# Mathematics

This approximation to S, which has the form of a Riemann sum, improves as the number of subintervals increases and as the length of the subintervals approaches 0. Specifically, as n S ∞ and as ∆x S 0, we obtain an integral for the surface area:

S = lim nS∞

a n

k = 1 2p f 1x*k221 + f ′1x*k22 ∆x

= L b

a 2p f 1×221 + f ′1×22 dx.

➤ Notice that f is assumed to be differentiable on 3a, b4; therefore, it satisfies the conditions of the Mean Value Theorem. Recall that a similar argument was used to derive the arc length formula in Section 6.5.

DEFINITION Area of a Surface of Revolution Let f be a nonnegative function with a continuous first derivative on the interval 3a, b4. The area of the surface generated when the graph of f on the interval 3a, b4 is revolved about the x-axis is

S = L b

a 2p f 1×221 + f ′1×22 dx.

QUICK CHECK 3 Let f 1×2 = c, where c 7 0. What surface is generated when the graph of f on 3a, b4 is revolved about the x-axis? Without using calculus, what is the area of the surface? M06_BRIG7345_02_SE_C06.6.indd 454 21/10/13 11:35 AM

6.6 Surface Area 455

Figure 6.65

y

1 3

2

4

x

y ! 2 x

(b)

!a a!h x

y ” a2 ! x2

y

a

EXAMPLE 1 Using the surface area formula The graph of f 1×2 = 21x on the interval 31, 34 is revolved about the x-axis. What is the area of the surface generated (Figure 6.65)?

SOLUTION Noting that f ′1×2 = 11x, the surface area formula gives S = L

b

a 2p f 1×221 + f ′1×22 dx

= 2pL 3

1 21xA1 + 1x dx Substitute for f and f ′.

= 4pL 3

1 2x + 1 dx Simplify.

= 8p 3 1x + 123>2 ` 3

1 =

16p 3

14 – 222. Integrate and simplify. Related Exercises 5–14

EXAMPLE 2 Surface area of a spherical cap A spherical cap is produced when a sphere of radius a is sliced by a horizontal plane that is a vertical distance h below the north pole of the sphere, where 0 … h … 2a (Figure 6.66a). We take the spherical cap to be that part of the sphere above the plane, so that h is the depth of the cap. Show that the area of a spherical cap of depth h cut from a sphere of radius a is 2pah.

SOLUTION To generate the spherical surface, we revolve the curve f 1×2 = 2a2 – x2 on the interval 3-a, a4 about the x-axis (Figure 6.66b). The spherical cap of height h cor- responds to that part of the sphere on the interval 3a -h, a4, for 0 … h … 2a. Noting that f ′1×2 = -x1a2 – x22-1>2, the surface area of the spherical cap of height h is

S = L b

a 2pf 1×221 + f ′1×22 dx

= 2pL a

a – h 2a2 – x221 + 1-x1a2 – x22-1>222 dx Substitute for f and f ′.

= 2pL a

a – h 2a2 – x2B a2a2 – x2 dx Simplify.

= 2pL a

a – h a dx = 2pah. Simplify and integrate.

It is worthwhile to check this result with three special cases. With h = 2a, we have a complete sphere, so S = 4pa2. The case h = a corresponds to a hemispherical cap, so S = 14pa22>2 = 2pa2. The case h = 0 corresponds to no spherical cap, so S = 0.

Related Exercises 5–14

➤ Notice that f is not differentiable at {a. Nevertheless, in this case, the surface area integral can be evaluated using methods you know.

➤ The surface area of a sphere of radius a is 4pa2.

Figure 6.66

h h

(a)

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456 Chapter 6 Applications of Integration

EXAMPLE 3 Painting a funnel The curved surface of a funnel is generated by

revolving the graph of y = f 1×2 = x3 + 1 12x

on the interval 31, 24 about the x-axis (Figure 6.67). Approximately what volume of paint is needed to cover the outside of the funnel with a layer of paint 0.05 cm thick? Assume that x and y are measured in centimeters.

