Calculus(II)

Calculus(II)
Department of Computer Science and Information Engineering
Chaoyang University of Technology
Taichung, Taiwan, Republic of China
Instructor: De-Yu Wang
E-mail: dywang@mail.cyut.edu.tw
Phone: (04)23323000 ext. 4538
Office: E738
z
✻
4
3
2
1
O
1 ✑
❳❳ 1 ❳❳❳ 2
✑ 3
2 ❳
✑ ❳❳❳ 4 5
3 ✑
❳3 y
4 ✑
✑
x
✑✰ ✑
y
3 ✻
< 1,−2 >
2 < 3,2 >
❆❑ r(2)
r(−2)
❆1
◆
❆ ✗
❆✑ ✑✑✑✑✑✸ ✲x
−2 −1 ❆ 1 2 3 4
−1 ❆
❯ r(0)
❆ ✕
−2 ❆❯
< 1,−2 >
♣ ✉ ✈ ✇ 1 2 3 4 5 6 7 7
February 10, 2012

CALCULUS
CALCULUS
Calculus
• Instructor:
1. Name: De-Yu Wang
2. Email: dywang@mail.cyut.edu.tw
3. Phone: (04)23323000 ext. 4538
4. Office: E738
• Textbook:
R. T. Smith and R. B. Minton, ”Calculus: Early Transcendental Functions,”
3e, 2006.
• Reference:
1. M. D. Weir, J. Hass and F. R. Giordano, “Thomas’ Calculus,” Eleventh
Edition, Greg Tobin, 2005.
2. J. Stewart, ”Early Transcendentals Calculus,” Five Edition, Thomson
Learning Inc., 2003.
3. R. Larson, R. Hostetler and B. H. Edwards ”Essential Calculus: Early
Transcendental Functions,” 2006.
• Grade:
1. Performance in class
2. Attendance condition
3. Weekly tests
4. Midterm
5. Final
De-Yu Wang CSIE CYUT
Calculus

CONTENTS
CONTENTS
Contents
1 LIMITS AND CONTINUITY 1
1.1 The Concept of Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Computation of Limits . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 Continuity and its Consequences . . . . . . . . . . . . . . . . . . . . . 9
1.4 Limits Involving Infinity . . . . . . . . . . . . . . . . . . . . . . . . . 13
2 DIFFERENTIATION 15
2.1 Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.2 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.3 Computation of Derivatives: the Power Rule . . . . . . . . . . . . . . 19
2.4 The Product and Quotient Rules . . . . . . . . . . . . . . . . . . . . 21
2.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.6 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.7 Derivatives of Trigonometric Functions . . . . . . . . . . . . . . . . . 28
2.8 Derivatives of Exponential and Logarithmic Functions . . . . . . . . . 32
3 APPLICATIONS OF DIFFERENTIATION 35
3.1 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . 36
3.2 Linearization and Newton’s Method . . . . . . . . . . . . . . . . . . . 37
3.3 Indeterminate Forms and L’Hôpital’s Rule . . . . . . . . . . . . . . . 41
3.4 Maximum and Minimum Values . . . . . . . . . . . . . . . . . . . . . 43
3.5 Monotonic Functions and the First Derivative Test . . . . . . . . . . 45
3.6 Concavity and the Second Derivative Test . . . . . . . . . . . . . . . 47
3.7 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
4 INTEGRATION 54
4.1 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.2 Area Under a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
4.3 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
4.4 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . 69
4.5 The Natural Logarithm as an Integral . . . . . . . . . . . . . . . . . . 72
4.6 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . 75
4.7 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
De-Yu Wang CSIE CYUT
i

CONTENTS
5 INTEGRATION TECHNIQUES 81
5.1 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . 82
5.2 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
5.3 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
5.4 Integrals Involving Powers of Trigonometric Functions . . . . . . . . . 89
5.5 Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . . . 94
6 DIFFERENTIAL EQUATIONS 97
6.1 Growth and Decay Problems . . . . . . . . . . . . . . . . . . . . . . . 98
6.2 Separable Differential Equations . . . . . . . . . . . . . . . . . . . . . 99
6.3 Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
7 INFINITE SERIES 105
7.1 Sequences of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . 106
7.2 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
7.3 The Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
7.4 Comparison Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
7.5 Alternating Series and Absolute Convergence . . . . . . . . . . . . . . 121
7.6 The Ratio Test and The Root Test . . . . . . . . . . . . . . . . . . . 125
7.7 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
7.8 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
8 PARAMETRIC EQUATIONS AND POLAR COORDINATES 136
8.1 Plane Curves and Parametric Equations . . . . . . . . . . . . . . . . 137
8.2 Calculus and Parametric Equations . . . . . . . . . . . . . . . . . . . 138
8.3 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
8.4 Calculus and Polar Coordinates . . . . . . . . . . . . . . . . . . . . . 144
9 VECTORS 149
9.1 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
9.2 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
9.3 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
9.4 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . 160
9.5 The Calculus of Vector-Valued Functions . . . . . . . . . . . . . . . . 163
9.6 Arc Length and Curvature . . . . . . . . . . . . . . . . . . . . . . . . 167
10 PARTIAL DIFFERENTIATION 170
10.1 Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . 171
10.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . 172
10.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
10.4 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
10.5 The Gradient and Directional Derivatives . . . . . . . . . . . . . . . . 180
10.6 Extrema of Functions of Several Variables . . . . . . . . . . . . . . . 182
De-Yu Wang CSIE CYUT
ii

CONTENTS
11 MULTIPLE INTEGRALS 186
11.1 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
11.2 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . 192
11.3 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
11.4 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
11.5 Triple Integrals in Cylindrical and Spherical Coordinates . . . . . . . 199
11.6 Change of Variables: Jacobians . . . . . . . . . . . . . . . . . . . . . 204
De-Yu Wang CSIE CYUT
iii

