Ambo University

Department of Mathematics

Lecture Note

Applied Mathematics II

June 4, 2016
Contents

1 Sequence and Series 3

1.1 Definition and types of sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1.1 Types of sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Convergence properties of sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.1 Monotonic and Bounded Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.3 Subsequence and limit points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.4 Definition of infinite series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.5 Convergence and divergence, properties of convergent series . . . . . . . . . . . . . . . . . . . . . 14
1.6 Nonnegative term series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.7 Tests of convergence (integral, comparison, ratio and root tests) . . . . . . . . . . . . . . . . . . . 17
1.7.1 The Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1.7.2 Comparison Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.7.3 Limit Comparison Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.8 Alternating series and alternating series test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
1.8.1 Alternating Series Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
1.9 Absolute and conditional convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.9.1 Ratio Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
1.9.2 Root Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
1.10 Generalized convergence tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2 Power Series 28
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.2 Definition of power series at any x0 and x0 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.3 Convergence and divergence, radius and interval of convergence . . . . . . . . . . . . . . . . . . 30
2.4 Representations of Functions as Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.5 Algebraic operations on convergent power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.6 Differentiation and integration of power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
2.7 Taylor series; Taylor polynomial and application . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2.7.1 Polynomial Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
2.7.2 Defined by Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
2.7.3 Indeterminate Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3 Differential Calculus of Function of Several Variables 40

3.1 Notations, examples, level curves and graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.1.1 Notations and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.1.2 Level curves and graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.2 Limits and continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.2.1 Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.2.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.3 Partial derivatives; tangent lines, higher order partial derivatives. . . . . . . . . . . . . . . . . . 45
3.3.1 A geometric interpretation of partial derivatives . . . . . . . . . . . . . . . . . . . . . . . 46
3.3.2 Higher order partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3.4 The chain rule, implicit differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.4.1 Chain rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.4.2 Implicit differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
3.4.3 Differentiability of functions of several variables . . . . . . . . . . . . . . . . . . . . . . . . 49
3.5 Directional derivatives and gradients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3.5.1 The Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.6 Tangent planes and Total differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
CONTENTS 2

3.6.1 Tangent Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.7 Tangent plane approximations of values of a function . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.8 Relative extrema of functions of two variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.9 Largest and smallest values of a function on a given set . . . . . . . . . . . . . . . . . . . . . . . 56
3.10 Extreme values under constraint conditions: Lagrange’s multiplier . . . . . . . . . . . . . . . . . 57

4 Multiple Integrals 59
4.1 Double integrals and their evaluation by iterated integrals . . . . . . . . . . . . . . . . . . . . . . 59
4.2 Iterated integrated . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
4.3 Change of variable in double integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
4.4 Double Integrals in Polar coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
4.5 Application of Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
4.5.1 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
4.5.2 Triple Integrals in Cartesian coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
4.6 Triple integrals in cylindrical and spherical coordinates . . . . . . . . . . . . . . . . . . . . . . . . 70
4.6.1 Triple Integrals in Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
4.6.2 Triple Integrals in Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
4.7 Application: Volume, center of mass of solid region . . . . . . . . . . . . . . . . . . . . . . . . . . 74
4.7.1 Some of the applications of multiple integrals . . . . . . . . . . . . . . . . . . . . . . . . . 75

Ambo University
DEPARTMENT OF MATHEMATICS Applied Mathematics II
Chapter 1

Sequence and Series

1.1 Definition and types of sequence

A sequence is a list of numbers
a1 , a2 , a3 , . . . , an , . . .
in a given order. A formal definition of sequence can be given as follows:
Definition 1.1.1. A sequence {an } is a function whose domain is the set of integers. The functional values
a1 , a2 , a3 , . . . , an , . . . are the terms of the sequence, and the term an is called the nth term of the sequence.
Note 1.1.1. • The sequence {a1 , a2 , a3 , . . . , an , . . . } is also denoted by {an }, {an }∞
n=1 .

• Sometimes it is convenient to begin a sequence with ak . In this case the sequence is {an }∞
n=k , and its
terms are ak , ak+1 , ak+2 , . . . , an , . . . .
Definition 1.1.2. A sequence is said to be a constant sequence if Sn = k(constant) for every n ∈ N.
Example 1.1.1. Let Sn = 2 for every n, is a constant sequence. That is, Sn = {2, 2, 2, 2, . . . }
Definition 1.1.3. A Recursively-defined sequence is a sequence where the first term(s) are given and the
next term is given in terms of the previous terms.
Example 1.1.2. List the first five terms of the sequence.
 ∞  
n
A. n+1 = 12 , 23 , 34 , 54 , 56 , . . .
n=1
 ∞  √

3 1
B. cos( nπ
6 ) = 1, 2 , 2 , 0, . . .
n=0
∞
√ √ √
  
C. n−3 = 0, 1, 2, 3, . . .
n=3

Exercise 1.1.1. List the first five terms of the sequence

A. {(−1)n n+1
3n }

B. { n cos(nπ)
2n−1 }
n
C. { (−π)
5n }

D. a1 = 2, a2 = 4, an+1 = 2an − an−1 for n ≥ 2.

Graphing a sequence
To graph the sequence {an } we plot the points (n, an ) as n ranges over all possible values on a graph.
∞
Example 1.1.3. Graph the sequence { n+1
n2 }n=1 . The first few points on the graph are then,

3 4 5
(1, 2), (2, ), (3, ), (4, ), . . . .
4 9 16
The graph is then,
Now let’s look at some special sequences, and their rules.

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
1.1 Definition and types of sequence 4

1.1.1 Types of sequences

Finite and infinite sequence
If the sequence goes on forever it is called an infinite sequence, otherwise it is a finite sequence.
Example 1.1.4. Some examples of finite and infinite sequence.
a) {1, 2, 3, 4, . . . } is a very simple sequence (and it is an infinite sequence)

b) {1, 3, 5, 7} is the sequence of the first 4 odd numbers (and is a finite sequence)
Arithmetic Sequence
In an arithmetic sequence the difference between one term and the next is a constant. In other words, you just
add some value each time . . . on to infinity. In general you could write an arithmetic sequence like this:

{a, a + d, a + 2d, a + 3d, . . . }

Where:
• a is the first term, and
• d is the difference between the terms (Usually called the ”common difference”)

And you can make the rule by:

xn = a + (n − 1)d
Example 1.1.5. Consider the sequence 1, 4, 7, 10, 13, 16, 19, 22, 25,. . .
This sequence has a difference of 3 between each number. Then its rule is xn = 3n − 2.

Geometric sequence
In a geometric sequence each term is found by multiplying the previous term by a constant.
Example 1.1.6. Consider the sequence 2, 4, 8, 16, 32, 64, 128, 256,. . .
This sequence has a factor of 2 between each number. Its Rule is xn = 2n .

In General you could write a geometric sequence like this:

{a, ar, ar2 , ar3 , . . . }.

Where:

• a is the first term, and

• r is the factor between the terms (called the ”common ratio”)
And the rule is: xn = ar(n−1) .
Note 1.1.2. r should not be 0. Because when r = 0 , you get the sequence {a, 0, 0, . . . } which is not geometric.

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
1.2 Convergence properties of sequences 5

1.2 Convergence properties of sequences

Let the sequence an = { n12 }, we see that if n is taken to be a large positive integer, then the nth term of the
sequence is small. That is as n becomes very large, { n12 } becomes very small or very close to zero. This sequence
behaves nicely in the sense that he terms approach some fixed real number( in this case zero) as n becomes
large. In this case we say that the sequence converges to zero. With this introduction, we define convergence of
a sequence.

Definition 1.2.1. A sequence {an } is said to converge to the real number A if and only if for each ε > 0 there
exists a positive real number N = N (ε) > 0 such that |an − A| < ε for all n > N. The sequence {an } is said to
be convergent if and only if there is a real number A such that {an } converges to A. The sequence {an } is said
to be divergent or to diverge if and only if it is not convergent.
Example 1.2.1. Show that the sequence { n12 }∞n=1 converges to 0.
Solution: Here, we must show that we can make n12 as close to 0 as desired, just by making n sufficiently large.
So, given any ε > 0, we must find N sufficiently large so that for every n > N ,

1 1
− 0 < ε, or 2 < ε (1.1)
n2 n

Since n2 and ε are positive, we can divide both sides of (1.1) by ε and multiply by n2 , to obtain
1
< n2 .
ε
Taking square roots gives us r
1
< n.
ε
q
1
Working backwards now, observe that if we choose N to be a real number with N = ε, then n > N implies
1
that n2 < ε, as desired.
So just how do we find the limits of sequences? We will use the following theorems and facts. The following
theorem is basically telling us that we take the limits of sequences much like we take the limit of functions. In
fact, in most cases we’ll not even really use this theorem by explicitly writing down a function. We will more
often just treat the limit as if it were a limit of a function.
Theorem 1.2.1. Given the sequence {an } if we have a function f (x) such that f (n) = an and lim f (x) = L,
x→∞
then lim an = L.
n→∞

The converse of Theorem 1.2.1 is false. Consider, for example, the sequence {sin nπ} = {0} This sequence
evidently converges to 0, since every term of the sequence is 0. But lim sin πx does not exist. In this section
x→∞
we continue our analysis of the convergence and divergence of sequences. Since sequences are functions, we
may add, subtract, multiply, and divide sequences just as we do for functions on Applied Mathematics I. Rules
for computing the limits of combination of sequences are analogous to the rules for limits of combinations of
functions. We present these rules now.
Theorem 1.2.2. Suppose lim an = L and lim bn = M and that c is a constant. Then
n→∞ n→∞

1. lim can = cL
n→∞

2. lim (an ± bn ) = L ± M
n→∞

3. lim an bn = LM
n→∞

4. lim ( abnn ) = L
M, provided that bn 6= 0 and and M 6= 0.
n→∞

5. lim (an )p = Lp , if p > 0 and an > 0.

n→∞

lim an
6. lim ean = en→∞
n→∞

Definition 1.2.2. Let {an }∞ ∞

n=m be a sequence. A number L is the limit of {an }n=m if for every ε > 0 there
is an integer N such that if n > N , then |an − L| < ε.
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DEPARTMENT OF MATHEMATICS Applied Mathematics II
1.2 Convergence properties of sequences 6

Theorem 1.2.3. If a sequence of real numbers {an }∞

n=m has a limit, then this limit is unique.
 
Example 1.2.2. Find the limit of the sequence enn .

Solution. The expression enn is an indeterminate form of type ∞/∞ as n → ∞, so L’Hopital’s rule is called
for. However, we cannot apply this rule directly to enn because the functions n and en are only defined at the
positive integers, and hence are not differentiable functions. To circumvent this problem, we will replace n by
x, and apply L’Hopital’s rule to the function exx . This yields

x 1
lim x
= lim x = 0
x→∞ e x→∞ e
using Theorem 1.2.1 we can conclude that
n
lim =0
n→∞ en
J
Example 1.2.3. Determine if the following sequences converge or diverge. If the sequence converges determine
its limit.
( )∞
3n2 −1
A. 10n+5n2
n=1

Solution. In this case all we need to do is recall the method that was developed in Applied Mathematics
I to deal with the limits of rational functions.[Please if you do not remember go to Applied Mathematics
I and read about Limit] To do a limit in this form all we need to do is factor from the numerator and
denominator the largest power of n, cancel and then take the limit.

3n2 − 1 n2 (3 − n12 ) 3 − n12 lim 3 − n12 3−0 3

n→∞
lim = lim 2 10 = lim 10 = 10 = = .
n→∞ 10n + 5n2 n→∞ n ( 0 + 5 5
n + 5) n +5 lim n + 5
n→∞
n→∞

So the sequence converges and its limit is 35 . J

( )∞
2n
e
B. n
n=1

Solution. We will need to be careful with this one. We will need to use L’Hospital’s Rule on this sequence.
The problem is that L’Hospital’s Rule only works on functions and not on sequences. Normally this would
2x
be a problem, but we’ve got Theorem 1.2.1 from above to help us out. Let’s define f (x) = ex and note
2n
that, f (n) = en . Theorem 1.2.1 says that all we need to do is take the limit of the function.

e2n e2x 2e2x

lim = lim f (x) = lim = lim =∞
n→∞ n x→∞ x→∞ x x→∞ 1

So, the sequence in this part diverges (to ∞ ). J

C. {(−1)n }

Solution. For this sequence all that we need to do is acknowledge that lim (−1)n doesn’t exist to get
n→∞
that the sequence is divergent. If youre not convinced that this limit doesnt exist write down the first few
terms of the sequence.   ∞
(−1)n
n=0
In order for a limit to exist the terms must be settling down towards a specific value and these clearly will
never do that. J

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
1.2 Convergence properties of sequences 7

Theorem 1.2.4 (Squeezing Theorem). If there exists some integer N such that an ≤ bn ≤ cn for all n ≥ N
and
lim an = lim cn = L, then lim bn = L.
n→∞ n→∞ n→∞
( )∞
cos
√n
Example 1.2.4. Find the limit of the sequences n
.
n=1

Solution. From the well known property of Cosine, we recall −1 ≤ cos(n) ≤ 1, ∀n.
−1 cos n 1
⇒ √ ≤ √ ≤ √ as n ≥ 1.
n n n
−1
Since √ n
→ 0 and √1n → 0 as n → 0, by Squeezing theorem cos
√n
n
→ 0.
Therefore, lim cos
√ n = 0.
n
J
n→∞

Definition 1.2.3. If the sequence {an } diverges but does not diverge to +∞ or does not diverge to −∞, then
the sequence {an } is said to oscillate or to be an oscillating sequence.
Example 1.2.5. {(−1)n } and {sin(nπ/4)} are oscillating sequences.

The next theorem is convenient for sequences that alternate in signs and note that it will only work if the
sequence has a limit of zero.
Theorem 1.2.5. If lim |an | = 0, then lim an = 0.
n→∞ n→∞
n
Example 1.2.6. Determine if the sequence { (−1) ∞
n }n=1 converge or diverge. If the sequence converges determine
its limit.
Solution. We will need to use Theorem 1.2.5 on this problem.

(−1)n 1
lim = lim =0
n→∞ n n→∞ n

Therefore, since the limit of the sequence terms with absolute value bars on them goes to zero we know by
Theorem 1.2.5 that,
(−1)n
lim =0
n→∞ n
Which also means that the sequence converges to a value of zero. J

Example 1.2.7. Show that if −1 < r < 1 then lim rn = 0.

n→∞

1 1
Solution. If r = 0, the result is trivial, so suppose otherwise. Then |r| > 1, and so |r| = 1 + p for some p > 0.
1
By Binomial Formula
1
= (1 + p)n = 1 + pn + positive terms ≥ pn
|r|n
Thus,
1
0 ≤ |r|n ≤
pn
1 1
Since lim = lim n1 = 0, it follows from the Squeezing Theorem that lim |r|n = 0, or equivalently,
n→∞ pn p n→∞ n→∞
lim |r | = 0. By Theorem 1.2.5, lim rn = 0.
n
J
n→∞ n→∞

Theorem 1.2.6. If lim an = L and the function f is continuous at L , then

n→∞

lim f (an ) = f ( lim an ) = f (L).

n→∞ n→∞
( )
1
Example 1.2.8. Show using the logarithmic function that a n , for any a > 0 converges to 1.

n
1 Binomial n
r n−r
Formula: (p + q)n =
P
r
p q
r=0
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DEPARTMENT OF MATHEMATICS Applied Mathematics II
1.2 Convergence properties of sequences 8

1
Solution. Let xn = a n . We have
1
⇒ ln(xn ) = ln a
n
⇒ lim ln(xn ) = 0,
n→∞

1
Using theorem 1.2.6, lim eln(xn ) = e0 = 1. Hence, lim a n = 1. J
n→∞

Exercise 1.2.1. 1. Find the limit of the given sequence and prove that the sequence converges to that limit.
√
A. {1/2n } C. {1 − (1/n)} E. {1 + ((−1)n / n)}
B. {1/n} D. {(−1)n /n2 } F. {n/(1 + n2 )}
 n 
a
2. Show that 1+ n converges to ea , for any number a.
 ∞
sin
√n
3. Find the limit of the sequence n
.
n=1
 ∞
sin2 n
4. Find the limit of the sequence n .
n=1
 ∞
ln n
5. Show that the sequence n converges to 0.
n=1
 ∞
2n3 +3
6. Determine the limit of 3n3 −4 .
n=1

1.2.1 Monotonic and Bounded Sequence

We now need to take a look at some more terminology and definitions for sequences.
Definition 1.2.4. Given any sequence {an } we have the following.
1. We call the sequence {an } is increasing if an ≤ an+1 for every n.
2. We call the sequence {an } is decreasing if an ≥ an+1 for every n.
3. We call the sequence {an } is strictly increasing if an < an+1 for every n.
4. We call the sequence {an } is strictly decreasing if an > an+1 for every n.
5. If {an } is an increasing sequence or {an } is a decreasing sequence we call it is monotonic sequence.
Testing for Monotonicity:
In order for a sequence to be strictly increasing all pairs of successive terms an and an+1 , must satisfy
an+1 − an > 0. More generally, a monotone sequence can be classified as follows

Difference between successive terms Classifications

an+1 − an > 0 Strictly increasing
an+1 − an < 0 strictly decreasing
an+1 − an ≥ 0 Increasing
an+1 − an ≤ 0 decreasing
1
Example 1.2.9. Show that the sequence defined by {an } = { n! } is monotonic strictly decreasing.
Solution.
1 1
an+1 − an = −
(n + 1)! n!
1 − (n + 1)
=
(n + 1)!
n
=− < 0, thus an+1 < an , ∀n ≥ 1.
(n + 1)!
Therefore, {an } is strictly decreasing sequence. J
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DEPARTMENT OF MATHEMATICS Applied Mathematics II
1.2 Convergence properties of sequences 9

Exercise 1.2.2. Use an+1 − an to show that the given sequence {an } is strictly increasing or strictly decreasing.
 ∞  ∞  ∞
1 n n
A. n B. 2n+1 C. 4n−1
n=1 n=1 n=1
 ∞  ∞  ∞
1
D. 1− n E. n − 2n F. n − n2
n=1 n=1 n=1

If an and an+1 are successive terms in a strictly increasing sequence, then an < an+1 . If the terms in the
sequence are all positive, then we can divide both sides of the inequality by an to obtain 1 < aan+1 n
. More
generally, monotone sequences with positive terms can be classified as follows:

Ratio of successive terms Conclusion

an+1 /an > 1 Strictly increasing
an+1 /an < 1 Strictly decreasing
an+1 /an ≥ 1 Increasing
an+1 /an ≤ 1 Decreasing
 
n
Example 1.2.10. Show that the sequence {an } = n+1 is strictly increasing by by examining the ratio of the
successive terms.
n n+1
an = , and an+1 =
n+1 n+2
Thus,
an+1 (n + 1)/(n + 2) n+1 n+1 n2 + 2n + 1
= = . = (1.2)
an (n)/(n + 1) n+2 n n2 + 2n
Since the numerator in (1.2) exceeds the denominator, it follows that an+1 /an > 1 for n ≥ 1. This proves that
the sequence is strictly increasing.
Exercise 1.2.3. Use an+1 /an to show that the given sequence {an } is strictly increasing or strictly decreasing.
 ∞  ∞  ∞
n 2n −n
A. 2n+1 B. 2n +1 C. ne
n=1 n=1 n=1
 ∞  ∞  ∞
10n nn
D. (2n)! E. n! F. n − n2
n=1 n=1 n=1

If f (n) = an is the nth term of a sequence, and if f is differentiable for x ≥ 1, then we have the following
results:

Derivative of f for x ≥ 1
Conclusion for the sequence an = f (n)
f 0 (x) > 0 Strictly increasing
f 0 (x) < 0 Strictly decreasing
f 0 (x) ≥ 0 Increasing
f 0 (x) ≤ 0 Decreasing
 
n
Example 1.2.11. Consider the sequence an = n+1 . Let

x
f (x) =
x+1

so that the nth term in the given sequence is an = f (n). The function f is increasing for x ≥ 1 since

(x + 1)(1) − x(1) 1
f 0 (x) = = > 0.
(x + 1)2 (x + 1)2

Thus,
an = f (n) < f (n + 1) = an+1
which proves that the given sequence is strictly increasing.

Exercise 1.2.4. Use differentiation to show that the sequence is strictly increasing or strictly decreasing.
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1.2 Convergence properties of sequences 10

 ∞  ∞  ∞
n 1 1
A. 2n+1 B. 3− n C. n+ln n
n=1 n=1 n=1
 ∞  ∞  ∞
−2n ln(n+2) −1
D. ne E. n+2 F. tan n
n=1 n=1 n=1

We now develop a method for determining the convergence of a sequence without direct application of the
definition of convergence. We begin by defining some special types of sequences.
Definition 1.2.5. Given any sequence {an } we have the following.
1. If there exists a number m such that m ≤ an for every n we say the sequence is bounded below. The
number m is sometimes called a lower bound for the sequence.

2. If there exists a number M such that M ≥ an for every n we say the sequence is bounded above. The
number M is sometimes called an upper bound for the sequence.
3. If the sequence is both bounded below and bounded above we call the sequence is bounded.
Definition 1.2.6. If A is a lower bound of a sequence {an } and if A has the property that for every lower
bound C of {an }, C ≤ A, then A is called the greatest lower bound of the sequence. Similarly, if B is an
upper bound of a sequence {an } and if B has the property that for every upper bound D of {an }, B ≤ D , then
B is called the least upper bound of the sequence.
The Axiom of Completeness
Every nonempty set of real numbers which has a lower bound has a great- est lower bound. Also, every set of
real numbers which has an upper bound has a least upper bound.

Theorem 1.2.7. Every bounded monotonic sequence is convergent.

Proof. We prove the theorem for the case when the monotonic sequence is increasing. Let the sequence be {an }.
Because {an } is bounded, there is an upper bound for the sequence. By the axiom of completeness , {an } has
a least upper bound, which we call B. Then if  is a positive number, B −  cannot be an upper bound of the
sequence because B −  < B and B is the least upper bound of the sequence. So for some positive integer N

B −  < aN (1.3)

Because B is the least upper bound of {an }, by Definition 1.2.5 it follows that

an ≤ B, for every positive integer n. (1.4)

Because {an } is an increasing sequence, we have from Definition 1.2.4(1)

an ≤ an+1 , for every positive integer n

and so
aN ≤ an , whenever n ≥ N (1.5)
From 1.3, 1.4, and 1.5 it follows that

B −  < aN ≤ an ≤ B < B +  whenever n ≥ N

from which we get

B −  < an < B + , whenever n ≥ N
which can be written as
|an − B| <  whenever n ≥ N.
This implies that lim an = B. Therefore, the sequence {an } is convergent.
n→∞

2n
Example 1.2.12. Investigate the convergence of the sequence n! .

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1.3 Subsequence and limit points 11

Multiplying both sides of (1.6) by an > 0

gives us an+1 ≤ an for all n and so, the 2n
n n!
sequence is decreasing. Next, since the se-
2 2
quence is decreasing, we have that
4 0.666667
2n 21 6 0.088889
|an | = ≤ = 2, 8 0.006349
n! 1!
10 0.000282
for n ≥ 1 (i.e., the sequence is bounded by 12 0.0000086
2). Since the sequence is both bounded and 14 1.88 × 10−7
monotonic, it must be convergent, by The- 16 3.13 × 10−9
orem 1.2.7. We display a number of terms 18 4.09 × 10−11
of the sequence in the accompanying table, 20 4.31 × 10−13
from which it appears that the sequence is
converging to approximately 0.

2n ∞
Solution. First, note that we do not know how to compute lim . This has the indeterminate form /∞ ,
n→∞ n!
n
but we cannot use L’Hopital’s Rule here directly or indirectly. (Why not?) The graph of 2n! suggests that the
sequence converges to 0. To confirm this suspicion, we first show that the sequence is monotonic. We have
2n+1
an+1 (n+1)! 2n+1 n!
= 2n = ·
an n!
(n + 1)! 2n
n
2(2 )n! 2
= n
= ≤ 1, for n ≥ 1 (1.6)
(n + 1)n!2 n+1
We can make a slightly stronger statement, though. Since we have established that the sequence is decreasing
and convergent, we have from our computations that

0 ≤ an ≤ a20 ≈ 4.31 × 10−13 , for n ≥ 20.