SOLUTION Note that f ′1×2 = 3×2 – 1 12×2

. Therefore, the surface area of the funnel

in cm2 is

S = L b

a 2pf 1×221 + f ′1×22 dx

= 2pL 2

1 ax3 + 1

12x bB1 + a3x2 – 112×2 b2 dx Substitute for f and f ′.

= 2pL 2

1 ax3 + 1

12x bB a3x2 + 112×2 b2 dx Expand and factor under square root.

= 2pL 2

1 ax3 + 1

12x b a3x2 + 1

12×2 b dx Simplify.

= 12,289

192 p. Evaluate integral.

Because the paint layer is 0.05 cm thick, the approximate volume of paint needed isa 12,289p 192

cm2b10.05 cm2 ≈ 10.1 cm3. Related Exercises 15–16

The derivation that led to the surface area integral may be used when a curve is revolved about the y-axis (rather than the x-axis). The result is the same integral with x replaced with y. For example, if the curve x = g1y2 on the interval 3c, d4 is revolved about the y-axis, the area of the surface generated is

S = L d

c 2pg1y221 + g′1y22 dy.

To use this integral, we must first describe the given curve as a differentiable function of y.

EXAMPLE 4 Change of perspective Consider the function y = ln a x + 2×2 – 1 2

b . Find the area of the surface generated when the part of the curve between the points 154, 02 and 1178 , ln 22 is revolved about the y-axis (Figure 6.68). SOLUTION We solve for x in terms of y in the following steps:

y = ln a x + 2×2 – 1 2

b ey =

x + 2×2 – 1 2

Exponentiate both sides.

2ey – x = 2×2 – 1 Rearrange terms. 4e2y – 4xey + x2 = x2 – 1 Square both sides.

x = g1y2 = ey + 1 4

e-y. Solve for x.Figure 6.68

y

ln 2

5 4

17 8

x

y ! ln( )! x2 ” 1x # 2

Figure 6.67

y

1 2

4

8 y ! x3 ” 112x

11111 x

M06_BRIG7345_02_SE_C06.6.indd 456 21/10/13 11:35 AM

6.7 Physical Applications 459

Figure 6.69 x ! a x ! b

Figure 6.70

x ! a x ! b

“x

Mass of kth subinterval: mk ! ! (xk*)”x

xk*

any linear function g1x2 = cx + d that is positive on the interval 3a, b4, where / is the slant height of the frustum. 37. Scaling surface area Let f be a nonnegative function with a con-

tinuous first derivative on 3a, b4 and suppose that g1x2 = c f 1×2 and h1x2 = f 1cx2, where c 7 0. When the curve y = f 1×2 on 3a, b4 is revolved about the x-axis, the area of the resulting surface is A. Evaluate the following integrals in terms of A and c.

a. L b

a 2pg1x22c2 + g′1×22 dx b. Lb>ca>c 2ph1x22c2 + h′1×22 dx

38. Surface plus cylinder Suppose f is a nonnegative function with a continuous first derivative on 3a, b4. Let L equal the length of the graph of f on 3a, b4 and let S be the area of the surface generated by revolving the graph of f on 3a, b4 about the x-axis. For a positive constant C, assume the curve y = f 1×2 + C is revolved about the x-axis. Show that the area of the resulting surface equals the sum of S and the surface area of a right circular cylinder of radius C and height L.

1. The surface area of the first cone 120015p2 is twice as great as the surface area of the second cone 110015p2. 2. The surface area is 63110p. 3. The surface is a cylinder of radius c and height b – a. The area of the curved surface is 2pc1b – a2.

the volume of the object is important. Animals typically generate heat at a rate proportional to their volume and lose heat at a rate pro- portional to their surface area. Therefore, animals with a low SAV ratio tend to retain heat, whereas animals with a high SAV ratio (such as children and hummingbirds) lose heat relatively quickly.

a. What is the SAV ratio of a cube with side lengths a? b. What is the SAV ratio of a ball with radius a? c. Use the result of Exercise 34 to find the SAV ratio of an ellip-

soid whose long axis has length 2a13 4, for a Ú 1, and whose other two axes have half the length of the long axis. (This scal- ing is used so that, for a given value of a, the volumes of the ellipsoid and the ball of radius a are equal.) The volume of a

general ellipsoid is V = 4p 3

ABC, where the axes have lengths 2A, 2B, and 2C.

d. Graph the SAV ratio of the ball of radius a Ú 1 as a function of a (part (b)) and graph the SAV ratio of the ellipsoid de- scribed in (part (c)) on the same set of axes. Which object has the smaller SAV ratio?

e. Among all ellipsoids of a fixed volume, which one would you choose for the shape of an animal if the goal is to minimize heat loss?

