Math 37 - Chapter 3

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1

1

CHAPTER 3

APPLICATIONS

OF THE DEFINITE

INTEGRAL

2

OUTLINE3.1 Area of a plane region

3.2 Volume of a solid of revolution3.3 Center of mass of a rod

3.4 Centroid of a plane region and a

Theorem of Pappus

3.5 Centroid of a solid of revolution

3.6 Length of an arc of a curve

3.7 Area of a surface of revolution

3.8 Work

3.9 Force due to Liquid Pressure

33

CHAPTER OBJECTIVES

At the end of the chapter, you should be able to use

the definite integral to find/compute for

1. the area of a plane region using vertical or horizontal

strips,

2. the volume of a solid of revolution using the

cylindrical or shell method,

3. the center of mass of a rod, plane region and of a solid

of revolution,

4. the length of an arc of a plane curve,

5. the area of a surface of revolution and

6. work and force due to liquid pressure. 4

3.1 Area of a plane region

Let f be a function of x which iscontinuous and non-negativeon the closedinterval [a, b].

Subdivide the closed interval [a, b] into

nsub-intervals by choosing (n- 1)

intermediate numbers

121 nx,...,x,x

.... bxxxxxa nn 1210

where

5

For example, if we want to subdivide

[a,b] into 2 sub-intervals, we only choose one

number x1between aand b.

a b| |

x1

|

If we want to subdivide [a,b] into 3 sub-

intervals, we choose two numbers x1 and x2

between aand b.

a b| || |x1 x2

66

x

y

y= f(x)

ba| |x1|

xi-1|

xi|

Denote the ith sub-

interval by Iiso that

101 x,xI

212 x,xI

323 x,xI

iii x,xI

1

nnn x,xI

1

For each ,

choose a number

n,...,,i 21

.Iii

xn-1|

1 i n

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7

Denote the length of the ith sub-interval

by

so that

xi

011 xxx

122 xxx

233 xxx

1

iii xxx

1

nnn xxx

88

Our objective is to

find a formula for

the area of the

region R enclosed

by the graph of f,

the x-axis and the

lines given by x = a

and x = b.x

y

y= f(x)

ba| |

R

9

x

y

y= f(x)

ba| |x1|

xi-1|

xi|

xn-1|

1 i n

1

f

if

nf On the ith sub-

interval, construct

a rectangle of length

and whose width is

.xi

if

10

The area of the ith rectangular region is

.xiif

Thus, the sum of areas of all nrectangular

regions is

xf 11

xf 22

...xf... ii

xf nn

or

.xfii

n

i

1

11

x

y

y= f(x)

ba| |

RThe area of the region

Ris given by

.xfl im ii

n

in

1

12

.dxxfxfl imb

aii

n

in

1

By definition,

Thus, the area of the region Ris given

by

.dxxfRAb

a

length of a

rectangularelement

width of a

rectangular

element

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13

Example 3.1.1 Find the area of the region

enclosed by the graphs of

.yx,x,xy 0and212

RA2

1

3

3

x

3

1

3

23

3

7

3

1

3

8

Thus, the area of the

region is 7/3 sq. units.2x1x

2xy

Solution:

dx 2x1

2

14

Example 3.1.2 Find the area of the region

enclosed by the graphs of

.yex,xl ny 0and

exxl nx 1

Thus, the area of the region Ris 1 sq. unit.

111 l neel ne1 0

11ee0y

xlnyex

Solution:

1x

RA dx xln1

e

15

Let g be a function of y which is

continuous and non-negative on the

closed interval [c, d].

Subdivide the closed interval [c, d]into


intermediate numbers

121 nyyy ,...,,

.... dyyyyyc nn 1210

where

1616

x

y

x= g(y)i

The area of the ith

rectangular region is

.yiig

The area of the

regionRis given

by

.ygl im ii

n

in

1

.dyygRAd

c

y= d

y= c

For each ,

choose a number

n,...,,i 21

.Iii

igx

1717

Example 3.1.3 Find the area of the region in

the first quadrant enclosed by the graphs of

.xy,y,xy 0and412

4

1

23

23

/

y /

Thus, the area of the region is 14/3 sq. units.

2xy

yx

41

23

3

2 /y

232314

3

2 //

183

2

3

14

0x

1y

4y

Solution:

RA dy y4

1

1818

Example 3.1.4 Find the area of the region in

the first quadrant enclosed by the graphs of

.ln 2 yxy and

20ye

Thus, the area of the region is (e2 - 1) sq. units.

xy ln 02 ee 2y

Solution:

RA dy ye2

0

yex

12 e

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1919

Theorem 3.1.1Let fand g be functions of

x which are continuous on [a, b] and

f(x) g(x)

for all xin [a, b].