Further, the limit L must also satisfy the inequality

0 ≤ L ≤ 4.31 × 10−13 .

We can confirm that the limit is 0, as follows. From (1.6),

2
L = lim an+1 = lim an ,
n→∞ n→∞ n+1
So that
2
L = lim lim an = 0 · L = 0.
n→∞ n + 1 n→∞
J
Theorem 1.2.8. Every convergent monotonic sequence is bounded.
Exercise 1.2.5. Determine if the following sequences are monotonic and/or bounded.
 ∞  ∞
2 n
A. − n D. n+1
n=1 n=1
 ∞  ∞
(−1)n
B. (−1)n+1 E. n
n=1 n=1
 ∞  ∞
2 nπ
C. n2 F. sin( 4 )
n=1 n=1

1.3 Subsequence and limit points

We now introduce the concept of a subsequence of a sequence.
Definition 1.3.1. Let {an } be a sequence , and let {nk } be a sequence of positive integers such that nk < nk+1
for each k, that is, {nk } = {nk }∞ ∞
k=1 is a strictly increasing sequence. Then the sequence {ank } = {ank }k=1 is
called a subsequence of the sequence {ank }.
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1.4 Definition of infinite series 12

Let’s have an example of a subsequence of a sequence.

Example 1.3.1. Consider the sequence { n1 }. If we let nk = 2k for each positive integer k, the corresponding
subsequence of { n1 } is { 2k
1 1
} = { 2n }. Furthermore, if we let {nk } be any strictly increasing sequence of positive
integers, then the sequence { nk } is a subsequence of the sequence { n1 }.
1

Next, lets have some results on convergence related to subsequences.

Note 1.3.1. If {an } is a sequence, then {an } is trivially a subsequence of itself.
Theorem 1.3.1. If the sequence {an } converges to A, then every subsequence of the sequence {an } also con-
verges to A.

Corollary 1.3.1. If the sequence {an } has two subsequences which converge to different limits, then the sequence
{an } is divergent.
Example 1.3.2. Consider the sequence {an } defined by the equation an = (−1)n for each positive integer n.
Then the subsequence a2n is defined by a2n = 1 for each positive integer n, and the subsequence a2n−1 is defined
by a2n−1 = −1 for each positive integer n. then, we have a2n → 1 and a2n−1 → −1. Hence Corollary 1.3.1
implies that the sequence {(−1)n } diverges.
Corollary 1.3.2. If the sequence {an } has a subsequence which is divergent, then the sequence {an } is divergent.
Example 1.3.3. Let an = sin( nπ 4 ) for each positive integer n, and let nk = 4k − 2 for each positive integer
k. Corollary 1.3.2 implies that the sequence {an } = sin( nπ
4 ) is divergent since this sequence has the divergent
subsequence
ank = {sin[(4k − 2)π/4]} = {(−1)k }.
Definition 1.3.2. Let {an } be a sequence. The number a is called a limit point of {an } if and only if there
is a subsequence of {an } converging to a.
Note that: limit point of a sequence and limit of a sequence are not the same. The following theorem points
out the relationship between the limit points of a sequence and the convergence of the sequence.
Theorem 1.3.2. Let {an } be a sequence. The point a is a limit point of {an } if and only if for each pair of
positive numbers ε and N there is a k > N such that

|ak − a| < ε.

Corollary 1.3.3. Let {an } be a sequence. The point a is a limit point of {an } if and only if for each ε > 0

|an − a| < ε

for infinitely many values of n.

   
Example 1.3.4. Consider the sequence (−1)n n+1
n . -1 and 1 are the limit points of the sequence (−1) n n+1
n .

Exercise 1.3.1. Find the limit points of the given sequences.

A. 1, 3, −1, 1, 3, −1, . . .
B. {cos(nπ/2)}
C. {(1 − 1/n)}
D. {(1 − 1/n)}{sin(nπ/2)}

E. {n}

1.4 Definition of infinite series

Definition 1.4.1. Given a sequence of numbers{an }, an expression of the form

a1 + a2 + a3 + · · · + an + · · ·

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1.4 Definition of infinite series 13

is an infinite series. The number an is the nth term of the series. The sequence {sn } defined by

s1 = a1
s2 = a1 + a2
s3 = a1 + a2 + a3
..
.
n
X
sn = a1 + a2 + a3 + · · · an = ak
k=1
..
.

is the sequence of partial sums of the series, the number sn being the nth partial sum.
Example 1.4.1. Find the first four elements of the sequence of partial sums sn and find a formula for sn in
terms of n.
Solution.
∞
X
A. n
n=1
s1 =1, s2 = 1 + 2 = 3, s3 = 1 + 2 + 3 = 6, s4 = 1 + 2 + 3 + 4 = 10
1
sn = n(n + 1)
2

∞
X 1
B.
n=1
(2n − 1)(2n + 1)
 
1 1 1
s1 = 1− =
2 3 3
   
1 1 1 1 1 1 2
s2 = 1− + − = (1 − ) =
2 3 3 5 2 5 5
     
1 1 1 1 1 1 1 1 3
s3 = 1− + − + − = (1 − ) =
2 3 3 5 5 7 2 7 7
       
1 1 1 1 1 1 1 1 1 1 4
s4 = 1− + − + − + − = (1 − ) =
2 3 3 5 5 7 7 9 2 9 9
       
1 1 1 1 1 1 1 1
sn = 1− + − + ··· + − + −
2 3 3 5 2n − 3 2n − 1 2n − 1 2n + 1
1 1 n
= (1 − )=
2 2n + 1 2n + 1

∞
X n
C. ln
n=1
n+1
∞ ∞
X n X
ln = [ln n − ln n + 1],
n=1
n + 1 n=1
s1 = ln 1 − ln 2 = − ln 2,
s2 = (ln 1 − ln 2) + (ln 2 − ln 3) = − ln 3
s3 = (ln 1 − ln 2) + (ln 2 − ln 3) + (ln 3 − ln 4) = − ln 4
s4 = (ln 1 − ln 2) + (ln 2 − ln 3) + (ln 3 − ln 4) + (ln 4 − ln 5) = − ln 5
sn = (ln 1 − ln 2) + (ln 2 − ln 3) + · · · + (ln(n − 1) − ln n) + (ln n − ln(n + 1)) = − ln(n + 1)

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1.5 Convergence and divergence, properties of convergent series 14

1.5 Convergence and divergence, properties of convergent series

Definition 1.5.1. Let {sn } be the sequence of partial sums of the series

a1 + a2 + · · · + an + · · ·

If the sequence {sn } converges to a limit S, then the series is said converge to S, and S is called the sum of
the series. We denote this by writing
∞
X
S= an .
n=1

If the sequence of partial sums diverges, then the series is said to diverge. A divergent series has no sum.
Remark 1.5.1. Sometimes it will be desirable to start the summation index in an infinite series at n = 0 rather
than n = 1, in which case we will view a0 as the zeroth term and s0 = a0 as the zeroth partial sum. It can
be proved that changing the starting value for the index has no effect on the convergence or divergence of an
infinite series.
Example 1.5.1. Determine whether the series converges. If the series converges, find its sum.
∞ ∞
X X 1
A. n B.
n=1 n=1
(2n − 1)(2n + 1)
∞
X n
C. ln
n=1
n+1

Exercise 1.5.1. Determine whether the series converges. If the series converges, find its sum.
∞ ∞
X 2n + 1 X 2
A. 2 (n + 1)2
B. n−1
n=1
n n=0
5
∞ ∞
X 2n−1 X 2
C. , D.
n=1
3n n=1
(4n − 3)(4n + 1)

Geometric Series
Geometric series play an important role in mathematical analysis. They also arise frequently in the field of
finance. The convergence or divergence of a geometric series is easily established.

Definition 1.5.2. A series of the form

∞
X
arn−1 = a + ar + ar2 + ar3 + · · · + arn−1 + · · · a 6= 0
n=1

is called a geometric series with common ratio r.

The following theorem tells us the conditions under which a geometric series is convergent.
Theorem 1.5.1. If |r| < 1 , then the geometric series
∞
X
arn−1 = a + ar + ar2 + ar3 + · · · + arn−1 + · · · a 6= 0
n=1

∞
a
arn−1 =
P
converges, and its sum is 1−r . The series diverges if |r| ≥ 1.
n=1

∞
 n−1
1 1 1 1 1 1
P
Example 1.5.2. 1. 1 + 2 + 4 + 8 + ··· = 2 = 1− 21
. Here a = 1 and r = 2. since |r| < 1, the
n=1
series converges.

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1.5 Convergence and divergence, properties of convergent series 15

∞
 n−1
e2 e3 e
. Here a = π and r = − πe
P
2. π − e + π − π2 + ··· = π − π
n=1
∞ n−1
π2

X e π
π − =  = .
π e π+e
n=1 1− π

e
The series converges since since − π < 1.

1 3
∞ √ √
( 2)n−1 . This series diverges to ∞ since a = 1 > 0 and r = 2 > 1.
P
3. 1 + 2 2 + 2 + 2 2 + · · · =
n=1

∞
(−1)n−1 . This series diverges since r = −1.
P
4. 1 − 1 + 1 − 1 + 1 − · · · =
n=1

5. Letx = 0.323232 · · · = 0.32; then

∞ n−1
322 323

32 X 32 1 32 1 32
x= + + + · · · = = 1 = 99
100 1002 1003 n=1
100 100 100 1 − 100

Telescoping Series
It is now time to look at the second of the three series in this section. In this portion we are going to look at a
series that is called a telescoping series. The name in this case comes from what happens with the partial sums
and is best shown in an example.
∞
1
P
Example 1.5.3. Show that the series n(n+1) , is convergent, and find its sum.
n=1

Solution. This is not a geometric series, so we go back to the definition of a convergent series and compute
the partial sums.
n
X 1 1 1 1 1
sn = = + + + ··· +
i=1
i(i + 1) 1 · 2 2 · 3 3 · 4 n(n + 1)
We can simplify this expression if we use the partial fraction decomposition
1 1 1
= −
i(i + 1) i i+1

Thus we have
n n
X 1 X 1 1
sn = = −
i=1
i(i + 1) n=1 i i+1
         
1 1 1 1 1 1 1 1 1
= 1− + − + − + ··· + − + −
2 2 3 3 4 n−1 n n n+1
1
=1−
n+1
And here,  
1
lim 1− = 1 − 0 = 1.
n→∞ n+1
Therefore, the given series is convergent and
∞
X 1
= 1.
n=1
n(n + 1)
J

Harmonic Series
The series
∞
X 1
n=1
n

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1.5 Convergence and divergence, properties of convergent series 16

Figure 1.1: A partial sum of the partial series

is called the Harmonic series. The harmonic series is divergent.

If sn is the partial sum of the Harmonic series, then
1 1 1
sn = 1 + + + · · · + = sum of the areas of rectangles shaded in Figure 1.1
2 3 n
1
> area under y = from x = 1 to x = n + 1
x
n+1
Z
dx
= = ln(n + 1)
x
1

Now lim ln(n + 1) = ∞. Therefore, lim sn = ∞ and

n→∞ n→∞

∞
X 1 1 1
= 1 + + + ··· diverges to infinity
n=1
n 2 3

Exercise 1.5.2. Determine whether the series converges. If the series converges, find its sum.
∞ ∞  
X 1 X 1 1
A. B. + n
n=1
n+2 n=1
2n 2
∞ ∞
X 3 X π
C. n
, D. tann
n=1
2 n=1
6
E. 0.272727 . . . D. 0.234234234 . . .

The next theorem tells us that the terms of a convergent series must ultimately approach zero.
∞
P
Theorem 1.5.2. If the series an is convergent, then lim an = 0.
n=1 n→∞

Remark 1.5.2. ThisP 1gives a necessary condition for the convergence of a series; it is not sufficient. For example
1
we have seen that n is divergent, even though n → 0 as n → ∞.
∞
P
Theorem 1.5.3 (The Test for Divergence). If lim an does not exist or if lim an 6= 0, then the series an
n→∞ n→∞ n=1
is divergent.
Example 1.5.4. Show that the following series are divergent
∞ ∞
X n X
A. B. (−1n )n sin(1/n)
n=2
2n −1 n=0

∞
n n
P
Solution. A. 2n−1 diverges to infinity since lim = 1/2 > 0
n=2 n→∞ 2n−1
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1.6 Nonnegative term series 17

∞
(−1n )n sin(1/n) diverges since
P
B.
n=0

1 sin n1 sin x
lim (−1n )n sin( ) = lim 1 = lim+ = 1 6= 0.
n→∞ n n→∞
n
x→0 x

The following properties of series are immediate consequences of the corresponding properties of the limits of
sequences.
∞
P ∞
P ∞
P ∞
P
Theorem 1.5.4. Let an and bn be two infinite series, and c is a constant. If an and bn converges
n=1 n=1 n=1 n=1
with sum a and b respectively, then

∞
X
A. (an ± bn ) converges to a ± b
n=1
X∞
B. can converges to ca
n=1
∞
X
C. If bn is divergent ,
n=1
∞
X
a) (an ± bn ) is divergent
n=1
X∞
b) cbn is divergent.
n=1

Example 1.5.5. Find the sum of the series

∞
X 1 + 2n+1
.
n=1
3n

Solution. The given series is the sum of two geometric series,

∞ ∞ 1
X 1 X 1 1 3
n
= n−1
= = 1/2
n=1
3 n=1
3 3 1 − 1/3

and
∞ ∞ 4
X 2n+1 X 4 2 n−1 3
= ( ) = = 4.
n=1
3n n=1
3 3 1 − 2/3
1 9
Thus its sum is 2 +4= 2 by Theorem 4.6.1(a). J

1.6 Nonnegative term series

If every term in a series is nonnegative, we may sometimes determine whether or not the series is convergent by
comparing it with a series of nonnegative terms that we know to be convergent or that we know to be divergent.

1.7 Tests of convergence (integral, comparison, ratio and root tests)

In this and the next two sections we will develop several tests for determining the convergence or divergence
of an infinite series by examining the nth term an of the series. These tests will confirm the convergence
of a series without yielding a value for its sum. From the practical point of view, however, this is all that
is required. Once it has been ascertained that a series is convergent, we can approximate its sum to any de-
gree of accuracy desired by adding up the terms of its nth partial sum sn , provided that n is chosen large enough.

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1.7 Tests of convergence (integral, comparison, ratio and root tests) 18

1.7.1 The Integral Test

∞
P
The Integral Test ties the convergence or divergence of an infinite series an to the convergence or divergence
n=1
R∞
of the improper integral f (x)dx, where f (n) = an .
1

Theorem 1.7.1 (The Integral Test). Suppose that f is a continuous, positive, and decreasing function on [1, ∞)
. If f (n) = an for n ≥ 1, then
X∞ Z∞
an and f (x)dx
n=1 1
either both converge or both diverge.
Example 1.7.1. Determine if the following series is convergent or divergent.
∞ ∞
X 1 X 2
A. B. ne−n
n=2
n ln n n=0
∞ ∞
X 1 X 1
C. p
, p = constant D. 2
n=1
n n=1
n

Solution. A.
ZM M
dx
lim = lim ln(ln x) = lim {ln(ln M ) − ln(ln 2)} = ∞
M →∞ x ln x M →∞ 2
M →∞
2
and the series diverges.
B.
ZM M
dx 1 2 1 1 2 1
lim = lim − e−x = lim { e−1 − e−M } = e−1
M →∞ xe−x2 M →∞ 2 1
M →∞ 2 2 2
2
and the series converges.
C. Consider
ZM ZM M
dx x1−p M 1−p − 1
= x−p dx = =
xp 1−p 1 1−p
1 1
where p 6= 1.
M 1−p −1
– If p < 1, lim 1−p = ∞, so that the integral and thus the series diverges.
M →∞
1−p
M −1 1
– If p > 1, lim 1−p = p−1 , so that the integral and thus the series converges.
M →∞
RM dx RM dx
– If p = 1, xp = x = ln M and lim ln M = ∞, so that the integral and thus the series diverges.
1 1 M →∞

D. Since p = 2 > 1, then the integral and thus the series converges.
J
Exercise 1.7.1. Use the integral test to determine the convergence of the following series.
∞ ∞
X 3 X 1
A. 7n2 e−n B. √
n=1 n=1
2n + 9
∞ 3 ∞
X [ln n] X 4n
C. D. 2+9
n=1
n n=1
n

P-series
∞
1
P
If k > 0 then np converges if p > 1 and diverges if p ≤ 1. Sometimes the series in this fact are called p-series
n=k
and so this fact is sometimes called the p-series test.
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Example 1.7.2. Determine if the following series is convergent or divergent.

∞ ∞
X 1 X 1
A. 7
B. √
n=4
n n=1
n
∞ ∞
X 1 X 1
C. √ D. 2
n=1
3
n n=1
n

Solution. A. In this case p = 7 > 1 and so by this fact the series is convergent.
1
B. For this series p = 2 ≤ 1 and so the series is divergent by the fact.
1
C. For this series p = 3 ≤ 1 and so the series is divergent by the fact.
D. For this series p = 2 > 1 and so the series is convergent by the fact.
J
It is important to note before leaving this section that in order to use the Integral Test the series terms
MUST be positive. If they are negative then the test doesn’t work. Also remember that the test only determines
the convergence of a series and does NOT give the value of the series.

1.7.2 Comparison Test

The first part says that if we have a series whose terms are smaller than those of a known convergent series,
then our series is also convergent. The second part says that if we start with a series whose terms are larger
than those of a known divergent series, then it too is divergent.
P P
Theorem 1.7.2 (The Comparison Test). Suppose that an and bn are series with positive terms.
P P
a. If bn is convergent and an ≤ bn for all n, then an is also convergent.
P P
b. If bn is divergent and an ≥ bn for all n, then an is also divergent.
Example 1.7.3. Determine if the following series is convergent or divergent.
∞ ∞
X n X n2 + 2
A. 2 − cos2 n
B.
n=1
n n=1
n4 + 5
∞ ∞
X 5 X 1
C. 2 + 4n + 3
D.
n=1
2n n=2
ln n

Solution. A. Since the cosine term in the denominator doesn’t get too large we can assume that the series
terms will behave like,
n 1
=
n2 n
which, as a series, will diverge. So, from this we can guess that the series will probably diverge and so
we’ll need to find a smaller series that will also diverge.
n n 1
> 2 = .
n2 − cos2 (n) n n
∞
1
P
Then, since n diverges (it’s harmonic and the p-series test) by the Comparison Test our original series
n=1
must also diverge.
B. The series terms should behave pretty much like

n2 1
= 2
n4 n
which will converge as a series. Therefore, we will need to find a larger series which also converges. So,

n2 + 2 n2 + 2
4
< .
n +5 n4
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DEPARTMENT OF MATHEMATICS Applied Mathematics II
1.7 Tests of convergence (integral, comparison, ratio and root tests) 20

Then,
∞ ∞ ∞ ∞ ∞
X n2 + 2 X n2 X 1 X 1 X 1
4
= 4
+ 4
= 2
+
n=1
n n=1
n n=1
n n=1
n n=1
n4
As shown, we can write the series as a sum of two series and both of these series are convergent by the
p-series test. Therefore, since each of these series are convergent we know that the sum,

n2 + 2
n4
is also a convergent series. Recall that the sum of two convergent series will also be convergent. Now,
since the terms of this series are larger than the terms of the original series we know that the original
series must also be convergent by the Comparison Test.
2
C. For
P 5large n the dominant term in the denominator is 2n so we compare the given series with the series
2n2 . Observe that
5 5
2
< 2
2n + 4n + 3 2n
because the left side has a bigger denominator. (In the notation of the Comparison Test, an is the left
side and bn is the right side.) We know that
∞ ∞
X 5 5X 1
=
n=1
2n2 2 n=1 n2

is convergent because it’s a constant times a p-series with p = 2 > 1. Therefore

∞
X 5
n=1
2n2

is convergent by part (i) of the Comparison Test.

∞
1 1 1
P
D. For n = 2, 3, 4, . . . , we have 0 < ln n < n. Thus ln n > n. Since n diverges to infinity(it is a harmonic
n=2
∞
1
P
series), so does ln n by harmonic series.
n=2
J
The terms of the series being tested must be smaller than those of a convergent series or larger than those
of a divergent series. If the terms are larger than the terms of a convergent series or smaller than those of a
divergent series, then the Comparison Test doesn’t apply. Consider, for instance, the series
∞
X 1
2n−1
n=1

The inequality
1 1
> n
−1 2n2
bn = ( 12 )n is convergent and an > bn . Nonethe-
P
is useless as far as the Comparison
P Test is concerned because
1
less, we have the feeling that 2n −1 ought to be convergent because it is very similar to the convergent
geometric series ( 12 )n . In such cases the following test can be used.
P

Exercise 1.7.2. Use the Comparison test to determine the convergence of the following series.
∞ ∞
X n X 5n
A. 3+1
B.
n=1
n n=1
7n + 1
∞ ∞
X 1 X n2
C. D.
n=1
n + n2 + 5
3
n=1
n3 − 5

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1.7 Tests of convergence (integral, comparison, ratio and root tests) 21

1.7.3 Limit Comparison Test

P P an
Theorem 1.7.3. Let an and bn be non negative series. Suppose lim = L, where L is a positive
n→∞ bn
number. If L is positive (i.e. L > 0 ) and is finite (i.e. L < ∞ ), then either both series converge or both series
diverge.
Example 1.7.4. Determine if the following series is convergent or divergent.
∞ ∞
X 1 X n+5
A. √ B. 3 − 2n + 3
n=1
1 + n n=1
n
∞ ∞
X 2n2 + 3n X 1
C. √ D. n−1
n=1
5 + n5 n=2
2

Solution. A. The terms of the series decrease √1 . Observe that

n

1√ √
1+ n n 1
L = lim = lim √ = lim √ = 1.
n→∞ √1 n→∞ 1+ n n→∞ (1/ n) + 1
n

∞ ∞
√1 1√
P P
Since the p−series n
diverges to ∞ (p = 1/2) so does the series 1+ n
by Limit Comparison Test.
n=1 n=1
∞
1
B. For large n the terms behave like n/n3 so let us compare the series with the p− series
P
n2 , which we
n=1
know converges.
n+5
n3 −2n+3 n3 + 5n2
L = lim 1 = lim = 1.
n→∞ n→∞ n3 − 2n + 3
n2
∞
n+5
P
Since L < ∞, the series n3 −2n+3 also converges by Limit Comparison Test.
n=1
√ 5
C. The dominant part of the numerator is 2n2 and the dominant part of the denominator is n5 = n 2 . This
suggests taking
2n2 + 3n 2n2 2
an = √ , bn = 5 = 1
5 + n5 n2 n2
1 5 3
an 2n2 + 3n n 2 2n 2 + 3n 2 2+ 3 2+0
lim = lim √ = lim √ = lim q n = √ =1
n→∞ bn n→∞ 5+n 2 5 n→∞ 2 5 + n 5 n→∞ 2 0+1
2 n55 + 1

bn = 2 1/n1/2 is divergent ( p-series with p = 12 < 1 ), the given series diverges by the Limit
P P
Since
Comparison Test.
D. We use the Limit Comparison Test with
1 1
an = , bn =
2n −1 2n
and obtain
1
an n 2n 1
lim = lim 2 1−1 = lim n = lim =1>0
n→∞ bn n→∞
2n
n→∞ 2 − 1 n→∞ 1 − 1/2n

1/2n is a convergent geometric series, the given series converges by the Limit
P
Since this limit exists and
Comparison Test.
J
Exercise 1.7.3. Determine if the following series is convergent or divergent.
∞ ∞
X 1 X n
A. 2+1
B. 2−3
n=1
n n=1
4n
∞ ∞
X n+2 X 3n
C. √ D.
n=1
(n + 1) n + 3 n=2
n5n

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
1.8 Alternating series and alternating series test 22

1.8 Alternating series and alternating series test

The convergence tests that we have looked at so far apply only to series with positive terms. In this section and
the next we learn how to deal with series whose terms are not necessarily positive. Of particular importance
are alternating series, whose terms alternate in sign. The last two tests that we looked at for series convergence
have required that all the terms in the series be positive. Of course there are many series out there that have
negative terms in them and so we now need to start looking at tests for these kinds of series.