Additional Exercises 36. Surface area of a frustum Show that the surface area of the frus-

tum of a cone generated by revolving the line segment between 1a, g1a22 and 1b, g1b22 about the x-axis is p1g1b2 + g1a22/, for 6.7 Physical Applications We continue this chapter on applications of integration with several problems from phys- ics and engineering. The physical themes in these problems are mass, work, force, and pressure. The common mathematical theme is the use of the slice-and-sum strategy, which always leads to a definite integral.

Density and Mass Density is the concentration of mass in an object and is usually measured in units of mass per volume 1for example, g>cm32. An object with uniform density satisfies the basic relationship

mass = density # volume. When the density of an object varies, this formula no longer holds, and we must appeal to calculus.

In this section, we introduce mass calculations for thin objects that can be viewed as line segments (such as wires or thin bars). The bar shown in Figure 6.69 has a density r that varies along its length. For one-dimensional objects, we use linear density with units of mass per length 1for example, g>cm2. What is the mass of such an object? QUICK CHECK 1 In Figure 6.69, suppose a = 0, b = 3, and the density of the rod in g>cm is r1x2 = 14 – x2. (a) Where is the rod lightest and heaviest? (b) What is the density at the middle of the bar?

We begin by dividing the bar, represented by the interval a … x … b, into n sub- intervals of equal length ∆x = 1b – a2>n (Figure 6.70). Let xk* be any point in the kth subinterval, for k = 1, c, n. The mass of the kth segment of the bar mk is approxi- mately the density at xk

* multiplied by the length of the interval, or mk ≈ r1xk*2∆x. So the approximate mass of the entire bar is

a n

k = 1 mk ≈ a

n

k = 1 r1xk*2∆x.

t

mk

➤ In Chapter 13, we return to mass calculations for two- and three- dimensional objects (plates and solids).

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460 Chapter 6 Applications of Integration

The exact mass is obtained by taking the limit as n S ∞ and as ∆x S 0, which produces a definite integral.

➤ Note that the units of the integral work out as they should: r has units of mass per length and dx has units of length, so r1x2 dx has units of mass.

➤ Another interpretation of the mass integral is that mass equals the average value of the density multiplied by the length of the bar b – a.

Figure 6.71

y

O xba

y ! F(x)

force

Force varies on [a, b]

DEFINITION Mass of a One-Dimensional Object Suppose a thin bar or wire is represented by the interval a … x … b with a density function r (with units of mass per length). The mass of the object is

m = L b

a r1x2 dx.

EXAMPLE 1 Mass from variable density A thin 2-m bar, represented by the interval 0 … x … 2, is made of an alloy whose density in units of kg>m is given by r1x2 =11 + x22. What is the mass of the bar? SOLUTION The mass of the bar in kilograms is

m = L b

a r1x2 dx = L20 11 + x22 dx = ax + x33 b ` 20 = 143 .

Related Exercises 9–16

QUICK CHECK 2 A thin bar occupies the interval 0 … x … 2 and has a density in kg>m of r1x2 = 11 + x22. Using the minimum value of the density, what is a lower bound for the mass of the object? Using the maximum value of the density, what is an upper bound for the mass of the object?

Work Work can be described as the change in energy when a force causes a displacement of an object. When you carry a basket of laundry up a flight of stairs or push a stalled car, you apply a force that results in the displacement of an object, and work is done. If a constant force F displaces an object a distance d in the direction of the force, the work done is the force multiplied by the distance:

work = force # distance. It is easiest to use metric units for force and work. A newton 1N2 is the force required to give a 1-kg mass an acceleration of 1 m>s2. A joule 1J2 is 1 newton-meter 1N@m2, the work done by a 1-N force over a distance of 1 m.

