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10. Application s of D efinite Integrals. Case Study. 10 .1 Finding Plane Areas by Integration. 10 .2 Volumes of Solids of Revolution. Chapter Summary. In some countries, there are very wide income gaps between the rich and the poor. - PowerPoint PPT Presentation
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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.
1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration
A. A. Area of the Region Bounded by a CurveArea of the Region Bounded by a Curve and the x-axisand the x-axis
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.
1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration
A. A. Area of the Region Bounded by a CurveArea of the Region Bounded by a Curve and the x-axisand the x-axis
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.
1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration
A. A. Area of the Region Bounded by a CurveArea of the Region Bounded by a Curve and the x-axisand the x-axis
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).
1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration
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 .
1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration
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)
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.
1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration
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)
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.
1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration
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 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
1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration
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)
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.
1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration
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.
1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration
CC. . Area of the Region Bounded by a Curve Area of the Region Bounded by a Curve and the y-axisand the y-axis
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.
1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration
CC. . Area of the Region Bounded by a Curve Area of the Region Bounded by a Curve and the y-axisand the y-axis
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.
1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration
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.
1010.1 .1 Finding Plane Areas by IntegrationFinding Plane Areas by Integration
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.
AA. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the x-axisx-axis
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.
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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.
AA. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the x-axisx-axis
h
r
We can also find the general formula for calculating the volume of a cylinder and a sphere using the disc method.
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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.
AA. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the x-axisx-axis
x
Solution:
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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.
AA. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the x-axisx-axis
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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.
AA. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the x-axisx-axis
dxxgdxxfVb
a
b
a 22 )(π)(π
dxxgxfb
a )()( π 22
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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.
AA. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the x-axisx-axis
Solution:
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
)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.
AA. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the x-axisx-axis
Solution:
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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:
BB. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the y-axisy-axis
4
π0 y
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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
BB. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the y-axisy-axis
29
2x
Solution:
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
)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
BB. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the y-axisy-axis
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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:
BB. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the y-axisy-axis
dyyvdyyuVd
c
d
c 22 )(π)(π
dyyvyud
c )()( π 22
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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:
BB. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the y-axisy-axis
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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.
BB. . Solids of Revolution Revolved about theSolids of Revolution Revolved about the y-axisy-axis
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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:
CC. . Solids of Revolution Revolved aboutSolids of Revolution Revolved about Lines Parallel to the Coordinate AxesLines Parallel to the Coordinate Axes
4
1
x
1
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
,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:
CC. . Solids of Revolution Revolved aboutSolids of Revolution Revolved about Lines Parallel to the Coordinate AxesLines Parallel to the Coordinate Axes
2
π0 y
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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.
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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.
DD. . Shell MethodShell Method
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
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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
DD. . Shell MethodShell Method
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.
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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.
DD. . Shell MethodShell Method
x
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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.
DD. . Shell MethodShell Method
1010..22 Volumes of Solids of RevolutionVolumes of Solids of Revolution
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
10.2 Volumes of Solids of Revolution
Chapter Chapter SummarySummary
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
10.2 Volumes of Solids of Revolution
Chapter Chapter SummarySummary
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
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