Definition 1.8.1. An alternating series is a series whose terms are alternately positive and negative.
Example 1.8.1.
∞
X (−1)n−1 1 1 1 1 1 1
A. = 1 − + − + − + + ···
n=1
n 2 3 4 5 6 7
∞
X n 1 2 3 4 5
B. (−1)n = − + − + − + ···
n=1
n+1 2 3 4 5 6

We see from these examples that the nth term of an alternating series is of the form

an = (−1)n−1 bn or an = (−1)n bn .

Where bn is a positive number (In fact, bn = |an |.)) The following test says that if the terms of an alternating
series decrease toward 0 in absolute value, then the series converges.

1.8.1 Alternating Series Test

an and either an = (−1)n bn or an = (−1)n+1 bn where bn ≥ 0, for all n. Then
P
Suppose that we have a series
if,
1. lim bn = 0
n→∞

2. bn+1 ≤ bn for all n.

then the series converges.
Example 1.8.2. Determine if the following series is convergent or divergent.
∞ ∞
X (−1)n+1 X (−1)n n2
A. B.
n=1
n n=1
n2 + 5
∞ √ ∞
X
n−3 n X cos(nπ)
C. (−1) D. √
n=0
n+4 n=1
n

Solution. A. First, identify the bn for the test.

∞ ∞
X (−1)n+1 X 1 1
= (−1)n+1 , bn =
n=1
n n=1
n n

Now, all that we need to do is run through the two conditions in the test.
1
lim bn = lim =0
n→∞ n→∞ n

and
1 1
bn =
> = bn+1 .
n n+1
Both conditions are met and so by the Alternating Series Test the series must converge.
B. First, identify the bn for the test.
∞ ∞
X (−1)n n2 X
n n
2
n2
= (−1) , bn =
n=1
n2 + 5 n=1
n2 + 5 n2 + 5

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
1.9 Absolute and conditional convergence 23

Lets check the conditions.

n2
lim bn = lim = 1.
n→∞ n→∞ n2 +5
So, the first condition isn’t met and so there is no reason to check the second. Since this condition isn’t
met we’ll need to use another test to check convergence. In these cases where the first condition isn’t met
it is usually best to use the divergence test.
(−1)n n2 n2
  
n
lim = lim (−1) lim
n→∞ n2 + 5 n→∞ n→∞ n2 + 5
 
= lim (−1)n (1)
n→∞

= lim (−1)n doesn’t exist

n→∞

This limit doesn’t exist and so by the Divergence Test this series diverges.
√
n
C. Let bn = n+4 . √
n
lim bn = lim =0
n→∞ n→∞ n+4
Let’s start with the following function and its derivative.
√
x 4−x
f (x) = , f 0 (x) = √
x+4 2 x(x + 4)2
Now, there are three critical points for this function, x = −4, x = 0, and x = 4. The first is outside the
bound of our series so we won’t need to worry about that one. Using the test points,
√
0 3 0 5
f (1) = f (5) =
50 810
and so we can see that the function in increasing on 0 ≤ x ≤ 4 and decreasing on x ≥ 4. Therefore,
f (n) = bn since we know as well that the bn are also increasing on 0 ≤ n ≤ 4 and decreasing on n ≥ 4.
The bn are then eventually decreasing and so the second condition is met. Both conditions are met and
so by the Alternating Series Test the series must be convergent.
D. To check for the convergence or divergence of this series use the fact cos(nπ) = (−1)n and then
∞ ∞
X cos(nπ) X (−1)n
√ = √
n=1
n n=1
n

1.9 Absolute and conditional convergence

We have convergence tests for series with positive terms and for alternating series. But what if the signs of the
terms switch back and forth irregularly? We will see that the idea of absolute convergence sometimes helps in
such cases.
P P
Definition 1.9.1. A series an is called absolutely convergent if the series of absolute values |an | is
convergent.
P
Notice that if an is a series with positive terms, then |an | = an and so absolute convergence is the same
as convergence in this case.
Example 1.9.1. The series
∞
X (−1)n−1 1 1 1
2
= 1 − 2 + 2 − 2 + ...
n=1
n 2 3 4
is absolutely convergent because
∞
X (−1)n−1 1 1 1 1
2
= 2 = 1 + 2 + 2 + 2 + ...
n=1
n n 2 3 4

is a convergent p-series ( p = 2).

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
1.9 Absolute and conditional convergence 24

Example 1.9.2. We know that the alternating harmonic series

∞
X (−1)n−1 1 1 1
= 1 − + − + ...
n=1
n 2 3 4

is convergent, but it is not absolutely convergent because the corresponding series of absolute values is
∞
X (−1)n−1 1 1 1 1
= = 1 − + − + ...
n=1
n n 2 3 4

which is the harmonic series ( p−series with p = 1) and is therefore divergent.

P
Definition 1.9.2. A series an is called conditionally convergent if it is convergent but not absolutely
convergent.
Example 1.9.2 above shows that the alternating harmonic series is conditionally convergent. Thus, it is
possible for a series to be convergent but not absolutely convergent. However, the next theorem shows that
absolute convergence implies convergence. We also have the following fact about absolute convergence.
P
Theorem 1.9.1. If an is absolutely convergent, then it is also convergent.

1.9.1 Ratio Test

In this section we are going to take a look at a test that we can use to see if a series is absolutely convergent
or not. Recall that if a series is absolutely convergent then we will also know that it’s convergent and so we
will more often than not use it to simply determine the convergence of a series. Before proceeding with the test
let’s do a quick reminder of factorials. This test will be particularly useful for series that contain factorials. If
n is an integer such that n ≥ 0, then n factorial is defined as,
n! = n(n − 1)(n − 2)(n − 3) . . . 3 · 2 · 1 if n ≥ 1.
0! = 1! = 1 by definition
Also, when dealing with factorials we need to be very careful with parenthesis. For (2n)! 6= 2n! as we can see
(2n)! = (2n)(2n − 1)(2n − 2)(2n − 3)(2n − 4) · · · · · 3 · 2 · 1
2n! = 2[(n)(n − 1)(n − 2) · · · · · 3 · 2 · 1]
P
Theorem 1.9.2 (Ratio Test). Suppose we have the series an . Define
an+1
r = lim
n→∞ an
Then,
1. If r < 1 the series is absolutely convergent (and hence convergent).
2. If r > 1 the series is divergent.
3. If r = 1 the series may be divergent, convergent, or absolutely convergent.
Example 1.9.3. Determine if the series is convergent or divergent.
∞ ∞
X n3 X n!
A. (−1)n B. (−1)n
n=1
3n n=1
5n
∞ ∞
X 1 X 1
C. (−1)n D. (−10)n
n=0
n2 + 1 n=1
42n+1 (n + 1)

Solution. A. We use the Ratio Test with

n3
an = (−1)n
3n
(n + 1)3
⇒ an+1 = (−1)n+1 n+1
3
3
(−1)n+1 (n+1)
3
(n + 1)3 3n

an+1 3n+1 1 n+1 1
r = lim = lim 3 = lim = lim = <1
n→∞ an n→∞ n
(−1) 3nn n→∞ 3n+1 n3 n→∞ 3 n 3
Thus, by the Ratio Test, the given series is absolutely convergent and therefore convergent.
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DEPARTMENT OF MATHEMATICS Applied Mathematics II
1.9 Absolute and conditional convergence 25

B. Here is the limit.

an+1 (n + 1)!5n 1 (n + 1)! n+1
r = lim = lim = lim = lim =∞>1
n→∞ an n→∞ 5n+1 n! n→∞ 5 n! n→∞ 5

So, by the Ratio Test this series diverges.

C. Let’s first get r,
1
an+1 (−1)(n+1) (n+1) 2 +1 n2 + 1
r = lim = lim = lim =1
n→∞ an n→∞ (−1)n n21+1 n→∞ (n + 1)2 + 1

So, as implied earlier we get r =1 which means the ratio test is no good for determining the convergence
of this series. We will need to resort to another test for this series. This series is an alternating series and
so lets check the two conditions from that test.
1
lim =0
n→∞ n2 + 1
1 1
bn = > = bn+1 .
n2 + 1 (n + 1)2 + 1
The two conditions are met and so by the Alternating Series Test this series is convergent.
D. Check that this series is convergent.
J
 
an+1 P
Raabe’s test. Let lim n 1 − an = L. Then the series an
n→∞

(a) converges (absolutely) if L > 1.

(b) diverges or converges conditionally if L < 1.
If L = 1 the test fails. This test is often used when the ratio tests fails.
Gauss’ test
If aan+1 =1− L cn P
n n + n2 , where |cn | < P for all n > N , then the series an

(a) converges (absolutely) if L > 1.

(b) diverges or converges conditionally if L ≤ 1.
This test is often used when Raabe’s test fails.

Exercise 1.9.1. Use the ratio test to determine the convergence of the following series.
∞ ∞
X 2n X 1
A. B.
n=1
n! n=1
n2 +1
∞ ∞
X X 5n
C. ne−n D.
n=0 n=0
2n + 3n

1.9.2 Root Test

This is the last test for series convergence that we’re going to be looking at. As with the Ratio Test this test will
also tell whether a series is absolutely convergent or not rather than simple convergence. This test is convenient
to apply when nth powers occur.
P
Theorem 1.9.3 (Root Test). Suppose we have the series an . Define
p 1
r = lim n |an | = lim |an | n
n→∞ n→∞

Then,
1. If r < 1 the series is absolutely convergent (and hence convergent).
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DEPARTMENT OF MATHEMATICS Applied Mathematics II
1.10 Generalized convergence tests 26

2. If r > 1 the series is divergent.

3. If r = 1 the series may be divergent, convergent, or absolutely convergent.
As with the ratio test, if we get r = 1 the root test will tell us nothing and we’ll need to use another test to
1
determine the convergence of the series. As an exercise for the reader please show that lim n n = 1.
n→∞

Example 1.9.4. Determine if the series is convergent or divergent.

∞ ∞
X nn X 2n + 3 n
A. 1+2n
B. ( )
n=1
3 n=1
3n + 2
∞ ∞
X 2n X 2n
C. D.
n=1
n n=1
n2

Solution. A. Here is the limit of the sequence

1
nn n
n
r = lim = lim 1 =∞>1
n→∞ 31+2n 3 n +2

So, by the Root Test this series is divergent

B. Now lets find the value of r
1
2n + 3
n n
2n + 3 2
r = lim = lim = <1
n→∞ 3n + 2 n→∞ 3n + 2 3

Hence, by the Root Test this series is convergent.

C. Similarly,
1
2n n
2
r = lim = lim 1
n→∞ n n→∞ nn
1
Now let’s compute lim n n = 1 Hence,
n→∞

1
2n n
2
r = lim = lim 1 =2>1
n→∞ n n→∞ nn
Therefore, by the root test the given series diverges.
J

Note: if we get r = 1 from the ratio test then the root test will also give r = 1 and so there isn’t any reason
to try the root test on anything that gives r = 1 on the ratio test.

1.10 Generalized convergence tests

By combining absolute convergence tests with tests for non negative series, we obtain convergence tests that
apply to any series, non negative or not.
P
Theorem 1.10.1 (Generalized convergence tests). Let be an a series.
∞
P ∞
P
a) Generalized comparison test: If |an | ≤ |bn | and |bn | converges, then an converges (absolutely).
n=1 n=1

an P
b) Generalized Limit comparison test: If lim bn = L, where L is a positive number, and if bn converges
n→∞
P
then an converges (absolutely).

an
c) Generalized ratio test: Suppose that an 6= 0 for n ≥ 1 and lim = r(possibly ∞) that
n→∞ an+1

∞
P
i) If r < 1, then an converges (absolutely).
n=1
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DEPARTMENT OF MATHEMATICS Applied Mathematics II
1.10 Generalized convergence tests 27

∞
P
ii) If r > 1, then an diverges
n=1
iii) If r = 1, no conclusion can be drawn.
p
d) Generalized root test: Suppose that lim n |an | = r (possibly ∞).
n→∞

∞
P
i) If r < 1, then an converges (absolutely).
n=1
P∞
ii) If r > 1, then an diverges
n=1
iii) If r = 1, no conclusion can be drawn.
∞
xn 2 3
= x+ x2 + x3 +· · · converges absolutely for |x| < 1. converges conditionally
P
Example 1.10.1. Show that n
n=1
for x = −1, and diverges for x = 1 and for |x| > 1.

Solution. If x = 0, the series obviously converges. If x 6= 0, then

xn+1 /(n + 1) n
lim n
= |x| lim = |x|
n→∞ x /n n→∞ n + 1

Therefore the Generalized Ratio Test implies that the given series converges for |x| < 1 and diverges |x| > 1. J
xn
Example 1.10.2. Show that lim = 0 for all x.
n→∞ n!

xn
Solution. If x = 0, then the limit is 0. If x 6= 0, let an = n! . Then

an+1 xn+1 /(n + 1)! 1

r = lim = lim = lim |x| = 0.
n→∞ an n→∞ xn /n! n→∞ n + 1

Since r < 1, the series converges for all x . J

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
Chapter 2

Power Series

2.1 Introduction
Until now, we have discussed a series with constant terms. Now, We extend our discussion of series to the case
where the terms of the series are functions of the variable x. Pay close attention, as the primary reason for
studying series is that we can use them to represent functions. This opens up numerous possibilities for us. For
instance,
• From approximating the values of transcendental functions to calculating derivatives and integrals of such
functions
• Developing the mathematical tools needed to investigate the convergence of Taylor and Maclaurin series.
• Defining functions as convergent series produces an explosion of new functions available to us, including
many important functions, such as the Bessel functions.

2.2 Definition of power series at any x0 and x0 = 0

Definition 2.2.1. Let x be a variable. A power series in x or a power series centered at the origin is
a series of the form
∞
X
an xn = a0 + a1 x + a2 x2 + a3 x3 + · · · + an xn + · · ·
n=0
where the an ’s are constants and are called the coefficients of the series.
More generally, a power series in (x − c) , where c is constant, is a series of the form
∞
X
an (x − c)n = a0 + a1 (x − c) + a2 (x − c)2 + a3 (x − c)3 + · · · + an (x − c)n + · · ·
n=0

Remark 2.2.1. 1) A power series in (x − c) is also called a power series centered at c, or a power
series about c . Thus, a power series in x is just a series centered at the origin.
2) To simplify the notation used for a power series, we have adopted the convention that (x − c)0 = 1, even
when x = c.
We can view a power series as a function defined by the rule
∞
X
f (x) = an (x − c)n
n=0

The domain of f is the set of all x for which the power series converges, and the range of f comprises the
sums of the series obtained by allowing x to take on all values in the domain of f . If a function f is defined in
∞
an (x − c)n .
P
this manner, we say that f is represented by the power series
n=0
Example 2.2.1. Determine the values of x for which the following power series converges.
∞ ∞ ∞
X n X (x − 3)n X
A. n+1
xn B. C. n!xn
n=0
3 n=1
n n=0

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
2.2 Definition of power series at any x0 and x0 = 0 29

Solution. A. Using the generalized Ratio Test, we have

an+1 (n + 1)xn+1 3n+1

lim = lim ·
n→∞ an n→∞ 3n+2 nxn
(n + 1)|x| |x| n+1
= lim = lim
n→∞ 3n 3 n→∞ n
|x|
= <1 = |x| ⇔ −3 < x < 3
3
So, the series converges absolutely for −3 < x < 3 and diverges for |x| > 3 (i.e., for x > 3 or x < −3).
Since the Ratio Test gives no conclusion for the endpoints x = ±3, we must test these separately. For
x = 3, we have the series
∞ ∞ ∞
X n n X n n Xn
x = 3 =
n=0
3n+1 n=0
3n+1 n=0
3
n
Since lim = ∞ 6= 0 the series diverges by the test for divergence. The series diverges when x = −3,
n→∞ 3
for the same reason. Thus, the power series converges for all x in the interval (−3, 3) and diverges for all
x outside this interval.
B. Using the generalized Ratio Test, we have

an+1 (x − 3)n+1 n
lim = lim ·
n→∞ an n→∞ n+1 (x − 3)n
n
= lim (x − 3)
n→∞ n + 1
n
= |x − 3| lim
n→∞ n + 1

= |x − 3| < 1 ⇐⇒ 2 < x < 4

Hence, the series converges absolutely for 2 < x < 4 and diverges for |x − 3| > 1 (i.e., for x < 2 or x > 4).
Since the Ratio Test gives no conclusion for the endpoints x = 2 and x = 4 , we must test these separately.
∞
(−1)n n1 which is convergent series by alternating series test.
P
. For x = 2, we have the series
n=1

∞
1
P
. For x = 4, we have the series n which is divergent harmonic series.
n=1
Therefore, the given power series converges for all x in the interval [2, 4) and diverges for all x outside
this interval.
C. For x 6= 0 and using the Generalized Ratio Test, we have

an+1 (n + 1)!xn+1
lim = lim
n→∞ an n→∞ n!xn

= lim (n + 1)x
n→∞

= |x| lim n + 1
n→∞
=∞

Hence, by generalized Ratio Test the series diverges when x 6= 0. Thus, given series converges for x = 0.
J
Exercise 2.2.1. Determine the values of x for which the following power series converges.
∞ ∞ ∞
X xn X xn X 2n xn
A. B. C. (−1)n+1
n=0
(n + 1)2n n=0
n! n=0
n3n

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2.3 Convergence and divergence, radius and interval of convergence 30

2.3 Convergence and divergence, radius and interval of convergence

The following theorem, which we state without proof, tells us that the domain of a power series is always an
interval with x = c as its center. In the extreme cases the domain consists of the infinite interval (−∞, ∞) or
just the point x = c, which may be regarded as a degenerate interval.
∞
an (x − c)n , exactly one of the following is true:
P
Theorem 2.3.1. Given a power series
n=0

1. The series converges only at x = c .

2. The series converges for all x.
3. There is a number R > 0 such that the series converges for |x − c| < R and diverges for |x − c| > R.

• The number R referred to in Theorem 2.3.1 is called the radius of convergence of the power series.
The radius of convergence is R = 0 in case (1) and ∞ in case (2).
• The set of all values for which the power series converges is called the interval of convergence of the
power series.

Theorem 2.3.1 tells us that the interval of convergence of a power series centred at c is one of the following

• just the single point c

• the interval (−∞, ∞)
• a finite interval centered at c

[c − R, c + R], or (c − R, c + R), or [c − R, c + R), or (c − R, c + R]

The radius of convergence, R, and found by using the ratio test on the power series: If

an+1 (x − c)n+1
 
an+1
ρ = lim = lim |x − c|
n→∞ an (x − c)n n→∞ an
∞
an (x − c)n converges absolutely when ρ < 1, that is, where
P
exists, then the series
n=0

an+1
|x − c| < R = 1/ lim
n→∞ an

The series diverges if |x − c| > R.

∞
an+1
an (x − c)n has radius of
P
Remark 2.3.1. Suppose that L = lim exists or ∞. Then the power series
n→∞ an n=0
convergence R = 1/L. (If L = ∞, then R = 0; if L = 0, then R = ∞.)
an xn converges absolutely on the interior of its interval of convergence.
P
Theorem 2.3.2. A power series
Example 2.3.1. Determine the center, radius and interval of convergence of the following power series
∞ ∞ ∞ ∞
X X xn X X (−1)n xn
A. xn B. C. n!xn D.
n=0 n=0
n! n=0 n=1
3n (n + 1)

Solution. A. By apply generalized ratio test , we have

an+1 xn+1
L = lim = lim
n→∞ an n→∞ xn
= lim |x|
n→∞
= |x|

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2.3 Convergence and divergence, radius and interval of convergence 31

Hence, the series absolutely converges if L = |x| < 1 and diverges if L = |x| > 1. The test is inconclusive
if |x| = 1 ( i.e if x = −1 and x = 1). so that we will have to investigate convergence at these points
separately. At these points the series becomes
∞
X
1n = 1 + 1 + 1 + · · · f or x = 1
n=0
∞
X
(−1)n = 1 − 1 + 1 − · · · f or x = −1
n=0

These both series diverge. Thus the given series converges in the interval (−1, 1).
Therefore, the interval of convergence for the given power series is (−1, 1), and the radius of convergence
is R = 1 .
B. Applying the generalized ratio test for absolute convergence, we have
an+1 x(n+1)! n!
L = lim = lim ·
n→∞ an n→∞ (n + 1)! xn

x
= lim
n→∞ n + 1

=0
Since L = 0 < 1 for all x,the series converges absolutely for all x.
Therefore, the interval of convergence is (−∞, +∞) and Radius of convergence is R= +∞
c. If x 6= 0, then the generalized ratio test for absolute convergence yields
an+1 (n + 1)!xn+1
L = lim = lim
n→∞ an n→∞ n!xn

= lim (n + 1)x
n→∞

= +∞
Hence, the series diverges for all non zero values of x. Therefore, the interval of convergence is single point
x = 0 and the radius of convergence is R = 0
D. Since |(−1)n | = |(−1)n+1 | = 1 and using generalized ratio test for absolute convergence we obtain
an+1 xn+1 3n (n + 1)
L = lim = lim ·
n→∞ an n→∞ 3n+1 xn
 
|x| n + 1
= lim ·
n→∞ 3 n+2
|x| n+1
= lim
3 n→∞ n + 2
|x|
=
3
Hence, by the generalized ratio test for absolute convergence implies that the given series converges
absolutely if |x| < 3 and diverges if |x| > 3. The test is inconclusive if |x| = 3 ( i.e if x = −3 and x = 3).
so that we will have to investigate convergence at these points separately.
Substituting x = −3 in the given series yields
∞ ∞
X (−1)n (−3)n X 1
n (n + 1)
=
n=1
3 n=1
n + 1

which is divergent harmonic series 1 + 21 + 13 + 1

4 + ···.
Substituting x = 3 in the given series yields
∞ ∞
X (−1)n (3)n X (−1)n 1 1 1
n (n + 1)
= = 1 − + − + ··· .
n=1
3 n=1
n + 1 2 3 4

which is the conditionally convergent alternating harmonic series.Thus,the interval of convergence for the
given series (−3, 3] and the radius convergence is R = 3
J
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2.4 Representations of Functions as Power Series 32

Exercise 2.3.1. 1. Determine the centre, radius and interval of convergence of the following power series
∞ ∞
X (x − 2)n X (−1)n 2n xn
A. B. ( √
n=1
n2 3n n=0
n+1
∞ ∞
X n(x + 2)n X (−1)n xn
C. D.
n=0
3n+1 n=0
3n (n + 1)

2. Show that if p is a positive integer, then the power series

∞
X (pn)! n
x
n=0
(n!)p

has a radius of convergence of 1/pp

2.4 Representations of Functions as Power Series

In this section we learn how to represent certain types of functions as sums of power series by manipulating
geometric series or by differentiating or integrating such a series. You might wonder why we would ever want
to express a known function as a sum of infinitely many terms. We will see later that this strategy is useful
for integrating functions that don’t have elementary antiderivatives, for solving differential equations, and for
approximating functions by polynomials.
Recall that the geometric series which is given by
∞
X a
axn = provided that |x| < 1
n=0
1−x

Now, if we take a = 1 the series becomes

∞
X 1
xn = provided that |x| < 1
n=0
1−x

∞
1
xn provided that |x| < 1.
P
So that we can represent the function f (x) = 1−x with power series
n=0
That means
∞
1 X
f (x) = = xn provided that |x| < 1 (2.1)
1 − x n=0

Remark 2.4.1. The radius of convergence of this power series is R = 1 and the interval of convergence is
|x| < 1.
Example 2.4.1. Find a power series representation for the following function and determine its interval of
convergence.