Chapter 7
INFINITE SERIES

7.1. SEQUENCES OF REAL NUMBERS
7.1 Sequences of Real Numbers
Definition 7.1.1. (a) Sequence {a n } ∞ n=n 0
: A function whose domain is the set of
integers.
(b) Sequence {a n } ∞ n=n 0
converges to L:
lim a n = L
n→∞
if for each ε > 0, there exist N > 0 such that |a n −L| < ε whenever n > N. If
the limit does not exist, then {a n } ∞ n=n 0
diverges.
{ } ∞ 1
Example 7.1.1. Write out the terms of the sequence {a n } ∞ n=1 = n 2 n=1
show that {a n } ∞ n=1
converges to 0.
and then
Solution:
{ } ∞ { 1 1
=
n 2 n=1
1 , 1 4 ,··· , 1 2,···}
n
Given any ε > 0, we must find N sufficiently large so that for every n > N,
∣ 1 ∣∣∣
∣n −0 < ε
2
1
n < ε 2
n 2 > 1 ε
n >
√
1
ε = N.
Theorem 7.1.1. Suppose that {a n } ∞ n=n 0
and {b n } ∞ n=n 0
both converge. Then
(a) lim
n→∞
(a n +b n ) = lim
n→∞
a n + lim
n→∞
b n ,
(b) lim(a n −b n ) = lim a n − lim b n ,
n→∞ n→∞ n→∞
( )(
(c) lim
n→∞
(a n b n ) =
a n
(d) lim =
n→∞ b n
lim
n→∞ a n
lim
n→∞ b n
lim
n→∞ a n
lim
n→∞ b n
)
,
( )
assuming lim b n ≠ 0 .
n→∞
De-Yu Wang CSIE CYUT 106

7.1. SEQUENCES OF REAL NUMBERS
Example 7.1.2. Evaluate lim
n→∞
5n+7
3n−5 .
Solution:
5n+7
lim
n→∞ 3n−5 = lim 5+ 7 n
n→∞ 3− 5 n
= 5 3
the sequence converges to 5 3 .
Example 7.1.3. Evaluate lim
n→∞
n 2 +1
2n−3 .
Solution:
n 2 +1
lim
n→∞ 2n−3 = lim n+ 1 n
n→∞ 2− 3 n
= ∞ the sequence diverges.
Theorem 7.1.2. Suppose that lim f(x) = L. Then, lim f(n) = L, also.
x→∞ n→∞
Remark. If lim f(n) = L, it need not be true that lim f(x) = L. For exapmle,
n→∞ x→∞
lim cos(2πn) = 1, lim
n→∞
cos(2πx)
x→∞
does not exist.
c n
5 10 15
2.0
✻
f(x)
✻
1
1.5
1.0
0.5
✲n
0.5
−0.5
−1
✲ x
2π 4π 6π
Example 7.1.4. Evaluate lim
n→∞
n+1
e n
Solution: Apply L’Hôpital’s rule for
x+1
lim
x→∞ e x
∴ lim
n→∞
n+1
e n
d
= lim dx (x+1)
x→∞ d
dx ex
= 0, also.
1
= lim
x→∞ e = 0 x
c n
✻
0.8
0.6
0.4
0.2
✲
n
5 10 15
De-Yu Wang CSIE CYUT 107

7.1. SEQUENCES OF REAL NUMBERS
Theorem 7.1.3. Squeeze theorem for sequences
If lim
n→∞
a n = L = lim
n→∞
b n ,
and there exists an integer N such that a n ≤ c n ≤ b n for all n > N, then
lim c n = L.
n→∞
Proof: Given any ε 1 , ε 2 > 0, we have
|a n −L| < ε 1 , |b n −L| < ε 2 , for n > N
L−ε 1 < a n < L+ε 1 , L−ε 2 < b n < L+ε 2
L−ε 1 < a n < c n < b n < L+ε 2
|c n −L| < max{ε 1 ,ε 2 }
Theorem 7.1.4. Absolute value theorem
If lim
n→∞
|a n | = 0, then lim
n→∞
a n = 0 also.
Proof: For all n,
−|a n | ≤ a n ≤ |a n | and lim |a n | = 0 = lim(−|a n |)
n→∞ n→∞
∴ lim a n = 0, also.
n→∞
{ } ∞ sinn
Example 7.1.5. Determine whether the sequence converges or diverges.
n 2 n=1
Solution:
−1 ≤ sinn ≤ 1, for all n,
−1
n ≤ sinn ≤ 1 2 n 2 n2, for all n ≥ 1,
lim
n→∞
−1
n 2 = 0 = lim
n→∞
1
n 2
∴ lim
n→∞
sinn
n 2 = 0, also.
Definition 7.1.2. For any integer n ≥ 1, the factorial,
n! = 1·2·3···n, and 0! = 1.
De-Yu Wang CSIE CYUT 108

