10.1Finding Plane Areas by Integration 10.2Volumes of Solids of Revolution Chapter Summary Case...

10.1 Finding Plane Areas by Integration

10.2 Volumes of Solids of Revolution

Chapter Summary

Case Study

Applications of Definite Integrals10

P. 2

To measure the inequality of income distribution in a society, we can make use of the Gini coefficient.

Case StudyCase Study

The ‘Lorenz Curve’ is sketched such that any point (x, y) on the curve indicates that the poorest x% families share y% of the total income of all the families in the society.

If the Gini coefficient is close to 1, then there is a very wide income gap between the rich and the poor; if it is close to 0, then the income is uniformly distributed among the families.

Yes, I know. How can we give a quantitative measure of the inequality of income distribution?

In some countries, there are very wide income gaps between the rich and the poor.

L

L

line under the area Total

line theand Curve Lorenz by the bounded Area t coefficien Gini

P. 3

In Chapter 9, we learnt that the definite integral of f (x) from a to b is defined as follows:

1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration

As shown in the figure, if f (x) 0 for a x b, then f (zi)x stands for thearea of a very thin rectangle drawn under the curve y f (x).

A. A. Area of the Region Bounded by a CurveArea of the Region Bounded by a Curve and the x-axisand the x-axis

n

abxxzfdxxf

n

iii

n

b

a

where,)(lim)(1

If f (x)0 in the interval a x b, then area

of the shaded region .

b

adxxfA )(

In this case, gives the area of the region bounded by the curve

y f (x), the x-axis, the lines x a and x b.

b

adxxf )(

P. 4

If f (x) takes both positive and negative values in the interval a x b, we can divide the shaded region into two parts: one part for f (x) 0 and the other for f (x) 0.



If f (x) 0 in the interval a x c and f (x) 0 in the interval c x b, then area of the shaded region A1 + A2

b

c

c

adxxfdxxf )()(

On the other hand, if f (x) 0 for a x b as shown in the figure, then f (zi)x is equal to the negative of the area of the thin rectangle.

If f (x) 0 in the interval a x b, then area

of the shaded region . .)(

b

adxxfA

P. 5

Example 10.1TFind the area of the region bounded by the curve y x2, the axes and the line x 3.



Solution:The required area

3

0

2dxx

3

0

3

3

x

9

P. 6

Example 10.2TFind the area of the region bounded by the curve y = x2 – 6x + 8, and the x-axis in the interval 0 x 4.



Solution:When x2 – 6x + 8 = 0,

4or 20)4)(2(

xxx

The required area

4

2

22

0

2 )86()86( dxxxdxxx

4

2

232

0

23

833

833

xx

xxx

x

3

4

3

20

8

P. 7

Once again, the area of the shaded region A can be approximated by a large number of thin, vertical rectangles of equal width ∆x. However, now the length of a rectangle is given by f (zi) – g(zi).


BB. . Area of the Region Bounded by y = f(x)Area of the Region Bounded by y = f(x) and y = g(x)and y = g(x)

In the figure, y = f (x) and y = g(x) are two continuous functions such that f (x) g(x) in the interval a x b.

n

iii

nxzgzf

1

)()(limregion shaded theof Area

where f (x) g(x) in the interval a x b.

,)()(region shaded theof Area dxxgxfAb

a

P. 8

Example 10.3TFind the area of the region bounded by the curves y = x2 and y = x3 .



Solution:Consider the points of intersection of the two curves.

)2(............ )1(............

3

2

xyxy

(2) – (1):

023 xx

1or 00)1(2

xxx

Hence the curves intersect when x = 0 and x = 1.

The required area 1

0

32 )( dxxx1

0

43

43

xx

4

1

3

1 12

1

P. 9

Example 10.4TFind the area of the region bounded by the curves y = x2, xy = 1 and the line y = 4x in the interval 0 x 1.



Solution:Solving y = x2 and xy = 1, the point of intersection is (1, 1). Solving xy = 1 and y = 4x, the point of intersection is .

2 ,

2

1

The required area

1

2

122

1

0

2 1)4( dxx

xdxxx

1

2

1

32

1

0

32

3ln

32

x

xx

x

24

72ln

24

11

6

12ln

P. 10

However, if y = f (x) and y = g(x) intersect in the interval as shown in the figure, we need to consider the shaded region on each side of the point of intersection independently.