Calculus enters the picture with variable forces. Suppose an object is moved along the x-axis by a variable force F that is directed along the x-axis (Figure 6.71). How much work is done in moving the object between x = a and x = b? Once again, we use the slice-and-sum strategy.

The interval 3a, b4 is divided into n subintervals of equal length ∆x = 1b – a2>n. We let xk

* be any point in the kth subinterval, for k = 1, c, n. On that subinterval, the force is approximately constant with a value of F 1xk*2. Therefore, the work done in mov- ing the object across the kth subinterval is approximately F 1xk*2∆x 1force # distance2. Summing the work done over each of the n subintervals, the total work over the interval 3a, b4 is approximately

W ≈ a n

k = 1 F 1xk*2∆x.

This approximation becomes exact when we take the limit as n S ∞ and ∆x S 0. The total work done is the integral of the force over the interval 3a, b4 (or, equivalently, the net area under the force curve in Figure 6.71).

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6.7 Physical Applications 461

An application of force and work that is easy to visualize is the stretching and com- pression of a spring. Suppose an object is attached to a spring on a frictionless horizontal surface; the object slides back and forth under the influence of the spring. We say that the spring is at equilibrium when it is neither compressed nor stretched. It is convenient to let x be the position of the object, where x = 0 is the equilibrium position (Figure 6.72).

Figure 6.72

Compressed

Equilibrium

Stretched

x ! 0

x ” 0

F ” 0

x ! 0

x ! 0x # 0

F # 0

Figure 6.73

F

x

F(x) ! kx

Spring is compressed.

Spring is stretched.

DEFINITION Work The work done by a variable force F moving an object along a line from x = a to x = b in the direction of the force is

W = L b

a F 1×2 dx.

QUICK CHECK 3 Explain why the sum of the work over n subintervals is only an approxi- mation of the total work.

According to Hooke’s law, the force required to keep the spring in a compressed or stretched position x units from the equilibrium position is F 1×2 = kx, where the positive spring constant k measures the stiffness of the spring. Note that to stretch the spring to a position x 7 0, a force F 7 0 (in the positive direction) is required. To compress the spring to a position x 6 0, a force F 6 0 (in the negative direction) is required (Figure 6.73). In other words, the force required to displace the spring is always in the direction of the displacement.

EXAMPLE 2 Compressing a spring Suppose a force of 10 N is required to stretch a spring 0.1 m from its equilibrium position and hold it in that position.

a. Assuming that the spring obeys Hooke’s law, find the spring constant k. b. How much work is needed to compress the spring 0.5 m from its equilibrium position? c. How much work is needed to stretch the spring 0.25 m from its equilibrium position? d. How much additional work is required to stretch the spring 0.25 m if it has already

been stretched 0.1 m from its equilibrium position?

SOLUTION

a. The fact that a force of 10 N is required to keep the spring stretched at x = 0.1 m means (by Hooke’s law) that F 10.12 = k10.1 m2 = 10 N. Solving for the spring constant, we find that k = 100 N>m. Therefore, Hooke’s law for this spring is F 1×2 = 100x.

b. The work in joules required to compress the spring from x = 0 to x = -0.5 is

W = L b

a F 1×2 dx = L -0.50 100x dx = 50×2 ` -0.50 = 12.5.

➤ Hooke’s law was proposed by the English scientist Robert Hooke (1635–1703), who also coined the biological term cell. Hooke’s law works well for springs made of many common materials. However, some springs obey more complicated spring laws (see Exercise 51).

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462 Chapter 6 Applications of Integration

Figure 6.74

y

x

!y

yn ” b

yn#1

yk

yk#1

y2

y1

yk*

y0 ” 0

h # yk*

y ” h (outflow level)

Work to lift kth layer ! A(yk*) !y !g \$ (h # yk*)

Work to lift all layers

! k”1

n

% A(yk*) !y !g \$ (h # yk*)

h

Volume of kth layer ! A(yk*) !y

Force on kth layer ! A(yk*) !y \$ ! \$ g

The kth layer occupying the interval 3yk – 1, yk4, for k = 1, c, n, is approximately yk* units above the bottom of the tank, where yk