1 2x2 x
A. g(x) = B. f (x) = C. h(x) =
1 + x3 1 + x3 5−x
1 1
Solution. A. By rewriting g(x) = 1+x3 = 1−(−x3 ) and substituting x = −x3 in equation 2.1 ,we have

1 1
g(x) = =
1 + x3 1 − (−x3 )
∞
X
= (−x3 )n
n=0
X∞
= (−1)n (x3n ) provided that | − x3 | < 1 ⇔ |x| < 1
n=0

∞
1
(−1)n (x3n ) with the interval
P
Therefore, the function g(x) = 1+x3 can be represented by power series
n=0
of convergence is |x| < 1.
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DEPARTMENT OF MATHEMATICS Applied Mathematics II
2.5 Algebraic operations on convergent power series 33

1 1
B. By rewriting f (x) = 2x2 1−(−x 3 ) and substituting x = −x
3
in equation 2.1 ,we have f (x) = 2x2 1−(−x3) =
∞ ∞
2x2 (−1)n x3n = (−1)n 2x3n+2 , provided that |x| < 1
P P
n=0 n=0

x x 1 x
C. By rewriting h(x) = 5−x = 5 · 1− x and substituting x = 5 in equation 2.1 ,we have
5

x x 1
g(x) = = ·
5−x 5 1 − x5
∞
x X  x n x
= provided that | | < 1 ⇔ |x| < 5
5 n=0 5 5

∞
x x n+1
P

Therefore, the function h(x) = 5−x can be represented by power series 5 with the interval of
n=0
convergence |x| < 5.
J

2.5 Algebraic operations on convergent power series

∞ ∞
an xn and bn xn be two power series with radius of convergence Ra and Rb respec-
P P
Theorem 2.5.1. Let
=0 n=0
tively, and let c be a constant. Then,
∞
can xn has radius of convergence Ra , and
P
i.
n=0

∞
X ∞
X
n
(can )x = c an xn
n=0 n=0

wherever the series on the right converges.

∞
(an + bn )xn has radius of convergence R at least as large as the smaller of Ra and Rb (i.e R ≥
P
ii.
n=0
min{Ra , Rb }), and
∞
X ∞
X ∞
X
(an + bn )xn = an xn + bn xn
n=0 n=0 n=0

wherever both series on the right converge.

∞  ∞ 
an xn · bn xn has radius of convergence R at least as large as the smaller of Ra and Rb (i.e R ≥
P P
iii.
n=0 n=0
min{Ra , Rb }), and
∞ ∞ ∞
! !
X X X
n n
an x · bn x = cn x n
n=0 n=0 n=0

where
n
X
cn = a0 bn + a1 bn−1 + ... + an b0 = aj bn−j
j=0

∞ ∞ ∞
cn xn is called Cauchy product of the series an xn and bn xn .
P P P
The series
n=0 n=0 n=0

1
Example 2.5.1. Find the power series representation of the function f (x) = by using Cauchy product
(1 − x)2
of series.
∞
1
xn holds for −1 < x < 1. We can determine a power series representation
P
Solution. We know that 1−x =
n=0
1
for (1−x)2 by taking the Cauchy product of this series with itself.

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2.6 Differentiation and integration of power series 34

Since an = bn = 1 for n = 0, 1, 2, ..., we have

n
X
cn = 1=n+1 and
n=0
∞
1 X
f (x) = = (n + 1)xn provided that |x| < 1
(1 − x)2 n=0

2.6 Differentiation and integration of power series

Theorem 2.6.1. Suppose that f (x) the sum of the power series on an interval I; that is,
∞
X
f (x) = an xn = a0 + a1 x + a2 x2 + a3 x3 + · · ·
n=0

then, if x is interior to I,
1. Differentiation of power series
∞
X ∞
X
f 0 (x) = Dx (an xn ) = n(an xn−1 )
n=0 n=1
= a1 + 2a2 x + 3a3 x2 + · · ·

2. Integration of power series

Z x ∞ Z x ∞
X X an n+1 1 1 1
f (t)dt = an tn dt = x = a0 x + a1 x2 + a2 x3 + a3 x4 + · · ·
0 n=0 0 n=0
n +1 2 3 4

Example 2.6.1. Apply Theorem 2.6.1 to the geometric series

1
= 1 + x + x2 + x3 + · · ·
1−x
to obtain formulas for two new series.
Solution. Differentiating term by term yields
1
= 1 + 2x + 3x2 + 4x3 + · · · , −1 < x < 1.
(1 − x)2

Integration term by term gives

Z x Z x Z x Z x Z x
1
dt = 1dt + tdt + t2 dt + t3 dt + · · ·
0 1−t 0 0 0 0

That is,
x2 x3
− ln(1 − x) = x +
+ + · · · , −1 < x < 1
2 3
If we replace x by −x in the later and multiply both sides by -11, we obtain

x2 x3 x4
ln(1 + x) = x − + − + ··· , −1 < x < 1
2 3 4
J
Example 2.6.2. Use the geometric series to find a power series representation for tan−1 x.

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
2.7 Taylor series; Taylor polynomial and application 35

Solution. Substitute −t2 for x in the geometric series. Since 0 ≤ t2 < 1 whenever −1 < t < 1, we obtain
1
= 1 − t2 + t4 − t6 + t8 − · · · (−1 < t < 1)
1 + t2
Now integrate from 0 to x, where |x| < 1:
Z x Z x
1
tan−1 x = 2
dt = (1 − t2 + t4 − t6 + t8 − · · · )dt
0 1+t 0
x3 x5 x7 x9
=x− + − + + ···
3 5 7 9
∞
X x2n+1
= (−1)n (−1 < x < 1)
n=0
2n + 1

J
2
Exercise 2.6.1. 1. Find a power series representation for 2xex given that a series representation for ex is
∞
xn
ex =
P
n! in (−∞, ∞).
n=0

2.7 Taylor series; Taylor polynomial and application

∞
an (x − c)n has a positive radius of convergence R, then the sum of the series defines a
P
If a power series
n=0
function f (x) on the interval (c − R, c + R). We say that the power series is a representation of f (x) on that
interval. What relationship exists between the function f (x) and the coefficients a0 , a1 , a2 , . . . of the power
series? The following theorem answers this question
Theorem 2.7.1 (Taylor’s Theorem). Suppose the series
∞
X
f (x) = an (x − c)n = a0 + a1 (x − c) + a2 (x − c)2 + a3 (x − c)3 + · · ·
n=0

converges to f (x) for c − R < x < c + R, where R > 0. Then

f (k) (c)
ak =
k!
for k = 0, 1, 2, 3, . . . .
Definition 2.7.1 (Taylor and Maclaurin series). If f (x) has derivatives of all orders at x = c(i.e.,if f (k) (c)
exists for k = 0, 1, 2, 3, . . . ), then the series
∞
X f (n) (c)
(x − c)n
n=0
n!
f 00 (c) f (3) (c)
= f (c) + f 0 (c)(x − c) + (x − c)2 + (x − c)3 + · · ·
2! 3!

is called the Taylor series of f about c (or the Taylor series of f in powers of (x − c). If c = 0,the Taylor
series becomes
f 00 (0) 2 f (3) (0) 3
f (x) = f (0) + f 0 (0)x + x + x + ···
2! 3!
and this series is called Maclaurin series.
Exercise 2.7.1.
Find Maclaurin series representation for ex , cos x, cosh x and sinh x.

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2.7 Taylor series; Taylor polynomial and application 36

2.7.1 Polynomial Approximation

The nth partial sum of the Taylor series centred at c,
n
X f (k) (c)
(x − c)k
k!
k=0
f 00 (c) f (3) (c) f (n) (c)
= f (c) + f 0 (c)(x − c) + (x − c)2 + (x − c)3 + · · · + (x − c)n
2! 3! n!

is called the nth -degree Taylor polynomial of f at c. If c = 0, we have the nth -degree Maclaurin polynomial
of f . First, notice that the partial sums of a Taylor series (like those for any power series) are simply polynomials.
We define
n
X f (k) (c)
Pn (x) = (x − c)k
k!
k=0
f 00 (c) f (n) (c)
= f (c) + f 0 (c)(x − c) + (x − c)2 + · · · + (x − c)n
2! n!
Theorem 2.7.2. If f has derivatives up to order n + 1 in an interval I containing c, then for each x in I,
there exists a number z between x and c such that
f 00 (c) f (3) (c) f (n) (c)
f (x) = f (c) + f 0 (c)(x − c) + (x − c)2 + (x − c)3 + · · · + (x − c)n + Rn (x) (2.2)
2! 3! n!
= Pn (x) + Rn (x)

where
f (n+1) (z)
Rn (x) = (x − c)n+1 .
(n + 1)!
n
f (k) (c)
− c)k which is a remainder (error)
P
Given f (x), c and n. Let Rn (x) = f (x) − Pn (x) = f (x) − k! (x
k=0
made in approximating f (x) by Pn (x).
Example 2.7.1. For f (x) = ex , find the Taylor polynomial of degree n expanded about x = 0. Then determine
a) Pn (1) b) P5 (1).
00
Solution. f (x) = ex , f 0 (x) = ex , f (x) = ex , f (3) (x) = ex and f (n) (x) = ex for all n. Thus,

f 0 (0) f 00 (0) 2 f (3) (0) 3 f (n) (0) n

Pn (x) = f (0) + x+ x + x + ··· + x
1! 2! 3! n!
n
x2 x3 xn X xk
=1+x+ + + ··· + =
2! 3! n! k!
k=0

As a result
n
1 1 1 1 X 1
Pn (1) = 1 + + + + ... + = and
1! 2! 3! n! k!
k=0
5
1 1 1 1 1 X 1
P5 (1) = 1 + + + + + =
1! 2! 3! 4! 5! 5!
k=0

J
Theorem 2.7.3 (Taylor’s Inequality). If |f (n+1) (x)| ≤ M for |x − x0 | ≤ d , then the remainder Rn (x) of the
Taylor series satisfies the inequality
M
Rn (x) ≤ |x − x0 |n+1 for |x − x0 | ≤ d.
(n + 1)!
We now turn our attention to finding the conditions under which the function has a power series represen-
tation.

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
2.7 Taylor series; Taylor polynomial and application 37

Theorem 2.7.4. Suppose that f has derivatives of all order on an interval I containing x0 and that Rn (x) is
the Taylor remainder of f at c. If
lim Rn (x) = 0
n→∞
for every x in I, then f (x) is represented by the Taylor series of f at c ; that is,
∞
X f (n) (c)
f (x) = (x − c)n
n=0
n!
Theorem 2.7.5. If x is any real number, then
|x|n
= 0.
lim
n→∞ n!
∞
xn x
P
Example 2.7.2. Show that the Maclaurin series n! of the function f (x) = e does represent f .
n=0

Solution. We use Taylor’s Theorem with c = 0 . Since f (n+1) (x) = ex , we see that
f (n+1) (z) n+1 ez
Rn (x) = x = xn+1
(n + 1)! (n + 1)!
where z is a number between 0 and x. If x > 0, then ez < ex , since the function f (x) = ex is increasing.
Therefore,
ex
0 < Rn (x) < xn+1
(n + 1)!
By Theorem 2.7.5,
ex xn+1
lim xn+1 = ex lim =0
n→∞ (n + 1)! n→∞ (n + 1)!

so the Squeeze Theorem implies that

lim Rn (x) = 0.
n→∞
If x < 0, then z < 0, and hence ez < e0 = 1. Therefore,
xn+1
0 < |Rn (x)| <
(n + 1)!
and once again, the Squeeze Theorem implies that
lim Rn (x) = 0.
n→∞

It follows from Theorem 2.7.4 that the Maclaurin series of represents the function f (x) = ex for all x 6= 0.
Finally, the series represents f at x = 0, since f (0) = e0 = 1, and this is also the value of the sum of the series
at 0. J
∞ n
x2n+1
P
Example 2.7.3. Show that the Maclaurin series (−1) (2n+1)! of the function f (x) = sin x does represent
n=0
f.
Solution. Using Taylor’s Theorem with c = 0, we have
f (n+1) (z) n+1
Rn (x) = x
(n + 1)!
where z is a number between 0 and x. But f (n+1) (x) is either ± sin x or ± cos x for any n(n = 0, 1, 2, . . . ).
Therefore, |f (n+1) (z)| ≤ 1 , so
f (n+1) (z)
|x|n+1
|Rn (x)| = |x|n+1 ≤
(n + 1)! (n + 1)!
By Theorem 2.7.5,
|x|n+1
lim =0
n→∞ (n + 1)!

so the Squeeze Theorem implies that

lim Rn (x) = 0.
n→∞
It follows from Theorem 2.7.4 that
∞
X n x2n+1
f (x) = (−1)
n=0
(2n + 1)!
as was to be shown. J
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DEPARTMENT OF MATHEMATICS Applied Mathematics II
2.7 Taylor series; Taylor polynomial and application 38

2.7.2 Defined by Integrals

Many functions that can be expressed as simple combinations of elementary functions cannot be antidifferenti-
ated by elementary techniques; their antiderivatives are not simple combinations of elementary functions. We
can, however, often find the Taylor series for the antiderivatives of such functions and hence approximate their
definite integrals.
Example 2.7.4. Find the Maclaurin series for
Z x
2
E(x) = e−t dt,
0

and use it to evaluate E(1) correct to 3 decimal places.

Solution. The Maclaurin series for E(x) is given by
Z x
t4 t6 t8

E(x) = 1 − t2 + − + − · · · dt
0 2! 3! 4!
3 5 x
t7 t9
 
t t
= t− + − + − ···
3 5 × 2! 7 × 3! 9 × 4! 0
∞
x3 x5 x7 x9 X x2n+1
=x− + − + − ··· = (−1)n ,
3 5 × 2! 7 × 3! 9 × 4! n=0
(2n + 1)n!

2
and is valid for all x because the series for e−t is valid for all t. Therefore,
1 1 1
E(1) = 1 −+ − + ···
3 5 × 2! 7 × 3!
1 1 1 (−1)n−1
≈1− + − + ···
3 5 × 2! 7 × 3! (2n − 1)(n − 1)!

We stopped with the nth term. Again, the alternating series test assures us that the error in this approximation
does not exceed the first omitted term, so it will be less than 0.0005, provided (2n + 1)n! > 2, 000. Since
13 × 6! = 9, 360, n = 6 will do. Thus,
1 1 1 1 1
E(1) ≈ 1 − + − + − ≈ 0.747,
3 10 42 216 1, 320
rounded to 3 decimal places. J

2.7.3 Indeterminate Forms

In this section we will see how Maclaurin polynomials could be used for evaluating the limits of indeterminate
forms. Here are two examples, using the series directly and keeping enough terms to allow cancellation of the
[0/0] factors.
Example 2.7.5. Evaluate

x − sin x (e2x − 1)(ln(1 + x3 ))

A. lim B. lim
x→0 x3 x→0 (1 − cos 3x)2

Solution. A.
 
x3 x5
x− x− 3! + 5! − ···
x − sin x
lim = lim
x→0 x3 x→0 x3
x3 x5
3! − 5! + ···
= lim
x→0 x3
1 x2 1 1
= lim − + ··· = =
x→0 3! 5! 3! 6

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2.7 Taylor series; Taylor polynomial and application 39

B.
  
(2x)2 (2x)3 3 x6
1 + (2x) + + + ··· − 1 x − 2 + ···
(e2x − 1)(ln(1 + x3 )) 2! 3!
lim = lim 2
(1 − cos 3x)2
x→0 x→0
 
(3x)2 (3x)4
1 − 1 − 2! + 4! − · · ·
2 + 2x + · · ·
= lim  2
x→0
9 2 34 4
2x − 4! x + ···
2 + 2x + · · · 2 8
= lim 2 = 9 2 =
(2) 81
x→0

9 4
2 − 34! x2 + · · ·

J
Exercise 2.7.2. Use Taylor series to find the following limits

sin x3 − x3 sin x tan−1 x − x

a) lim b) lim c) lim
x→0 x9 x→0 x x→0 x3
√
ex − 1 ln 1 + x − sin 2x
d) lim e) lim
x→0 x x→0 x

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
Chapter 3

Differential Calculus of Function of

Several Variables

3.1 Notations, examples, level curves and graphs

3.1.1 Notations and Examples
There are many familiar formulas in which a given variable depends on two or more other variables. For example,
the area A of a triangle depends on the base length b, and height h by the formula A = 21 bh : the volume V
of a rectangular box depends on the length l, the width w, and the height h by the formula V = lwh and the
arithmetic average x of n real numbers, depends on those numbers by the formula
x1 + x2 + · · · + xn
x=
n
Thus, we say that
• A is a function of the two variables b and h
• V is a function of the three variables l, w, and h;
• x is a function of the n variables .
Definition 3.1.1. A function f of two variables, x and y, is a rule that assigns a unique real number f (x, y)
to each point (x, y) in some set D in the xy-plane. The set D is the domain of f and its range is the set of
values that f takes on, that is, {f (x, y)|(x, y) ∈ D}.
√
Example 3.1.1. Let f (x, y) = 3x2 y −1. Find f (1, 4), f (0, 9), f (t2 , t), f (ab, 9b), and the natural domain of f .

Solution. By substitution
√
f (1, 4) = 3(1)2 4 − 1 = 5
√
f (0, 9) = 3(0)2 9 − 1 = −1
√ √
f (t2 , t) = 3(t2 )2 t − 1 = 3t4 t − 1
√ √
f (ab, 9b) = 3(ab)2 9b − 1 = 9a2 b2 b − 1

J
Domain= {(x, y) ∈ R2 : y ≥ 0}
The domain of functions of several variables consists of all points in space or (in plane for functions of two
variables) for which the formula is meaningful.
p
Example 3.1.2. f (x, y) = 9 − 4x2 − y 2 The domain of f is the region in the xy plane bounded by the ellipse
4x2 + y 2 = 9, because the square root is defined only for nonnegative numbers.
p
Example 3.1.3. g(x, y, z) = x2 + y 2 + z 2 The domain of g consists of all points (x, y, z) in space, because
x2 + y 2 + z 2 ≥ 0 for all (x, y, z).
The following are some of functions of several variables with there domains and representations
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DEPARTMENT OF MATHEMATICS Applied Mathematics II
3.1 Notations, examples, level curves and graphs 41

1. f (x, y) = xy, for x ≥ 0 and y ≥ 0, Area of rectangle.

2. f (x, y, z) = xyz, for x ≥ 0, y ≥ 0 and z ≥ 0, Area of rectangular parallelepiped.
3. f (x, y, z) = 2xy + 2yx + 2xz for surface area of a rectangular parallelepiped for x ≥ 0, y ≥ 0 and z ≥ 0.

Combinations of Functions of Several Variables

The sum, product, and quotient of two functions f and g of several variables are defined as follows.
• (f + g)(x, y) = f (x, y) + g(x, y)
• (f − g)(x, y) = f (x, y) − g(x, y)

• (f g)(x, y) = f (x, y)g(x, y)

 
• fg (x, y) = fg(x,y)
(x,y)

The formula for functions of thee variables is analogous. The domain of f + g, f − g , and f g consists of all
points simultaneously in the domain of f and in the domain of g, whereas the domain of fg consists of all points
simultaneously in the domain of f and in the domain of g at which g does not assume the value 0. A function
f of two variables x and y is a polynomial function if it is a sum of function of the form cxm y n , where c is
a number and m and n are nonnegative integers. A rational function is, as with functions of one variable,
the quotient of two polynomials functions. Similar terminology is used for polynomial and rational function of
three variables.

3.1.2 Level curves and graphs

The graph of a function of two variables is the collection of points (x, y, f (x, y)) for which (x.y) is in the domain
of f . It is customary to let z = f (x, y) the graph of f consists of all points such that z = f (x, y).
In sketching the graph of the function f of two variables, it is often helpful to determine the intersection of the
graph of f with planes of the form z = c. We call each such intersection the trace of the graph of f in the
plane z = c. Thus the trace of the graph of f in the plane z = c is the collection of points (x, y, c) such that
c = f (x, y).
Definition 3.1.2. The set of points (x, y) in the xy plane such that f (x, y) = c is called a level curve of f .

Example 3.1.4. Describe the graph of the function in an xyz coordinate system and the level curves associated
with them.
a. f (x, y) = 1 − x − 21 y.

Solution. By definition, the graph of the given function is the graph of the equation z = 1 − x − 12 y
which is a plane. A triangular portion of the plane can be sketched by plotting the intersection with the
coordinates axes and joining them ((0, 0, 1), (0, 2, 0) and (1, 0, 0)) with line segment. For any value of c
the level curve f (x, y) = c is the line with equation x − 21 y = 1 − c. J

b. f (x, y) = 1 − x2 − y 2

Solution. By definition, the graph of the given function is the graph of the equation

z = 1 − x2 − y 2 . (3.1)

In this case the level curve f (x, y) = c is a circle if c ≤ 1. J

Exercise 3.1.1. Identify the type of the level curve f (x, y) = c if,
a) f (x, y) = x2 + 4y 2 ; c = 1, 4

b) f (x, y) = x2 − y; c = −2, 2
c) f (x, y) = x2 − y 2 ; c = −1, 0, 1

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
3.2 Limits and continuity 42

(0, 0, 1)
y
(0, 2, 0)
y
y
(1, 0, 0)

Figure 3.2:

Figure 3.1:

3.2 Limits and continuity

3.2.1 Limit
If D is a set of points in 2 space, then a point (a, b) is called an interior point of D if there is some circular disk
with positive radius, centered at (a, b), and containing only points in D.
Definition 3.2.1. Let f be defined throughout a set containing a disk centered at (a, b) except possibly at (a, b)
itself, and let L be a number. Then L is the limit of f at (a, b) if for every  > 0 there is a δ > 0 such that
p
if 0 < (x − a)2 + (y − b)2 < δ, then |f (x, y) − L| < 
In this case we write
lim f (x, y) = L
(x,y)→(a,b)

and say that lim f (x, y) exists. Similarly let f be defined throughout a set containing a ball centered at
(x,y)→(a,b)
(a, b, c) except possibly at (a, b, c) it self. Then L is the limit of f at (a, b, c) if for every  > 0 there exists a
δ > 0 such that p
if 0 < (x − a)2 + (y − b)2 + (z − c)2 < δ, then |f (x, y, z) − L| < 
In this case we write
lim f (x, y, z) = L
(x,y,z)→(a,b,c)

and say that lim f (x, y, z) exists.

(x,y,z)→(a,b,c)

Example 3.2.1. Show that lim x = a and lim y=b

(x,y)→(a,b) (x,y)→(a,b)

Solution. Let  > 0, Observe that

p p
(x − a)2 ≤ (x − a)2 + (y − b)2

Therefore, if we let δ = ε it follows that

p
0< (x − a)2 + (y − b)2 < δ,
p
then |x − a| = (x − a)2 < δ < ε. This proves that lim x = a . The second limit is established
(x,y)→(a,b)
analogously. J
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DEPARTMENT OF MATHEMATICS Applied Mathematics II
3.2 Limits and continuity 43

Exercise 3.2.1. Show that lim y = b.

(x,y,z)→(a,b,c)

Theorem 3.2.1. If lim f (x, y) and lim g(x, y) exist, then

(x,y)→(a,b) (x,y)→(a,b)

i. lim (f ± g) (x, y) = lim f (x, y) ± lim g(x, y).