Then the area of the region bounded by

the graph of the graphs of fand gand

the lines given by x = aand x= bis given

by

. b

a dxxgxfA

2020

xgy

xfy

ax bx

xgxf

b

adxxgxfA

21

Example 3.1.5 Set-up the definite integral that

gives the area of the region R bounded by the

graphs of and the y-axis.93 y,y x

Solution:

(2,9)f(x) = 9

g(x) = 3x

dxxgxfRA ba

dx 9(2

0)x

3

22

Solution:


gives the area of the region R bounded by the

graphs of and .12 xy 42 xy

4212 xx

0322 xx

013 xx

13 xx or

POI: (3,10) and (-1,2)

23

dxxgxfRA ba

dx 42 x3

-112 x

3-1

12 xxg

42 xxf

2424

Theorem 3.1.2Let fand g be functions of

y which are continuous on [c, d] and

f(y) g(y)

for all yin [c, d].

Then the area of the region bounded by

the graph of the graphs of fand gand

the lines given by y = cand y= dis given

by

.dyygyfA d

c

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2525

dy

cy

.dyygyfAd

c

ygyf

yfx ygx

26

Solution:


gives the area of the region bounded by the

graphs of and y= x3.12

xy

312 yy

022 yy

012 yy

12 yy or

POI: (5,2) and (2,-1)

27

y2= x - 1

y=x - 3

(2,-1)

(5,2)

x=y + 3

x = y2+1

dyygyfAd

c

dyyy 13 22

1

.dyyy 2

1-

22

28

3.2 Volume of a Solid of Revolution

Soli d of Revolution

A soli d of revoluti on is obtained when

a region in the xy-plane is rotated about a

fixed line in the plane.

The fixed line is called the axis of

symmetry or axis of revolution of the

resulting solid.

We shall only consider vertical or

horizontal line as axis of revolution.

29

Rotate a tri angular regionabout one of

its legs to obtain a conical soli d.

30

Rotate a rectangular regionabout one of its

sides to obtain a ri ght-cir cular cyli ndrical

solid.

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31

Rotate a parabolic regionabout its axis of

symmetry to obtain a parabolic cyli ndri cal

solid.32

33

Methods of finding the volume of a solid of

revolution

1. Circular Disk Method

- a rectangular element isperpendicular to the axis ofrevolution.

2. Cylindrical Shell

- a rectangular element is parallelto the axis of revolution.

3434

3.2.1 Circular Disk Method

We derive the formula for finding

the volume of a solid formed when a

plane region is revolved about the x-axis

using the circular disk method.

The method may be applied even

when the axis of revolution is no longer

the x-axis but with a slight modification.

35

We recall that

the volume of a solid

enclosed by a right

circular cylinder of

radius rand altitude

his

V=r2h.

r

h

36

Let the function f be continuous on

the closed interval [a,b]and assume that

f(x) 0 for all xin [a,b].

Let Rbe the

region bounded by

the graph of y = f(x),

the x-axisand the

lines given by x= a

and x= b.

xfy

ax bx

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3737

Suppose Ris revolved around the x-axis.

38

xfy

ax bx

38

i

When the ith rectangular

region is revolved about

the x-axis, a circular disk

is formed.Its volume is

Vi=r2h = xf ii

2

The sum of volumes of

the ndisks is

i

n

i

V1

.xfii

n

i

2

1

3939

Thus, the volume of the solid formed

when the region Ris revolved about

the x-axis is

xfl imV iin

in

2

1

or equivalently,

.dxxfV ba 2

40

Example 3.2.1

xxf

Consider the region bounded by the graphs of

, and the x-axis.xy 4x

If the region is

revolved about the x-

axis, a solid isformed.

Using the Disk

Method, the radius of

the disk is given by

xxf

Solution:

4141

Using the Disk

Method, the

volume of the

solid generated

is

dxxfVb

a 2

dxxV24

0

xdx4

0

4

0

2

2 x 8

242

cu. units

42

Example 3.2.2


in the first quadrant.