* is any point in 3yk – 1, yk4. The cross-sectional area of the kth layer at yk

*, denoted A1yk*2, is determined by the shape of the tank; the solution depends on being able to find A for all values of y. Because the volume of the kth layer is approximately A1yk*2∆y, the force on the kth layer (its weight) is

mass t

Fk = mg ≈ A1yk*2∆y # r # g. t 6

volume density

To reach the level y = h, the kth layer is lifted an approximate distance 1h – yk*2 (Figure 6.74). So the work in lifting the kth layer to a height h is approximately

Wk = A1yk*2∆yrg # 1h – yk*2. force distance

Summing the work required to lift all the layers to a height h, the total work is

W ≈ a n

k = 1 Wk = a

n

k = 1 A1yk*2rg1h – yk*2∆y.

This approximation becomes more accurate as the width of the layers ∆y tends to zero and the number of layers tends to infinity. In this limit, we obtain a definite integral from y = 0 to y = b. The total work required to empty the tank is

W = lim nS ∞ a

n

k = 1 A1yk*2rg1h – yk*2∆y = Lb0 rgA1y21h – y2 dy.

This derivation assumes that the bottom of the tank is at y = 0, in which case the distance that the slice at level y must be lifted is D1y2 = h – y. If you choose a different location for the origin, the function D will be different. Here is a general procedure for any choice of origin.

➤ The choice of a coordinate system is somewhat arbitrary and may depend on the geometry of the problem. You can let the y-axis point upward or downward, and there are usually several logical choices for the location of y = 0. You should experiment with different coordinate systems.

c. The work in joules required to stretch the spring from x = 0 to x = 0.25 is

W = L b

a F 1×2 dx = L0.250 100x dx = 50×2 ` 0.250 = 3.125.

d. The work in joules required to stretch the spring from x = 0.1 to x = 0.35 is

W = L b

a F 1×2 dx = L0.350.1 100x dx = 50×2 ` 0.350.1 = 5.625.

Comparing parts (c) and (d), we see that more work is required to stretch the spring 0.25 m starting at x = 0.1 than starting at x = 0. Related Exercises 17–26

Lifting Problems Another common work problem arises when the motion is vertical and the force is the gravitational force. The gravitational force exerted on an object with mass m is F = mg, where g ≈ 9.8 m>s2 is the acceleration due to gravity near the surface of Earth. The work in joules required to lift an object of mass m a vertical distance of y meters is

work = force # distance = mgy. This type of problem becomes interesting when the object being lifted is a body of

water, a rope, or a chain. In these situations, different parts of the object are lifted different distances—so integration is necessary. Here is a typical situation and the strategy used.

Suppose a fluid such as water is pumped out of a tank to a height h above the bottom of the tank. How much work is required, assuming the tank is full of water? Three key observations lead to the solution.

Water from different levels of the tank is lifted different vertical distances, requiring different amounts of work.

Two equal volumes of water from the same horizontal plane are lifted the same distance and require the same amount of work.

A volume V of water has mass rV, where r = 1 g>cm3 = 1000 kg>m3 is the density of water.

To solve this problem, we let the y-axis point upward with y = 0 at the bottom of the tank. The body of water that must be lifted extends from y = 0 to y = b (which may be the top of the tank). The level to which the water must be raised is y = h, where h Ú b (Figure 6.74). We now slice the water into n horizontal layers, each having thickness ∆y.

➤ Notice again that the units in the integral are consistent. If F has units of N and x has units of m, then W has units of F dx, or N@m, which are the units of work 11 N@m = 1 J2.

QUICK CHECK 4 In Example 2, explain why more work is needed in part (d) than in part (c), even though the displacement is the same.

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6.7 Physical Applications 463

PROCEDURE Solving Lifting Problems 1. Draw a y-axis in the vertical direction (parallel to gravity) and choose a con-

venient origin. Assume the interval 3a, b4 corresponds to the vertical extent of the fluid.

2. For a … y … b, find the cross-sectional area A1y2 of the horizontal slices and the distance D1y2 the slices must be lifted.

3. The work required to lift the water is

W = L b

a rgA1y2D1y2 dy.

