# Mathematics

Can this area be found using a different approach? Sometimes it helps to use geometry. Notice that the region R can be formed by taking the entire region under the curve y = x2>3 on the interval 30, 84 and then removing a triangle whose base is the interval 34, 84 (Figure 6.21). The area of the region R1 under the curve y = x2>3 is

L 8

0 x2>3 dx = 3

5 x5>3 ` 8

0 =

96 5

.

The triangle R2 has a base of length 4 and a height of 4, so its area is 1 2 # 4 # 4 = 8. There-

fore, the area of R is 965 – 8 = 56 5 , which agrees with the first calculation.

➤ To find the point of intersection in Example 4, solve y3>2 = y + 4 by first squaring both sides of the equation.

Figure 6.21

0 84

y

4

R

x

0 84

y

4

R1

x 0 84

y

4

R2

x

y ! x2/3

y ! x ” 4

Area of region R ! (area of region R1) ” (area of region R2)

4

4

Related Exercises 33–38

QUICK CHECK 4 An alternative way to determine the area of the region in Example 3 (Figure 6.18) is to compute 18 + 120 1x + 6 – x32 dx. Why?

SECTION 6.2 EXERCISES Review Questions 1. Draw the graphs of two functions f and g that are continuous and

intersect exactly twice on 1- ∞ , ∞2. Explain how to use integra- tion to find the area of the region bounded by the two curves.

2. Draw the graphs of two functions f and g that are continuous and intersect exactly three times on 1- ∞ , ∞2. How is integration used to find the area of the region bounded by the two curves?

3. Make a sketch to show a case in which the area bounded by two curves is most easily found by integrating with respect to x.

4. Make a sketch to show a case in which the area bounded by two curves is most easily found by integrating with respect to y.

M06_BRIG7345_02_SE_C06.2.indd 416 21/10/13 11:32 AM

6.2 Regions Between Curves 417

7. y

x

y ! 3 ” x

y ! 2x

O

(Hint: Find the intersection point by inspection.)

8.

O x

y ! 4 cos2 x

y ! sec 2 x

4

y

9–14. Regions between curves Sketch the region and find its area.

9. The region bounded by y = 21x + 12, y = 31x + 12, and x = 4 10. The region bounded by y = cos x and y = sin x between

x = p>4 and x = 5p>4 11. The region bounded by y = ex, y = e-2x, and x = ln 4

12. The region bounded by y = 2x and y = x2 + 3x – 6

13. The region bounded by y = 2

1 + x2 and y = 1

14. The region bounded by y = 241x and y = 3×2 15–22. Compound regions Sketch each region (if a figure is not given) and then find its total area.

15. The region bounded by y = sin x, y = cos x, and the x-axis between x = 0 and x = p>2

xq

1

y

y ! sin x

y ! cos x

16. The regions between y = sin x and y = sin 2x, for 0 … x … p

x!q

1

!1

y

y ” sin x

y ” sin 2x

17. The region bounded by y = x, y = 1>x, y = 0, and x = 2 18. The regions in the first quadrant on the interval 30, 24 bounded by

y = 4x – x2 and y = 4x – 4.

19. The region bounded by y = 2 – ! x ! and y = x2

20. The regions bounded by y = x3 and y = 9x

21. The region bounded by y = ! x – 3 ! and y = x>2 22. The regions bounded by y = x213 – x2 and y = 12 – 4x

23–26. Integrating with respect to y Sketch each region (if a figure is not given) and find its area by integrating with respect to y.

23. The region bounded by y = A x2 + 1, y = 11 – x, and y = 0.

x

y ! 1 ” xy ! ” # 12 x

y

24. The region bounded by x = cos y and x = -sin 2y

!1

!

1 x

y

2 ”

4 ”

4 ”

x # !sin 2y

x # cos y

25. The region bounded by x = y2 – 3y + 12 and x = -2y2 – 6y + 30

4

!4

x ” !2y2 ! 6y # 30

x ” y2 ! 3y # 12

10 3020 x

y

2

!2

26. Both regions bounded by x = y3 – 4y2 + 3y and x = y2 – y

x ! y3 ” 4y2 # 3y

x ! y2 ” y

4

“1

5 10 x

y

3

2

1

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422 Chapter 6 Applications of Integration

QUICK CHECK 1 Why is the volume, as given by the general slicing method, equal to the average value of the area function A on 3a, b4 multiplied by b – a? EXAMPLE 1 Volume of a “parabolic cube” Let R be the region in the first quadrant bounded by the coordinate axes and the curve y = 1 – x2. A solid has a base R, and cross sections through the solid perpendicular to the base and parallel to the y-axis are squares (Figure 6.25a). Find the volume of the solid.

