# Mathematics

74. Shifting sines Consider the functions f 1×2 = a sin 2x and g1x2 = 1sin x2>a, where a 7 0 is a real number. a. Graph the two functions on the interval 30, p>24, for a = 12, 1,

and 2. b. Show that the curves have an intersection point x* 1other than

x = 02 on 30, p>24 that satisfies cos x* = 1>12a22, provided a 7 1>12.

c. Find the area of the region between the two curves on 30, x*4 when a = 1.

d. Show that as a S 1>12 +. the area of the region between the two curves on 30, x*4 approaches zero.

QUICK CHECK ANSWERS 1. If g1x2 = 0 and f 1×2 Ú 0, then the area between the curves is 1ba 1 f 1×2 – 02 dx = 1ba f 1×2 dx, which is the area between y = f 1×2 and the x-axis. 2. 1ba f 1×2 dx is the area of the region between the graph of f and the x-axis. 1ba g1x2 dx is the area of the region between the graph of g and the x-axis. The difference of the two integrals is the area of the region between the graphs of f and g. 3. a. 120 1x dx + 142 11x – x + 22 dx b. 120 1y + 2 – y22 dy 4. The area of the triangle to the left of the y-axis is 18. The area of the region to the right of the y-axis is given by the integral.

T

T

T

that Area ∆PQR = 2 # Area ∆P′Q′R′ in the following cases. (In fact, the property holds for any three points on any parabola.) (Source: Mathematics Magazine 81, 2, Apr 2008)

y y ! x2!P

!R

!Q

P”

P

Q”

Q

R

R”

x

a. P1-a, a22, Q1a, a22, and R10, 02, where a is a positive real number

b. P1-a, a22, Q1b, b22, and R10, 02, where a and b are positive real numbers

c. P1-a, a22, Q1b, b22, and R is any point between P and Q on the curve

67. Minimum area Graph the curves y = 1x + 121x – 22 and y = ax + 1 for various values of a. For what value of a is the area of the region between the two curves a minimum?

68. An area function Graph the curves y = a2x3 and y = 1x for various values of a 7 0. Note how the area A1a2 between the curves varies with a. Find and graph the area function A1a2. For what value of a is A1a2 = 16?

69. Area of a curve defined implicitly Determine the area of the shaded region bounded by the curve x2 = y411 – y32 (see figure).

!1 x1

1

!1

y

x2 ” y4(1 ! y3)

70. Rewrite first Find the area of the region bounded by the curve

x = 1 2y

– A 14y2 – 1 and the line x = 1 in the first quadrant. (Hint: Express y in terms of x.)

T

T

6.3 Volume by Slicing We have seen that integration is used to compute the area of two-dimensional regions bounded by curves. Integrals are also used to find the volume of three-dimensional regions (or solids). Once again, the slice-and-sum method is the key to solving these problems.

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6.3 Volume by Slicing 421

General Slicing Method Consider a solid object that extends in the x-direction from x = a to x = b. Imagine cut- ting through the solid, perpendicular to the x-axis at a particular point x, and suppose the area of the cross section created by the cut is given by a known integrable function A (Figure 6.22).

To find the volume of this solid, we first divide 3a, b4 into n subintervals of length ∆x = 1b – a2>n. The endpoints of the subintervals are the grid points x0 = a, x1, x2, c, xn = b. We now make vertical cuts through the solid perpendicular to the x-axis at each grid point, which produces n slices of thickness ∆x. (Imagine cutting a loaf of bread to create n slices of equal width.) On each subinterval, an arbitrary point xk

* is identified. The kth slice through the solid has a thickness ∆x, and we take A1xk*2 as a representative cross-sectional area of the slice. Therefore, the volume of the kth slice is approximately A1xk*2∆x (Figure 6.23). Summing the volumes of the slices, the approximate volume of the solid is

V ≈ a n

k = 1 A1xk*2∆x.

As the number of slices increases 1n S ∞2 and the thickness of each slice goes to zero 1∆x S 02, the exact volume V is obtained in terms of a definite integral (Figure 6.24): V = lim

nS ∞ a n

k = 1 A1x*k2∆x = Lba A1x2 dx.

Figure 6.22

x

xb

Cross section with area A(x)

a

Figure 6.23

xb

a

!x

Cross-sectional area ” A (xk*)

Volume of kth slice ! A(xk*) !x

xk*

Figure 6.24

Increase the number of slices.

n ! !

Volume ” lim k”1

n

# A(xk*)$x

a

b

” ! A(x)dx n!!

We summarize the important general slicing method, which is also the basis of other vol- ume formulas to follow.