7.1. SEQUENCES OF REAL NUMBERS
Definition 7.1.3. Monotonic sequence {a n } ∞ n=1
(a) The sequence {a n } ∞ n=1 is increasing if a 1 ≤ a 2 ≤ ··· ≤ a n ≤ a n+1··· .
(b) The sequence {a n } ∞ n=1 is decreasing if a 1 ≥ a 2 ≥ ··· ≥ a n ≥ a n+1··· .
Definition 7.1.4. Bounded sequence {a n } ∞ n=n 0
:
If there is a number M > 0 for which |a n | ≤ M, for all n.
Theorem 7.1.5. Every bounded, monotonic sequence converges.
{ 2
n
Example 7.1.6. Prove that the sequence {a n } ∞ n=1 =
n!} ∞
n=1
converges.
Proof: We cannot apply L’Hôpital’s rule to n!
(a) {a n } ∞ n=1 is monotonic (decreasing).
a n+1
a n
=
2 n+1
(n+1)!
2 n
= 2
n+1
n!
a n+1 ≤ a n , for n ≥ 1,
(b) {a n } ∞ n=1 is bounded.
≤ 1, for n ≥ 1
0 < 2n
n! = 2·2·2···2·2
n·(n−1)···2·1 ≤ 2 = 2, for n ≥ 1
1
|a n | ≤ 2, for n ≥ 1.
∴ {a n } ∞ n=1 converges.
Exercise 7.1.1. Write out the terms of the squence
(a) a n = 2n−1
n 2
(b) a n = 3 n!
4
(c) a n = √ n+1
(d) a n = (−1) n n
3n−1
Exercise 7.1.2. Find the limit of the squence
(a) a n = n
2n+1
(b) a n =
4
√ n+1
De-Yu Wang CSIE CYUT 109

7.2. INFINITE SERIES
Exercise 7.1.3. Determine whether the sequence converges or diverges of
(a) a n = ne −n
(b) a n = n2 +3
n 3 +1
(c)
(d)
{ n!
} ∞
2 n n=1
{
(−1)
n+2} nn+3
∞
n=1
Exercise 7.1.4. Determine whether the sequence is increasng, decreasing or neither.
(a)
(b)
{ } e
n ∞
n
n=1
{ 2
n ∞
(n+1)!}
n=1
(c) a n = n+3
n+2
Exercise 7.1.5. Show that the sequence is bounded
(a)
{ 3n 2 −2
n 2 +1
} ∞
n=1
(b)
{ 6n−1
n+3
} ∞
n=1
7.2 Infinite Series
Definition 7.2.1. Infinite Series
∞∑
a k = lim
k=1
n→∞
n∑
k=1
a k = lim
n→∞
S n = S,
where S n = a 1 +a 2 +···+a n is the nth partial sum.
De-Yu Wang CSIE CYUT 110

7.2. INFINITE SERIES
Summary. Convergence tests for infinte series
Page Test Series Conditions of Comment
Convergence Divergence
∞∑
111 kth-Term a k
lim k ≠ 0 Cannot be
k→∞
k=k 0 used to show
convergence.
112
∞∑
Geometric ar k |r| < 1 |r| ≥ 1 S = ark0
1−r
series k=k 0
114 Integral
f is continuous and decreasing
a k = f(k) ≥ 0
∞∑
∫ ∞ ∫ ∞
a k f(x)dx f(x)dx 0 ∫ < R n < ∞
k=1
1
1
converges diverges f(x)dx
116 p-series
118 Comparison
(a k ,b k > 0)
119 Limit Comparison
(a k ,b k > 0)
122 Alternating
Series
124 Absolute
Convergence
125 Ratio
126 Root Test
∞∑
k=1
∞∑
k=1
∞∑
k=1
1
k p p > 1 p ≤ 1
a k
a k
∞∑
(−1) k+1 a k
k=1
∞∑
k=1
a k
0 ≤ a k ≤ b k and
∞∑
b k converges
k=1
∞
∑
k=1
a k and
lim
∞∑
k→∞
k=1
b k
lim a k = 0and0 <
k→∞
a k+1 ≤ a k
∞
∑
k=1
|a k |
0 ≤ b k ≤ a k and
∞∑
b k diverges
k=1
a k
b k
= L > 0
both converge or
both diverge
n
|S − S n | ≤
a n+1
converges
∣ lim
a k+1∣∣∣
k→∞∣
= L
a k
∞∑
a k (k!, b k ) L < 1 L > 1 L = 1 (no
√
conclusion)
k
lim |ak | = L
k→∞
∞∑
a k (k k ,b k ) L < 1 L > 1 L = 1 (no
conclusion)
k=1
k=1
Theorem 7.2.1. kth-term test for divergence
∞∑
(a) If a k converges, then lim a k = 0.
k→∞
k=1
(b) If lim
k→∞
a k ≠ 0, then
∞∑
a k diverges.
k=1
Remark. This test cannot be used to show convergence.
De-Yu Wang CSIE CYUT 111

7.2. INFINITE SERIES
Proof: Assume that
∞∑
k=1
a k = lim
n→∞
S n = L
a n =
n∑ ∑n−1
a k − a k = S n −S n−1
k=1
k=1
lim a n = lim(S n −S n−1 ) = lim S n − lim S n−1 = L−L = 0
n→∞ n→∞ n→∞ n→∞
Example 7.2.1. Use the kth-term test to determine if the series
k
Solution: lim a k = lim
k→∞ k→∞ k +1
= 1 ≠ 0, ∴ this series diverges.
Example 7.2.2. Use the kth-term test to determine if the series
Solution: lim a k = lim
k→∞ k = 0.
Remark. This does not say that the series converges.
k→∞
1
∞∑
k=1
∞∑
k=1
k
k +1 diverges.
1
k diverges.
Theorem 7.2.2. For a ≠ 0, the geometric seies
and diverges if |r| ≥ 1.
Proof:
∞∑
k=k 0
ar k converges to ark 0
1−r
if |r| < 1
S n = ar k 0
+ ar k 0+1
+ ar k 0+2
+ ··· + ar n
− rS n = ar k 0+1
+ ar k 0+2
+ ar k 0+3
+ ··· + ar n+1
(1−r)S n = ar k 0
− ar n+1
S n = ark 0
−ar n+1
1−r
S = lim
n→∞
S n = lim
n→∞
ar k 0
−ar n+1
1−r
⎧
⎨ ar k 0
if |r| < 1;
= 1−r
⎩
does not exist if |r| ≥ 1
De-Yu Wang CSIE CYUT 112