SOLUTION Focus on a cross section through the solid at a point x, where 0 … x … 1. That cross section is a square with sides of length 1 – x2. Therefore, the area of a typical cross section is A1x2 = 11 – x222. Using the general slicing method, the volume of the solid is

V = L 1

0 A1x2 dx General slicing method

= L 1

0 11 – x222 dx Substitute for A1x2.

= L 1

0 11 – 2×2 + x42 dx Expand integrand.

= 8 15

. Evaluate.

The actual solid with a square cross section is shown in Figure 6.25b.

Figure 6.25

x

x R

y

y ! 1 ” x2

1 ” x2

1 ” x2

(x, 1 ” x2)

Base of solid

Square slice

1

1

(a)

R

y ! 1 ” x2

(b)

y

x

Related Exercises 7–16

EXAMPLE 2 Volume of a “parabolic hemisphere” A solid has a base that is bounded by the curves y = x2 and y = 2 – x2 in the xy-plane. Cross sections through the solid perpendicular to the base and parallel to the y-axis are semicircular disks. Find the volume of the solid.

SOLUTION Because a typical cross section perpendicular to the x-axis is a semicircular disk (Figure 6.26), the area of a cross section is 12 pr

2, where r is the radius of the cross section.

Figure 6.26

!1

Radius of slice

” ((2 ! x2) ! x2)

” 1 ! x2

y ” x2

y ” 2 ! x2

Base of solid

Semicircular slice

y 1 2

x 1

M06_BRIG7345_02_SE_C06.3.indd 422 21/10/13 4:45 PM

6.3 Volume by Slicing 423

The key observation is that this radius is one-half of the distance between the upper bound- ing curve y = 2 – x2 and the lower bounding curve y = x2. So the radius at the point x is

r = 1 2

112 – x22 – x22 = 1 – x2. This means that the area of the semicircular cross section at the point x is

A1x2 = 1 2

pr2 = p

2 11 – x222.

The intersection points of the two bounding curves satisfy 2 – x2 = x2, which has solu- tions x = {1. Therefore, the cross sections lie between x = -1 and x = 1. Integrating the cross-sectional areas, the volume of the solid is

V = L 1

-1 A1x2 dx General slicing method

= L 1

-1

p

2 11 – x222 dx Substitute for A1x2.

= p

2 L 1

-1 11 – 2×2 + x42 dx Expand integrand.

= 8p 15

. Evaluate.

Related Exercises 7–16

The Disk Method We now consider a specific type of solid known as a solid of revolution. Suppose f is a continuous function with f 1×2 Ú 0 on an interval 3a, b4. Let R be the region bounded by the graph of f , the x-axis, and the lines x = a and x = b (Figure 6.27). Now revolve R around the x-axis. As R revolves once about the x-axis, it sweeps out a three-dimensional solid of revolution (Figure 6.28). The goal is to find the volume of this solid, and it may be done using the general slicing method.

Figure 6.27

y ! f (x)

a

R

b

y

xO

Figure 6.28

x

y

O a

R

b x

y

O a b

x

y

O a b x

y

O a b

Revolving the region R generates a solid of revolution.

QUICK CHECK 2 In Example 2, what is the cross-sectional area function A1x2 if cross sections perpendicular to the base are squares rather than semicircles?

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424 Chapter 6 Applications of Integration

Disk Method about the x-Axis Let f be continuous with f 1×2 Ú 0 on the interval 3a, b4. If the region R bounded by the graph of f , the x-axis, and the lines x = a and x = b is revolved about the x-axis, the volume of the resulting solid of revolution is

V = L b

a p f 1×22 dx.

QUICK CHECK 3 What solid results when the region R is revolved about the x-axis if (a) R is a square with vertices 10, 02, 10, 22, 12, 02, and 12, 22, and (b) R is a triangle with vertices 10, 02, 10, 22, and 12, 02?