In Example 10.3T, the function f (x) – g(x) does not change sign in the interval a x b, so the area of the region bounded by the two curves can

be simply calculated by using the formula . )()( dxxgxfb

a

P. 11

Example 10.5TFind the area of the region bounded by the curves and y = sin x in the interval 0 x



xy cos3Solution:

At the point of intersection, .cos3sin xx

The coordinates are .2

3 ,

3

On the left hand side of the intersection, is positive. xx sincos3 On the right hand side of the intersection, is negative. xx sincos3 The required area

3

30

)cos3(sin)sincos3( dxxxdxxx

3

30 ]sin3cos[]cossin3[ xxxx

)]2(1[)12( 4

3tan x

3

x

P. 12

Using similar arguments as we have studied in the previous section, we can find the area of the region bounded by the curve x = u(y), the y-axis and the lines y = c and y = d.


CC. . Area of the Region Bounded by a Curve Area of the Region Bounded by a Curve and the y-axisand the y-axis

Sometimes, the equation of a curve may be written in the form x = u(y), such as x = y2.

d

c

n

ii

ndyyuyzuA )()(limregion shaded theof Area

1

d

e

e

cdyyudyyuA )()(region shaded theof Area

d

c

n

ii

ndyyuyzuA )()(limregion shaded theof Area

1

P. 13

Example 10.6TFind the area of the region bounded by the curvey = x3 – 2, the y-axis and the line y = 6.



Solution:For y = x3

– 2,

3 2 yx

The required area dyy 6

23 2

6

2

3

4

)2(4

3

y

12

P. 14

Example 10.7TFind the area of the region bounded by the curve x = –y2 +2y + 3 and the y-axis in the interval –1 y 6.



Solution:The required area

6

3

23

1

2 )32()32( dyyydyyy

6

3

233

1

23

33

33

yyy

yyy

)27(3

32

3

113

P. 15

As shown in the figure, x = u(y) and x = v(y) are two continuous functions such that u(y) v(y) in the interval c y d.


DD. . Area of the Region Bounded by x = u(x) Area of the Region Bounded by x = u(x) and x = v(y)and x = v(y)

where u(y) v(y) in the interval c y d.

,)()(region shaded theof Area dyyvyuAd

c

P. 16

Example 10.8T

Solution:

In the figure, y 4x – 3 is a tangent to the curve y x4.(a) Find the point of contact.(b) Find the area of the region bounded by the curve y

x4, the line y 4x – 3 and the x-axis.


DD. . Area of the Region Bounded by x = u(x) Area of the Region Bounded by x = u(x) and x = v(y)and x = v(y)

(a) Slope of y = 4x – 3 is 4.

For y = x4, 34x

dx

dy

When 4x3 = 4,

113

xx

The point of contact is (1, 1).

(b) The required area

1

04

1

4

3dyy

y

1

0

4

52

5

4

4

3

8

yy

y

40

3

P. 17

Consider the region bounded by the curve y + 1, the x-axis, the y-axis and the line x = 4.

1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution

AA. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the x-axisx-axis

2

2x

If it is revolved about the x-axis, then it will generate a solid as shown in the figure.

We will learn how to find the volume V of the solid of revolution using integration.

In this case, we call it a solid of revolution and the x-axis the axis of revolution.

P. 18

Consider the curve of a non-negative function y = f (x) defined in the interval a x b.


The area is approximated by n thin rectangles of width

x = and length f (zi),

where i = 1, 2, ... , n. n

ab

If the first rectangle from the left is revolved about the x-axis to form a disc, the volume of the disc is .)(π 2

1 xzf If n is getting larger indefinitely, the width x tends to zero. Then the sum of the volumes of the discs will approach the actual volume of the solid of revolution.

dxxfVb

a 2)(π

This method of finding the volume of the solid of revolution by integration is called the disc method.


P. 19

As shown in the figure, the shaded region enclosed by the line y x (where r, h > 0) and x h is revolved about the x-axis.


h

r

We can also find the general formula for calculating the volume of a cylinder and a sphere using the disc method.


Since the equation of the function is y x witha = 0 and b = h,

the required volume

h

r

dxxh

rh

0

2

π hr2π3

1

P. 20

Example 10.9TFind the volume of the solid generated by revolving the region enclosed by the curve y , the x-axis and the line x 6 about the x-axis.


x

Solution:


The required volume

6

0

2)( dxx

6

0

2

2

x

18

P. 21

Example 10.10TFind the volume of the solid generated by revolving the region enclosed by the curve y2 x2(x + 1) about the x-axis.



Solution:When x2(x + 1) = 0,

x = –1 or 0

The required volume

0

1

2 )1( dxxx

0

1

23 )( dxxx0

1

34

34

xx

12

P. 22

Now, consider two continuous functions f (x) and g(x), where f (x) g(x) > 0 in the interval a x b.