7.2. INFINITE SERIES
Example 7.2.3. Determine if the geometric series
Find the sum of the series if it converges.
∞∑
k=3
( ) k
1 2
converges or diverges.
3 3
Solution:
r = 2 3 , |r| = 2 3 < 1.
S = ar3 1 ·(2) 3
1−r = 3 3
1− 2 3
= 8
27
∴ ∞
∑
k=3
( ) k
1 2
converges.
3 3
Example 7.2.4. Determine if the geometric series
(
1
− 3 k
converges or di-
3 2)
verges. Find the sum of the series if it converges.
∞∑
k=1
Solution:
r = − 3 2 , |r| = 3 2 > 1 ∴ ∞
∑
k=1
(
1
− 3 k
diverges.
3 2)
Exercise 7.2.1. Use the kth-term test to determine if the series diverges.
(a)
(b)
∞∑
k=1
∞∑
k=0
2
k
4k
k +2
(c)
(d)
∞∑
(−1) k+1 2k
k +1
∞∑
( 1
2 − 1 )
k k +1
k=0
k=0
Exercise 7.2.2. Determine if the geometric series converges or diverges. Find the
sum of the series if it converges.
(a)
(b)
∞∑
k=2
∞∑
k=1
(
5 − 1 ) k
2
( 1
3
4
) k
(c)
(d)
∞∑
k=0
∞∑
k=3
1 ( ) 4
−1 k
3
1
5
(
− 7 ) k
2
Exercise 7.2.3. Prove Theorem 7.2.2 and Theorem 7.2.1.
De-Yu Wang CSIE CYUT 113

7.3. THE INTEGRAL TEST
7.3 The Integral Test
Theorem 7.3.1. Integral Test
(a) If f(k) = a k for all k = 1,2,··· , f is continuous and decreasing, and f(x) ≥ 0
∫ ∞ ∞∑
for x ≥ 1, then f(x)dx and a k either both converge or both diverge.
1
k=1
∞∑
(b) If f(k) converges. Then, the remainder R n satisfies
k=1
∞∑
0 ≤ R n = a k ≤
∫ ∞
k=n+1
n
f(x)dx.
Proof: (a) 0 ≤ a 2 +a 3 +···+a n = S n −a 1 ≤
(i) If
(ii) If
∫ ∞
1
y
✻
y = f(x)
(2,a
2 )
✲x
1 2 3 4 5 6
(5,a
5 )
f(x)dx converges to L
S n −a 1 ≤
∫ n
lim (S n −a 1 ) ≤ lim
n→∞
∫ ∞
1
∫ n
lim
n→∞
1
1
f(x)dx
n→∞
∫ n
1
f(x)dx =
∫ ∞
lim S n ≤ lim(L+a 1 ) = L+a 1 ,
n→∞ n→∞
f(x)dx diverges
f(x)dx =
∫ ∞
1
∫ n
1
1
∫ n
1
f(x)dx ≤ S n−1
f(x)dx ≤ a 1 +a 2 +···+a n−1 = S n−1
f(x)dx = L
f(x)dx = ∞ ≤ lim
n→∞
S n−1
lim S n−1 > ∞,
n→∞
y
✻
y = f(x)
(1,a
1 )
converges, too.
(4,a
4 )
✲x
1 2 3 4 5 6
diverges, too.
De-Yu Wang CSIE CYUT 114

7.3. THE INTEGRAL TEST
(iii) If
∞∑
a k converges to L
k=1
∫ n
(iv) If
∫ n
lim
n→∞
1
f(x)dx =
∞∑
a k diverges
k=1
(b) R n =
1
∫ ∞
1
∫ ∞
1
S n −a 1 ≤
f(x)dx ≤ S n−1
f(x)dx ≤ lim
n→∞
S n−1 = L
f(x)dx ≤ L,
∫ n
lim (S n −a 1 ) = ∞ ≤ lim
n→∞
∞∑
∫ ∞
1
k=n+1
1
f(x)dx ≥ ∞,
a k ≤
∫ ∞
n
f(x)dx
n→∞
∫ n
f(x)dx
1
f(x)dx =
converges, too.
∫ ∞
1
diverges, too.
y
✻
y = f(x)
f(x)dx
(n+1,a n+1 )
nn+1
✲x
∞∑ 3
Example 7.3.1. Usetheintegraltesttodetermineiftheseries
(2+k) converges 2 k=0
or diverges.
3
Solution: f(x) = is continuous and decreasing, and f(x) ≥ 0 for x ≥ 0
(2+x)
2
∫ ∞ ∫
3
R
R
3 −3
dx = lim dx = lim
(2+x)
2 R→∞ (2+x)
2 R→∞ (2+x) ∣
0
∴
0
= −3 lim
R→∞
∞∑
k=0
[
3
(2+k) 2
1
2+R − 1 2
0
]
= 3 2 , converges
converges, too.
De-Yu Wang CSIE CYUT 115

7.3. THE INTEGRAL TEST
Example 7.3.2. Estimate the error for the integral test in using the partial sum S 100
∞∑ 3
to approximate the sum of the series
(2+k) 2.
Solution:
0 ≤ R 100 ≤
∫ ∞
100
k=0
∫
3
R
dx = lim
(2+x)
2 R→∞
100
= −3 lim
R→∞
3
dx = lim
(2+x)
2 R→∞
[
1
2+R − 1 ]
102
= 3
102 .
−3
(2+x) ∣
R
100
Example 7.3.3.
Determine the number of terms needed to obtain an approximation to the sum of the
∞∑ 3
series
(2+k) correct to within 2 10−6 with error estimate for the integral test.
k=0
Solution:
0 ≤ R n ≤
∫ ∞
n
∫
3
R
dx = lim
(2+x)
2 R→∞
n
= −3 lim
R→∞
3
dx = lim
(2+x)
2 R→∞
[
1
2+R − 1 ]
n+2
R n ≤ 3
n+2 ≤ 10−6 , ∴ n ≥ 3×10 6 −2.
−3
(2+x) ∣
R
n
= 3
n+2 ≤ 10−6 ,
Theorem 7.3.2. p series
∞∑ 1
If p-series converges if p > 1 and diverges if p ≤ 1.
kp k=1
Proof: f(x) = 1 x p = x−p is continuous and decreasing, and f(x) ≥ 0 for x ≥ 1
(a) For p ≠ 1
∫ ∞ ∫ R
x −p dx = lim x −p x −p+1
dx = lim
1
R→∞
1
R→∞ −p+1∣
1
=
−p+1 lim
[
R −p+1 −1 ]
R→∞
⎧
⎨ 1
= p−1 , if p > 1
⎩
∞, if p < 1
R
1
De-Yu Wang CSIE CYUT 116