With a solid of revolution, the cross-sectional area function has a special form because all cross sections perpendicular to the x-axis are circular disks with radius f 1×2 (Figure 6.29). Therefore, the cross section at the point x, where a … x … b, has area

A1x2 = p1radius22 = pf 1×22. By the general slicing method, the volume of the solid is

V = L b

a A1x2 dx = Lba p f 1×22 dx.

Because each slice through the solid is a circular disk, the resulting method is called the disk method.

Figure 6.29

x

y

O x

f (x)

Cross sections of a solid of revolution are circular disks of radius f (x) and area ! f (x)2.

EXAMPLE 3 Disk method at work Let R be the region bounded by the curve f 1×2 = 1x + 122, the x-axis, and the lines x = 0 and x = 2 (Figure 6.30a). Find the volume of the solid of revolution obtained by revolving R about the x-axis.

SOLUTION When the region R is revolved about the x-axis, it generates a solid of revolu- tion (Figure 6.30b). A cross section perpendicular to the x-axis at the point 0 … x … 2 is a circular disk of radius f 1×2. Therefore, a typical cross section has area

A1x2 = p f 1×22 = p11x + 12222. Integrating these cross-sectional areas between x = 0 and x = 2 gives the volume of the solid:

V = L 2

0 A1x2 dx = L20 p 11x + 12222 dx Substitute for A1x2.

= L 2

0 p1x + 124 dx Simplify.

= p u5

5 ` 3 1

= 242 p

5 . Let u = x + 1 and evaluate.

Related Exercises 17–26

Washer Method A slight variation on the disk method enables us to compute the volume of more ex- otic solids of revolution. Suppose that R is the region bounded by the graphs of f and g between x = a and x = b, where f 1×2 Ú g1x2 Ú 0 (Figure 6.31). If R is revolved about the x-axis to generate a solid of revolution, the resulting solid generally has a hole through it.

Once again we apply the general slicing method. In this case, a cross section through the solid perpendicular to the x-axis is a circular washer with an outer radius of R = f 1×2

Figure 6.30

y

x

x

f (x) ! (x ” 1)2

Radius ! (x ” 1)2

R

9

6

3

2x

2

Cross-sectional area ! !((x ” 1)2)2

x

y

0

0

(a)

(b)

M06_BRIG7345_02_SE_C06.3.indd 424 21/10/13 11:32 AM

6.3 Volume by Slicing 425

and an inner radius of r = g1x2, where a … x … b. The area of the cross section is the area of the entire disk minus the area of the hole, or

A1x2 = p1R2 – r22 = p1 f 1×22 – g1x222 (Figure 6.32). The general slicing method gives the area of the solid.

Washer Method about the x-Axis Let f and g be continuous functions with f 1×2 Ú g1x2 Ú 0 on 3a, b4. Let R be the region bounded by y = f 1×2, y = g1x2, and the lines x = a and x = b. When R is revolved about the x-axis, the volume of the resulting solid of revolution is

V = L b

a p1 f 1×22 – g1x222 dx.

Figure 6.31

y

xa b

b a

x x

y ! f (x)

y ! g(x)

Revolving region R about the x-axis…

… produces a solid with a hole.

y

O O

R

QUICK CHECK 4 Show that when g1x2 = 0 in the washer method, the result is the disk method.

EXAMPLE 4 Volume by the washer method The region R is bounded by the graphs of f 1×2 = 1x and g1x2 = x2 between x = 0 and x = 1. What is the volume of the solid that results when R is revolved about the x-axis?

SOLUTION The region R is bounded by the graphs of f and g with f 1×2 Ú g1x2 on 30, 14, so the washer method is applicable (Figure 6.33). The area of a typical cross section at the point x is

A1x2 = p1 f 1×22 – g1x222 = p111x22 – 1×2222 = p1x – x42. Therefore, the volume of the solid is

V = L 1

0 p1x – x42 dx Washer method

= pa x2 2

– x 5

5 b ` 1

0 =

3p 10

. Fundamental Theorem

➤ The washer method is really two applications of the disk method. We compute the volume of the entire solid without the hole (by the disk method) and then subtract the volume of the hole (also computed by the disk method).