If the region bounded by the curves y = f (x) and y = g(x) is revolved about the x-axis, then a hollow solid of revolution will be formed.


dxxgdxxfVb

a

b

a 22 )(π)(π

dxxgxfb

a )()( π 22


The volume enclosed by the outer surface of this hollow solid is given

by , while the volume of the empty space is equal to

.

dxxfb

a2)(π

dxxgb

a2)(π

P. 23

Example 10.11TFind the volume of the solid generated when the region bounded by the curves y x2 + 2 and y –x2 + 4 is revolved about the x-axis.


Solution:


)2(......... 4)1(............ 2

2

2

xyxy

Substituting (1) into (2),42 22 xx

1or 112

xx

The curves intersect at (1, 3) and (1, 3).

The required volume

1

1

2222 ])2()4[( dxxx

1

1

2 )1212( dxx

11

3 ]412 xx

)]8(8 16

P. 24

Example 10.12TFind the volume of the solid generated when the region bounded by the curves y = ex and y = e–x in the interval0 ≤ x ≤ ln 2 is revolved about the x-axis.


Solution:


The required volume

2ln

0

22 ])()[( dxee xx

2ln

0

22 )( dxee xx

2ln

0

22

22

xx ee

1

8

17

8

9

P. 25

Using the method, we can derive the formula for the volume of the solid generated by revolving about the y-axis.

If u(y) 0 in the interval c y d, then when the shaded region is revolved about the y-axis, the volume of the solid of revolution can be found by using the following formula:

BB. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the y-axisy-axis

dyyuVd

c 2)(π


P. 26

Example 10.13TFind the volume of the solid generated when the region between the y-axis and the curve x = tan y, where is revolved about the y-axis.Solution:


4

π0 y


The required volume

40

2tan ydy

40

2 )1(sec dyy

40][tan

yy

4

4

2

P. 27

Example 10.14TThe region bounded by the curve y , the line

3x – 4y – 8 0 and the y-axis with x > 0 is revolved aboutthe y-axis, find the volume of the solid generated


29

2x

Solution:


)2(....................... 0843

)1(............ 2

92

yx

xy

Substituting (1) into (2),

082

9432

x

x

2916)83(

22 x

x

0804817 2 xx0)4)(2017( xx

(rejected)17

20or 4 x

The point of intersection is (4, 1).

,2

9For 2x

y

22

22

2182

9

yx

yx

0843For yx

3

84

yx

P. 28

Example 10.14T



The required volume

dyydy

y)218(

3

84 22

dyydyyy )218()44(

9

1 22

3

1

31

2

23

3

21842

39

1

yyyy

y

3

5236

3

8

3

19

9

1

3

104

The region bounded by the curve y , the line

3x – 4y – 8 0 and the y-axis with x > 0 is revolved aboutthe y-axis, find the volume of the solid generated

29

2x

Solution:

P. 29

Now let us consider two continuous functions u(y) and v(y), where u(y) v(y) > 0 in the interval c y d.

If the region bounded by the curves x = u(y) and x = v(y) is revolved about the y-axis, then the volume V of the hollow solid formed is given by:


dyyvdyyuVd

c

d

c 22 )(π)(π

dyyvyud

c )()( π 22


P. 30

Example 10.15TA container is obtained by revolving the region bounded by the curves y 10 – x2, y 8 – x2 and the line y –2 about the y-axis. Find the volume of the material needed to make the container.Solution:



Rewrite the equations of the curves y 10 – x2 and y 8 – x2 as x2 10 – y and x2 8 – y respectively.

The required volume

8

2

10

2)8()10( dyydyy8

2

210

2

2

28

210

y

yy

y

507222

P. 31

Example 10.16TFind the volume of the solid generated when the region bounded by the curve y 2 – (x + 1)2 and the line y 1 is revolved about the y-axis.



Solution:Vertex (1, 2),

2)1(2 xy

yx 2)1( 2

yx 21

yx 21

Since the x-coordinates of the point on the left side of vertex less than –1, the equation of it is ,where 1 y 2.

yx 21

The equation of the point on the right side of vertex is ,

where 1 y 2.yx 21

The required volume

dyy

y

])21(

)21[(

2

2

1

2

2

124 dyy

2

1

23

)2(3

8

y

3

8

P. 32

Sometimes, the solid of revolution may be formed by revolving about a line parallel to the x-axis or the y-axis. In this case, we need to modify the formulas obtained in Sections 10.2 A and B in order to find the volume of the solid of revolution.