2xy 4yand

Set up the definite integral

which gives the volume of the

solid formed when the given


1. x= 0 2. x= 2 3. x= -1

4. y= 0 5. y= 4 6. y= -1

2xy

(2,4)

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43

2xy

1. x= 0

yxr

V

(2,4)

dyh

2y dy0

4

44

2xy

2. x= 2

yx

21 VVV

dyh

2r

dyh

yr 2

1V 2

2 dy0

4

xr 2

2V 2

2 y dy0

4

45

3. x= -1

2xy

yx

x= -1 x= -1 x= -1

dyh

xr 1

yr 1

dyh

1r

21 VVV

1V 2

1 y dy0

4

2V 2

1 dy0

4

46

2xy

4. y= 0

21 VVV

dxh

4r

yr

dxh

2xr

1V 2

4 dx0

2

2V 22

x0

2dx

4747

2xy

5. y= 4

yr 4

dxh

V dx0

2

24 x

224 x

48

2xy

6. y= -1

21 VVV

dxh

5r1 yr

dxh

12

xr

1V

25 dx

0

2

2

V

221x

0

2dx

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4949

3.2.2 Cylindrical Shell Method

A cylindrical shell is asolid contained between

two cylinders having the

same altitude and axis.

The Cylindrical Shell

method involves taking

rectangular elements

which are parallel to the

axis of revolution.

r1

r2

h

hrV 2

hrhrV2

1

2

2

hrrV 21

2

2

5050

Theorem 3.2.2Let the

function f be continuous

on the closed interval [a, b]

and assume that f(x) 0

for all xin [a, b].

Let Rbe the region

bounded by the graph of

y= f(x), the x-axis and the

lines given by x= aand

x= b.

Suppose Ris revolved

around the y-axis.

xfy

ax bx

5151

xfy

ax bxi

When the ith rectangular


the y-axis, a circular shell

is formed.

Its volume is

hrrVi

2

1

2

2

iii f-xx 2

1

2

iiiii fxx- xx 11

iii fx 2Let .

2

1 iiixx

xf iii 2

5252

The sum of volumes of the nshells is

.xfxf iiin

i

iii

n

i

2211

Thus, the volume of the solid formed

when the region Ris revolved about the y-

axis is

or equivalently,

xflimV iiin

in

21

.dxxxfVb

a

2

Distance from the

axis of revolution

Length of a

rectangular element

Width of a

rectangular element

5353

Example 3.2.3

2xy


, and they-axis in the first quadrant.2xy 4y

Using Shell Method, set up thedefinite integral which gives thevolume of the solid formedwhen the given region isrevolved about

1. x= 0 4. y= 0

2. x= -1 5. y= 53. x= 3 6. y = -2

yx

Solution:

54

24 xh

1. x= 0

xd

.dxxxV 2

0

242

x= 02xy

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55

2. x= -1

24 xh

1xd

.dxxxV 2

0

2412

x= -12

xy

5656

3. x= 3

24 xh

xd 3

.dxxxV 2

0

2432

x= 3

2

xy

57

yh

4. y= 0

yd

.dyyyV 4

02

y= 0

2xy

5858

5. y= 5

yd 5

y= 5

yh

.dyyyV 4

052

2xy

59

6. y= -2

2yd

y= -2

yh

.dyyyV 4

022

2xy

60

3.3 Center of mass of a rod

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0

Consider a horizontal rod, of

negligible weight and thickness, placedon the x-axis.

On the rod is a system of nparticles

located at points x1, x2, , xn.

m1

m2 mi mnx1 x2 xi xn

62

The total mass of the systemis

M= m1

+ m2

+ + mn

.mn

i i

1

The moment of mass of the ith

particle with respect to the originis

The total moment of mass of the

system with respect to the originis

.xmn

i

ii

1

.xmii

nnxm...xmxmM 22110

63

The center of mass of the systemis

at

.

m

xm

n

i

i

n

i

ii

1

1

.xmxmn

i

ii

n

i

i

11

M

Mx 0

64

Example 3.3.1 Find the center of mass of the

system of 5 particles with masses 3,4,2,1 and 6

kgms located at -2,-1,0,3 and 4 respectively.

Solution:

5

1

5

1

i

i

i

ii

m

xm

x 61243

4333031323

.4

3

16

12

The center of mass of the system is 0.75 to the right

of the origin.

65

A rod is said to be homogeneous if the

mass of a part of it is proportional to the length

of that part.

im li

ii klm

Linear density

The center of mass of a homogeneous rod is at

its center.66

Consider a non-homogeneousrod, that is

the linear density varies along the rod. Let L

meters be the length of the rod and place the

rod on the x-axisso that an endpoint of the rod

coincides with the origin and the other

endpoint is at L .

0 L

Suppose the linear density at point x is(x) where is continuous on [0, L ].