Figure 6.32

x

(x, f (x))

(x, g(x))

Cross-sectional area ! !(f (x)2 ” g(x)2)

y

O

b a

x

r

R

Cross-sectional area ! !R2 ” !r2

M06_BRIG7345_02_SE_C06.3.indd 425 21/10/13 11:33 AM

402 Chapter 6 Applications of Integration

SOLUTION

a. The velocity function (Figure 6.5a) is positive for 0 6 t 6 p, which means the block moves in the positive (upward) direction. At t = p, the block comes to rest momen- tarily; for p 6 t 6 2p, the block moves in the negative (downward) direction. We let s1t2 be the position at time t Ú 0 with the initial position s102 = -14 m. Method 1: Using antiderivatives Because the position is an antiderivative of the velocity, we have

s1t2 = Lv1t2 dt = L 14 sin t dt = – 14 cos t + C. To determine the arbitrary constant C, we substitute the initial condition s102 = -14 into the expression for s1t2:

– 1 4

= – 1 4

cos 0 + C.

Solving for C, we find that C = 0. Therefore, the position for any time t Ú 0 is

s1t2 = – 1 4

cos t.

Method 2: Using Theorem 6.1 Alternatively, we may use the relationship

s1t2 = s102 + L t0 v1x2 dx. Substituting v1x2 = 14 sin x and s102 = -14, the position function is

s1t2 = – 1 4

+ L t

0 1 4

sin x dx

s102 v1x2 = – 1

4 – a 1

4 cos xb ` t

0 Evaluate integral.

= – 1 4

– 1 4

1cos t – 12 Simplify. = – 1

4 cos t. Simplify.

b. The graph of the position function is shown in Figure 6.5b. We see that s102 = -14 m, as prescribed.

c. The block initially moves in the positive s direction (upward), reaching the origin 1s = 02 when s1t2 = -14 cos t = 0. So the block arrives at the origin for the first time when t = p>2.

d. The block moves in the positive direction and reaches its high point for the first time when t = p; the position at that moment is s1p2 = 14 m. The block then reverses direction and moves in the negative (downward) direction, reaching its low point at t = 2p. This motion repeats every 2p seconds.

Related Exercises 15–24

QUICK CHECK 4 Without doing further calculations, what are the displacement and distance traveled by the block in Example 2 over the interval 30, 2p4?

➤ It is worth repeating that to find the displacement, we need to know only the velocity. To find the position, we must know both the velocity and the initial position s102.

e

Figure 6.5

(a)

(b)

0 t

0.25

!0.25

2!! 3!

0 t

0.25

!0.25

v

2!! 3!

s

v(t) ” E sin t

Velocity function

Position function

s(t) ” !E cos t

M06_BRIG7345_02_SE_C06.1.indd 402 21/10/13 11:30 AM

6.1 Velocity and Net Change 403

EXAMPLE 3 Skydiving Suppose a skydiver leaps from a hovering helicopter and falls in a straight line. He reaches a terminal velocity of 80 m>s at t = 0 and falls for 19 sec- onds, at which time he opens his parachute. The velocity decreases linearly to 6 m>s over a two-second period and then remains constant until he reaches the ground at t = 40 s. The motion is described by the velocity function

v1t2 = c 80 if 0 … t 6 19783 – 37t if 19 … t 6 21 6 if 21 … t … 40.

Determine the height above the ground from which the skydiver jumped.

SOLUTION We let the position of the skydiver increase downward with the origin 1s = 02 corresponding to the position of the helicopter. The velocity is positive, so the distance traveled by the skydiver equals the displacement, which is

L 40

0 0 v1t2 0 dt = L190 80 dt + L2119 1783 – 37t2 dt + L4021 6 dt

= 80t ` 19 0

+ a783t – 37t2 2

b ` 21 19

+ 6t ` 40 21

Fundamental Theorem

= 1720. Evaluate and simplify.

The skydiver jumped from 1720 m above the ground. Notice that the displacement of the skydiver is the area under the velocity curve (Figure 6.6).

➤ The terminal velocity of an object depends on its density, shape, size, and the medium through which it falls. Estimates for human beings in free fall in the lower atmosphere vary from 120 mi>hr 154 m>s2 to 180 mi>hr 180 m>s2.

Figure 6.6

t40302010

v

80

40

Distance traveled ! ! “v(t)” dt ! ! 80 dt ” ! (783 # 37t) dt ” ! 6 dt

! (783 # 37t) dt 19

21

0

19

! 80 dt 0

19

0

40

v (t) ! 783 # 37t

v (t) ! 6

v (t) ! 80

21

40

19

21

21

40! 6 dt

0

Related Exercises 25–26

QUICK CHECK 5 Suppose (unrealistically) in Example 3 that the velocity of the skydiver is 80 m>s, for 0 … t 6 20, and then it changes instantaneously to 6 m>s, for 20 … t … 40. Sketch the velocity function and, without integrating, find the distance the skydiver falls in 40 s.