➤ The factors in this volume integral have meaning: A1x2 is the cross-sectional area of a slice and dx represents its thickness. Summing (integrating) the volumes of the slices A1x2 dx gives the volume of the solid.

General Slicing Method Suppose a solid object extends from x = a to x = b and the cross section of the solid perpendicular to the x-axis has an area given by a function A that is integrable on 3a, b4. The volume of the solid is

V = L b

a A1x2 dx.

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426 Chapter 6 Applications of Integration

Related Exercises 27–34

QUICK CHECK 5 Suppose the region in Example 4 is revolved about the line y = -1 in- stead of the x-axis. (a) What is the inner radius of a typical washer? (b) What is the outer radius of a typical washer?

Revolving about the y-Axis Everything you learned about revolving regions about the x-axis applies to revolving re- gions about the y-axis. Consider a region R bounded by the curve x = p1y2 on the right, the curve x = q1y2 on the left, and the horizontal lines y = c and y = d (Figure 6.34a).

To find the volume of the solid generated when R is revolved about the y-axis, we use the general slicing method—now with respect to the y-axis (Figure 6.34b). The area of a typical cross section is A1y2 = p1p1y22 – q1y222, where c … y … d. As before, inte- grating these cross-sectional areas of the solid gives the volume.

➤ Ignoring the factor of p, the integrand in the washer method integral is f 1×22 – g1x22, which is not equal to 1 f 1×2 – g1x222.

Figure 6.33

Interval of integration

g(x) ! x2

g(x)

Area of washer face ! !( f (x)2 ” g(x)2) ! !(x ” x4)

y

x

1 1

1x

R

f (x) ! x

f (x)

Revolving the region about the x-axis… … produces a bowl-

shaped solid. y

x

00 1

x

Figure 6.34

O

O

Interval of integration

q(y)

p(y)

y

x

y

x

d

y

c

d

cx ! p(y)

x ! q(y) x ! p(y)

x ! q(y)

y

Inner radius ! q(y)

Outer radius ! p(y)

(b)(a)

R

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6.3 Volume by Slicing 427

EXAMPLE 5 Which solid has greater volume? Let R be the region in the first quad- rant bounded by the graphs of x = y3 and x = 4y. Which is greater, the volume of the solid generated when R is revolved about the x-axis or the y-axis?

SOLUTION Solving y3 = 4y, or equivalently, y1y2 – 42 = 0, we find that the bounding curves of R intersect at the points 10, 02 and 18, 22. When the region R (Figure 6.35a) is revolved about the y-axis, it generates a funnel with a curved inner surface (Figure 6.35b). Washer-shaped cross sections perpendicular to the y-axis extend from y = 0 to y = 2. The outer radius of the cross section at the point y is determined by the line x = p1y2 = 4y. The inner radius of the cross section at the point y is determined by the curve x = q1y2 = y3. Applying the washer method, the volume of this solid is

V = L 2

0 p1p1y22 – q1y222 dy Washer method

= L 2

0 p116y2 – y62 dy Substitute for p and q.

= pa 16 3

y3 – y7

7 b ` 2

0 Fundamental Theorem

= 512p

21 . Evaluate.

➤ The disk>washer method about the y-axis is the disk>washer method about the x-axis with x replaced with y.

Disk and Washer Methods about the y-Axis Let p and q be continuous functions with p1y2 Ú q1y2 Ú 0 on 3c, d4. Let R be the region bounded by x = p1y2, x = q1y2, and the lines y = c and y = d. When R is revolved about the y-axis, the volume of the resulting solid of revolution is given by

V = L d

c p1p1y22 – q1y222 dy.

If q1y2 = 0, the disk method results: V = L

d

c pp1y22 dy.

Figure 6.35 (a)

2

8

Interval of integration

0

y

x

y

R

8

Inner radius ! y3

Outer radius ! 4y

q(y) ! y3

(8, 2)

p(y) ! 4y

Area of washer face ! !(p(y)2 ” q(y)2) ! !(16y2 ” y6)

x

y

0

(b)

y 2

When the region R is revolved about the x-axis, it generates a different funnel (Figure 6.36). Vertical slices through the solid between x = 0 and x = 8 produce wash- ers. The outer radius of the washer at the point x is determined by the curve x = y3, or

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428 Chapter 6 Applications of Integration

y = f 1×2 = x1>3. The inner radius is determined by x = 4y, or y = g1x2 = x>4. The volume of the resulting solid is

V = L 8

0 p1 f 1×22 – g1x222 dx Washer method

= L 8

0 pax2>3 – x2

16 b dx Substitute for f and g.

= pa 3 5

x5>3 – x3 48

b ` 8 0

Fundamental Theorem

= 128p

15 . Evaluate.

