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THE KENYA METHODIST UNIVERSITY
DEPARTMENT OF PURE AND
APPLIED SCIENCES
MATHS 104: CALCULUS II
COURSE INSTRUCTORALICE LUNANI MURWAYI (MSc – Mathematics)
DEPARTMENT OF PURE AND APPLIED SCIENCESP.O. Box 267
MERU
MATHS 104: CALCULUS II
INTRODUCTION
This course will introduce the students to the concepts and skills of Integral calculus essential for
application finding areas, length of arc, Volume and Surface area of Solids of revolution.
COURSE OBJECTIVES
During the course, the undergraduate student will be able to:.
BROAD AREAS OF COVERAGE
1. The antiderivatives
2. Use basic formula to integration of algebraic functions
Power
Sum and difference
Constant time a function
3. Integration of exponential functions and trigonometric functions using basic formulae
4. Integration using substitutions
5. Integration by parts
6. Integration using Trigonometric Identities
7. Integration using Partial Fractions
8. Definite integrals and the fundamental theorem of Calculus
9. Area under a curve
10. Improper Integrals
11. Length of an arc of a curve
12. Volume and Surface Area of Solid of Revolution
13. Trapezoidal, Mid ordinate and Simpson’s rule
TEACHING AND LEARNING METHOD’S
Lecture method in which definitions are stated and illustrate, theorems are stated, proved
and problems solved as examples.
Questions and answer method.
Tutorials in which problems are solved test the understanding.
Take away assignments.
STUDENTS ASSESSMENT
Assignment
Course Examination
Individual student’s marks of these assignments will contribute to the final examination mark
and grade for the course.
COURSE OUTLINE FOR THE TRIMESTER
WEEK 1
The antiderivatives
Use basic formula to integration of algebraic functions
Power
Sum and difference
Constant time a function
WEEK 2
Integration of exponential functions and trigonometric functions using basic formulae
WEEK 3
Integration using substitutions
WEEK 4
Integration by parts
WEEK 5
Integration using Trigonometric Identities
WEEK 6
Integration using Partial Fractions
WEEK 7
Definite integrals and the fundamental theorem of Calculus
WEEK 8
Area under a curve
WEEK 9
Improper Integrals
WEEK 10
Length of an arc of a curve
WEEK 11
Volume and Surface Area of Solid of Revolution
WEEK 12
Trapezoidal, Mid ordinate and Simpson’s rule
WEEK 14 AND 15
Assignments
Cat 1-------------------------------15%
Cat 2-------------------------------15%
Total Cats-------------------------30%
Course Examinations---------------------70%
Grant Total---------------------------------100%
LEARNING RESOURCES
Larry, J. G, et al: Calculus and it Applications, Prentice – Hall International Press, London, 1993
Swokowski, E. W.: Calculus with Analytic Geometry, Alternate, Edition PWS Publishers, 1983
Thomas, G. G and Finney, R. L.: Calculus and Analytic Geometry, Narosa Publishing House, 6th Edition, 1998
TABLE OF CONTENTS
CHAPTER ONE: INTEGRAL CALCULUS..................................................................................7
CHAPTER TWO: EXTENSION FORMULAE...........................................................................15
CHAPTER THREE: INTEGRATION USING SUBSTITUTIONS.............................................18
CHAPTER FOUR: INTEGRATION OF TRIGONOMETRIC FUNCTIONS..........................23
CHAPTER FIVE: PARTIAL FRACTIONS.................................................................................36
CHAPTER SIX : INTEGRATION BY PARTS...........................................................................44
CHAPTER SEVEN:FACTORIZATION BY COMPLETING THE SQUARE...........................48
CHAPTER EIGHT : INTEGRATION THE INVERSE TRIGONOMETRIC FUNCTIONS......50
CHAPTER NINE : CHANGE OF VARIABLE...........................................................................61
CHAPTER TEN: DEFINITE INTEGRALS.................................................................................65
CHAPTER ELEVEN: : AREA UNDER A CURVE....................................................................71
CHAPTER TWELVE: NUMERICAL INTEGRATION..............................................................85
CHAPTER THIRTEEN :VOLUME AND SURFACE AREA OF SOLIDS OF REVOLUTION.......................................................................................................................................................97
CHAPTER FOURTEEN : LENGTH OF ARC OF A CURVE..................................................112
CHAPTER ONE: INTEGRAL CALCULUS In Differential calculus, you studied the different methods of differentiation and its
applications. Now in integral calculus we shall learn the different methods of integration and its
applications
1.1: CHAPTER OBJECTIVES
By the end of the chapter, the student should be able to:
1. Explain the meaning of integration
2. Explain the meaning of the constant of integration
3. State and use the rule for Integrating
a. A constant function
b. An algebraic function
c. A constant times a function
d. Sum of functions
e. Functions of the form
f. Exponential functions
1.2: MEANING OF INTEGRATION
The process of integration is defined as the reverse or converse process of differentiation
For example
, we say that the integration of . This is written in symbols as
, we say the integral of
The sign ’’means the integral of ’’ and is called the integral sign. Means
integration of the function with respect to . Indicates the variable of
integration. C is the constant of integration.