(x,y)→(a,b) (x,y)→(a,b) (x,y)→(a,b)

ii. lim (f g) (x, y) = lim f (x, y) lim g(x, y)

(x,y)→(a,b) (x,y)→(a,b) (x,y)→(a,b)

  lim f (x,y)
f (x,y)→(a,b)
iii. lim g (x, y) = lim g(x,y) provided that lim g(x, y) 6= 0.
(x,y)→(a,b) (x,y)→(a,b) (x,y)→(a,b)

iv. The limit of a polynomial always exists and is found simply by substitution. lim Pn (x, y) = Pn (a, b)
(x,y)→(a,b)

Example 3.2.2. Show that

x3 + y 3 7 x3 + y 3
lim = and lim = 0.
(x,y)→(−1,2) x2 + y 2 5 (x,y)→(0,0) x2 + y 2

Solution. Since lim x = −1 and lim y = 2 The product formula yields lim x3 = −1 and
(x,y)→(−1,2) (x,y)→(−1,2) (x,y)→(−1,2)
lim y3 = 8
(x,y)→(−1,2)
lim x2 = 1 and lim y 2 = 4 Hence the sum and the quotient formulas combined to yield
(x,y)→(−1,2) (x,y)→(−1,2)

lim (x3 + y 3 )
x3 + y 3 (x,y)→(−1,2)
lim =
(x,y)→(−1,2) x2 + y 2 lim (x2 + y 2 )
(x,y)→(−1,2)

lim x3 + lim y3
(x,y)→(−1,2) (x,y)→(−1,2)
=
lim x2 + lim y2
(x,y)→(−1,2) (x,y)→(−1,2)
−1 + 8 7
= = .
1+4 5
We cannot use quite the same procedure to verify the second limit, since lim (x2 + y 2 ) = 0. So, the
(x,y)→(0,0)
quotient formula does not apply. To verify the second limit we will show first that

x3
lim =0 (3.2)
(x,y)→(0,0) x2 + y 2

For this limit we observe that

x3 x3
0≤ ≤ 2 = |x|
x2 +y 2 x
Since lim x = 0 by Example 3.2.1, a version of the squeezing theorem for functions of two variable yields
(x,y)→(0,0)
3.2. A similar argument shows that
y3
lim =0 (3.3)
(x,y)→(0,0) x2 + y 2

we can use the sum formula to combine the limits in 3.2 and 3.3:
x3 + y 3 x3 y3
lim 2 2
= lim 2 2
+ lim =0+0=0
(x,y)→(0,0) x + y (x,y)→(0,0) x + y (x,y)→(0,0) x + y 2
2

J
2x2 y+3xy
Exercise 3.2.2. 1. Evaluate lim 2 .
(x,y)→(2,1) 5xy +3y

2. Evaluate lim ln xy .
(x,y)→(e,1)

Disproving limits:
• For a limit to exist, the function must approach that limit for every possible path of (x, y) approaching
(a, b). Thus, it is usually very hard to prove a limit exists, and easier to show a limit does not exist.
Ambo University
DEPARTMENT OF MATHEMATICS Applied Mathematics II
3.2 Limits and continuity 44

• So, if a function f (x, y) approaches L1 as (x, y) approaches (a, b) along a path P1 and f (x, y) approaches
L2 6= L1 as (x, y) approaches (a, b) along a different path P2 , then lim f (x, y) does not exist.
(x,y)→(a,b)

• Some simple paths to try are the lines along x = a, y = b, or any other line through the point.
y
Example 3.2.3. Show the following limit does not exist: lim
(x,y)→(1,0) x+y−1

Solution. The point (x, y) may approach (1, 0) in infinitely many paths and if the limit exits it is independent
of the path followed. Among these paths we can take the line x = 1 and the line y = x − 1.
Case 1: when (x, y) may approach (1, 0) along x = 1.
y y
lim = lim = 1.
(x,y)→(1,0) x + y − 1 y→0 y

Case2: when (x, y) may approach (1, 0) along the line y = x − 1

y x−1 1
lim = lim = .
(x,y)→(1,0) x + y − 1 x→1 x + x − 1 − 1 2
y
Thus the limit is not the same in different paths and by uniqueness of limit lim doesn’t exist. J
(x,y)→(1,0) x+y−1

x2 −y 2
Example 3.2.4. Show that lim 2 2 does not exist.
(x,y)→(0,0) x +y

x2 −y 2 x2
Solution. Let f (x, y) = x2 +y 2 . First let’s approach (0, 0) along the x-axis. Then y = 0 gives f (x, 0) = x2 =1
for all x 6= 0,so
f (x, y) → 1 as (x, y) → (0, 0) along the x-axis.
y2
We now approach along the y-axis by putting x = 0. The f (0, y) = y2 = −1 for all y 6= 0, so

f (x, y) → −1 as (x, y) → (0, 0) along the y-axis.

Since f has two different limits a long two different lines, the given limit does not exist. J
Exercise 3.2.3. Show the following limit does not exist
xy
1. lim 2 2
(x,y)→(0,0) x +y

xy 2
2. lim x 2 +y 4
(x,y)→(0,0)

3.2.2 Continuity
Definition 3.2.2. a) A function f of two variables is continuous at (a, b) if

lim f (x, y) = f (a, b)

(x,y)→(a,b)

b) A function f of three variables is continuous at (a, b, c) if

lim f (x, y, z) = f (a, b, c)

(x,y,z)→(a,b,c)

c) A function of several variables is continuous if it is continuous at each point in its domain.

x2 −y 2
Example 3.2.5. Where is the function f (x, y) = x2 +y 2 continuous?

Solution. The function f is discontinuous at (0, 0) because it is not defined there. Since f is a rational function,
it is continuous on its domain, which is the set D = {(x, y)|(x, y) 6= (0, 0)}. J
Example 3.2.6. Let
x2 −y 2
(
x2 +y 2 if (x, y) 6= (0, 0)
g(x) =
0 if (x, y) = (0, 0)
g is defined at (0, 0) but g is still discontinuous there because lim g(x, y) does not exist.
(x,y)→(0,0)

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
3.3 Partial derivatives; tangent lines, higher order partial derivatives. 45

3.3 Partial derivatives; tangent lines, higher order partial deriva-

tives.
Definition 3.3.1. Let f be a function of two variables and let (x0 , y0 ) be in the domain of f . The partial
derivative of f with respect to x at (x0 , y0 ) is defined by
f (x0 + h, y0 ) − f (x0 , y0 )
fx (x0 , y0 ) = lim
h→0 h
provided that the limit exists. The partial derivative of f with respect to y at is defined by
f (x0 , y0 + h) − f (x0 , y0 )
fy (x0 , y0 ) = lim
h→0 h
provided that the limit exists.
The functions fx and fy that arise through partial differentiation and are defined by
f (x + h, y) − f (x, y) f (x, y + h) − f (x, y)
fx (x, y) = lim and fy (x, y) = lim
h→0 h h→0 h
∂f ∂f
are called the partial derivatives of f and are frequently denoted ∂x and ∂y .
∂f ∂f
Example 3.3.1. Let f (x, y) = x2 + 3xy + y − 1. Find the values of ∂x = fx and ∂y = fy at the point (4, −5).
∂f
Solution. To find ∂x = fx , we treat y as a constant and differentiate f with respect to x.
∂f ∂f 2
= (x + 3xy + y − 1) = 2x + 3y
∂x ∂x
∂f ∂f
The value of ∂x at (4, −5) is 2(4) + 3(−5) = −7. To find ∂y , we treat x as a constant and differentiate f with
respect to y:
∂f ∂ 2
= (x + 3xy + y − 1) = 3x + 1.
∂y ∂y
∂f
The value of ∂y at (4, −5) is 3(4) + 1 = 13. J
Example 3.3.2. Find the partial derivatives of
f (x, y) = 2x3 y 2 + 2y + 4x.
Solution. Treating y as a constant and differentiating with respect to x, we obtain
fx (x, y) = 6x2 y 2 + 4.
Treating x as a constant and differentiating with respect to y, we obtain
fx (x, y) = 4x3 y + 2.
J
Exercise 3.3.1. a) Let f (x, y) = 24xy − 6x2 y . Find fx and fy at (1, 2).
∂z ∂z
b) Let z = x2 cosy. Find ∂x and ∂y .
Theorem 3.3.1. If f and g have partial derivatives, then
(f ± g)x = fx + gx and (f ± g)y = fy + gy
(f g)x = fx g + f gx and (f g)y = fy g + f gy
   
f fx g − f gx f fy g − f gy
= 2
and =
g x g g y g2
Example 3.3.3. Let
x3 y − xy 3
f (x, y) = .
x2 + y 2
Find fx and fy
Solution. Since f is a rational function, fx (x, y) and fy (x, y) are defined for all x and y such that the denom-
inator x2 + y 2 is not 0, that is, for all points except the origin. To find fx (x, y) we hold y constant and use the
quotient rule for partial derivatives to differentiate with respect to x:
(3x2 y − y 3 )(x2 + y 2 ) − (x3 y − xy 3 )2x x4 y + 4x2 y 3 − y 5
fx (x, y) = 2 2 2
=
(x + y ) (x2 + y 2 )2
J
Ambo University
DEPARTMENT OF MATHEMATICS Applied Mathematics II
3.3 Partial derivatives; tangent lines, higher order partial derivatives. 46

3.3.1 A geometric interpretation of partial derivatives

When we hold y equal to a constant y = y0 , z = f (x, y) becomes the function z = f (x, y0 ) of x, whose graph is
the intersection of the surface z = f (x, y) with the vertical plane y = y0 (Figure 4). The x-derivative fx (x0 , y0 )
is the slope in the positive x-direction of the tangent line to this curve at x = x0 . Similarly, when we hold x

equal to a constant x0 , z = f (x, y) becomes the function z = f (x0 , y) of y, whose graph is the intersection of the
surface with the plane x = x0 (Figure 5), and the y-derivative fy (x0 , y0 ) is the slope in the positive y-direction
of the tangent line to this curve at y = y0 .

3.3.2 Higher order partial derivatives

If f is a function of two variables, then its partial derivatives fx and fy are also functions of two variables, so we
can consider their partial derivatives (fx )x , (fx )y , (fy )x and (fy )y which are called the second partial derivatives
of f . If z = f (x, y), we use the following notation:
 
∂ ∂f ∂2f ∂2z
1. (fx )x = fxx = f11 = ∂x ∂x = ∂x2 = ∂x2
 
∂ ∂f ∂2f ∂2z
2. (fx )y = fxy = f12 = ∂y ∂x = ∂y∂x = ∂y∂x
 
∂ ∂f ∂2f ∂2z
3. (fy )x = fyx = f21 = ∂x ∂y = ∂x∂y = ∂x∂y
 
∂ ∂f ∂2f ∂2z
4. (fy )y = fyy = f22 = ∂y ∂y = ∂y 2 = ∂y 2

Example 3.3.4. Find all the second order derivatives for f (x, y) = cos(2x) − x2 e5y + 3y 2

Solution. We should first need the first order derivatives

fx (x, y) = −2 sin(2x) − 2xe5y
fy (x, y) = −5x2 e5y + 6y
Now, let’s get the second order derivatives.
fxx (x, y) = −4 cos(2x) − 2e5y
fxy (x, y) = −10xe5y
fyx (x, y) = −10xe5y
fyy (x, y) = −25x2 e5y + 6
J
Example 3.3.5. Let f (x, y) = sin(xy 2 ). Find all the second partial derivatives of f .
Solution. The first partial derivatives are given by
fx (x, y) = y 2 cos(xy 2 ) and fy (x, y) = 2xy cos(xy 2 )
We obtain the second partials by computing the partial derivatives of the first partials:
fxx (x, y) = −4y 4 sin(xy 2 )
fxy (x, y) = 2y cos(xy 2 ) − 2xy 3 sin(xy 2 )
fyx (x, y) = 2y cos(xy 2 ) − 2xy 3 sin(xy 2 )
fyy (x, y) = 2y cos(xy 2 ) − 4x2 y 2 sin(xy 2 ).

J
Ambo University
DEPARTMENT OF MATHEMATICS Applied Mathematics II
3.4 The chain rule, implicit differentiation 47

Exercise 3.3.2. Let (

x3 y−xy 3
x2 +y 2 for (x, y) 6= (0, 0)
f (x, y) =
0 for (x, y) = (0, 0)
Show that fxy (0, 0) 6= fyx (0, 0)
Theorem 3.3.2. Let f be a function of two variables, and assume that fxy (x, y) and fyx (x, y) are continuous
at (a, b). Then
fxy (a, b) = fyx (a, b).
2 2
Exercise 3.3.3. 1. Verify the Theorem 3.3.2 for f (x, y) = xe−x y
.
2. Compute all four second-order partial derivatives for
(a) f (x, y) = 3x2 + 4y 2 − 2x2 y + 7xy 2 − 9
(b) f (x, y) = cos(x) sin(y) − exy
3. Compute
(a) fxyx and fxyxy for f (x, y) = cos(x) sin(y) − exy .
√
(b) fx , fxy and fxyz for f (x, y, z) = xyz + x2 y 3 z 4 .

3.4 The chain rule, implicit differentiation

3.4.1 Chain rule
The chain rule for functions of one variable provides a formula for differentiating the composite of two functions
f and g. If u = f (x) and y = g(u) = g(f (x)), the chain rule can be expressed in the Leibniz notation as
dy dy du
=
dx du dx
Composites involving functions of several variables have their own versions of the chain rule, which involve
derivatives and partial derivatives.
Theorem 3.4.1 (The chain rule(case 1)). Suppose that z = f (x, y) is a differentiable function of x and y,where
x = g(t) and y = h(t) are both differentiable functions of t. Then z is a differentiable function of t and
dz ∂z dx ∂z dy
= + (3.4)
dt ∂x dt ∂y dt
dz
Example 3.4.1. Let z = x2 ey , x = sin t, and y = t3 . Find dt .
3 3
∂z dy
Solution. Using 3.4, we find that dz
dt = ∂z dx
∂x dt + ∂y dt = 2xey cos t+x2 ey (3t2 ) = 2(sin t)e(t ) cos t+3(sin2 t)e(t ) t2
J
Example 3.4.2. The pressure P (in kilopascal), volume V (in liter), and temperature T (in Kelvin) of a mole
of an ideal gas are related by the equation P V = 8.31T . Find the rate at which the pressure is changing when
the temperature is 300k and increasing at a rate of 0.1k/s and the volume is 100L and increasing at a rate of
0.2L/s.
Solution. If t represents the time elapsed in seconds, then at given instant we have T = 300, dT
dt = 0.1, V =
100, dV
dt = 0.2. Since P = 8.31T
V the chain rule gives
dP ∂P dT ∂P dV 8.31 dT 8.31 dV 8.31 8.31(300)
= + = − 2 = (0.1) − (0.2) = −0.04155
dt ∂T dt ∂V dt V dt V dt 100 1002
The pressure is decreasing at a rate of about 0.042kpa/s. J
Theorem 3.4.2 (The chain rule(case 2)). Suppose that z = f (x, y) is a differentiable function of x and y,
where x = g1 (s, t) and y = g2 (s, t). Then z = f (g1 (s, t), g2 (s, t)), and
∂z ∂z ∂x ∂z ∂y
= + (3.5)
∂s ∂x ∂s ∂y ∂s
∂z ∂z ∂x ∂z ∂y
= + (3.6)
∂t ∂x ∂t ∂y ∂t

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
3.4 The chain rule, implicit differentiation 48

∂x s
∂s
∂z
∂z ∂z ∂x ∂z ∂y ∂x x
= + ∂x
∂s ∂x ∂s ∂y ∂s ∂t
t
z ∂y s
∂z ∂z ∂x ∂z ∂y ∂s
= + y
∂t ∂x ∂t ∂y ∂t ∂z
∂y ∂y
∂t t

∂z ∂z
Example 3.4.3. Let z = x ln y, x = u2 + v 2 , and y = u2 − v 2 . Find ∂u and ∂v .

Solution. Using 3.5, we find that

 
∂z ∂z ∂x ∂z ∂y x
= + = (ln y)(2u) + (2u)
∂u ∂x ∂u ∂y ∂u y
u2 + v 2
= 2u ln(u2 − v 2 ) + 2u 2
u − v2
J
Example 3.4.4. Suppose that w = f (x, y), x = g(u, v), y = h(u, v), u = j(t), and v = k(t). Find a formula for
dw
dt .

Example 3.4.5. Let z = f (u − v, v − u). show that

∂z ∂z
+ =0
∂u ∂v
Example 3.4.6. Suppose the volume V of a canonical sand pile grows at a rate of 4 cubic inches per second
and the radius r of the circular base grows at rate of e−r inches per second. Find the rate at which the height
h of the cone is growing at the instant when the volume is 60 cubic inches and the radius is 6 inches.
dw
Exercise 3.4.1. 1. Let w = x cos yz 2 , x = sin t, y = t2 , and z = et . Find dt
∂u
2. If u = x4 y + y 2 z 2 ,where x = rset , y = rs2 e(−t) ,and z = r2 s sin t find the value of ∂s when r = 2, s = 1, t =
0.
√
3. Let w = x + y 2 z 3 , x = 1 + u2 v 2 , y = uv, and z = 3u. Find ∂w ∂w
∂u and ∂v .

3.4.2 Implicit differentiation

Through the chain rule we can more completely describe the process of implicit differentiation which you have
studied in applied I.
dw
Example 3.4.7. Suppose that w = f (x, y) and y = g(x). Find a formula for dx

Solution. The appropriate diagram is as follows: It follows that

dw ∂w ∂w dy
= + (3.7)
dx ∂x ∂y dx

(In this formula, ∂w

∂x refers to the partial derivative of z = w(x, y) with respect to x, whereas
dw
dx refers to the
derivative of z = f (x, g(x)) with respect to x.) J
Now suppose that f is a function of two variables that has partial derivatives, and assume that the equation
f (x, y) = 0 defines a differentiable function y = g(x) of x, so that f (x, g(x)) = 0. If w = f (x, y), then by
assumption,
dw d d
= (f (x, g(x))) = (0)
dx dx dx
Therefore by 3.7,
dw ∂w ∂w dy
0= = +
dx ∂x ∂y dx
∂w dy
Finally, if ∂y 6= 0, then by solving for dx , we obtain
−∂w
dy ∂x
= ∂w
(3.8)
dx ∂y

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3.5 Directional derivatives and gradients 49

dy
Example 3.4.8. Let x3 + y 3 = 2xy. Find dx .

Solution. Let w = x3 + y 3 − 2xy. Then

∂w ∂w
= 3x2 − 2y and = 3y 2 − 2x
∂x ∂y
and 3.8 implies that
−∂w
dy ∂x −(3x2 − 2y)
= ∂w
=
dx ∂y
3y 2 − 2x
J

3.4.3 Differentiability of functions of several variables

Recall that in the case of a function of a single variable, a function f (x) is differentiable only if it is continuous;
but that continuity does not guarantee differentiability. Intuitively, continuity of f (x) requires that its graph be
a continuous curve; and differentiability requires also that there is always a unique tangent vector to the graph
of f (x). In other words, a function f (x) is differentiable if and only if its graph is a smooth continuous curve
with no sharp corners (a sharp corner would be a place where there would be two possible tangent vectors).
If we try to extend this graphical picture of differentiability to functions of two or more variables, it would
be natural to think of a differentiable function of several variables as one whose graph is a smooth continuous
surface, with no sharp peaks or folds. Because for such a surface it would always be possible to associate a
unique tangent plane at a given point. However, “differentiability” in this sense turns out to be a much stronger
condition than the mere existence of partial derivatives. For the existence of partial derivatives at a point x0
requires only a smooth approach to the point f (x0 ) along the direction of the coordinate axes. We have seen
examples of functions that are discontinuous even though

lim (x, 0) = lim (0, y) both exist.

x→0 y→0

2
For example, the function f (x, y) = (x−y) ∂f ∂f
x2 +y 2 has this property, and in fact both ∂x and ∂y exist and are
continuous functions at the point (0, 0). With this sort of phenomenon in mind we give the following definition
of differentiability.
Theorem 3.4.3. Let f be a function having partial derivatives throughout a set containing a disk centered at
(x0 , y0 ). If fx and fy are continuous at (x0 , y0 ) then there are functions ε1 and ε2 of two variables such that

f (x, y) − f (x0 , y0 ) = fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) + ε1 (x, y)(x − x0 ) + ε2 (x, y)(y − y0 ) for (x, y) in D

where lim ε1 (x, y) = 0 and lim ε2 (x, y) = 0.

(x,y)→(x0 ,y0 ) (x,y)→(x0 ,y0 )

Definition 3.4.1. A function f of two variables is differentiable at (x0 , y0 ) if there exists a disk D centered
at (x0 , y0 ) and functions ε1 and ε2 of two variables such that

f (x, y)−f (x0 , y0 ) = fx (x0 , y0 )(x−x0 )+fy (x0 , y0 )(y−y0 )+ε1 (x, y)(x−x0 )+ε2 (x, y)(y−y0 ) for (x, y) in D (3.9)

where lim ε1 (x, y) = 0 and lim ε2 (x, y) = 0.

(x,y)→(x0 ,y0 ) (x,y)→(x0 ,y0 )

3.5 Directional derivatives and gradients

Definition 3.5.1. Let f be a function defined on a set containing a disk D centered at (x0 , y0 ) and let u =
a1 i + a2 j be a unit vector. Then the directional derivative of f at (x0 , y0 ) in the direction of u, denoted
Du f (x0 , y0 ), is defined by

f (x0 + ha1 , y0 + ha2 ) − f (x0 , y0 )

Du f (x0 , y0 ) = lim
h→0 h
provided this limit exists.
Theorem 3.5.1. Let f be differentiable at (x0 , y0 ). Then f has a directional derivative at (x0 , y0 ) in every
direction. Moreover, if u = a1 i + a2 j is a unit vector, then

Du f (x0 , y0 ) = fx (x0 , y0 )a1 + fy (x0 , y0 )a2 . (3.10)

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3.5 Directional derivatives and gradients 50

Example 3.5.1. Let f (x, y) = 6 − 3x2 − y 2 , and let u = √1 i − √1 j. Find Du f (1, 2).
2 2

Solution. Notice that u is a unit vector. First we calculate the partial derivatives of f :

fx (x, y) = −6x and fy (x, y) = −2y.

Therefore by 3.11
   
1 1
Du f (1, 2) = fx (1, 2) √ + fy (1, 2) √
2 2
√
   
1 −1
= (−6) √ + (−4) √ =− 2
2 2
J
Remark 3.5.1. The directional derivative in the direction of an arbitrary non-zero vector a is defined to be
1
Du f (x0 , y0 ), where u = kak a.

Example 3.5.2. Let f (x, y) = xy 2 and let a = i − 2j. Find the directional derivative at (−3, 1) in the direction
of a.
p √
Solution. In this case kak = 12 + (−2)2 = 5, so we will find Du f (−3, 1), where

1 1 2
u= a= √ i− √ j
kak 5 5

Since fx (x, y) = y 2 and fy (x, y) = 2xy it follows from 3.11 that

   
1 1
Du f (−3, 1) = fx (−3, 1) √ + fy (−3, 1) √
5 5
13 √
   
1 −2
= (1) √ + (−6) √ = 5.
5 5 5
J
Note 1. Let u = a1 i + a2 j + a3 k be a unit vector in space. The directional derivative Du f (x0 , y0 , z0 ) is defined
by
f (x0 + ha1 , y0 + ha2 , z0 + ha3 ) − f (x0 , y0 , z0 )
Du f (x0 , y0 , z0 ) = lim
h→0 h
provided that the limit exists. Moreover if f is differentiable at (x0 , y0 , z0 ). Then

Du f (x0 , y0 , z0 ) = fx (x0 , y0 , z0 )a1 + fy (x0 , y0 , z0 )a2 + fz (x0 , y0 , z0 )a3 . (3.11)

3.5.1 The Gradient

Definition 3.5.2. a. Let f be a function of two variables that has partial derivatives at (x0 , y0 ). Then the
gradient of f at (x0 , y0 ), denoted grad f (x0 , y0 ) or ∇f (x0 , y0 ) is defined by

grad f (x0 , y0 ) = ∇f (x0 , y0 ) = fx (x0 , y0 )i + fy (x0 , y0 )j

b. Let f be a function of three variables that has partial derivatives at (x0 , y0 , z0 ). Then the gradient of f at
(x0 , y0 , z0 ), which is denoted grad (x0 , y0 , z0 ) or ∇f (x0 , y0 , z0 ) is defined by

grad f (x0 , y0 , z0 ) = ∇f (x0 , y0 , z0 ) = fx (x0 , y0 , z0 )i + fy (x0 , y0 , z0 )j + fz (x0 , y0 , z0 )k

Example 3.5.3. Find the gradient of the function f (x, y) = x2 y 3 − 4y at the point (2, −1).
Solution. We first compute the partial derivatives at (2, −1).

fx (x, y) = 2xy 2 and fy (x, y) = 3x2 y 2 − 4.

Hence fx (2, −1) = −4 and fy (2, −1) = 8. Therefore, ∇f (2, −1) = −4i + 8j. J
Note 2.
Du f (x, y) = ∇f (x, y) · u (3.12)
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Example 3.5.4. If f (x, y, z) = x sin yz,

a. Find the gradient of f and
b. Find the directional derivative of f at (1, 3, 0) in the direction of v = i + 2j − k.