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Consider a partition of [0, L ] into nsub-

intervals and let the ith subinterval be

.x,xI iii 1

Let i be a number in Ii .

The length of the ith subinterval is

.xmiii

.xi

An approximation to the mass of the part of the

rod in the ith subinterval is

68

0 Li

x1ix

i

.xmiii

An approximation to the mass of the rod is

i

n

i

m1

xii

n

i

1

The mass of the rod is

xl imMii

n

in

1

dxxL

0

69

An approximation to the moment

of mass of the part of the rod in the ith

subinterval with respect to the originis

iixm

iii x

An approximation to the momentof mass of the rod with respect to the

originis

ii

n

i

xm1

.xiii

n

i

1

.xiii

70

xl imMiii

n

in

1

0 .dxxxL

0

The moment of mass of the rod

with respect to the originis

The center of mass of the rod is at

.

dxx

dxxx

L

L

0

0

M

Mx 0

71

Example 3.3.2 Find the center of mass of a rod that

is 20 cm. long and the linear density on the rod at a

point xcentimeters from one end is (3x+ 2)

gm/cm.

Solution:

dxx

dxxxx

L

L

0

0

dxx

dxxx

23

23

2

2

0

0

0

0

dxx

dxxx

23

23

2

22

0

0

0

0

20

0

20

0

xx

xx

2

2

3 2

23

20220

2

3

2020

2

23

20220

2

3

2020202

32

420

8

105

72

Example 3.3.3 A rod is 12 cm. long and the linear

density at a point on the rod is a linear function of

the distance of the point from the left endpoint of

the rod. If the linear density at the left end is 3

gm/cm and at the right end is 4 gm/cm, find the

center of mass of the rod.

Solution:

Let xbe the distance from the left endpoint

of a point on the rod and (x)be the linear

density at x. Then for some constants aand b,

.baxx

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Since (0) = 3 and (12) = 4, and

(x) =ax+ b,

30 3 b

3 axx

412 4312 a

12

1a

312

1 xx

74

dxx

dxxxx

L

L

0

0

dxx

dxxx

312

1

312

1

1

1

2

0

2

0

dxx

dxxx

312

1

312

1

1

21

2

0

2

0

12

0

12

0

xx

xx

324

1

2

3

36

1

2

23

75

1231224

1

122

312

36

1

2

23

1231224

1

122

312

36

1 2

3

2

1

1223

31

2

7

184

2

7

22

7

44

76

3.4 Centroid of a Plane Region

Consider a system of nparticles located at

the points (x1, y1), (x2,y2) , (xn,yn) in the xy

plane and let their masses be m1, m2, .., mn,

respectively. Imagine the particles as supported

by a sheet of negligible weight and thickness.

(x1, y1) (x2, y2)

(x3, y3)

(x4, y4)

(x5, y5)

77

Let the ith particle, located at (xi,yi) have

mass mi.

The total mass of the system is

.mMn

i

i

1

The moment of mass of the system with

respect to the x-axisis

.xmMn

i

iiy

1

.ymMn

i

iix

1

The moment of mass of the system with

respect to the y-axisis

78

The center of mass of the system is at

y,xwhere

andM

xm

M

Mx

n

i

iiy

1

.M

ym

M

M

y

n

i

ii

x

1

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system of 4 particles with masses 4,2,3,6 and which

are located at the points (-2,-3), (1,2), (3,-2) and

(4,-3).

Solution:

4

1i

imM 156324

4

1i

iix ymM 36232234 4

4

1i

iiy xmM 46331224 27

80

The center of mass of the system is

at

y,x

where

.M

My x

15

4

and5

9

15

27

M

Mx y

81

A lamina is said to be homogeneousif

the mass of a portion of it is proportional

to the area of the part. Thus, if M is the

mass of the region R and A is its area,

then there is a constant ksuch that

M= kA .

Area density

82

So if we have a homogeneous

rectangular region, its mass Mi is

given by

Mi=k(lw).

Also, its centr oid or center of mass

is at its center.

83

Let Rbe the region

bounded above and

below, respectively by

the graphs of

and the lines given by

xgy,xfy

ax .bxand

xgy

ax bx

xfy

84

The area of the region is given by

.dxxgxfAb

a

Its mass is given by

.dxxgxfkMb

a

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85

b,a

xgy

ax bx

xfy

i

As we did in the

first 2 sections,

we subdivide

[a, b] into n

subintervals and

we choose a

number i in

each subinterval.