7.3. THE INTEGRAL TEST
(b) For p = 1
∫ ∞
1
1
dx = lim
x R→∞
= lim
R→∞
∫ R
1
1 x
[ln|R|−1] = ∞.
dx = lim
R→∞ ln|x||R 1
Example 7.3.4. Determine convergence or divergence of the p-series
∞∑ ∞∑ ∞∑
k −11/10 , k −10/11 1
, and
k .
k=1
k=1
Solution: (a)
(b)
(c)
∞∑
k −11/10 =
k=1
k=1
∞∑
k=1
1
k 11/10
∵ p = 11
∞
10 > 1 ∴ ∑
k −11/10 converges to
∞∑
k −10/11 =
k=1
∞∑
k=1
1
k
∞∑
k=1
1
k 10/11
k=1
∵ p = 10
∞
11 < 1 ∴ ∑
k −10/11 diverges.
∵ p = 1
∴
∞∑
k=1
k=1
1
k diverges.
11
10
1
= 10.
−1
Exercise 7.3.1. Use the integral test to determine if the series converges or diverges.
If the series converges:
(1) Estimate the error in using the partial sum S 100 to approximate the sum of the
series.
(2) Determine the number of terms needed to obtain an approximation to the sum
of the series correct to within 10 −6 with error estimate for the integral test.
(a)
(b)
∞∑
k=3
∞∑
k=2
k +1
k 2 +2k +3
2
klnk
(c)
(d)
∞∑
k=3
∞∑
k=3
1
(1+2k) 2
e 1/k
k 2
De-Yu Wang CSIE CYUT 117

7.4. COMPARISON TESTS
Exercise 7.3.2. Determine convergence or divergence of the p-series
(a)
(b)
∞∑
k=1
∞∑
k=1
4
5√
k
k −9/10
(c)
(d)
∞∑
k=1
∞∑
k=6
k −2/3
4
√
k
3
Exercise 7.3.3. Prove Theorem 7.3.1 and Theorem 7.3.2
7.4 Comparison Tests
Theorem 7.4.1. Comparison test
Suppose that 0 ≤ a k ≤ b k , for all k
∞∑
∞∑
(a) If b k converges, then a k converges, too.
k=1
k=1
∞∑
∞∑
(b) If a k diverges, then b k diverges, too.
k=1
k=1
∞∑
Proof for (a): If b k = B
k=1
0 ≤ S n = a 1 +a 2 +···+a n ≤ b 1 +b 2 +···+b n ≤
{S n } ∞ n=1 is bounded, (∵ 0 ≤ S n ≤ B)
∞∑
b k = B
k=1
Proof for (b): If
∴
S n = a 1 +a 2 +···+a n ≤ a 1 +a 2 +···+a n +a n+1 = S n+1
{S n } ∞ n=1 is increasing, (∵ S n ≤ S n+1 )
∴ lim
n→∞
S n = S converges.
∞∑
k=1
∞∑
a k = ∞
k=1
b k = lim
n→∞
(b 1 +b 2 +···+b n ) ≥ lim
n→∞
(a 1 +a 2 +···+a n ) =
∞∑
b k =∞ diverges, too.
k=1
∞∑
a k = ∞.
k=1
De-Yu Wang CSIE CYUT 118

7.4. COMPARISON TESTS
Example 7.4.1. Use the comparison test to determine if the series
2
3k 2 +1 converges
or diverges.
∞∑
k=1
Solution:
2
3k 2 +1 < 2
3k2, for k ≥ 1
∞∑ 2
∵ converges (∵ p = 2),
3k 2 k=1
∞∑ 2
∴ converges, too.
3k 2 +1
k=1
Example 7.4.2. Use the comparison test to determine if the series
converges or diverges.
∞∑ 2
k=1
3 3√ k 2 −1
Solution:
2
3 3√ k 2 −1 > 2
3 3√ k , for k ≥ 1
2
∞∑ 2
∵
3 3√ k = ∑ ∞ (
2
diverges ∵ p = 2 )
,
k=1
2 k=1
3k 2 3 3
∞∑ 2
∴
3 3√ diverges, too.
k 2 −1
k=1
Theorem 7.4.2. Limit Comparison test
a k
Suppose that a k ,b k > 0 and lim = L > 0 for k large. Then, either
k→∞ b k
∞∑
b k both converge or both diverge.
k=1
a k
Proof: If lim = L > 0, we can make
k→∞ b k
a k
∣ −L
b k
∣ < ε = L , for all k > N
2
∞∑
a k and
k=1
L
2