1.3: THE CONSTANT OF INTEGRATION
The differentiation of a constant is always equal to zero; therefore several functions have the
same derivative because any constant will disappear in the process of differentiation. After
differentiation we always add a constant c or k as we can see in the two examples. Generally,
when gives all the anti-derivatives of , Here c is the constant
of integration. An integral like in which the constant c is not known/ cannot be
determined is called an indefinite integral
1.3: THE INTEGRAND
In the integral sign cannot be divorced from if the integral is with respect to . The
function in is called the integrand of the integral. For example in , is
called the integrand of the integral.
1.4: THE GENERAL RULE FOR INTEGRATING ALGEBRAIC FUNCTIONS.
Examples
Find the integrals of the following functions
1.5: INTEGRATION OF A CONSTANT TIMES A FUNCTION
EXAMPLE
Integrate 2x5
Solution
1.6: INTEGRATION OF A CONSTANT
Example
Evaluate the integral of 4 with respect to x
Solution
1.7: INTEGRATION OF THE SUM OF FUNCTIONS
The integral of sum of two functions is the sum of their integrals. This extends to the
sum of any finite number of functions.
Example 2
Find
Solution
We expand and then integrate
Example 3
Find
Solution
First we separate the terms by dividing each term by x and then integrate
Example 4
Find
Solution
We simplify the integrant and the integrate
1.8: INTEGRATION OF
Recall
For all values when using definite integrals
1.9: INTEGRATION OF EXPONENTIAL FUNCTIONS
Exponential functions are functions in which the variable is the index or the power .These
are functions of the form ex and ax
1)
Generally,
Example
2. Find
Solution
3. Integrate
Solution
EXERCISE
Integrate the following functions with respect to
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14) If find x.
15) If , find the
values of x
CHAPTER TWO: EXTENSION FORMULAE
2.1: CHAPTER OBJECTIVES
By the end of the chapter, the student should be able to:
1) Write the extension formulae corresponding to the basic formulae for integrating
algebraic and exponential functions.
2) Use the extension formulae to integrate expressions /functions of first degree in x.
2.2: EXTENSION FORMULAE FOR INTEGRATING ALGEBRAIC
FUNCTIONS CONTAINING FIRST DEGREE EXPRESSIONS
This is the extension of the integration rules seen in chapter one. It is one of
the techniques of integration in which the variable x in the basic formulae is
replaced by a first degree expression of the form (ax +b). It entails finding
the integrals like:
, where real numbers
And
is a linear function of .
Generally
EXAMPLES
2.3: INTEGRATING FUNCTIONS OF THE FORM
Generally,
Examples
2.4: INTEGRATING EXPONENTIAL FUNCTIONS USING EXTENDED
FORMULAE
These are functions of the form and .