Solution. a. The gradient of f is

∇f (x, y, z) = hfx (x, y, z), fy (x, y, z), fz (x, y, z)i = hsin yz, xz cos yz, xy cos yzi = h0, 0, 3i

b. The unit vector in the direction of v = i + 2j − k is u = √1 i + √2 j − √1 k. Therefore,

6 6 6

  r
1 2 1 −1 3
Du f (1, 3, 0) = ∇f (1, 3, 0) · u = 3k · √ i + √ j − √ k = 3 √ =−
6 6 6 6 2
J
Theorem 3.5.2. Suppose f is differentiable function of two or three variables. The maximum value of the
directional derivative Du f (x) is |∇f (x)| and it occurs when u has the same direction as the gradient vector
∇f (x).
Proof. We know that Du f = ∇f · u. Thus, Du f = ∇f · u = |∇f |kuk cos θ = k∇f k cos θ where θ is the angle
between ∇f and u. The maximum value of cos θ is 1 and this occurs when θ = 0. Therefore the maximum
value of Du f is k∇f k and it occurs when θ = 0, that is, when u has the same direction ∇f .

Example 3.5.5. a. If f (x, y) = xey , find the rate of change of f at the point P (2, 0) in the direction from
P to Q(1/2, 2).
b. In what direction does f have the maximum rate of change? What is this maximum rate of change?
Solution. a. We first compute the gradient vector:

∇f (x, y) = hfx , fy i = hey , xey i ⇒ f (2, 0) = h1, 2i.

The unit vector in the direction of P~Q = ( −3 −3 4

2 , 2) is a u = h 5 , 5 i, so the rate of change of f in the
direction from P to Q is
   
−3 4 −3 4
Du f (2, 0) = ∇f (2, 0) · u = h1, 2i · h , i=1 +2 =1
5 5 5 5

b. By the above Theorem f increases fastest in the direction

√ of the gradient vector ∇f (2, 0) = h1, 2i. The
maximum rate of change is h∇f (2, 0)i = kh1, 2ik = 5.
J

3.6 Tangent planes and Total differential

3.6.1 Tangent Planes

Example 3.6.1. Assuming that the curve x2 − xy + 3y 2 = 5 is smooth, find a unit vector that is perpendicular
to the curve at (1, −1).

Definition 3.6.1. Let f be differentiable at a point (x0 , y0 , z0 ) on a level surface S of f . If gradf (x0 , y0 , z0 ) 6= 0,
then the plane through (x0 , y0 , z0 ) whose normal is gradf (x0 , y0 , z0 ) is the plane tangent to S at (x0 , y0 , z0 ),
and gradf (x0 , y0 , z0 ) is normal to S
Since
gradf (x0 , y0 , z0 ) = fx (x0 , y0 , z0 )~i + fy (x0 , y0 , z0 )~j + fz (x0 , y0 , z0 )~k
and Since gradf (x0 , y0 , z0 ) is normal to the tangent plane at (x0 , y0 , z0 ), an equation of tangent plane is

fx (x0 , y0 , z0 )(x − x0 ) + fy (x0 , y0 , z0 )(y − y0 ) + fz (x0 , y0 , z0 )(z − z0 ) = 0

√
Example 3.6.2. Find an equation of the plane tangent to the sphere x2 + y 2 + z 2 = 4 at the point (−1, 1, 2).

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3.6 Tangent planes and Total differential 52

The Gradient as a Normal Vector grad f (x0 , y0 )

y C
Suppose a smooth curve C is a level curve of a func-
tion f that is differentiable at a point (x0 , y0 ) on C. as f (x, y) = c
u
a point (x, y) moves along C, the value f (x, y) is con-
stant by hypothesis and hence does not change. This
suggests that at (x0 , y0 ) the rate of change of f in the (x0 , y0 )
direction of a unit vector u tangent to C is 0, that is,
Du f (x0 , y0 ) = 0 (Figure 3.3). However, 3.12 implies
that Du f (x0 , y0 ) = 0 for any unit vector u perpendic-
ular to gradf (x0 , y0 ). combining these results, we can
reasonably conjecture that gradf (x0 , y0 ) is perpendicu-
lar, or equivalently, normal to C at (x0 , y0 )(Figure 3.3)
Theorem 3.6.1. Let f be differentiable at (x0 , y0 ),
and let f (x0 , y0 ) = c. Also let C be a level curve
f (x, y) = c that passes through (x0 , y0 ). If C is smooth Du f (x0 , y0 ) = 0
and gradf (x0 , y0 ) 6= 0, then gradf (x0 , y0 ) is normal to
C at (x0 , y0 ).
Figure 3.3:

tangent plane gradf (x0 , y0 , z0 )

Now let f be a function of three variables that is differen-

tiable at a point (x0 , y0 , z0 ), and let f (x0 , y0 , z0 ) = c. If
C is any smooth curve on the level surface f (x, y, z) = c
and if C passes through (x0 , y0 , z0 ), then an argument tangent vector
similar to the one used in above Theorem shows that if
gradf (x0 , y0 , z0 ) 6= 0, then gradf (x0 , y0 , z0 ) is perpen-
dicular to the tangent vector of C at x0 , y0 , z0 (Figure
3.4). Because gradf (x0 , y0 , z0 ) is perpendicular to the
tangent vector to any such curve C at (x0 , y0 , z0 ), all
such tangents lie in a single plane called a tangent
plane.
(x0 , y0 , z0 )
f (x, y, z) = c

Figure 3.4:

2 2 2
Solution.
√ The sphere is the level surface f (x, y, z) = 4, where f (x, y, z) = x + y + z . The partials of f at
(−1, 1, 2) are given by
√ √ √ √
fx (−1, 1, 2) = −2, fy (−1, 1, 2) = 2, fz (−1, 1, 2) = 2 2
√
Therefore an equation of the plane tangent at (−1, 1, 2) is
√ √
−2(x − (−1)) + 2(y − 1) + 2 2(z − 2) = 0

or equivalently, √
−x + y + 2z = 4
J
Suppose f is a function of two variables that is differentiable at (x, y0 ), and let

g(x, y, z) = f (x, y) − z

Then the graph of f is the level surface g(x, y, z) = 0. Accordingly, we define the plane tangent to the graph of
f at (x0 , y0 , f (x0 , y0 )) to be the plane tangent to the level surface g(x, y, z) at (x0 , y0 , f (x0 , y0 )). Since

gradg(x0 , y, z0 ) = fx (x0 , y0 )~i + fy (x0 , y0 )~j − ~k

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3.6 Tangent planes and Total differential 53

an equation of the tangent plane of f at (x0 , y0 , f (x0 , y0 )) is

fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) − [z − f (x0 , y0 )] = 0 (3.13)
or equivalently,
z = f (x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) = 0 (3.14)
and the vector gradg(x0 , y, z0 ) = fx (x0 , y0 )~i + fy (x0 , y0 )~j − ~k is said to be normal to the graph of f at
(x0 , y0 , f (x0 , y0 )).
Example 3.6.3. Find the tangent plane to the elliptic paraboloid z = 2x2 + y 2 at the point (1, 1, 3).
Solution. Let f (x, y) = 2x2 + y 2 . Then fx (x, y) = 4x and fy (x, y) = 2y. which implies
fx (1, 1) = 4, fy (1, 1) = 2
Then (by definition) the equation of the tangent plane at (1, 1, 3) is
z − 3 = 4(x − 1) + 2(y − 1) ⇒ z = 4x + 2y − 3
J
Example 3.6.4. Let f (x, y) = 6 − 3x2 − y 2 . Find a vector normal to the graph of f at (1, 2, −1), and find an
equation of the plane tangent to the graph of f at (1, 2, −1).

Differentials
If f is a function of two variables, we can replace (x0 , y0 ) by any point (x, y) in the domain of f at which f is
differentiable and the linear approximation is transformed into

f (x + h, y + k) − f (x, y) ≈ fx (x, y)h + fy (x, y)k (3.15)

The number fx (x, y)h + fy (x, y)k on the right side of (3.15) is usually called the differential (or total
differential) of f (at (x, y) with increments h and k) and is denoted df . Thus
f = fx (x, y)h + fy (x, y)k. (3.16)
Of course, df depends on x, y, h and k, even though they are not indicated in the notation df . If g1 (x, y) = x
and g2 (x, y) = y, then the differential dg1 is denoted by dx, and the differential dg2 is denoted by dy. Since
(g1 )x (x, y) = 1, (g1 )y (x, y) = 0, (g2 )x (x, y) = 0, (g2 )y (x, y) = 1
We have dx = dg1 = 1.h + 0.k = h and dy = dg2 = 0.h + 1.k = k Therefore we can write (3.16) as
∂f ∂f
df = fx (x, y)dx + fy (x, y)dy or df = dx + dy
∂x ∂y
Example 3.6.5. Let f (x, y) = xy 2 + y sin x. Find df .
∂f ∂f
Solution. Since ∂x = y 2 + y cos x and ∂x = 2xy + sin x. Thus df = (y 2 + y cos x)dx + (2xy + sin x)dy J
Note 3. If f is a function of three variables that is differentiable at (x0 , y0 , z0 ), then the differential df is defined
by df = fx (x, y, z)h + fy (x, y, z)k + fz (x, y, zl The more usual form for df is
df = fx (x, y, z)dx + fy (x, y, z) + fz (x, y, z)dz
∂f ∂f ∂f
df = dx + dy + dz
∂x ∂y ∂z
Example 3.6.6. Let f (x, y, z) = x2 ln(y − z). Find df .
∂f x2 ∂f −x2 x2 2 2
Solution. Since ∂x = 2x ln(y − z), ∂f
∂y = y−z , ∂z = y−z = z−y .
x
Thus, df = 2x ln(y − z)dx + y−z x
dy + z−y dz. J

Example 3.6.7. Find the tangent plane to the elliptic paraboloid z = 2x2 + y 2 at the point (1, 1, 3).
Solution. Let f (x, y) = 2x2 + y 2 . Then fx (x, y) = 4x and fy (x, y) = 2y. which implies
fx (1, 1) = 4, fy (1, 1) = 2
Then (by definition) the equation of the tangent plane at (1, 1, 3) is
z − 3 = 4(x − 1) + 2(y − 1) ⇒ z = 4x + 2y − 3
J
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3.7 Tangent plane approximations of values of a function 54

3.7 Tangent plane approximations of values of a function

In example 3.6.7 the tangent line L(x, y) = 4x + 2y − 3 is a good approximation to f (x, y) when (x, y) is near
(1, 1). The function L is called the linearization of f at (1, 1) and the approximation

f (x, y) ≈ 4x + 2y − 3

is called the linear approximation or tangent plane approximation of f at (1, 1).

Definition 3.7.1. An equation of the tangent plane to the graph of a function f of two variables at the point
(a, b, f (a, b)) is z = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b) is called the linearization of f at (a, b) and the
approximation
f (x, y) ≈ f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b)
is called the linear approximation or the tangent plane approximation of f at (a, b).
Theorem 3.7.1. If the partial derivatives fx and fy exist near (a, b) and are continuous at (a, b), then f is
differentiable at (a, b).
Example 3.7.1. Show that f (x, y) = xex y is differentiable at (1, 0) and find its linearization there. Then use
it to approximate f (1.1, −0.1).
Solution. The partial derivatives are fx (x, y) = ex y + xyex y, fy (x, y) = x2 ex y, fx (1, 0) = 1 and fy (1, 0) = 1.
Both fx and fy are continuous functions, so f is differentiable by the above theorem. The linearization is
L(x, y) = f (1, 0) + fx (1, 0)(x − 1) + fy (1, 0)(y − 0) = x + y. The corresponding linear approximation is
xex y ≈ x + y.So,
f (1.1, −0.1) ≈ 1.1 − 0.1 = 1
Compare this with the actual value of f (1.1, −0.1) = (1.1e)( − 0.1) ≈ 0.98542. J

3.8 Relative extrema of functions of two variables

Definition 3.8.1. Let f be a function of two variables, and let R be a set contained in the domain of f .
Then f has an absolute maximum (respectively, an absolute minimum value) on R at (x0 , y0 ) if f (x, y) ≤
f (x0 , y0 )(respectively, f (x, y) ≥ f (x0 , y0 )) for all (x, y) in R. If R is the domain of f , we say that f has an
absolute maximum value (respectively, an absolute minimum value)at (x0 , y0 ). Furthermore, f has a relative
maximum value (respectively, a relative value) at (x0 , y0 ) if there is a disk D centered at (x0 , y0 ) and contained
in the domain of f such that f (x, y) ≤ f (x0 , y0 ) (respectively, f (x, y) ≥ f (x0 , y0 )) for all (x, y) in D.

Relative Maximums

Relative Minimums

Theorem 3.8.1. Let f have a relative extreme value at (x0 , y0 ). If f has partial derivatives at (x0 , y0 ), then
fx (x0 , y0 ) = fy (x0 , y0 ) = 0.
Definition 3.8.2. The point (a, b) is a critical point (or a stationary point) of f (x, y). Provided one of the
following is true,

1. ∇f (a, b) = 0 (this is equivalent to saying that fx (a, b) = 0 and fy (a, b) = 0 )

2. fx (a, b) and /or fy (a, b) doesn’t exist.
Example 3.8.1. Let f (x, y) = 3 − x2 + 2x − y 2 − 4y. Find all critical points of f .

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Solution. The partial derivatives of f exist at every point in the domain of f , so relative extreme values can
occur only at points at which both partial derivatives are 0. The partial derivatives are

fx (x, y) = −2x + 2 and fy (x, y) = −2y − 4.

Therefore fx (x, y) = 0 only if x = 1,and fy (x, y) = 0 only if y = −2. This means that fx (x, y) = fy (x, y) only
if (x, y) = (1, −2), and thus (1, −2) is the only critical point of f . J
p
Example 3.8.2. Let f (x, y) = (x2 + y 2 ). Determine all critical points and all relative extreme values of f .
x y
Solution. We find that fx (x, y) = √ and fy (x, y) = x2 +y 2 . Since the partial derivatives exist at all
x2 +y 2
points except the origin, and since the origin is in the domain of f , the origin is a critical point of f . Because
fx (x, y) = 0 only if x = 0 and fy (x, y) = 0 only if y = 0, there is no point (x, y) such that fx (x, y) = fy (x, y) = 0.
Consequently (0, 0) is the only critical point of f . Since f (0, 0) = 0 and f (x, y) ≤ 0 for all (x, y), it follows that
0 is the only (relative) minimum value of f , and there is no relative maximum value. J
Theorem 3.8.2. If the point (a, b) is a relative extrema of the function f (x, y), then (a, b) is also a critical
point of f (x, y) and ∇f (a, b) = 0.
Note that this does NOT say that all critical points are relative extrema. It only says that relative extrema
will be critical points of the function. To see this let’s consider the function

f (x, y) = xy.

The two first order partial derivatives are,

fx (x, y) = y, fy (x, y) = x

The only point that will make both of these derivatives zero at the same time is (0, 0) and also (0, 0) is a
critical point for the function. But if we start at the origin and move into either of the quadrants where both
x and y are the same sign the function increases. However, if we start at the origin and move into either of the
quadrants where x and y have the opposite sign then the function decreases. In other words, no matter what
region you take about the origin there will be points larger than f (0, 0) = 0 and points smaller than
f (0, 0) = 0. Therefore, there is no way that (0, 0) can be a relative extrema. Critical points that exhibit this
kind of behaviour are called saddle points.
Exercise 3.8.1. 1. Let f (x, y) = y 2 − x2 . Show that the origin is the only critical point but that f (0, 0) is
not a relative extreme value of f .

2. Let f (x, y) = sin x + sin y for 0 ≤ x ≤ π2 , 0 ≤ y ≤ π2 find all critical points and determine whether each
critical point yields a relative maximum value, a relative minimum value or a saddle point
Definition 3.8.3. If f is a function for which fx (x0 , y0 ) = fy (x0 , y0 ) = 0, we say that f has a saddle point
at (x0 , y0 ) if there is a disk centered at (x0 , y0 ) such that the following condition holds: f assume its maximum
value on one diameter of the disk only at (x0 , y0 ), and assumes its minimum value on another diameter of the
disk only at (x0 , y0 ). For example, if f (x, y) = y 2 − x2 then f has a saddle point at the origin.
Theorem 3.8.3 (Second Partial Test). Assume f has a critical point at (x0 , y0 ) and that f has continuous
second partial derivatives in a disk centered at (x0 , y0 ). Let

D(x0 , y0 ) = fxx (x0 , y0 )fyy (x0 , y0 ) − [fxy (x0 , y0 )]2

a. If D(x0 , y0 ) > 0 and fxx (x0 , y0 ) < 0 (or fyy (x0 , y0 ) < 0), then f has a relative maximum value at (x0 , y0 ).
b. If D(x0 , y0 ) > 0 and fxx (x0 , y0 ) > 0 (or fyy (x0 , y0 ) > 0), then f has a relative maximum value at (x0 , y0 ).
c. If D(x0 , y0 ) < 0 , then f has a saddle point at (x0 , y0 ).
Finally, if D(x0 , y0 ) = 0, then f may or may not have a relative extreme value at (x0 , y0 ). The expression
D(x0 , y0 ) in the second partial test is called the discriminant of f at (x0 , y0 ). It can also be given in
determinant form:
fxx (x0 , y0 ) fxy (x0 , y0 )
D(x0 , y0 ) =
fxy (x0 , y0 ) fyy (x0 , y0 )

Definition 3.8.4. A critical point (x0 , y0 ) is said to be degenerate if D(x0 , y0 ) = 0.

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3.9 Largest and smallest values of a function on a given set 56

Example 3.8.3. Let f (x, y) = x2 − 2xy + 13 y 3 − 3y Using the second partial derivative test, determine at which
points f has relative extreme values and at which points f has saddle points.
Solution. We find that fx (x, y) = 2x − 2y and fy (x, y) = −2x + y 2 − 3 Observe that fx (x, y) = 0 if x = y and
fy (x, y) = 0 if −2x + y 2 − 3 = 0. Thus (x, y) is a critical point if x = y and −2x + y 2 − 3 = 0. By substituting
y for x we can transform the second equation into y 2 − 2y − 3 = 0. The two solutions of this equation are
y = 3 and y = −1. Thus the critical points of f are (3, 3) and (−1, 1). For the second partials of f we find that
fxx (x, y) = 2, fyy (x, y) = 2y, and fxy (x, y) = −2. Therefore

D(3, 3) = fxx (3, 3)fyy (3, 3) − [fxy (3, 3)]2 = (2)(6) − 22 = 8 > 0

and
D(−1, 1) = fxx (−1, −1)fyy (−1, −1) − [fxy (−1, −1)]2 = (2)(−2) − (−2)2 = −8 < 0.
Since D(3, 3) > 0 and fxx (3, 3) = 2 > 0, f has a relative minimum value at (3, 3). Since D(−1, −1) < 0, f has
a saddle point at (−1, −1).
J

3.9 Largest and smallest values of a function on a given set

Definition 3.9.1. 1. A region in R2 is called closed if it includes its boundary. A region is called open if
it doesn’t include any of its boundary points.
2. A region in R2 is called bounded if it can be completely contained in a disk. In other words, a region
will be bounded if it is finite.

Theorem 3.9.1 (Extreme value theorem). If f is continuous in some closed, bounded set D in R2 , then there
are points in D, (x1 , y1 ) and (x2 , y2 ) so that f (x1 , y1 ) is the absolute maximum and (x2 , y2 ) is the absolute
minimum of the function in D.
Steps to find extreme value
1. Find all the critical points of f that lie in the region D and compute the values of f at these points.

2. Find the extreme values of f on the boundary.

3. The largest and smallest values found in the first two steps are the absolute minimum and the absolute
maximum of the function.

y
Let f (x, y) = xy−x2 , and let R be
l4
Example 3.9.1. the square region shown in Figure 1
below. Find extreme values of f l1 l3
0 l2 1 x

Solution. By the Extreme value Theorem f has extreme values on R. Since fx (x, y) = y − 2x and fy (x, y) = x
it follows that fx (x, y) = 0 if y = 2x and fy (x, y) = 0 if x = 0. Thus the only critical point of f is (0, 0), which
happens to be a boundary point of R. Therefore the extreme values of f on R must occur on the boundary
of R, which is composed of the four line segments l1 , l2 , l3 and l4 . On l1 , x = 0 and 0 ≤ y ≤ 1 and since
f (0, y) = 0, the maximum and the minimum values of f on l1 are both 0. On l2 , y = 0 and 0 ≤ x ≤ 1, and since
f (x, 0) = −x2 the maximum value of f on l2 is 0 and the minimum value is −1. On l3 , x = 1 and 0 ≤ y ≤ 1,
and since f (1, y) = y − 1 the maximum value of f On l3 is 0 and the minimum value is -1. On l4 , y = 1 and
0 ≤ x ≤ 1, and since f (x, 1) = x − x2 the maximum value of f on l4 is 1/4 and the minimum value is 0. By
comparing the extreme values of f on l1 , l2 , l3 and l4 , we conclude that the maximum value of f on R is 1/4
and the minimum value is -1. J

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3.10 Extreme values under constraint conditions: Lagrange’s multiplier 57

3.10 Extreme values under constraint conditions: Lagrange’s

multiplier
Let us consider the problem of finding an extreme value of a function f of two variables subject to a constraint
of the form g(x, y) = c; that is we seek an extreme value of f on the level curve g(x, y) = c (rather than on the
entire domain of f ). If f has an extreme value on the level curve at the point (x0 , y0 ), then under certain
conditions there exists a number λ such that

∇f (x0 , y0 ) = λ∇g(x0 , y0 ).

Theorem 3.10.1. Let f and g be differentiable at (x0 , y0 ). Let C be the level curve g(x, y) = k that contains
(x0 , y0 ), Assume that C is smooth, and that (x0 , y0 ) is not an end point of the curve. If ∇f (x0 , y0 ) and if f has
an extreme value on C at (x0 , y0 ), then there is a number λ such that

∇f (x0 , y0 ) = λ∇g(x0 , y0 )

is called a Lagrange multiplier for f and g.

The method of determining extreme values by means of Lagrange multipliers proceed as follows:
1. Assume that f has an extreme value on the level curve g(x, y) = k
2. Solve the equations Constraint g(x, y) = k

fx (x, y) = λgx (x, y)
∇f (x, y) = λ∇g(x, y) =
fy (x, y) = λgy (x, y)

3. Calculate the values of f at each point (x, y) that arises in step 2, and at each end point (if any) of
the curve. If f has maximum value in the level curve g(x, y) = k it would be the largest of the values
computed; if f has a minimum value on the level curve, it will be the smallest of the values computed.
Example 3.10.1. Let f (x, y) = x2 + y 3 , Find the extreme values of f on the ellipse x2 + 2y 2 = 1. Solution:
Let g(x, y) = x2 + 2y 2 So the constraint is g(x, y) = x2 + 2y 2 = 1. Since ∇f (x, y) = 2xi + 12y 2 j and
∇g(x, y) = 2xi + 4yj The equation we will use to find x and y are constraint x2 + 2y 2 = 1

2x = λ2x
∇f (x0 , y0 ) = λ∇g(x0 , y0 ) =
12y 2 = λ4y

By (3) either x = 0 or λ = 1 . If x = 0, then the (2) implies 2y 2 = 1 so that y = √12 or y = − √12 . If λ = 1

then (3) becomes 12y 2 = 4y, which means that y = 0 or y = 31 . By (2) If y = 0, then x2 + 2(0)2 = 1 , so x = 1
 2 √ √
or x = −1. If y = 3 , then x + 2 13
1 2
= 1 , so x = 37 or x = − 37 Thus the only possible extreme values
√ √ √
of occur at (0, √12 ), (0, − √12 ), (1, 0), (−1, 0), ( 37 , 31 ), and ( −3 7 , 31 ). f (0, √12 ) = 2, f (1, 0) = 1 = f (−1, 0),
√ √ √
f ( 37 , 13 ) = ( −3 7 , 13 ) = 25
27 . We conclude that the maximum value 2 of f occurs at (0, √12 ) and the minimum
√
value of f = − 2 occurs at (0, − √12 ).

The Lagrange Method for Functions of Three variables

Next we consider the problem of finding extreme values of a function of three variables subject to a constraint
of the form g(x, y, z) = c. By an argument similar to that used for functions of two variables, It is possible to
show that if f has such an extreme value at (x0 , y0 , z0 ), then ∇f (x0 , y0 , z0 ) and ∇g(x0 , y0 , z0 ), if not 0, are
both normal to the level surface g(x, y, z) = c at (x0 , y0 , z0 ), and hence are parallel to each other. Thus there
is a number λ, again called a lagrange multiplier, such that

∇f (x0 , y0 , z0 ) = λ∇g(x0 , y0 , z0 ).