Acceleration Because the acceleration of an object moving along a line is given by a1t2 = v′1t2, the relationship between velocity and acceleration is the same as the relationship between

M06_BRIG7345_02_SE_C06.1.indd 403 21/10/13 11:30 AM

404 Chapter 6 Applications of Integration

position and velocity. Given the acceleration of an object, the change in velocity over an interval 3a, b4 is

change in velocity = v1b2 – v1a2 = Lba v′1t2 dt = Lba a1t2 dt. Furthermore, if we know the acceleration and initial velocity v102, then the velocity at future times can also be found.

➤ Theorem 6.2 is a consequence of the Fundamental Theorem of Calculus.

THEOREM 6.2 Velocity from Acceleration Given the acceleration a1t2 of an object moving along a line and its initial velocity v102, the velocity of the object for future times t Ú 0 is

v1t2 = v102 + L t0 a1x2 dx. EXAMPLE 4 Motion in a gravitational field An artillery shell is fired directly upward with an initial velocity of 300 m>s from a point 30 m above the ground (Figure 6.7). Assume that only the force of gravity acts on the shell and it produces an acceleration of 9.8 m>s2. Find the velocity of the shell while it is in the air. SOLUTION We let the positive direction be upward with the origin 1s = 02 correspond- ing to the ground. The initial velocity of the shell is v102 = 300 m>s. The acceleration due to gravity is downward; therefore, a1t2 = -9.8 m>s2. Integrating the acceleration, the velocity is

v1t2 = v102 + L t0 a1x2 dx = 300 + L t0 1-9.82 dx = 300 – 9.8t. 300 m>s -9.8 m>s2 The velocity decreases from its initial value of 300 m>s, reaching zero at the high point of the trajectory when v1t2 = 300 – 9.8t = 0, or at t ≈ 30.6 s (Figure 6.8). At this point, the velocity becomes negative, and the shell begins its descent to Earth.

Knowing the velocity function, you could now find the position function using the methods of Example 3.

Related Exercises 27–37

Net Change and Future Value Everything we have said about velocity, position, and displacement carries over to more general situations. Suppose you are interested in some quantity Q that changes over time; Q may represent the amount of water in a reservoir, the population of a cell culture, or the amount of a resource that is consumed or produced. If you are given the rate Q′ at which Q changes, then integration allows you to calculate either the net change in the quantity Q or the future value of Q.

We argue just as we did for velocity and position: Because Q1t2 is an antiderivative of Q′1t2, the Fundamental Theorem of Calculus tells us that

L b

a Q′1t2 dt = Q1b2 – Q1a2 = net change in Q over 3a, b4.

Geometrically, the net change in Q over the time interval 3a, b4 is the net area under the graph of Q′ over 3a, b4. We interpret the product Q′1t2 dt as a change in Q over a small increment of time. Integrating Q′1t2 accumulates, or adds up, these small changes over the interval 3a, b4. The result is the net change in Q between t = a and t = b. We see that accumulating the rate of change of a quantity over the interval gives the net change in that quantity over the interval.

Figure 6.8

t6020 40

!300

!100

!200

v

300

200

100

v (t) ” 300 ! 9.8t

High point of trajectory at t ! 30.6 s

0

g ! 9.8 m/s2 v(0) ! 300 m/s

s(0) ! 30 m

s

Figure 6.7

ee

➤ Note that the units in the integral are consistent. For example, if Q′ has units of gallons>second, and t and x have units of seconds, then Q′1×2 dx has units of 1gallons>second21seconds2 = gallons, which are the units of Q.

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6.1 Velocity and Net Change 405

Alternatively, suppose we are given both the rate of change Q′ and the initial value Q102. Integrating over the interval 30, t4, where t Ú 0, we have

L t

0 Q′1×2 dx = Q1t2 – Q102.

Rearranging this equation, we write the value of Q at any future time t Ú 0 as

Q1t2 = Q102 + L t0 Q′1×2 dx. e e e

future initial

net change

value value over 30, t4