The extended formulae for the functions of the form and are:
1)
2)
Example
EXERCISE
Integrate the following functions with respect to
1)
2)
3)
4)
5)
6)
7) ) 8).
CHAPTER THREE: INTEGRATION USING SUBSTITUTIONSWhen the integrand consists of two functions one of which is the derivative of the other function,
we use the substitution technique of integration. The main function is u and the derivative u’ or
so that u’ dx becomes du. The substitution transforms the integral into one of the basic
formula of integration in the new variable u only. We then integrate using the basic formula.
3.1: CHAPTER OBJECTIVES
By the end of the chapter, the student should be able to:
1) Recognize a function u and its derivative u’ or in the integrand
2) Adjust the derivative so that the integrand consists of one function u and its derivative
3) Substitute u and u’ or in the integrand and integrate
3.2: THE SUBSTITUTION METHOD OF INTEGRATION
This involves the changing of the variable so as to simplify integration. There are different types
of integrands to consider like:
Type 1
Examples
1)
Solution
Do this problem by making the substitution
2)
Solution
3)
Solution
4)
Solution
Type 3
EXERCISE
Evaluate the following
1)
2)
3)
CHAPTER FOUR: INTEGRATION OF TRIGONOMETRIC FUNCTIONS
4.1: CHAPTER OBJECTIVES
By the end of the chapter, the student should be able to:
1) Integrate trigonometric functions using the six basic formulae
2) Integrate trigonometric functions with first degree expressions using the Extension
functions
3) Use trigonometric Identities to integrate trigonometric functions
4) Use Substitution to integrate trigonometric functions
4.2: INTEGRATION OF TRIGONOMETRIC FUNCTIONS USING THE
BASIC FOMULAE
The following is a list of the basic trigonometric formula. They are also in the Advanced
Mathematical tables.
1)
2)
3)
4)
5)
6)
Examples
1) Evaluate
Solution
2) Find
Solution
3) Find
Solution
EXERCISE
Evaluate
1)
2)
4.3:INTEGRATION OF TRIGONOMETRIC FUNCTIONS USING
SUBSTITUTIONS
Example
Find
Solution
Example
Find
Solution
NB.
Example
Exercise
1) Show that
2) Use a suitable substitution to show that:
3)
4)
5) Use a suitable substitution to show that
a)
b)
c)
4.4: INTEGRATION OF TRIGONOMETRIC FUNCTIONS USING
TRIGONOMETRIC IDENTITIES
The Basic Trigonometric Identities
Examples
4.5: INTEGRATION OF TRIGONOMETRIC FUNCTIONS USING
MULTIPLE ANGLES
Multiple angle Identities
EXERCISE
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
SOLUTIONS
CHAPTER FIVE: PARTIAL FRACTIONS5.1: CHAPTER OBJECTIVES
By the end of the chapter, the student should be able to:
1. Decompose compound fractions into partial fractions
2. Use appropriate method to integrate the partial fractions
5.2: INTRODUCTION TO PARTIAL FRACTIONS
In this section we show how to disintegrate compound fractions byexpressing them as a sum of simpler fractions, called partial fractions. To
illustrate the method, observe that by taking the fractions and
to a common denominator we obtain
If we now reverse the procedure, we see how to integrate the function on the right side ofthis equation:
To see how the method of partial fractions works in general, let’s consider a rationalFunction
Where and are polynomials. It’s possible to express as a sum of simpler fractionsprovided that the degree of is less than the degree of . Such a rational function is calledproper. Recall that if
Where , then the degree of is and we write deg = .
If is improper, that is, , then we must take the preliminary stepof dividing into (by long division) until a remainder is obtained such that
. The division statement is
Where and are also polynomials.As the following example illustrates, sometimes this preliminary step is all that is required.
EXAMPLE 1
Find
SOLUTION Since the degree of the numerator is greater than the degree of the denominator,we first perform the long division.
This enables us to write
5.3: Denominator with linear factors examples.
(1).Evaluate
Examples
5.4: DENOMINATOR WITH A QUADRATIC FACTOR
But
Each numerator of a partial fraction must be assumed to be a polynomial of degree 1 less than
that of the corresponding denominator.