To find the extreme values of f subject to the constraint g(x, y, z) = c, we follow the same approach as in
steps 1-3 for functions of two variables:
1. Assume that f has an extreme value on the level surface g(x, y, z) = c

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3.10 Extreme values under constraint conditions: Lagrange’s multiplier 58

2. Solve the equations

Constraint g(x, y, z) = c

 fx (x, y, z) = λgx (x, y, z)
∇f (x, y, z) = λ∇g(x, y, z) fy (x, y, z) = λgy (x, y, z)
fz (x, y, z) = λgz (x, y, z)


3. Calculate the values of f at each point (x, y, z) that arises in step 2, and at each end point (if any) of
the curve. If f has maximum value in the level curve g(x, y, z) = c it would be the largest of the values
computed; if f has a minimum value on the level curve, it will be the smallest of the values computed.

Example 3.10.2. Let V (x, y, z) = xyz for x ≥ 0, y ≥ 0, and z ≥ 0. Find the maximum value of V subject to
the constraint 2x + 2y + z = 108.
Solution. Let g(x, y, z) = 2x + 2y + z, so the constraint is g(x, y, z) = 2x + 2y + z = 108. Because

∇V (x, y, z) = yz~i + xz~j + xy~k and ∇g(x, y, z) = 2~i + 2~j + ~k

the equations we will use to find x, y and z are

Constraint 2x + 2y + z = 108 (3.17)


 yz = 2λ
∇f (x, y, z) = λ∇g(x, y, z) xz = 2λ (3.18)
xy = λ


First we solve for λ in terms of x, y and z in 3.18

yz xz
λ= = = xy (3.19)
2 2
Since V (x, y, z) = 0 if x, y or z is 0, and since 0 is obviously not the maximum value of V subject to (3.17), we
can assume that x, y and z are different from 0. Then (3.19) tells us that x = y and z = 2y. Substituting for x
and z in (3.17) yields
2y + 2y + 2y = 108, so that y = 18
Thus x = 18 and z = 36, and therefore V (18, 18, 36) is the only possible extreme value of V subject to the
constraint. Since we are assuming that V has a maximum value subject to the constraint, we conclude that
V (18, 18, 36) = 11, 664 is that value. J
Exercise 3.10.1. 1. Let f (x, y) = 3x2 + 2y 2 − 4y + 1. Find the extreme values of f on the disk x2 + y 2 ≤ 16.
√
2. Find the minimum distance from a point on the surface xy + 2xz = 5 5 to the origin.

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Chapter 4

Multiple Integrals

4.1 Double integrals and their evaluation by iterated integrals

Recall that the definition of definite integral of function of one variable.
Definition 4.1.1. For any function f defined on the interval [a, b] and kP k (the norm of the partition) defined
as the maximum of all the intervals on [a, b] (i.e., kP k = max{∆xi }), the definite integral of f on [a, b] is:
Z b n
X
f (x) dx = lim f (ci )∆xi ,
a kP k→0
i=1

Provided the limit exists and is the same for all values of the evaluation points ci ∈ [xi−1 , xi ] for i = 1, 2, . . . , n.
In this case, we say f is integrable on [a, b].
In this section we want to integrate a function of two variables, f (x, y). With functions of one variable we
integrated over an interval ( i.e. a one-dimensional space) and so it makes some sense then that when
integrating a function of two variables we will integrate over a region of R2 (two dimensional space).
We will start out by assuming that the region in R2 is a rectangle which we will denote as follows,

R = [a, b] × [c, d]

This means that the ranges for x and y are a ≤ x ≤ b, c ≤ y ≤ d. Also, we will initially assume that
f (x, y) ≥ 0 although this doesn’t really have to be the case. Let’s start out with the graph of the surface S
give by graphing f (x, y) over the rectangle R.

Now, just like with functions of one variable let’s not worry about integrals quite yet. Let’s first ask what the
volume of the region under S (and above the xy-plane of course) is. We will first approximate the volume
much as we approximated the area above. We will first divide up a ≤ x ≤ b into n subintervals and divide up
c ≤ y ≤ d into m subintervals. This will divide up R into a series of smaller rectangles and from each of these
we will choose a point (x∗i , yj∗ ). Here is a sketch of this set up.

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4.2 Iterated integrated 60

Now, over each of these smaller rectangles we will construct a box whose height is given by f (x∗i , yj∗ ). Here is a
sketch of that.

Each of the rectangles has a base area of ∆A and a height of f (xi ∗ , yj ∗ ) so the volume of each of these boxes
is f (xi ∗ , yj ∗ )∆A . The volume under the surface S is then approximately,
n X
X m
V = f (xi ∗ , yj ∗ )∆A
i=1 j=1

We will have a double sum since we will need to add up volumes in both the x and y directions.
To get a better estimation of the volume we will take n and m larger and larger and to get the exact volume
we will need to take the limit as both n and m go to infinity. In other words,
n X
X m
V = lim f (xi ∗ , yj ∗ )∆A
n,m→∞
i=1 j=1

Now, this should look familiar. This looks a lot like the definition of the integral of a function of single
variable. In fact this is also the definition of a double integral, or more exactly an integral of a function of two
variables over a rectangle. Here is the official definition of a double integral of a function of two variables over
a rectangular region R as well as the notation that we’ll use for it.

Definition 4.1.2 (Double Integral of a Function of Two Variables over a Rectangular Region). For any function
f (x, y) defined on the rectangle R = {(x, y)|a ≤ x ≤ b, c ≤ y ≤ d}, divide a ≤ x ≤ b into n subintervals and
divide c ≤ y ≤RRd into m subintervals each of the rectangles has a base area ∆A, the double integral of f over R
is denoted by f (x, y)dA defined as:
R
ZZ n X
X m
f (x, y)dA = lim f (xi ∗ , yj ∗ )∆A
n,m→∞
R i=1 j=1

provided the limit exists and is the same for all choices of the evaluation points (xi ∗ , yj ∗ ) ∈ R for i = 1, 2, , n.
In this case, we say f is integrable over R. If f is non-negative and integrable on R, then the volume V of the
solid region between the graph of f and R is given by
ZZ
V = f (x, y)dA.
R

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4.2 Iterated integrated 61

Figure 4.1: Vertical simple region Figure 4.2: Horizontal simple region

4.2 Iterated integrated

Definition 4.2.1. A. A plane region R is said to be vertically simple if there are two continuous functions
g1 and g2 on an interval [a, b] such that g1 (x) ≤ g2 (x) for a ≤ x ≤ b and such that R is the region between
the graph of g1 and g2 on [a, b]. In this case we say that R is the vertically simple region between the
graphs of g1 and g2 on [a, b] see figure 4.1.
B. A plane region R is horizontally simple if there are two continuous functions h1 and h2 on an interval
[c, d] such that h1 (y) ≤ h2 (y) for c ≤ y ≤ d and such that R is the region between the graph h1 and h2 on
[c, d]. In this case we say that R is the horizontally simple region between the graphs of h1 and h2 on [c, d]
see figure 4.2.
C. A plane region R is simple if it is both vertically simple and horizontally simple.
Example 4.2.1. Let R be the region between the graphs of y = x2 and y = x + 6.
Show that r is simple.
Solution. First let us find the intersection points of the graphs of y = x2 and y = x + 6. Observe that y = x2
and y = x + 6 if and Only if x2 = x + 6. Which means that x = −2 or x = 4. The graph intersects at the point
(−2, 4) and (3, 9). Therefore, if g1 (x) = x2 and g2 (x) = x + 6 , then g1 ≤ g2 on [−2, 3] , so R is the vertically
simple region between the graphs of g1 and g2 . To prove R is horizontally simple, we notice that R is composed
of two portions, one for which 0 ≤ y ≤ 4 and the other for which 4 ≤ y ≤ 9. Thus (x, y) is in R provided that
√ √
either 0 ≤ y ≤ 4 and − y ≤ x ≤ y. Or consequently
 √
− y, for 0 ≤ y ≤ 4
h1 (y) =
y − 6, 4 ≤ y ≤ 9
√
h2 (y) = y. Then, h1 ≤ h2 on [0, 9], so R is horizontally simple region between the graphs of h1 and h2 on
[0, 9]. Since R is both vertically and horizontally simple, R is simple by definition. J
To compute a double integral, we can think of taking thin slices of the solid parallel to the yz plane and
integrating the function while holding x constant to find A(x) (i.e., perform a partial integration), and then
integrating the volume of all the thin slices to find the total volume. This is called an iterated integral.
Likewise, we could take thin slices parallel to the xz plane to find A(y), and then integrate the volume of
those slices to find total volume. In either case, we get the same answer.
Theorem 4.2.1. Let f be continuous on a region R in the plane
a If R is vertically simple region between the graph of g1 and g2 on [a, b], then f is integrable on R, and

ZZ Zb gZ2 (x)
f (x, y)dA = f (x, y) dx dy
R a g1 (x)

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4.2 Iterated integrated 62

b If R is horizontally simple region between the graph of h1 and h2 on [c, d] , then f is integrable on R, and

ZZ Zb hZ2 (x)
f (x, y)dA = f (x, y) dy dx
R a h1 (x)

Example 4.2.2. Let R be a rectangular region bounded by the lines x = −1, x = 2, y = 0 and y = 2. Find
ZZ
x2 ydA
R

Solution. The graph of R is vertically simple region between the graphs of y = 0 and y = 2 for −1 ≤ x ≤ 2.
Therefore,
ZZ Z2 Z2
f (x, y)dA = x2 y dy dx
R −1 0

R2
To evaluate the iterated integral, we first compute x2 y dy for each x in −1 ≤ x ≤ 2 We obtain
0

Z2 Z2
2 2 1
x y dy = x y dy = x2 ( y 2 )(2 − 0) = 2x2
2
0 0

Because x is held constant when we integrate with respect to y. We conclude

ZZ Z2 Z2 Z2
f (x, y)dA = x2 y dy dx = 2x2 dx = 6
R −1 0 −1

J
6x + 2y 2 dA, where R is the region enclosed by the parabola x = y 2 and the line
RR
Example 4.2.3. Evaluate
R
x + y = 2.

Solution. The upper boundary changes form at x = 1. The left boundary is the same throughout R. The
right boundary is the same throughout R. Therefore choose horizontal strips.

Z1 2−y
Z Z1
2
 2 x=y2
I= (6x + 2y ) dx dy = 3x + 2xy 2 x=2−y dy
−2 y 2 −2

Z1
(3(2 − y)2 + 2(2 − y)y 2 ) − (3y 4 + 2y 4 ) dy


=
−2
Z1
(12 − 12y + 3y 2 ) + (4y 2 − 2y 3 ) − 5y 4 dy


=
−2
Z1
12 − 12y + 7y 2 − 2y 3 − 5y 4 dy


=
−2
7 1
=[12y − 6y 2 + y 3 − y 4 − y 5 ]1−2
3 2
7 1 56
=(12 − 6 + − − 1) − (−24 − 24 − − 8 + 32)
3 2 3
99
=
2
J

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
4.3 Change of variable in double integrals 63

The area A of a plane region R by

ZZ
A= 1dA
R

where R is the region between the graphs of two continuous functions g1 and g2 on [a, b] such that g1 ≤ g2

ZZ Zb gZ2 (x) Zb
A= 1dA = 1 dy dx = (g1 − g2 ) dx.
R a g1 (x) a

√
Example 4.2.4. Find the area of the region in the xy-plane bounded by the curves y = x3 and y = x.
Solution.
√ 
ZZ Z x 
Z1   Z1
√

2 3
 2
2 2
3


A= 1dA = dy dx = x − x dx = x 2 = (1 − 0) = .

3 
 3 1 3 3
R 0 x 0

J
Theorem 4.2.2 (Linear Combinations of Double Integrals). Let the function f (x, y) and g(x, y) be integrable
over the region R, and let c be any constant. Then the following holds:
RR RR
i cf (x, y)dA = c f (x, y)dA
R R
RR RR RR
ii (f (x, y) + g(x, y)dA = f (x, y))dA + f (x, y)dA
R R R

iii . if R = R1 ∪ R2 , where R1 and R2 are non overlapping regions, then .

ZZ ZZ ZZ
f (x, y)dA = f (x, y)dA + f (x, y)dA
R R1 R2

4.3 Change of variable in double integrals

In evaluating a multiple integral over a region R, it is often convenient to use the coordinate other than
rectangular, such as the polar coordinates, cylindrical coordinates, spherical coordinates, etc.
If we let (u, v) be curvilinear coordinates of points in a plane, there will be a set of transformation
x = f (x, y), y = g(x, y) mapping points (x, y) of the xy-plane into points (u, v) of the uv-plane. In such case
0
the region R of the xy-plane is mapped into a region R of the uv-plane. We the have
ZZ ZZ
∂(x, y)
F (x, y) dx dy = G(u, v)| |dudv (4.1)
R R ∂(u, v)

where G(u, v) = F {f (u, v), g(u, v)} and

∂x ∂x
∂(x, y) ∂u ∂v
= ∂y ∂y
∂(u, v) ∂u ∂v

is the Jacobian of x and y with respect to u and v. The result 4.1 corresponds to change of variables for
double integrals.
∂(x,y)
Example 4.3.1. If u = x2 − y 2 and v = 2xy, find ∂(u,v) in terms of u and v.

Solution.
∂u ∂u
∂(u, v)
= ∂x
∂v
∂y
∂v = 4(x2 + y 2 )
∂(x, y) ∂x ∂y

From the identity (x2 + y 2 )2 = (x2 −√y 2 )2 + (2xy)2 we have

(x2 + y 2 )2 = u2 + v 2 and x2 + y 2 = u2 + v 2 .
∂(x,y) 1
Then ∂(u,v) = ∂(u,v) = 4(x21+y2 ) = 4√u12 +v2 J
∂(x,y)

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4.4 Double Integrals in Polar coordinates 64

∂(x,y)
Example 4.3.2. a Let x = r cos θ and y = r sin θ. Find the value of ∂(r,θ) .
RR p
b Evaluate x2 + y 2 dx dy, where R is the region in the xy-plane bounded by x2 +y 2 = 4 and x2 +y 2 = 9.
R

∂(x,y) xr xθ cos θ −r sin θ

Solution. a ∂(r,θ) = = = r cos2 θ + r sin2 θ = r.
yr yθ sin θ r cos θ

b By transforming the rectangular coordinates (x, y) in to polar coordinates (r, θ) and using 4.1 we have

Z2πZ3 Z2π 3 Z2π

r3
ZZ p ZZ
∂(x, y) 2 19 38π
x2 + y2 dx dy = r dr dθ = r dr dθ = dθ = dθ = .
∂(r, θ) 3 2 3 3
R R0 0 2 0 0

J
Exercise 4.3.1. 1. Evaluate the following iterated integrals.
3
R1 Rx
R1 Rx y
a. 1 + x2 dy dx c. e x dy dx
0 0 0 0
Rπ sin
Rx
ex sin y dx dy
RR
b. y dy dx d.
0 0

2. Compute the double integral of the function f (x, y) = x2 + y 2 over the region bounded by the curves
y = 1 − x2 and y = x2 − 1 in the xy-plane.
3. Compute the double integral of the function f (x, y) = 6 − x + 2y over the region bounded by the curves
x = y 2 and y = 2x in the xy-plane.
of the solid region D bounded above by the paraboloid z = 4 − x2 − y 2 and below by the
4. Find the volume V RR
xy-plane(hint V = (4 − x2 − y 2 )dA)
R

4.4 Double Integrals in Polar coordinates

To this point we have seen quite a few double integrals. However, in every case we’ve seen to this point the
region R could be easily described in terms of simple functions in Cartesian coordinates. In this section we
want to look at some regions that are much easier to describe in terms of polar coordinates. For instance, we
might have a region that is a disk, ring, or a portion of a disk or ring. In these cases using Cartesian
coordinates could be somewhat cumbersome.
So, if we could convert our double integral formula into one involving polar coordinates we would be in pretty
good shape. The problem is that we can’t just convert the dx and the dy into a dr and a dθ . In computing
double integrals to this point we have been using the fact that dA = dx dy and this really does require
Cartesian coordinates to use. Once we’ve moved into polar coordinates dA 6= dr dθ and so we’re going to need
to determine just what dA is under polar coordinates. After some manipulations it is found that
dA = r dr dθ.Here is a sketch of some region using polar coordinates.

So, our general region will be defined by inequalities,

α≤θ≤β
h1 (θ) ≤ r ≤ h2 (θ)

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4.4 Double Integrals in Polar coordinates 65

Theorem 4.4.1. Suppose that h1 and h2 are continuous on [α, β], where 0 ≤ α ≤ β ≤ 2π, and that 0 ≤ h1 (θ) ≤
h2 (θ) for α ≤ θ ≤ β, Let R be the region between the polar graphs of r = h1 (θ) and r = h2 (θ) for α ≤ θ ≤ β
and if f is continuous on R, then

ZZ Zβ hZ2 (θ)
f (x, y)dA = f (r cos θ, r sin θ)r dr dθ
R α h! (θ)

In the event f is non-negative on R, the volume V of the region between the graph of f and R is given by

Zβ hZ2 (θ)
V = f (r cos θ, r sin θ)r dr dθ
α h1 (θ)

and the area A of R is given by

Zβ hZ2 (θ)
A= r dr dθ
α h1 (θ)

Example 4.4.1. Evaluate the integral by converting into polar coordinates,

ZZ
2xydA
R

, R is the portion of the region between the circles of radius 2 and radius 5 centred at the origin that lies in the
first quadrant.
Solution. First let’s get R in terms of polar coordinates. The circle of radius 2 is given by r = 2 and the circle
of radius 5 is given by r = 5 . We want the region between them so we will have the following inequality for r.

2≤r≤5

Also, since we only want the portion that is in the first quadrant we get the following range of θ’s. 0 ≤ θ ≤ π/2.
Now that we’ve got these we can do the integral.
π
ZZ Z2 Z5
2xydA = 2r cos θr sin θr dr dθ
R 0 2
π
Z2 Z5
= 2r3 cos θ sin θ dr dθ
0 2
π
2
5
r4
Z  
= 2 cos θ sin θ dθ
4 2
0
π
Z2
609
= cosθsinθ dθ
2
0
609
=
4
J
Example 4.4.2. Let D be the solid region bounded above by the paraboloid z = 4 − x2 − y 2 and below by the
xy plane. Find the volume V of D.

Solution. The region R over which the integral is to be taken is bounded by the circle (Intersection of the
surface with plane), whose equation in polar coordinate is r = 2: therefore D can be described as the region

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4.5 Application of Double Integrals 66

between the graph of the paraboloid and the disk, As a result

ZZ Z2πZ2
2 2
4 − (r cos θ)2 − (r sin θ)2 r dr dθ



4−x −y dA =
R 0 0
Z2πZ2
4r − r3 dr dθ


=
0 0
Z2π 2
r4
= (2r2 − ) dθ
4 0
0
Z2π
=4 dθ = 8π
0

J
Exercise 4.4.1. 1 Determine the volume of the region that lies under the sphere, x2 + y 2 + z 2 = 1 above
the plane z = 0 and inside the cylinder x2 + y 2 = 5.

2 Determine the area of the region that lies inside r = 3 + sin θ and outside r = 2.
3 Find the volume of the region that lies inside the z = x2 + y 2 and below the plane z = 16.

4.5 Application of Double Integrals

Double integrals have different applications in solving practical problems. Some of its applications are used
i to find area of a plane figure and volume of certain solids(We have already seen.)
ii to find surface area of three dimensional surfaces(We will see in the next subsection), e.t.c.

4.5.1 Surface Area

Definition 4.5.1. Let R be a vertically or horizontally simple region, and let f have continuous partial deriva-
tives on R. If G is the graph of f on R, then the surface area S of G is defined by
ZZ q
S= [fx ]2 + [fy ]2 + 1dA. (4.2)
R

Example 4.5.1. Find the surface area S of the portion of the paraboloid

z = 2 − x2 − y 2

that lies above the xy-plane.

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4.5 Application of Double Integrals 67

Solution. The given surface lies over the region R in the xy-plane bounded by the circle x2 + y 2 = 2. If
f (x, y) = 2 − x2 − y 2 , then fx (x, y) = −2x and fy (x, y) = −2y. By equation 4.2,
ZZ q ZZ p
S= 2 2
[fx ] + [fy ] + 1dA = 4x2 + 4y 2 + 1dA
R R

This double integral can easily be evaluated by using polar coordinates.

√
ZZ p Z2 Z 2p
13
S= 4(x2 + y 2 ) + 1dA = 4r2 + 1r dr dθ = π
3
R 0 0

J
p
Example 4.5.2. Find the surface area S of the frustum of the cone Z = x2 + y 2 with minimum and maximum
radii 1 and 2 respectively.
2 2 2 2
Solution. Thep given surface lies over the xregion R in the xy-planey bounded by the annulus 1 ≤ x + y ≤ 2 .
If f (x, y) = x + y , then fx (x, y) = √ 2 2 and fy (x, y) = √ 2 2 .
2 2
x +y x +y
So that
ZZ s
x2 y2
ZZ q
S= [fx ]2 + [fy ]2 + 1dA = + 2 + 1dA
+y 2 x2
x + y2
R
ZRZ √ √ ZZ √ √ √
= 2dA = 2 dA = 2A(R) = 2π(22 − 12 ) = 3 2π
R R

J
√
Exercise 4.5.1. 1. Find the surface area of that portion of the surface z = 4 − x2 that lies above the
rectangle R in the xy plane whose coordinates satisfy 0 ≤ x ≤ 10 and 0 ≤ y ≤ 40
2. Find the surface area of the portion of the paraboloid x2 + y 2 = z below the plane z = 1.
3. Find the area of the portion of the plane Z = x + 3y that lies inside the elliptical cylinder with equation
x2 y2
4 + 9 = 1.

These are not the only applications of double integrals. The other applications related center of gravity of a
body, mass of a body and moment of inertia are mentioned at the end of the next chapter.

4.5.2 Triple Integrals in Cartesian coordinates

Now that we know how to integrate over a two-dimensional region we need to move on to integrating over a
three-dimensional region. We used a double integral to integrate over a two dimensional region and so it
shouldn’t be too surprising that we’ll use a triple integral to integrate over a three dimensional region. The
notation for the general triple integrals is,
ZZZ
f (x, y, z)dV.
D

Let’s start simple by integrating over the box, D = [a, b] × [c, d] × [r, s]. Note that when using this notation we
list the x’s first, the y’s second and the z’s third. The triple integral in this case is,
Zr Z d Z b
f (x, y, z) dx dy dz.
c a
s

Note that we integrated with respect to x first, then y, and finally z here, but in fact there is no reason to the
integrals in this order. There are 6 different possible orders to do the integral in and which order you do the
integral in will depend upon the function and the order that you feel will be the easiest. We will get the same
answer regardless of the order however. This can be stated as a theorem:
Theorem 4.5.1 (Fubini’s Theorem). Let D be the rectangular box dened by the inequalities

a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s
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DEPARTMENT OF MATHEMATICS Applied Mathematics II
4.5 Application of Double Integrals 68

If f is continuous on the region D, then

ZZZ Zb Z d Z s
f (x, y, z)dV = f (x, y, z) dz dy dx
c r
D a

Moreover, the iterated integral on the right can be replaced with any of the ve other iterated integrals that
result by altering the order of integration.
Example 4.5.3. Evaluate the following integral,
ZZZ
8xyzdV,
D

where D = [2, 3] × [1, 2] × [0, 1]

Solution. Just to make the point that order doesn’t matter let’s use a different order from that listed above.
We’ll do the integral in the following order.

ZZZ Z2 Z 2 Z 0
xyzdV = xyz dz dx dy
3 1
D 1
Z2 Z 2 1
= 4xyz 2 dx dy
3 0
1
Z2 Z 2 1
= 4xy dx dy
3 0
1
Z2 2
= 2x2 y dx dy
3
1
Z2 2
= 10y dx dy = 15
3
1

J
Theorem 4.5.2. Let D be the solid region between the graphs of two continuous functions F1 and F2 on
vertically or horizontally simple region R in the plane, and let f be continuous on D. Then
ZZZ ZZ Z F2 (x,y)
f (x, y, z)dV = ( f (x, y, z) dz)dA
F1 (x,y)
D R

F2 R
(x,y)
We evaluate f (x, y, z) dz by integrating with respect to z while both x and y are held fixed, thus obtaining
F1 (x,y)
a number depending on x and y.