7.4. COMPARISON TESTS
(a) If
∞∑
b k converges to B, then
k=1
∞∑
a k =
N∑
a k +
∞∑
a k <
N∑
a k + 3L 2
∞∑
b k <
N∑
a k + 3L 2
B, converges,
k=1
too.
k=1
k=N+1
k=1
k=N+1
k=1
∞∑
(b) If b k diverges, then
k=1
∞∑
a k =
N∑ ∞∑
a k + a k >
k=1 k=1 k=N+1
N∑
a k + L 2
k=1
∞∑
k=N+1
b k
} {{ }
∞
, diverges, too.
∞∑
(c) If a k converges to A, then
k=1
∞∑
b k =
N∑ ∞∑
b k + b k <
k=1 k=1 k=N+1 k=1
N∑
b k + 2 L
∞∑
a k <
k=N+1
N∑
k=1
b k + 2 A, converges, too.
L
∞∑
(d) If a k diverges, then
k=1
∞∑
b k =
N∑ ∞∑
b k + b k >
k=1 k=1 k=N+1
N∑
k=1
b k + 2
3L
∞∑
k=N+1
a k
} {{ }
∞
, diverges, too.
Example 7.4.3. Use the limit comparison test to determine if the series
converges or diverges.
∞∑
k=3
1
k 3 −5k
Solution: Let a k =
1
k 3 −5k > 0, b k = 1 > 0 for k large.
k3 a k
lim = lim
k→∞ b k
k 3 −5k
1
k 3
k→∞
1
1
= lim
k→∞ 1− 5 = 1 > 0
k 2
Since
∞∑
k=3
1
k is a convergent p-series (p = 3 > 1), ∑ ∞
1
3 k 3 −5k
k=3
is also convergent.
De-Yu Wang CSIE CYUT 120

7.5. ALTERNATING SERIES AND ABSOLUTE CONVERGENCE
Example 7.4.4. Use the limit comparison test to determine if the series
∞∑ k 2 −2k +7
converges or diverges.
k 5 +5k 4 −3k 3 +2k −1
k=1
Solution: Let
a k =
a k
lim = lim
k→∞ b k
k 2 −2k +7
k 5 +5k 4 −3k 3 +2k −1 = 1− 2 + 7
k k 2
k 3 +5k 2 −3k + 2 − 1 , b k = 1 k
k k 3
2
1− 2 + 7
k k 2
k→∞
k 3 +5k 2 −3k + 2 k − 1
k 2
1
k 3
1− 2
= lim
+ 7
k k 2
k→∞ 1+ 5 − 3
k
+ 2 − 1
k 2 k 3
k 5 = 1 > 0
∞∑ 1
Since
k is a convergent p-series (p = 3 > 1), ∑ ∞
k 2 −2k +7
3 k 5 +5k 4 −3k 3 +2k −1
k=1
k=1
convergent.
is also
Exercise 7.4.1. Use the comparison test to determine if the series converges or
diverges.
(a)
(b)
∞∑
k=1
∞∑
k=1
2k
k 3 +4
2
3k 2 +1
(c)
(d)
∞∑
k=3
∞∑
k=2
2k
k 4 +5
2
3k 2 +k
Exercise 7.4.2. Use the limit comparison test to determine if the series converges
or diverges.
(a)
∞∑
k=1
k 3 +2k +3
k 4 +2k 2 +4
(b)
∞∑
k=3
k 2 +3
k 5 +k 2 +4
Exercise 7.4.3. Prove Theorem 7.4.1 and Theorem 7.4.2
7.5 Alternating Series and Absolute Convergence
Definition 7.5.1. Alternating Series
Any series of the form
∞∑
(−1) k+1 a k = a 1 −a 2 +a 3 −a 4 +a 5 −a 6 +··· ,
k=1
where a k > 0, for all k.
De-Yu Wang CSIE CYUT 121

7.5. ALTERNATING SERIES AND ABSOLUTE CONVERGENCE
Theorem 7.5.1. Alternating Series Test
Suppose that lim a k = 0 and 0 < a k+1 ≤ a k , for all k ≥ 1. Then, the alternating
k→∞
∞∑
series (−1) k+1 a k converges.
k=1
Proof: (a) The even-index partial sums S 2n
{S 2n } ∞ n=1 is monotonic(increasing).
S 2 = a 1 −a 2 > 0
S 4 = S 2 +(a 3 −a 4 ) ≥ S 2 , ∵ a 3 ≥ a 4
.
S 2n = S 2n−2 +(a 2n−1 −a 2n−2 ) ≥ S 2n−2
S 2n ≥ S 2n−2 ≥ ··· ≥ S 4 ≥ S 2 > 0.
{S 2n } ∞ n=1 is bounded.
0 < S 2n = a 1 +(−a 2 +a 3 ) +(−a
} {{ } 4 +a 5 ) +···+(−a
} {{ } 2n−2 +a 2n−1 ) −a
} {{ } }{{} 2n
≤ 0 ≤ 0
≤ 0 ≤ 0
≤ a 1
∴ lim
n→∞
S 2n converges.
(b) The odd-index partial sums S 2n+1
lim S 2n+1 = lim(S 2n +a 2n+1 ) = lim S 2n + lim a 2n+1 = lim S 2n
n→∞ n→∞ n→∞ n→∞ n→∞
converges, too.
Example 7.5.1. Determine if the alternating series
diverges.
k=0
∞∑
k=0
(−1) k+1 k 2
k 2 +3
k 2
∞
Solution: lim a k = lim
k→∞ k→∞ k 2 +3 = 1 ≠ 0, ∴ ∑
(−1) k+1 k 2
k 2 +3 diverges.
Example 7.5.2. Determine if the alternating series
diverges.
Solution: lim a k = lim
k→∞ k 2 +3 = 0 and a k+1 =
∞∑
∴ (−1) k+1 1
k 2 +3 converges.
k=0
k→∞
1
∞∑
k=0
(−1) k+1 1
k 2 +3
1
(k +1) 2 +3 < 1
k 2 +3 = a k,
converges or
converges or
De-Yu Wang CSIE CYUT 122