Relating the numerators
5.5: DENOMINATOR WITH REPEATED LINEAR FACTORS
EXAMPLES
Exercise
Show that
1)
2)
3)
4)
CHAPTER SIX : INTEGRATION BY PARTS6.1: CHAPTER OBJECTIVES
By the end of the chapter, the student should be able to:
1. Derive the formula for integration by parts
2. Use the formula to integrate the product of two functions
6.2: INTEGRATION BY PARTS
This method applies to integrals of the form in which can be differentiated
repeatedly to become zero, while is integrated repeatedly without much difficulty.
Integration by parts is equivalent to the product rule and it states that:
Integrate with respect to x
EXAMPLES
Exercise
Use integration by parts.
1) 2).
3). 4).
5). 6).
7). xdx 1sin 8). xdxex sin
9). 10).
CHAPTER SEVEN:FACTORIZATION BY COMPLETING THE SQUARE7.1: CHAPTER OBJECTIVES
By the end of the chapter, the student should be able to:
1) Factorize Quadratic expressions by completing the square
2) Use substitution to integrate the factorized function
7.2:INTEGRALS INVOLVING
By completing the square of where a
where u is a function of x evaluate by integration
Examples
1.
Completing square
lets u= differentiating both side dx = cos ydy
2.Evaluate
CHAPTER EIGHT : INTEGRATION THE INVERSE TRIGONOMETRIC FUNCTIONS8.1: CHAPTER OBJECTIVES
By the end of the chapter ,the student should be able to:
1) Define and find the inverse of a given function
2) Draw graphs of inverse trigonometric functions
3) Combine Differentiation and integration to prove identities involving
Inverse trigonometric functions.
8.2: THE INVERSE TRIGONOMETRIC FUNCTIONS
If is a function whose inverse if then
Definition:
The inverse sure function is denoted by is defined by
if and only if where .
The domain and range are very important
One to one function – inverse can be easily obtained
The Trigonometric functions are not one to one and therefore have no inverse. However, if there
domains are restricted then there is an inverse.
The inverse of the sine function
The graph of can be draw from the equation
2
1
0
-1
-2
2
1
0
-1
-2
f x
2
1
0
-1
-2
4
3
2
1
0
1
-1-1
-2
f x
21
32
342
262
10062
142
232
32
1
)(sin 1
xx
2
2
x
y
1-1
The inverse of the Cosine function
Definition:
The inverse of cosine function denoted by is defined by
x
yy = cos(x)
x
yy = arccos(x)
The inverse of the Tangent function
Definition:
The inverse of tangent function is defined by . If where
8.3. DIFFERENTIATION AND INTEGRATION
Exercise
Show that
a)
b)
c)
CHAPTER NINE : CHANGE OF VARIABLE9.1: CHAPTER OBJECTIVES
By the end of the lesson, the student should be able to:
1) Evaluate integrals by changing the variable to
2) Evaluate integrals by changing the variable to
9.2: THE CHANGE OF VARIABLE It’s used in functions with even powers of sin x and cosxExample
Find
Solution
The change of variable makes the integration easier or possible. It is important to simplify the expression as much as possible using the trigonometric identities.
Exercise
1) 2).
3). 4).
9.3: THE CHANGE OF VARIABLE
Recall trigonometric identities and change of variables
Examples
1. Find
Solution
Exercise 1)
2)
3)
CHAPTER TEN: DEFINITE INTEGRALS
10.1: CHAPTER OBJECTIVES
By the end of the chapter, the student should be able to:
1. State the fundamental theories of Calculus
2. Use the theories in Solving problems involving Calculus
3. Calculate definite integrals.
10.2:FUNDAMENTAL THEORIES OF CALCULUS1. let be continuous on closed interval
Theorem If a function F is defined by For all in then F is an antiderivative on 2. If s continuous on and is an antiderivative of on then
F F =
x
y
Substitution in definite integrals Example
1)
Examples Evaluate
Exercise
1)
2)
3)
4)
5) Find the area between and the
Solutions
Exercise on definite integralsEvaluate the following integrals.