This time, we choose ias the abscissa

of the midpoint of each sub-interval.8686

Then we construct a

rectangle whose width

is and whose

length is .

xi

ii gf

The area of the ith


.xgfiii

Thus, the mass of the

ith rectangular region

is

. xgfk iii

b,a

xgy

ax bx

xfy

i

8787

b,a

xgy

ax bx

xfy

The centroid of the ith


at

.

2

ii gf

i,

2

ii gf

i,

The moment of mass

mi,ywith respect to the

yaxis of the ith


the product of its mass

and the distance of its

centroid from the y-axis. Thus,

iy,im xgfk iii

.xgfk iiii

8888

b,a

xgy

ax bx

xfy

2

ii gf

i,

The moment of mass

mi,xwith respect to the

x-axis of the ith


the product of its

mass and the distance

of its centroid fromthe x-axis. Thus,

2

ii gf

x,im xgfk iii

.xgfk iii 2

22

89

The total moment of massMi,ywith respect to

the yaxisof the nrectangular regions is given

by

y,i

n

i

y,i mM

1

.xgfk iiiin

i

1

The moment of massMywith respect to the y

axisof region Ris given by

xgfklim iiiin

in

1

y,i

n

in

y,in

y mlimMlimM

1

xgflimk iiiin

in

1

dxxgxfxkb

a

90

.xgfk iiin

i

2

22

1

The total moment of massMi,xwith respect to

the xaxisof the nrectangular regions is given

by

x,i

n

i

x,i mM

1

The moment of massMxwith respect to the x

axisof region Ris given by

xgfklim iiin

in

2

22

1

x,i

n

in

x,in

x mlimMlimM

1

dxxgxfk ba

22

2

xgflimk

iii

n

in 2

22

1

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91

If is the centroid of the plane region R

whose mass is M, then

y,x

M

Mx y M

My xand .Since

,dxxgxfkMb

a

,dxxgxfxkMb

ay

dxxgxfkM ba

x

22

2

92

M

My x

dxxgxfk

dxxgxfk

b

a

b

a

22

2

dxxgxf

dxxgxf

b

a

b

a

2

22

M

M

x

y

dxxgxfk

dxxgxfxk

b

a

b

a

dxxgxf

dxxgxfx

b

a

b

a

9393

Summary:

xgy

ax bx

xfy

dxxgxf

dxxgxfxx

b

a

b

a

dxxgxf

dxxgxfy

b

a

b

a

2

22

y,x

94

xxf

Example 3.4.2

Find the centroid of the region bounded by the

graphs of , and thex-axis.xy 4x

4x 0xg

Solution:

95

dxxgxf

dxxgxfxx

b

a

b

a

dxx

dxxx

0

0

4

0

4

0

dxx

dxx

/

/

214

0

234

0

4

0

23

4

0

25

23

25

/

x

/

x

/

/

4

0

23

4

0

25

3

2

5

2

/

/

x

x

3

42

5

42

23

25

/

/

3

165

64

.5

12

16

3

5

64

9696

dxx

dxx

02

0

4

0

224

0

dxx

xdx

/214

0

4

0

2

3

162

2

4

0

2

x

dxxgxf

dxxgxfy

b

a

b

a

2

22

3

162

2

42

3

32

8 .

4

3

32

38

8/11/2019 Math 37 - Chapter 3

17/2817

9797

xxf

(2.4, 0.75)

The centroid of the region bounded by the

graphs of , and thex-axis is atxy 4x

(2.4, 0.75).

98

24 xxf

Solution:

Example 3.4.3


graphs of and .2

4 xy 2xy

2xxg

022 xx

242 xx

012 xx

.xx 1or2

(-2,0)

(1,3)

9999

dxxgxf

dxxgxfxx

b

a

b

a

dxxx

dxxxx

24

24

21

2

21

2

dxxx

dxxxx

21

2

21

2

2

2

dxxx

dxxxx

21

2

231

2

2

2

1

2

23

1

2

342

232

34

xx

x

xxx

2

2

3

222

2

1

3

12

3

2

4

22

3

1

4

11

23

342

100

23

84

2

1

3

12

3

842

3

1

4

11

3

8

62

1

3

1

2

3

82

3

1

4

11

18

1

9

2

4

1

2

94

1

2

15

4

1

101

dxxx

dxxx

242

24

21

2

2221

2

dxxgxf

dxxgxfy

b

a

b

a

2

22

dxxx

dxxxxx

21

2

2421

2

22

44816

dxxx

dxxxx

21

2

421

2

22

4912

102102

dxxx

dxxxx

21

2

421

2

22

4912

1

2

23

1

2

25

3

2322

25

312

xx

x

xx

xx

2

2

3

222

2

1

3

122

225

2232122

5

1312

23

25

3

23

84

2

1

3

122

85

322424

5

17

425

12

9

5

108

2

152

5

3315

.