7.5. ALTERNATING SERIES AND ABSOLUTE CONVERGENCE
Theorem 7.5.2. If the alternating series
∞∑
(−1) k+1 a k converges to some number
S, then the error in approximating S by the nth partial sum S n satisfies
|S −S n | ≤ a n+1 .
Proof: (a) n = 2m (even)
S −S 2m = a 2m+1 +(−a 2m+2 +a 2m+3 ) +(−a
} {{ } 2m+4 +a 2m+5 ) +···
} {{ }
≤ 0
≤ 0
(b) n = 2m+1 (odd)
≤ a 2m+1 = a n+1
k=1
S −S 2m+1 = −a 2m+2 +(a 2m+3 −a 2m+4 ) +(a
} {{ } 2m+5 −a 2m+6 ) +···
} {{ }
≥ 0
≥ 0
∴ |S −S n | ≤ a n+1
≥ −a 2m+2 = −a n+1
Example 7.5.3. Determine how many terms needed to estimate the sum of the
∞∑
alternating series (−1) k+1 4 k correct to within 2 10−8 .
Solution:
k=1
|S −S n | ≤ a n+1 ≤ 10 −8
4
(n+1) 2 ≤ 10−8
n+1 ≥ 2×10 4
n ≥ 19999
Definition 7.5.2. Absolute and conditional convergence
(a)
(b)
∞∑
∞∑
a k is absolutely convergent if |a k | converges.
k=1
k=1
∞∑
∞∑
a k is conditionally convergent if a k converges but
k=1
k=1
∞∑
|a k | diverges.
k=1
De-Yu Wang CSIE CYUT 123

7.5. ALTERNATING SERIES AND ABSOLUTE CONVERGENCE
Theorem 7.5.3. Absolute Convergence
∞∑
∞∑
If |a k | converges, then a k converges.
k=1
k=1
∞∑
Proof: If |a k | converges to L
k=1
∞∑
a k = a 1 +a 2 +a 3 +···
k=1
∞∑
≤ |a 1 |+|a 2 |+|a 3 |+··· = |a k | = L, converges, too.
k=1
Example 7.5.4. Use the absolute convergence to determine if the series
is absolutely convergent, conditionally convergent or divergent.
∞∑
(−1) k 2 k 2
k=1
Solution:
∞∑
|a k | =
∴
k=1
∞∑
a k =
k=1
∞∑
2 ∣ ∣∣∣
∣ (−1)k =
k 2
k=1
∞∑
(−1) k 2 k 2
k=1
∞∑
k=1
2
k2, is a convergent p-series (p = 2 > 1)
is absolutely convergent.
Example 7.5.5. Use the absolute convergence to determine if the series
∞∑
(−1) k 2
3√ is absolutely convergent, conditionally convergent or divergent.
k
2
k=1
Solution:
∴
∞∑
k=0
∞∑
|a k | =
k=1
∞∑
2
∞
∣ (−1)k 3√
k
2∣ = ∑
k=1
lim a 2
k = lim
k→∞ k→∞
and a k+1 =
(−1) k 2
k 2 3
2
k 2 3
(k +1) 2 3
= 0
< 2
k 2 3
= a k ,
2
k=1
k 2 3
is conditionally convergent.
, is a divergent p-series
(
p = 2 )
3
De-Yu Wang CSIE CYUT 124

7.6. THE RATIO TEST AND THE ROOT TEST
Exercise 7.5.1. Determine if the alternating series is absolutely convergent, conditionally
convergent or divergent. If the series converges, then determine how many
terms needed to estimate the sum of the alternating series correct to within 10 −6 .
(a)
(b)
∞∑
(−1) k+1 2 k 3
k=0
∞∑
(−1) k+1 √ 3
k
k=2
(c)
(d)
∞∑
(−1)
k=7
k+12k −1
k 3
∞∑
(−1) k+1 k 2
k +1
k=1
Exercise 7.5.2. Prove Theorem 7.5.1, Theorem 7.5.3 and Theorem 7.5.2.
7.6 The Ratio Test and The Root Test
Theorem 7.6.1. Ratio Test
∞∑
∣ ∣ ∣∣∣ a k+1∣∣∣
Given a k with a k ≠ 0 for all k, suppose that lim = L. Then,
k→∞ a k
k=1
(a) if L < 1, the series converges absolutely,
(b) if L > 1, the series diverges and
(c) if L = 1, there is no conclusion.
∣ ∣ ∣∣∣ a k+1∣∣∣
Proof for (a): For L < 1, pick any number r with lim = L < r < 1.
k→∞ a k
Then, we can make
∣ a k+1∣∣∣
∣ < r, for k > N
a k
|a k+1 | < r|a k |
|a k+2 | < r|a k+1 | < r 2 |a k |
.
|a k+m | < r|a k+m−1 | < r m |a k |
∴
∞∑ N∑ ∞∑
|a k | = |a k |+ |a k |
k=1
∞∑
k=1
a k
<
k=1
N∑
|a k |+
k=N+1
∞∑
k=1 k=N+1
converges absolutely.
r k−N−1 |a N+1 | =
N∑
k=1
|a k |+ |a N+1|
1−r
< ∞, converges.
De-Yu Wang CSIE CYUT 125