1)
2)
3)
4)
5)
6)
CHAPTER ELEVEN: AREA UNDER A CURVE
11.1: CHAPTER OBJECTIVES
By the end of the chapter the student should be able to:
1. Find the area under a curve
2. Calculate the area between two curves
12.2 :AREA UNDER THE CURVE
Suppose we wish to find the area under the curve from to .1. Divide the area into thin vertical strips; find their area and them up to find the
approximate area under the curve.2. Consider one strip PP’Q, Q where P(x, y).The length of the strip = y and the width can be
considered as a small increase in x ( ). The area of the strip = y ( )3. If A is the area of the region up to PP’ then the area of PPQ”Q can be considered as a
small increase in
Area of strip =
x=a x=b
y f x
P’ Q’
y
x
pQ
The accuracy of the area increases as decreases.
Total area=
from (1)
From (2) and(3)
The area of the curve is the integration of the function. Since the boundary values of defining the total area are from Then the total area is given by
is called the definite integral of over the integral
PROPERTIES
1)
2) direction interchanges the
intervals and hence the sign.
3)
11.3: FUNDAMENTAL THEORIES OF CULCULUS1. Let be continuous on a closed interval [a,b]
If a function F is defined by for all x in [a, b], then F is an antiderivative
on [a ,b].2. If is continuous on [a,b ] and F is an antiderivative of on [a,b] then,
Example1) Find the area under the curve enclosed between .line and where
are positive.
Solution
Sketch the graph
y
x
Area of shaded region
=
= = =
y
x
y mx
2 (a) Find the area between the curve and the -axis from to
SolutionSketch the graph
b) Find the area of the region bound by and the from to x=2
SolutionSketch the graph to identify the region
2 4y x
y
x
3.Find the area of the region enclose by the curve
SolutionFactorize the function
Recall curve sketch in Calculus 1
Curve cuts the x-axis when y = 0 ;
Curve cuts the x-axis at three points Find the turning points by differentiating and determine their nature by differentiating the 2nd
time.
24y x
y
x-2
2
When min. point at
When max. at
y
x-2 2
3 4y x x
Interpretation: The area under the x-axis is equal to that above the x-axis.
4. Find the area bonded by the curve and the
SolutionSketch the graphFind the y-intercepts and the x-intercepts
y
x
2
-2
4
24x y
When the function is in terms of y, integrate with respect to y and the intervals are in the form y= a and y = b.
EXERCISEFind the area bounded by given curve and given vertical line
1)2)3)
4)
5)
6)
7)
8)9) Find the area bounded by the co-ordinate axis and the line .10) Find the area bounded by and co-ordinate axis11) Find the area between the curve and the
12) Find the area between and the x-axis
11.4: AREA BETWEEN TWO CURVESRecall area under a curve as below.
y f x
a b
c
d
x g y
Suppose are continuous in and that
then the graph of lies above on the interval .
Divide the shaded area into vertical strips each of width
Consider one inch strip as shown Area of strip =
If sum of vertical strips
Defining the area between two curves If through out interval
1y f x
2y f x
x= a x= b
Area between and from to is
Examples: 1.Find the area of the region bounded by the parabola and the line Make a sketch
SolutionSketch the graphs of the functions
.These are the x-co-ordinates of the points of intersectionOf the two functions. They are the intervals.
2.Find the area of the region bound by the graphs
Solution Sketch the graphs to identify the region
2y x
21 2y x
If area is bounded by and the curves for all in cd
Then
2
24
21x y
2
22
2 4
2 4
y x
or x y
ExerciseFind the area of the regions bounded by the following curves
Exercise on definite integralsEvaluate the following integrals.