8/11/2019 Math 37 - Chapter 3

18/2818

103103

(-1/18, 2.4).

The centroid of the region bounded by the

graphs of and is at24 xy 2xy

24 xxf

2xxg

104104

ygxcy

dy

yfx

dyygyf

dxygyfx

d

c

d

c

2

22

dyygyf

dyygyfyy

d

c

d

c

y,x

105

12 xy

Solution:


graphs of and .

Example 3.4.4

12 xy 5x

5x

(5,-2)

(5,2)

1

2

y

ygx 5yf

106106

dyygyf

dxygyfx

d

c

d

c

2

22 dyy

dxy

152

15

22

2

2222

2

dyygyf

dyygyfyy

c

c

c

c

dyy

dxyy

15

15

22

2

22

2

The centroid of the region

is at where,y,x

107

Theorem of Pappus

When a plane region R is revolved

about a line which does not intersect the

region except possibly at its boundaries,

the volume of the solid formed is equal to

the area of the region times the distance

traveled by the centroidof the region.

108

y,x

Let Abe the area of the

region.

d

The distance traveled by the

centroid of the region when

the region is revolved about

the given line is 2d.

The volume of the solid of

revolution is A*(2d).

8/11/2019 Math 37 - Chapter 3

19/2819

109

Use the Theorem of Pappusto find the volume of

the solid formed when the region bounded by the

graphs of , and thex-axis

is revolved about the indicated line.

a. y= 0 b. y= 6 c. x= 0 d. x= -2

Example 3.4.5

xy 4x

110

From Example 3.4.2,

the centroid of the

region is at (2.4, 0.75).

Solution:

xxf

4x

0xgAlso from Example

3.4.2, the area of the

region is 16/3 sq. units.

111

(2.4, 0.75) 750.yd

3

16V 7502 .

4

32

3

16

.unitscubic8

a. Abouty= 0.

112112

b. Abouty= 6.

4

216

4

3 yd

3

16V

4

212

.unitscubic56

(2.4, 0.75)

6y

113

(2.4, 0.75)

c. Aboutx= 0.

5

1242 .xd

3

16V

5

122

.unitscubic5

128

114114

d. Aboutx= -2.

5

222

5

122 xd

3

16V

5

222

.unitscubic15

704

(2.4, 0.75)

2x

8/11/2019 Math 37 - Chapter 3

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115

3.5 Center of Mass of a Solid of

Revolution

In this section, we only consider

finding the center of mass of a solid of

revolution formed when a region whose

boundary includes the x-axis is revolved

about the x-axis or when a region whose

boundary includes the y-axis is revolved

about the y-axis.

116116

Remarks:

In a solid of revolution, the center of mass

lies along the axis of revolution, that is,

i. if the solid of revolutionSis generated by

revolving a region Rabout the x-axis, then

the centroid of Sis located at . 00,,x

ii. if Sis generated by revolving a region R

about the y-axis, then the centroid of Sis

located at . 00 ,y,

117117

x

y

z

x

y

z

x

y

00,,x xfy

b

a

b

a

dxxf

dxxfxx

2

2

118118

x

y

00 ,y, ygx

d

c

d

c

dyyg

dyygyy

2

2

x

y

z x

y

z

119


solid formed when the region bounded by the

graphs of

and thex-axis in the first quadrant is revolved

about thex-axis.

xl ny ex,

120

0y

xlnyex

Solution:

The centroid of the

solid is at

where,

00,,x

e

e

dxxln

dxxlnxx

1

2

1

2

8/11/2019 Math 37 - Chapter 3

21/2821

121

dxxlnx2

udv

vduuv

dxxxlnxx

xln1

222

222

xdxlnx

xlnx

2

22

xdxdv,xlnu 2

2

12

2xv,dx

xxlndu

122

xdxdv,xlnu

2

12x

v,dxxdu

xdxlnx

udv vduuv

dxxxx

xln1

22

22

xdxxlnx

2

1

2

2

Cxxlnx

22

1

2

22

Cxxlnx

42

22

123

dxxlnx2

xdxlnxxlnx

2

22

xdxlnx Cxxlnx

42

22

Cxxlnxxlnx

dxxlnx

4222222

2

Cxxlnxxlnx

dxxlnx 422

22222

124

e

e xxlnxxlnxdxxlnx

1

2222

1

2

422

4

1

2

1

2

1

422

22222 lnlneelneelne

0 01 1

4

1

4

1

4

22

ee

125

dxxln2

udv dxdv,xlnu 2

xv,dxx

xlndu 1

2 vduuv

dxxxlnxxxln

12

2

xdxlnxlnx 22

Cxxlnxxlnx 22

Cxxl nxxl nx 222

126126

ee xxl nxxl nxxl n 1221

22

212122 2221

l nl neel neel nexl ne

0 0

222221

eeeexl ne

1 1

The center of mass of the solid is at where, 00,,x

e

e

dxxln

dxxlnx

x

1

21

2

241

24

12

2

e

e

e

e

242.

8/11/2019 Math 37 - Chapter 3

22/2822

127

Example 3.5.2 Find the center of mass of

the solid formed when the region boundedby the graphs of

,

and they-axis in the first quadrant is

revolved about they-axis.

2xy 4y

128

2xy

Solution:

The centroid of the region

is at where,00 ,y,

d

c

d

c

dxyg

dxygyy

2

2

4

0

2

4

0

2

dxy

dxyy

4

0

4

0

2

ydy

dyy

.3

8

8

3

64

4

0

2

4

0

3

2

3

y

y

129

3.6 Length of Arc of a Curve

Let the function f and its derivative f

be continuous on the closed interval [a, b].

af,a

xfy

bf,b

130

af,a

xfy

bf,b

Subdivide the closed interval [a, b] into


intermediate numbers .121 nx,...,x,x

x1

|

xi-1

|

xi

|

xn-1

|

x0

|

xn

|

131

The distance between (xi-1, f(xi-1)) and

(xi, f(xi)) is

21

2

1 iiiii xxxfxfd

221 xxfxf iii

22

2

1 1 xx

xfxfi

i

ii

2

2

1 1 xx

xfxfi

i

ii

132

x

x

xfxfd

i

i

ii

i

2

11

By the Mean-Value Theorem, there exists

an iin [xi-1, xi] such that

1

1

ii

ii

ixx

xfxff

x

xfxf

fi

ii

i

1

8/11/2019 Math 37 - Chapter 3

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133

xfdiii

2

1

An approximation to the length of the curve

from (a, f(a)) to (b, f(b)) is

.xfdii

n

i

i

n

i

2

11

1

134

b

adxx'fL

21

The length of the curve from (a, f(a)) to

(b, f(b)) is

.xfl imdl imLii

n

in

i

n

in

2

11

1

135

Example 3.6.1 Find the length of the curve

given by from the point where to the

point where

xey 1x.x 2

Solution:

b

a dxx'fL2

1

2

1

2

1 dxex

2

1

21 dxe x

xx ex'fexf

136

dxe x2

1

dxy

ydy

12

xey2

1

xey22

1

dxeydy x222

dxeydy x2

dxyydy 12

y1

2 y

ydy

12

2

y

dyy

dyy 1

11

2

dy

yy 12

1

12

11

Cyl nyl ny 12

11

2

1

137

Cy

yl nydxe

x

1

1

2

11

2

Ce

el nydxe

x

x

x

11

11

2

11

2

22

2

1

21 dxe x

2

1

2

2

11

11

2

1

x

x

e

el ny

138

11

11

2

11

11

11

2

12

2

2

4

4

e

el n

e

el n

11

11

2

1

11

11

2

11

2

2

4

4

e

el n

e

el n

11

11

11

11

2

11

2

2

4

4

e

e

e

el n

8/11/2019 Math 37 - Chapter 3

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139

Solution:

2

2

44

x

xx'fxxf


given by on the interval24 xy .,22

b

adxx'fL

21

2

2

2

24

1 dxx

x

2

2 2

2

41 dxx

x

140

2

2 24

12 dx

x

2

0 2

0

2 24

124

12 dxx

dxx

(Improper of type DB and DA)

b

baa

dxx

l imdxx

l im0 22

0

22 4

12

4

12

bxb

a

x

a

si nArcl imsinArcl im

022

0

22

22

141

2

2

02 aa

sinArcsi nArcl im

022

2

sinArcsinArcl im bb

1-1

x

y

2

2

0

0

2

2

2 aa

sinArcl im

2

2

2 bb

si nArcl im

2

2 2

2

2

142

Let the function g and its derivative gbe

continuous on the closed interval . Then the

length of the curve from the point

to the point is given by

d,c ygx c,cg

d,dg

d

cdyy'gL

21

143


given by from the point where to the

point where

1xy 1y

.y 2

Solution:

2

111

yy'g

yygxxy

d

cdyy'gL

21

2

1

2

2

11 dy

y

L

2

1

4

11 dy

y

The evaluation of the integral is left to you as an exercise.144

3.7 Area of a surface of

revolution

When a region

is revolved about a

line, a solid is

formed.