CC. . Solids of Revolution Revolved aboutSolids of Revolution Revolved about Lines Parallel to the Coordinate AxesLines Parallel to the Coordinate Axes

For example, as shown in the figure, the region bounded by the curve y = f (x) and the line y = h (where h > 0) in the interval a x b is revolved about the line y = h to form the solid of revolution. If we translate the curve y = f (x) and the line y = h downwards by h units, then we can see that it is just the same as revolving the curve y = f (x) – h about the x-axis ! Thus, we can conclude that, the volume V of the solid of revolution is given by

dxhxfVb

a 2)(π


P. 33

Example 10.17TFind the volume of the solid generated when the region bounded by the

curve y , the lines y = 2 and x is revolved about the line y 2.

Solution:


4

1

x

1


,1

For x

y when y 2, x

12

The required volume

2

1

41

2

21

dxx

2

1

41 2

441

dxxx

2

1

4

14ln4

1

xx

x

)]2ln83(2ln4[

)2ln43(

2

1x

P. 34

Example 10.18TFind the volume of the solid generated when the region bounded by thecurve x cos y and the lines x y + 1 and x 2 in the interval is revolved about the line x 2.Solution:


2

π0 y


For x y + 1,

when x 2, y 1

The required volume

1

0

220

2 )21()2(cos dyydyy

1

0

220

2 )1()4cos4(cos dyydyyy1

0

32

0 3

)1(4cos4

2

2cos1

ydyy

y

3sin4

4

2sin

2

9 2

0

yy

y3

4

13

4

9 2

P. 35

In the disc method mentioned previously, the volumes of solids of revolution are found by approximating with circular discs. But in some situations, it may be easier to use the shell method, in which thin cylindrical shells are used to approximate the volume of the solid of revolution.

DD. . Shell MethodShell Method

For example, consider the region bounded by the curve y = f (x), the x-axis, the lines x = a and x = b as shown in the figure, where f (x)0 and 0 a b.

A hollow solid is obtained when the region is revolved about the y-axis.


P. 36

Now, if we approximate the region by a number of vertical rectangles and all these rectangles are revolved about the y-axis, each of them will generate a thin cylindrical shell of thickness x, height f (x) and radius x as shown in the figure.


If x is very small, the volume of such a shell is approximately equal to the volume of a thin rectangular sheet of thickness x, height f (x) and length equal to the base circumference of the shell, that is, 2x.

Hence the volume of a shell is approximately equal to 2x f (x)x and thus the total volume of the solid is

.)(π2 dxxxfVb

a


P. 37

Similarly, if the region bounded by the function x = g(y), the lines y = c and y = d is revolved about the x-axis, then


Note:1. In the above discussion, f (x) and g(y) are assumed to be

non-negative within the interval of integration. If not, theformulas should be modified as

.)(2 dyyygVd

c

.)(π2 and )(π2 b

a

b

adyyfyVdxxfxV

2. In the shell method, when the region is revolved about the y-axis, the volume is integrated along x; when it is revolved about the

x-axis, the volume is integrated along y. It is opposite to the formulas we learnt in the disc method. Students should pay attention to this.


P. 38

Example 10.19TFind the volume of the solid generated when the region bounded by thecurve y , the x-axis and the line x 9 is revolved about y-axis.


x


Solution:The required volume

9

02 dxxx

9

0

2

5

5

22

x

5

972

P. 39

Example 10.20TFind the volume of the solid generated when the region bounded by thecurve y = x3, the x-axis and the line x = 1 is revolved about the line x =

2.



Solution:The required volume

1

0

3)2(2 dxxx1

0

54

522

xx

5

3

P. 40

10.1 Finding Plane Areas by Integration

Chapter Chapter SummarySummary

1. For a b c,

e

d

f

edyyudyyu )()(area Shaded

c

b

b

adxxfxgdxxgxf )()()()(area Shaded

c

b

b

adxxfdxxf )()(area Shaded

f

e

e

ddyyuyvdyyvyu )()()()(area Shaded

2. For a b c,

3. For d e f,

4. For d e f,

P. 41



1. Hollow Solid of revolution about the x-axisIn the figure, by the disc method,

2. Hollow Solid of revolution about the y-axisIn the figure, by the disc method,

When the region bounded by the curve y = f (x), the lines x = a, x = b and y = h is revolved about y = h, the volume of solid of revolution is given by

. )()( π volume 22 dxxgxfb

a

. )()( π volume 22 dyyvyud

c

.)(π volume 2dxhxfb

a

P. 42



In the figure, by the shell method,

1.

2. Similarly, when the region bounded by the curve x = g(y), the lines y = c and y = d is revolved about the x-axis,

.)(π2 volume dxxxfb

a

.)(π2 volume dyyygd

c

10.1Finding Plane Areas by Integration 10.2Volumes of Solids of Revolution Chapter Summary Case...

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