7)
8)
9)
10)
11)
12)
13)
CHAPTER TWELVE: NUMERICAL INTEGRATION 12.1: CHAPTER OBJECTIVES By the end of the chapter, the student should be able to:
1. Derive the Trapezoidal rule2. Use the Trapezoidal rule to find approximate area under a curve3. Derive the Simpson’s rule4. Find the approximate area using the Simpson’s rule
12.2: TRAPEZOIDAL RULE.
Used to approx. areas under the curves whose integrals can’t be found easily.It estimates the integral by dividing the area into trapeziums instead of rectangles consider the area under the curve.
from to
Divide the area into is trapeziums of equal width h such that
y f x
xn= bx0= a x1 x2
y0 y1 y3
Consider the shaded trapezium
Total area (TA) = sum of area of all trapeziums
Example
Using trapezoidal rule with
Solution
The more the number of trapezia the region is divided into, and approximate area exact value
This means as increase decrease and the Total area approaches exact value. Taking large enough we can make the difference between the exact values of the area (the error ) as small as possible.However, in practice, its not possible as we can’t tell how large n should be:-If is the continuous on where is 2nd derivative of it can be shown that.
Where c is a value between Thus as the error
Use the inequality
Maximum absolute value 0f evaluated at a and b .It gives us the upper
bound of the magnitude of error error.max .In practice, the exact value of can’t be obtained. We therefore estimate an upper bound value for it instead.
If M is the upper bound for the value then the inequality becomes
Definition: If is continuous and m is any upper bound for the values of on the interval
the error in the trapezoidal approx of the integral of from to statistics the inequality
That is if M is a real number such that for all x in then the error in using
the trapezoidal rule is not greater than
Examples
1. Find the upper bound on the error (max error) in the approximation for
SolutionFrom previous example
The max value of on occurs at :
.
2. Approximate by using Trapeziodal rule with n=10. Estimate the max error
in the approximation.
Solution
Max value of on occurs at 1x
12.3: SIMPSON’S RULE-This involves Approximating area under curves using parabolas.Consider the area under curve from to
Any 3 non-collinear points can’t be joined/ fitted into a parabola Divide the area into n parabolas
Consider the following co-ordinates axis
x
y f x
y
x=a x=b
Let the equation of parabola be of the form
Since the points satisfy the equation
xx=-h x=h
0y 1y 2y
y
Solve for A,B and CAdd (i) and (iii)
Simpson’s rule results from applying the formula for AP to successive pieces of the curve between and
To approx. we sum all the areas
Where n is even
x
y f x
y
x=a x=b
0y ny1ny
1y
Error Estimation for Simpson’s rule.If is continuous and is any upper bound for the values of that is,
for all in the error in the Simpson’s rule approximation of the
satisfies the inequality
Examples
1. Approximate Using Simpson’s rule with .What is the max error in
the estimate?
Solution
Using Simpson’s
Estimating the error.
Max. Occurs at any point
Both Simpson’s and Trapezoidal rule can be used to estimate
,Where is not known but the data is given .Suppose its found experimentally
that 2 physical variables are related as shown.
Assuming that is continuous estimate using (a) Trapezoidal rule.
(b) Simpson’s rule.
Solutionb). Simpson’s rule.
ExerciseUse Trapezoidal & Simpson’s rules together with max-errors of Estate to approximate the following definite integrals for the stated value of n.
a) c).
b) d).
c) e).
CHAPTER THIRTEEN :VOLUME AND SURFACE AREA OF SOLIDS OF REVOLUTION
13.1: CHAPTER OBJECTIVES By the end of the topic the student should be able to:
1.Use the formula for the volume of the solid of revolution to the volume of common solids of revolution
2.Find the surface area of common solids of revolution using the formula for the surface area of solid of revolution.
13.2: VOLUMES OF SOLID OF REVOLUTION
Consider the region bounded by ,lines and
Revolve this region about the .If a region in a plane is rotated about a line in the plane, the resulting solid is called a solid of revolution and solid is said to be generated by the region. The line about which the rotation occurs is called the axis of revolution.Consider the region bounded by
Definition:
x=a x=b
y f x
2
-2
55
The volume obtained by rotating the curve about the between the
ordinates is given by
Means rotation about or any line parallel to . But
ExampleIf ,find the volume of the solid of revolution under this graph from to
about the .