When an arc is

revolved about a

line, a surface is

formed.

8/11/2019 Math 37 - Chapter 3

25/2825

145

.rrlSA 21

l

1r

2r

From Euclidean Geometry, the surface

area of a frustum of a cone is

The frustum is generated by revolving

the red segment about the blue one.

146

Let the function f be continuous on

the closed interval [a, b], differentiable on

the open interval (a, b) and f(x) 0 foreach xin [a, b].

Let Cbe the arc of the graph of fon

[a, b]. We wish to find the area of the

surface formed when Cis revolved about

the x-axis.

147

x

y

y=f(x)

ba| | x

y

ba| |

x1|

xi-1|

xi|

xn-1|

148

The length of the ith segment on the curve is

xfdiii

21

11With ixfr and, 2 ixfr

,xfl ii 2

1

The area of the ith frustum is

iiiii

xfxfxfSA 12

1

21rrlSA

149

,xfxfiii 21

2

xff iii 2

12

2

1 ii

i

xfxfxf

xfxf iii 212

The area of the surface of revolution is

approximately equal to

.xff iiin

i

2

1

12 150

b

adxx'fxfSA

212

Distance of a point of the

curve from the axis of

revolution

Element of arclength

xffl imSA iiin

in

2

1

12

The area of the surface of revolution is equal to

b

adsldSA 2

8/11/2019 Math 37 - Chapter 3

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151

Example 3.7.a Find the area of the surface

formed when the arc of from the point

where to the point where is

revolved about the given indicated line.

xey

1x 2

x

a. y= 0 d. x= 0

b. y= 9 e. x= 5

c. y= -2 f. x= -1

Solution:

.dxeL x 2

1

21

From Example 3.6.1, the length of the

arc is given by

152

x

y

xey

.dxeL x 2

1

21

a. y= 0

b

adxx'fxfSA

212

2

1

212 dxee xx

153

x

y

xey

.dxeL x 2

1

21

b. y= 9

2

1

21?2 dxeSA x

2

12192 dxee xx

154

x

y

xey

.dxeL x 2

1

21

c. y= -2

2

1

21?2 dxeSA x

2

1

2122 dxee xx

155

x

y

xey

.dxeL x 2

1

21

d. x= 0

2

1

212 dxex x

2

1

21?2 dxeSA x

156

x

y

xey

.dxeL x 2

1

21

e. x= 5

2

1

21?2 dxeSA x

x = 5

2

1

2152 dxex x

8/11/2019 Math 37 - Chapter 3

27/2827

157157

x

y

xey

f. x= -1

.dxeL x 2

1

21

2

1

21?2 dxeSA x

x= -1

2

1

2112 dxex x

158

Example 3.7.b Find the area of the surface




xey1x 2x

a. y= 0 d. x= 0

b. y= 9 e. x= 5

c. y= -2 f. x= -1

Solution:

lnxy e y x

1dy dx

y

1dx

dy y

159

2

2

11

e

eL dy

y

The length of the arc is given by

2

2

11

e

eL dy

y

xey

,1when x ey

,2when x 2ey

160160

Example 3.7.c Find the area of the surface




1y

x1y 2y

a. x= 0 d. y= 0

b. x= 3 e. y= 3

c. x= -1 f. y= -1

Solution:1

xy

2

1dx

dy y

161161

b. x= 3

x

y

yx

1

a. x= 0

c. x= -1

2

1 4

11?2 dy

ySA

x= 3x= -1

2

1 4

1

1 dyyL

d

cdyy'gygSA

212

2

1 4

11

12 dy

yy

2

1 4

11

132 dy

yy

2

1 4

11?2 dy

ySA

2

1 4

11

112 dy

yy

162162

d. y= 3

x

y

yx

1

c. y= 0

e. y= -1

2

1 4

11?2 dy

ySA

y = 3

y= -1

2

1 4

1

1 dyyL

d

cdyy'gySA

212

2

1 4

112 dy

yy

2

1 4

1132 dy

yy

2

1 4

11?2 dy

ySA

2

1 4

1112 dy

yy

8/11/2019 Math 37 - Chapter 3

28/28

163163

The End

of Chapter

Math 37 - Chapter 3

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