The kth layer occupying the interval 3yk – 1, yk4, for k = 1, c, n, is approximately yk* units above the bottom of the tank, where yk

* is any point in 3yk – 1, yk4. The cross-sectional area of the kth layer at yk

*, denoted A1yk*2, is determined by the shape of the tank; the solution depends on being able to find A for all values of y. Because the volume of the kth layer is approximately A1yk*2∆y, the force on the kth layer (its weight) is

mass t

Fk = mg ≈ A1yk*2∆y # r # g. t 6 volume density

To reach the level y = h, the kth layer is lifted an approximate distance 1h – yk*2 (Figure 6.74). So the work in lifting the kth layer to a height h is approximately

Wk = A1yk*2∆yrg # 1h – yk*2. force distance

Summing the work required to lift all the layers to a height h, the total work is

W ≈ a n

k = 1 Wk = a

n

k = 1 A1yk*2rg1h – yk*2∆y.

This approximation becomes more accurate as the width of the layers ∆y tends to zero and the number of layers tends to infinity. In this limit, we obtain a definite integral from y = 0 to y = b. The total work required to empty the tank is

W = lim nS ∞ a

n

k = 1 A1yk*2rg1h – yk*2∆y = Lb0 rgA1y21h – y2 dy.

This derivation assumes that the bottom of the tank is at y = 0, in which case the distance that the slice at level y must be lifted is D1y2 = h – y. If you choose a different location for the origin, the function D will be different. Here is a general procedure for any choice of origin.

➤ The choice of a coordinate system is somewhat arbitrary and may depend on the geometry of the problem. You can let the y-axis point upward or downward, and there are usually several logical choices for the location of y = 0. You should experiment with different coordinate systems.

t t

We now use this procedure to solve two pumping problems.

EXAMPLE 3 Pumping water How much work is needed to pump all the water out of a cylindrical tank with a height of 10 m and a radius of 5 m? The water is pumped to an outflow pipe 15 m above the bottom of the tank.

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464 Chapter 6 Applications of Integration

SOLUTION In this problem, we choose a different origin by letting y = 0 and y = -5 correspond to the center and the bottom of the tank, respectively. For -5 … y … 0, a hori- zontal layer of gasoline located at a depth y is a rectangle with a length of 10 and width of 2225 – y2 (Figure 6.76). Therefore, the cross-sectional area of the layer at depth y is

A1y2 = 20225 – y2. ➤ Again, there are several choices for the

location of the origin. The location in this example makes A1y2 easy to compute.

SOLUTION Figure 6.75 shows the cylindrical tank filled to capacity and the outflow 15 m above the bottom of the tank. We let y = 0 represent the bottom of the tank and y = 10 represent the top of the tank. In this case, all horizontal slices are circular disks of radius r = 5 m. Therefore, for 0 … y … 10, the cross-sectional area is

A1y2 = pr2 = p52 = 25p. Note that the water is pumped to a level h = 15 m above the bottom of the tank, so the lifting distance is D1y2 = 15 – y. The resulting work integral is

W = L 10

0 rgA1y2 D1y2 dy = 25prgL100 115 – y2 dy.

25p 15 – y

Substituting r = 1000 kg>m3 and g = 9.8 m>s2, the total work in joules is W = 25prgL

10

0 115 – y2 dy

= 25p11000219.82a15y – 1 2

y2b ` 10 0

r g

≈ 7.7 * 107.

The work required to pump the water out of the tank is approximately 77 million joules. Related Exercises 27–37

QUICK CHECK 5 In the previous example, how would the integral change if the outflow pipe were at the top of the tank?

EXAMPLE 4 Pumping gasoline A cylindrical tank with a length of 10 m and a radius of 5 m is on its side and half-full of gasoline (Figure 6.76). How much work is required to empty the tank through an outlet pipe at the top of the tank? 1The density of gasoline is r ≈ 737 kg>m3.2

➤ Recall that g ≈ 9.8 m>s2. You should verify that the units are consistent in this calculation: The units of r, g, A1y2, D1y2, and dy are kg>m3, m>s2, m2, m, and m, respectively. The resulting units of W are kg m2>s2, or J. A more convenient unit for large amounts of work and energy is the kilowatt-hour, which is 3.6 million joules.

e e eFigure 6.75

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