Solution Sketch the graph showing he region and the solid after rotation
Definition Volume obtained by rotating the curve about the between the ordinates and and the .
2 1y x
-11
x
y
Example. The region bound by the between and is evolved about the
.find the volume of the solid of revolution. SolutionSketch the graph
c
d
x g y
y = 1
y= 8
3y x
x
y
Definition.If the region bounded by the two curves from and is revolved about the
Its volume is given by.
Example1. If the region bound by is revolved about . Find the volume of the solid of revolution.
x = a x = b
1f x
Solution
Rewrite the equations in terms of y,that is
NOTEIf the region is rotated about:
1. Volume=
2. : Volume =
1 12y x
2 2y x
x = 1
Examples 2. Find the volume obtained by revolving the region bounded by and lines and about the line
Solution
3. The region in the first quadrant bounded by is revolved about .Find the
volume of the solid of revolution.
Solution
The equations are rewritten in terms of x
4. The region bounded by the graphs is revolved About the line y = 3. Find the volume of the solid of revolution.
Solution
1
1
2x y
1 31 2x y
EXERCISE Sketch the regions bounded by the following curves. Find the volumes generated about the given lines.Exercise
1). 2).
3). 4).
5). 6).
7). 8).
9). 10). 11). 12).
13). 14).
15).
1 12y x
2 2y x
x = 1
13.4: THE SURFACE OF SOLID OF REVOLUTION Definition: surface area.
1) The area of the surface swept by revolving the curve from to about the
2) If the axis of revolution is y-axis from y = c and y = d
3) Using parametric equation.If the curve that sweeps out the surface is given in parametric form with and as functions of a variable t that varies from to then
Examples 1)Find the area of the surface obtained by revolving the curve
Solution
2) The line segment from to is revolved about the y axis generated
a cone. Find its surface area.
2
y x
x
y
Solution
3)Find the area of the sphere generated by revolving the circle about.
The top semi circle is enough to generate the required area.Parametrically the equation of the circle is given by:
REMARKS
1) If revolved about
y
x
y
x
a
a
a
2) If revolved about
EXERCISEFind the area of the surface generated by revolving the given curve about the given lines.
1)2)
3)
4)
5)
6)
7)8)
CHAPTER FOURTEEN : LENGTH OF ARC OF A CURVE
14.1: CHAPTER OBJECTIVES
By the end of the chapter the student should be able to:1. Derive the formula for finding the length of an arc /curve if the the equation of the
curve is known2. Accurately calculate the length of a curve using the formula.
14.2: THE LENGTH OF A CURVE IN CARTESIAN EQUATION
The length can be approximated by short line segments. The more the line segments the more accurate the approximation. Suppose that the curve whose length we want to find is
between and .The arc the polygonal path APQB.The length of the arc is defined to be the limit of the length of successively finer polygonal approximations. Divide the curve into n parts and connect the successive end points with line segments. Consider a typical segment . The length of
by Pythagoras for n segments the length of the curve from and
is approx. by the sum
The approximation improves as n increases and length of segment approaches the exact value.
x = a x = b
AB y f x
P
Q
14.3: MEAN VALUE THEOREM.
If a function is continuous on the closed interval and is differentiable on the
open interval , then there exists a number C in such that
the derivative of at C)
Suppose that has a derivative that is continuous at every point then by the mean value theorem their exists a point on the curve between such that
From (1)
Examples.
Find the length of the curve
14.4: THE LENGTH OF A CURVE IN PARAMETRIC EQUATIONSuppose
Examples
1. Find the distance traveled between and by a particle whose position at time
is given by and
Solution
At a point on the curve where fails to exist may exist. Then the arc length of
is;
2.
Solution
Sketch the curve
EXERCISEFind the length of the curves.
a). e).
2 3y x
8-1 x
y
b). f).
c). g).
d). h).
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