as in the ﬁgure). Let y be the distance from the “southern” end
of the needle to the nearest line to the north. (If the needle’s southern end lies on a line, let
y 0. If the needle happens to lie eastwest, let the “western” end be the “southern” end.) Let
be the angle that the needle makes with a ray extending eastward from the “southern” end.
Then 0 y L and 0
. Note that the needle intersects one of the lines only when
y h sin . Now, the total set of possibilities for the needle can be identiﬁed with the rectangular region 0 y L, 0
, and the proportion of times that the needle intersects a
line is the ratio h sin ¨ L y
L
h area under y h sin
area of rectangle
This ratio is the probability that the needle intersects a line. Find the probability that the
needle will intersect a line if h L. What if h L 2?
12. If the needle in Problem 11 has length h
π
2 FIGURE FOR PROBLEM 11 π ¨ L, it’s possible for the needle to intersect more
than one line.
(a) If L 4, ﬁnd the probability that a needle of length 7 will intersect at least one line.
[Hint: Proceed as in Problem 11. Deﬁne y as before; then the total set of possibilities for
the needle can be identiﬁed with the same rectangular region 0 y L, 0
.
What portion of the rectangle corresponds to the needle intersecting a line?]
(b) If L 4, ﬁnd the probability that a needle of length 7 will intersect two lines.
(c) If 2L h 3L, ﬁnd a general formula for the probability that the needle intersects
three lines. 621 ...
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This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas at Austin.
 Fall '09
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