We see that revolving the region about the y-axis produces a solid of greater volume.

Figure 6.36

2

1

80

y

x

R

x

(8, 2)

Interval of integration

Inner radius ! x/4

Outer radius ! x1/3 f (x) ! x1/3

g(x) ! x 4

Area of washer face ! !(f (x)2 ” g(x)2)

! !(x2/3 ” x 2

16 )

x

y

2

1

x 8

0

Related Exercises 35–44

QUICK CHECK 6 The region in the first quadrant bounded by y = x and y = x3 is revolved about the y-axis. Give the integral for the volume of the solid that is generated.

The disk and washer methods may be generalized to handle situations in which a region R is revolved about a line parallel to one of the coordinate axes. The next example discusses three such cases.

EXAMPLE 6 Revolving about other lines Let f 1×2 = 1x + 1 and g1x2 = x2 + 1. a. Find the volume of the solid generated when the region R1 bounded by the graph of f

and the line y = 2 on the interval 30, 14 is revolved about the line y = 2. b. Find the volume of the solid generated when the region R2 bounded by the graphs of f

and g on the interval 30, 14 is revolved about the line y = -1. c. Find the volume of the solid generated when the region R2 bounded by the graphs of f

and g on the interval 30, 14 is revolved about the line x = 2. SOLUTION

a. Figure 6.37a shows the region R1 and the axis of revolution. Applying the disk method, we see that a disk located at a point x has a radius of 2 – f 1×2 = 2 – 11x + 12 = 1 – 1x. Therefore, the volume of the solid generated when R1 is revolved about y = 2 is

L 1

0 p11 – 1×22 dx = pL10 11 – 21x + x2 dx = p6 .

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6.3 Volume by Slicing 429

3. The region bounded by the curves y = 2x and y = x2 is revolved about the x-axis. Give an integral for the volume of the solid that is generated.

4. The region bounded by the curves y = 2x and y = x2 is revolved about the y-axis. Give an integral for the volume of the solid that is generated.

b. When the graph of f is revolved about y = -1, it sweeps out a solid of revolution whose radius at a point x is f 1×2 + 1 = 1x + 2. Similarly, when the graph of g is revolved about y = -1, it sweeps out a solid of revolution whose radius at a point x is g1x2 + 1 = x2 + 2 (Figure 6.37b). Using the washer method, the volume of the solid generated when R2 is revolved about y = -1 is

L 1

0 p111x + 222 – 1×2 + 2222 dx

= pL 1

0 1-x4 – 4×2 + x + 41×2 dx

= 49p 30

.

c. When the region R2 is revolved about the line x = 2, we use the washer method and integrate in the y-direction. First note that the graph of f is described by y = 1x + 1, or equivalently, x = 1y – 122, for y Ú 1. Also, the graph of g is described by y = x2 + 1, or equivalently, x = 1y – 1 for y Ú 1 (Figure 6.37c). When the graph of f is revolved about the line x = 2, the radius of a typical disk at a point y is 2 – 1y – 122. Similarly, when the graph of g is revolved about x = 2, the radius of a typical disk at a point y is 2 – 1y – 1. Finally, observe that the extent of the region R2 in the y-direction is the interval 1 … y … 2.

Applying the washer method, simplifying the integrand, and integrating powers of y, the volume of the solid of revolution is

L 2

1 p112 – 1y – 12222 – 12 – 1y – 1222 dy = 31p

30 .

Figure 6.37

0

Radius: ! 2 ” ( x # 1) ! 1 ” x

Interval of integration

y

x

R1 2

1

1x

y ! x # 1

(a)

!1

y ” x2 # 1

y ” !1

Outer radius ” x # 2

y ” x # 1

Inner radius ” x2 # 2

y

x

2

1x

1

Interval of integration

(b)

0

R2 x ! (y ” 1)2

x ! 2

Interval of integration y

y

x

2

1

21

x ! y ” 1

Inner radius ! 2 ” y ” 1

Outer radius ! 2 ” (y ” 1)2

(c)

0

R2

Related Exercises 45–52

SECTION 6.3 EXERCISES Review Questions 1. Suppose a cut is made through a solid object perpendicular to the

x-axis at a particular point x. Explain the meaning of A1x2. 2. A solid has a circular base and cross sections perpendicular to the

base are squares. What method should be used to find the volume of the solid?

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430 Chapter 6 Applications of Integration

11. The solid with a semicircular base of radius 5 whose cross sec- tions perpendicular to the base and parallel to the diameter are squares

12. The solid whose base is the region bounded by y = x2 and the line y = 1, and whose cross sections perpendicular to the base and parallel to the x-axis are squares

square cross section

base

y ! x2

y

x