If R is the vertically simple region between the graphs of g1 (x) and g2 (x) on [a, b], we evaluate the double
integral over R by using

ZZZ Zb Z g2 (x) Z F2 (x,y)

f (x, y, z)dV = (( f (x, y, z) dz) dy) dx.
g1 (x) F1 (x,y)
D a

In contrast, if R is the horizontally simple region between the graphs of h1 (y) and h2 (y) on [c, d], we
evaluate the double integral over R by using
ZZZ Zc Z h2 (y) Z F2 (x,y)
f (x, y, z)dV = (( f (x, y, z) dz) dx) dy.
h1 (y) F1 (x,y)
D d

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4.5 Application of Double Integrals 69

Example 4.5.4. Let R be the triangular region in the xy plane between the graphs of y = 0 and y = x for
0 ≤ x ≤ 1 , and let D be the solid region between the graphs of the surfaces z = −y 2 and z = x2 for all (x, y)
in R. Evaluate ZZZ
(x + 1)dV.
D

Solution. By the above discussions we have,

ZZZ Z1 Z x Z x2
(x + 1)dV = [ ( (x + 1) dz) dy] dx
0 −y 2
D 0
Z1 Z x x2
= (x + 1)z dy dx
0 −y 2
0
Z1 Z x
= (x + 1)(x2 + y 2 ) dy dx
0
0
Z1 x
xy 3 ) y3
= (x3 y + + x2 y + dx
3 3 0
0
Z1
4 3
= (x4 + x3 ) dx =
3 5
0

J
Evaluation of triple integrals as they are may sometimes be difficult or even possible. In this case changing
one coordinate system to another may facilitate the process. If (u, v, w) are curvlinear coordinates in three
dimensions, there will be a set of transformation equations x = f (u, v, w), y = g(u, v, w) and z = h(u, v, w)
and we can write:
ZZZ ZZZ
∂(x, y, z)
F (x, y, z) dx dy dz = G(u, v, w) dudvdw (4.3)
∂(u, v, w)
D D0

where G(u, v, w) = F {f (u, v, w), g(u, v, w), h(u, v, w)} and

∂x ∂x ∂x
∂(x, y, z) ∂u ∂v ∂w
∂y ∂y ∂y
= ∂u ∂v ∂w
∂(u, v, w) ∂z ∂z ∂z
∂u ∂v ∂w

is the Jacobian of x, y and z with respect to u, v and w. The result (4.3)corresponds to change of variables for
triple integrals.
Exercise 4.5.2. 1. Evaluate the following
√ πiterated integrals.
R1 Rx x+y R 6 Ry Ry
(1 + y 2 z cos xz) dx dz dy
R
A. (z − 2x − y) dz dy dx B.
−1 0 x−y 0 0 0
π π
R 3 R1 Ry
ln 2 R2 R2 sin
Rz
C. (z 2 + 1)ey dz dy dx D. (sin y) dx dy dz
0 0 0 0 0 0
√1 π
R13 Re x 2 cos
Rz cos
R yz
z(ln x)2 dz dx dy
R R
E. F. (x cos(yz) dx dy dz
−15 1 0 −π
2
− cos z − cos yz

ey dV , where D is the solid region bounded by the planes y = 1, z = 0, y = x,

RRR
2. Evaluate the integral
D
y = −x and z = y.
RRR
3. Evaluate the integral zydV , where D is the solid region in the first octant bounded above by the planes
pD
z = 1 and below by the cone z = x2 + y 2 .
Now we are going to see cylindrical and spherical coordinates as special cases of the topic change of variables
in triple integrals.
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4.6 Triple integrals in cylindrical and spherical coordinates 70

4.6 Triple integrals in cylindrical and spherical coordinates

4.6.1 Triple Integrals in Cylindrical Coordinates
Let (x, y, z) be the rectangular coordinates of a point P in space. If (r, θ) is a set of polar coordinates for the
point (x, y), then we call (r, θ, z) a set of cylindrical coordinates for P as shown in the figure 4.3.

P (x, y, z) = P (r, θ, z)
Given the rectangular coordinates (x, y, z) of
a point P , we can determine a set of cylindri-
cal coordinates for P with aid of the formulas
x2 + y 2 = r2 and tan θ = xy (if x 6= 0). Con-
versely, from any set (r, θ, z) of cylindrical
coordinates of a point P we can determine
the rectangular coordinates (x, y, z) of P by r
the formulas x = r cos θ and y = r sin θ. θ
y

Figure 4.3:

Let see equations of common solids in rectangular and cylindrical coordinates.

Surface Rectangular Cylindrical
Cylinder x2 + y 2 = a2 r=a
Sphere x2 + y 2 + z 2 = a2 r2 + z 2 = a2
Double circular cone x2 + y 2 = a2 z 2 r = az or z = a cot φ
Circular paraboloid x2 + y 2 = az r2 = az
As you have seen in previous section,
ZZZ ZZ Z F2 (x,y)
f (x, y, z)dV = ( f (x, y, z) dz)dA
F1 (x,y)
D R

for a solid region between the graphs of two continuous functions F1 and F2 on a vertically or Horizontally
simple region R in the xy-plane and for a continuous function f on D.
Theorem 4.6.1. Let D be a solid region between the graphs of F1 and F2 on R, where R is the plane region
between the polar graphs of h1 and h2 on [α, β], with 0 ≤ β − α ≤ 2π and 0 ≤ h1 (θ) ≤ h2 (θ) for α ≤ θ ≤ β. If
f is continuous on D, then
ZZZ Zα Z h2 (θ) Z F2 (r cos θ,r sin θ)
f (x, y, z)dV = f (r cos θ, r sin θ, z)r dz dr dθ.
h1 (θ) F1 (r cos θ,r sin θ)
D β

Integration by means of cylindrical coordinates is especially effective when expressions containing x2 + y 2

appear in the integrand or in the limits of integration and the region over which the integration is taken is
easily described by polar coordinates.
Example 4.6.1. Let D be the solid region bounded above by the plane y + z = 4, below by the xy- plane and
on the sides by the cylinder x2 + y 2 = 16 as shown in the figure 4.4. Evaluate
ZZZ p
x2 + y 2 dV.
R

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4.6 Triple integrals in cylindrical and spherical coordinates 71

Figure 4.4:

Solution. Observe that D is a solid region between the graphs of z = 0 and z = 4 − y on R, where R is the disk
x2 + y 2 ≤ 16. In polar coordinate R is the region between the polar graphs of r = 0 and r = 4 for 0 ≤ θ ≤ 2π.
Consequently, in cylindrical coordinates D is the solid region between the graphs z = 0 and z = 4 − r sin θ for
(r, θ) in R. Then by Theorem 4.6.1 we have

ZZZ p Z2πZ4 4−r

Z sin θ Z2πZ4 4−r sin θ
x2 + y 2 dV = r · r dz dr dθ = r2 z dr dθ
0
R 0 0 0 0 0
Z2πZ4 Z2π 4
4 3 r4

2 3
= (4r − r sin θ) dr dθ = r − sin θ dθ
3 4 0
0 0 0
Z2π    2π
256 256 512
= − 64 sin θ dθ = θ + 64 cos θ = π
3 3 0 3
0

J
√
R3 9−x2 R1
y 2 dz dy dx.
R
Example 4.6.2. Evaluate
√
−3 − 9−x2 (x2 +y 2 )2
√ √
Solution. The limit of integrations −3 and 3 in the first integral and − 9 − x2 and 9 − x2 on the second
integral tells us that those two integrals are taken over the region of the disk x2 + y 2 ≤ 9 or r ≤ 3, in the xy-
plane. It follows that
√
Z3 Z9−x2 Z1 Z0 Z3 Z1
2
y dz dy dx = (r sin θ)2 r dz dr dθ
√
−3 − 9−x2 (x2 +y 2 )2 2π 0 r 4

Z0 Z3  1 Z0 Z3  
2
= 3
r sin θz dr dθ = r − r sin2 θ dr dθ]
3 7

r4
2π 0 2π 0
Z2π 4 8 3
 Z 2π
r r 6399
= − sin2 θ dθ = − sin2 θ dθ
4 8 0 8 0
0
Z 2π  2π
6399 6399 sin 2θ 6399
=− (1 − cos 2θ) dθ = − θ− =− π
16 0 16 2 0 8
J
Exercise 4.6.1. Evaluate the following integrals by changing into cylindrical coordinates.
RRR 2
1. (x + y 2 )dV , where D is the solid region bounded by the cylinder x2 + y 2 = 1 and the planes z = 0
D
and z = 4.

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DEPARTMENT OF MATHEMATICS Applied Mathematics II
4.6 Triple integrals in cylindrical and spherical coordinates 72

(xz)dV , where D is the portion of the ball portionof theballx2 + y 2 + z 2 ≤ 4 in the first octant.
RRR
2.
D

(y 2 z)dV , where D is the solid region bounded above by the sphere x2 + y 2 + z 2 = 4.

RRR
3.
D

4.6.2 Triple Integrals in Spherical Coordinates

Given a point P with rectangular coordinates (x, y, z). Let ρ = |OP | be the distance between the origin O and
−−→
P , and let φ be the angle measured (downward) from the positive z− axis to OP as shown in the figure 4.5.
Also let θ be the same angle as in the case of cylindrical coordinates, namely the angle from the positive x−
axis to OQ, where Q is the projection of P onto the xy plane(θ is measured in counter clockwise direction as
shown on the xy-plane from the side of the positive z− axis).

P (x, y, z) = P (ρ, φ, θ)
Then the point P is said to have spherical
coordinates ρ, θ and φ, and we write P = ρ
(ρ, θ, φ) as well as P = (x, y, z). Where 0 ≤
φ
r ≤ ∞, 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π. From
trigonometry we found that r = ρ sin φ and
z = ρ cos φ . These equations along with the
polar coordinated formulae x = r cos θ and r
y = r sin θ. θ
y

Figure 4.5:

Yield the following formulae for converting from spherical coordinates to rectangular coordinates:
x = r cos θ = ρ sin φ cos θ, y p
= r sin θ = ρ sin φ sin θ and z = ρcosφ. As you can easily verify that
x2 + y 2 + z 2 = ρ2 ⇒ ρ = x2 + y 2 + z 2 , tan θ = xy , cos φ = √ 2 z 2 2 , (x2 + y 2 + z 2 6= 0).
x +y +z

Theorem 4.6.2 (Triple integrals in spherical coordinates). If f (x, y, z) is continuous on a solid region D, then
ZZZ ZZZ
f (x, y, z)dV = f (ρ sin φ cos θ, ρ sin φ sin θ, ρcosφ)ρ2 sin φ dρ dφ dθ.
D D0

z 2 dV , where D is a solid region x2 + y 2 + z 2 ≤ 1.

RRR
Example 4.6.3. Evaluate
D

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4.6 Triple integrals in cylindrical and spherical coordinates 73

Solution. By using spherical coordinates

ZZZ Z2πZπ Z1
2
z dV = (ρ cos φ)2 ρ2 sin φ dρ dφ dθ
D 0 0 0
Z2π Zπ  Z1 Zπ  1
ρ5
 
4 2 2
= ρ dρ (cos φ) ρ sin φ dφ dθ = 2π cos2 φ sin φ dφ
5 0
0 0 0 0
Zπ  
2π 2 2π 2 4
= cos φ sin φ dφ = = π
5 5 3 15
0

J
RRR p
Example 4.6.4. Use spherical coordinates and evaluate x2 + y 2 + z 2 dV , where D is the ball x2 +y 2 +z 2 ≤
D
a2 (a 6= 0)

Solution. Using the transformation equations x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ, ρ2 = x2 + y 2 + z 2
and dV = ρ2 sin φ dρ dφdθ, we have

ZZZ p Z2πZπ Za
x2 + y2 + z 2 dV = (ρ · ρ2 sin φ dρ dφ dθ
D 0 0 0
Z2π Zπ  Za   Z2π Zπ  4 a  
3 ρ
= ρ dρ sin φ dφ dθ = sin φ dφ dθ
4 0
0 0 0 0 0
Z2π Zπ 
a4 sin φ
 
= dφ dθ
4
0 0
Z2π π Z2π
a4 a4
=− (cos φ) dθ = dθ = πa4
4 0 2
0 0

J
(x2 + y 2 )dV , where D is a solid region bounded above by
RRR
Exercise 4.6.2. 1. Evaluate the triple integral
q D

the unit sphere x2 + y 2 + z 2 = 1 and below by the cone z = − 13 (x2 + y 2 ) using spherical coordinates.
dV
where D is the bounded solid region between the cylinder x2 + y 2 = 4 and the
RRR
2. Evaluate x2 +y 2 +z 2 ,
D
nappes of the cone x2 + y 2 = z 2 using spherical coordinates.
dV 2 2 2
RRR
3. Evaluate 3 , where D is the spherical shell 1 ≤ x + y + z ≤ e using spherical coordinates.
2 2 2
(x +y +z ) 2
D

4. Use spherical coordinates to evaluate

√ √
4−x2 −y 2
Z2 Z4−x2 Z p
z2 x2 + y 2 + z 2 dz dy dx.
√
−2 − 4−x2 0

5. Use either cylindrical or spherical coordinates to evaluate the integral

√ √ 2 2
Z1 Z1−x2 1−x Z −y √ 2 2 2 3
e−( x +y +z ) 2 dz dy dx.
−1 0 0

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4.7 Application: Volume, center of mass of solid region 74

4.7 Application: Volume, center of mass of solid region

Suppose f (x, y, z) = 1. The volume of the solid D is defined to be
ZZZ
V = dV.

Example 4.7.1. Use triple integral to find the volume of the solid enclosed between the cylinder x2 + y 2 = 16
and the planes z = 0 and y + z = 4.

Solution. The solid D and its projection R on the the xy- are shown in Figure 4.6. The lower surface of the
solid is the plane z = 0, and the upper surface is the plane y + z = 4, or equivalently, z = 4 − y. Thus,

ZZZ Z Z  4−y
Z  ZZ
V = dv = dz dA = (4 − y)dA
D R 0 R
Z2πZ4
= (4 − r sin θ)r dr dθ (by using cylindrical coordinates)
0 0
Z2π Z4 Z2π 4
r3
 
2
= (4r − r sin θ) dr dθ = 2r2 − θ dθ
sin 0
0 0 0
Z2π    2π
64 64
= 32 −
sin θ dθ = 32θ + cos θ
3 3 0
0
 
64 64
= 64π + − = 64π
3 3

Figure 4.6:

Example 4.7.2. Find the volume V of the region in the first octant bounded by the planes z = 10 + x + y,
y = 2 − x, y = x, z = 0 and x = 0.
Solution. The region R of the solid bounded in the xy-plane is as shown in the fig—. The line y = 2 − x and
y = x intersect at (1, 1). Hence, x ≤ y ≤ 2 − x as 0 ≤ x ≤ 1. Therefore,

ZZZ Z 10+x+y
Z1 2−x Z Z1  2−x
Z  10+x+y
Z  
V = dv = dx dy dz = dz dy dx
D x=0 y=x z=0 x=0 y=x z=0
Z1  2−x Z1 1
x3
Z 
34
= (10x + x + y) dy dx = {20 − 20x − 2x2 } dx = 2{11x − 5x2 − } = .
3 0 4
0 x 0

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4.7 Application: Volume, center of mass of solid region 75

4.7.1 Some of the applications of multiple integrals

Some of the applications of multiple integral are used to
• evaluate the area of a region enclosed by a plane curves

• calculate area of curved surfaces

• find the volume of a solid region
• calculate a mass of a lamina

• calculate a mass of a solid figure

• calculate center of gravity of a plane lamina and a solid
• calculate moment of inertia of a plane lamina and a solid
Now let see some of its applications:

1. Calculation of Mass
(a) For a plane lamina, if the surface density at the point P (x, y) be ρ = f (x, y), then the elementary
mass at P is equal to ρ dx dy. therefore, the total mass of the lamina is
ZZ
ρ dx dy

with integrals embracing the whole area of the lamina. In polar coordinates, taking ρ = φ(r, θ) at
the point P (r, θ), the total mass of the lamina is
ZZ
ρ dr dθ

(b) For a solid if the density at a point P (x, y, z) be ρ = f (x, y, z), then the total mass of the solid is
ZZ
ρ dx dy dz

with appropriate limit of integration.

2. Center of Gravity
(a) To find the center of gravity (x, y) of a plane lamina, take the element of mass ρ dx dy at the point
P (x, y), then RR RR
xρ dx dy yρ dx dy
x = RR , y = RR
ρ dx dy ρ dx dy
(b) To find Center of Gravity (x, y, z) of a solid, take an element of mass ρ dx dy dz enclosing the point
P (x, y, z). Then,
RRR RRR RRR
xρ dx dy dz yρ dx dy dz zρ dx dy dz
x= RRR , y= RRR , z = RRR
ρ dx dy dz ρ dx dy dz ρ dx dy dz
x
Example 4.7.3. Find the mass of the tetrahedron bounded by the coordinate planes and the plane a +
y z
b + c = 1, with variable density ρ(x, y, z) = µxyz.

Solution.
RRR Elementary mass at point P (x, y, z) is µxyz dx dy dz. Therefore, the whole mass is equal to
µxyz dx dy dz,
 the integrals
 embracing the whole volume of OABC(Fig
  4.7). The limits for Z are
y
from 0 to Z = c 1 − xa − b . The limits for y are from 0 to y = b 1 − xa and the limits of x are from 0
to a. Hence the required mass is

Ambo University
DEPARTMENT OF MATHEMATICS Applied Mathematics II
4.7 Application: Volume, center of mass of solid region 76

Figure 4.7:

 
y
c 1− x
a−b
 
b(1− x
a) b(1− x
a)
Za Za c 1− x y
a−b
xyz 2
Z Z Z
µxyz dz dy dx = µ dy dx
2 z=0
0 0 0 0 0
b(1− x
a)
Za
c2
Z
x y
=µ xy · (1 − − )2 dy dx
2 a b
0 0
x
Za b(1−
Z a) 2
µc2 x y2 y3
   
x
= x 1− y−2 1− + 2 dy dx
α a a b b
0 0
Za 
2
2
x y2 x y3 y4
 
µc
= x 1− −2 1− + 2 dx
α a 2 a 3b 4b
0

2 Za  4 4 4 
b2 2b2 b2
  
µc x x x
= 1− − 1− + 1− dx
α 2 a 3 a 4 a
0

2 2 Za  4
µb c x µa2 b2 c2
= x 1− dx = .
24 a 720
0

J
Example 4.7.4. Find the volume and centroid (center of gravity) of the uniform “ice-Cream Cone” D
bounded by the cone φ = π6 and the sphere ρ = 2a cos φ of radius a. The sphere and the part of the cone
within it are shown in figure–.
π
Solution. The ice-cream cone is described by the inequalities 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ 6 and 0 ≤ ρ ≤ 2a cos φ.
Its volume is given by
π π
Z2πZ6 2aZcos φ Z2πZ6
8
V = ρ2 sin φ dρ dφ dθ = a3 sin φ dφdθ
3
0 0 0 0 0
  π6
16 3 1 4 7
= a π − cos φ = πa3
3 4 0 12

Now for the centroid, it is clear by symmetry that x = y = 0. We may also assume that D has a constant
7
density δ, so that the mass of D is numerically equal to volume times its density equal to 12 πa3 δ. Because
Z = ρ cos φ, the Z-coordinates of the centroid of D is
Ambo University
DEPARTMENT OF MATHEMATICS Applied Mathematics II
4.7 Application: Volume, center of mass of solid region 77

RRR RRR
ρzdV zdV ZZZ
D D 1
Z = RRR = RRR = zdV
ρdV dV V
D D D
π
Z2πZ 6 2aZcos φ
12
= ρ3 cos φ sin φ dρ dφ dθ
7πa3
0 0 0
π
Z2πZ6   π6
48 5 96 1 6 37
= a cos φ sin φ dφ dθ = a − cos φ = a
7 7 6 0 28
0 0
 
Hence the centroid of the ice-cream is located at the point 0, 0, 37
28 a . J

3. Moment of Inertia

If a particle of mass m of a body be at a dis-

tance r from a given line, them mr2 is called
the moment of inertia of the particle about
the given line and the sum of similar expres-
sions
P taken for all the particles of the body,
i.e mr2 is called the moment of inertial of
the body about the given line. (fig4.8) If M
be the total mass of the body and we write
its moment of inertia is equal to M k 2 , then k
is called the radius of gyration of the body
about the axis.

Figure 4.8:

(a) Moment of inertia of a plane lamina Consider an elementary mass ρ dx dy at the point P (x, y) of a
plane area A so that it’s Moment of Inertia about x- axis is equal to ρ dx dyy 2 .
∴ M.I of the lamina about the x− axis, i.e. Ix is given by
ZZ
Ix = ρy 2 dx dy.
R

Similarly, M.I of the lamina about the y− axis, i.e. Iy is given by

ZZ
Iy = ρx2 dx dy.
R

Also, M.I of the lamina about an axis perpendicular to xy-plane, i.e. Iz is given by
ZZ
Iz = ρ(x2 + b2 ) dx dy.
R

(b) Moment of inertia of a solid Consider elementary mass ρ dx dy dzpenclosing a point P (x, y, z) of a
solid from volume V . Distance of P from the x- axis is equal to y 2 + z 2 . Therefore, M.I. of this
element about the x− axis is equal to ρ dx dy dz(y 2 + z 2 ). Thus M.I of this solid about the x− axis
is ZZZ
Ix = ρ(y 2 + z 2 ) dx dy dz.
D
Ambo University
DEPARTMENT OF MATHEMATICS Applied Mathematics II
4.7 Application: Volume, center of mass of solid region 78

Similarly, M.I about the y− axis is

ZZZ
Iy = ρ(x2 + z 2 ) dx dy dz.
D

M.I about the z− axis is ZZZ

Iz = ρ(x2 + y 2 ) dx dy dz.
D

Example 4.7.5. A solid ball D with constant density δ is bounded by the spherical surface with equation
ρ = a. Use spherical coordinates to compute its volume V and its moment of inertia Iz around the z−
axis.

Solution. The points of the ball are described by the inequalities 0 ≤ ρ ≤ a, 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π.
Volume of the ball is given by

ZZZ Z2πZπ Za Z2πZπ

1 3
V = dV = ρ2 sin φ dρ dφ dθ = a sin φ dφ dθ
3
D 0 0 0 0 0
Z2π Z2π
1 3  π 2 3 4 3
= a − cos φ 0
dθ = a dθ = πa .
3 3 3
0 0

The distance from a typical point (ρ, φ, θ) of the sphere to the z- axis is r = ρ sin φ, so the moment of
inertial of the sphere around that axis is

ZZZ Z2πZπ Za
Iz = 2
r δdV = δρ4 sin3 φ dρ dφ dθ
D 0 0 0
Z2πZa Zπ
1 3 2 2
= δ5 sin φ dφ dθ = πδa5 sin3 φ dφ = ma2
5 5 5
0 0 0

where m = 34 πa3 δ. J
Exercise 4.7.1. (a) Find the volume bounded by the cylinder x2 + y 2 = 4 and the planes y + z = 4 and
z = 0.
(b) Find the mass of a lamina in the form of the cardioid r = a(1 + cos θ).
(c) Find the centroid of a loop of the lemniscate r2 = a2 cos 2θ.
(d) If the density at any point of the solid octant of the ellipsoid ( xa )2 + ( yb )2 + ( zc )2 = 1 various as xyz,
find the coordinates of the center of gravity of the solid.
(e) Using double integrals, find the moment of inertial about the x− axis of the area enclosed by the lines
x = 0, y = 0, xa + yb = 1.
(f ) Find the moment of inertia about the z-axis of a homogeneous tetrahedron bounded by the planes
x = 0, y = 0, z = x + y and z = 1.
(g) A lamina is shaped like the circular sector R in the first quadrat bounded by the coordinate axes and
an arc of the unit circle x2 + y 2 = 1. Find the mass and center of gravity (x, y) of the lamina its
density function is ρ(x, y) = x2 + y 2 .

Ambo University
DEPARTMENT OF MATHEMATICS Applied Mathematics II