7.6. THE RATIO TEST AND THE ROOT TEST
Proof for (b): For L > 1, we have lim
k→∞
∣ ∣∣∣ a k+1
a k
∣ ∣∣∣
= L > 1.
Then, we can make
∣ a k+1∣∣∣
∣ > 1, for k > N
a k
|a k+1 | > |a k | ⇒ lim
k→∞
|a k | ≠ 0
By the kth-term test,
∞∑
a k diverges.
k=1
Example 7.6.1. Use the ratio test to determine if the alternating series
is absolutely convergent, conditionally convergent or divergent.
∞∑
(−1) k 2 k!
k=0
∣ ∣∣∣ a k+1
Solution: lim
k→∞
∞∑
∴ (−1) k 2 k!
k=0
a k
∣ ∣∣∣
= lim
2
(k+1)!
k→∞
2
k!
is absolutely convergent.
1
= lim
k→∞ k +1 = 0 < 1
Theorem 7.6.2. Root Test
∞∑ ∣ ∣∣
Given a k , suppose that lim
√ k |a k | ∣ = L. Then,
k→∞
k=1
(a) if L < 1, the series converges absolutely,
(b) if L > 1, the series diverges and
(c) if L = 1, there is no conclusion.
Proof for (a): For L < 1, pick any number r with lim
k→∞
k √ |a k | = L < r < 1.
Then, we can make
√
k
|ak | < r, for k > N
|a k | < r k
|a k+1 | < r k+1
.
|a m | < r m
De-Yu Wang CSIE CYUT 126

7.6. THE RATIO TEST AND THE ROOT TEST
∴
∞∑ N∑ ∞∑
|a k | = |a k |+ |a k |
k=1
∞∑
k=1
a k
<
k=1
N∑
|a k |+
k=N+1
∞∑
k=1 k=N+1
r k =
converges absolutely.
N∑
k=1
|a k |+ rN+1
1−r
< ∞, converges.
Proof for (b): For L > 1, we have lim
√ k
|a k | = L > 1.
k→∞
Then, we can make
√
k
|ak | = L > 1, for k > N
|a k | > L k
|a k+1 | > L k+1 ⇒ lim
k→∞
|a k | ≠ 0
By the kth-term test,
∞∑
a k diverges.
k=1
Example 7.6.2. Use the root test to determine if the series
convergent, conditionally convergent or divergent.
Solution:
∞∑
k=1
(−1) k 3 is absolutely
kk ∴
lim
k→∞
∞∑
k=1
√ ∣∣∣∣ ∣
√
k
k
3
√ ∣∣∣
k
3
|ak | = lim = lim
k→∞ k k k→∞ k = 0 < 1
(−1) k 3 is absolutely convergent.
kk Exercise 7.6.1. Use the ratio test to determine if the alternating series is absolutely
convergent, conditionally convergent or divergent.
(a)
(b)
(c)
∞∑
(−1) k 2 k
k=0
∞∑
(−1) k+1k!
4 k
k=1
∞∑
k=1
2
k!
(d)
(e)
(f)
∞∑
k=4
∞∑
k=1
∞∑
k=3
(−1) k10k
k!
(−1) k k!
e k
(−1) kk2 3 k
2 k
De-Yu Wang CSIE CYUT 127

7.7. POWER SERIES
Exercise 7.6.2. Use the root test to determine if the series is absolutely convergent,
conditionally convergent or divergent.
(a)
(b)
∞∑ 2
(c)
(k +3) k
∞∑
( ) k 6k
(d)
5k +1
k=1
k=1
∞∑
e 3k
k 3k
k=2
∞∑
k=1
( 4
3k +2
) k
Exercise 7.6.3. Prove Theorem 7.6.1 and Theorem 7.6.2.
7.7 Power Series
Definition 7.7.1. Power Series
∞∑
b k (x−c) k = b 0 +b 1 (x−c)+b 2 (x−c) 2 +b 3 (x−c) 3 +···
k=0
Theorem 7.7.1. Convergence of a Power Series
∞∑
For a power series centered at c, b k (x−c) k , precisely one of the following is ture.
The series converges
(a) only for x = c, r = 0;
k=0
(b) for x ∈ (c−r, c+r), 0 < r < ∞;
(c) for all x ∈ (−∞,∞), r = ∞.
where r is the radius of convergence of the power series.
2
Example 7.7.1. Find a power series representation of about c = 0. Also
1−x
determine the radius of convergence and interval of convergence.
Solution:
∞∑
∵ ab k = a , for |b| < 1
1−b
∴
k=0
∞
2
1−x = ∑
2·x k for |x| < 1,
k=0
the radius of convergence r = 1 and the interval of convergence x ∈ (−1,1).
De-Yu Wang CSIE CYUT 128

7.7. POWER SERIES
Example 7.7.2. Determine the radius and interval of convergence for the geometric
∞∑ 10 k
power series
k! (x−1)k .
k=0
Solution: From the ratio test, we have
∣ ∣∣∣∣∣∣∣ 10 ∣ k+1 ∣ ∣∣∣∣∣∣∣ lim
a k+1∣∣∣
(k +1)! (x−1)k+1 1
∣ = lim = 10|x−1| lim
a k k→∞ 10 k
k→∞ k +1 = 0 < 1
k! (x−1)k
k→∞
∴
∞∑
k=0
10 k
k! (x−1)k converges absolutly for all x ∈ (−∞,∞) and r = ∞.
Example 7.7.3. Determine the radius and interval of convergence for the geometric
∞∑ x k
power series
k4 k.
k=1
Solution: From the ratio test, we have
lim
k→∞
For x = 4,
∞∑
k=1
For x = −4,
∞∑
k=1
∣ a k+1∣∣∣
∣ = lim
a k
x k
k4 = ∑ ∞ k
k=1
x k
k4 = ∑ ∞ k
k=1
∴ r = 4, x ∈ [−4,4).
∣ ∣∣∣∣∣∣∣ x k+1
(k +1)4 k+1
k→∞ x k
k4 k
4 k
k4 = ∑ ∞ k
(−4) k
k4 k =
k=1
∣ ∣∣∣∣∣∣∣
= |x|
4 lim
k→∞
k
k +1 = |x|
4 < 1.
1
, is a divergent p-series (p = 1).
k
∞∑ (−1) k
, is a convergent alternating series.
k
k=1
Theorem 7.7.2. If the function given by
f(x) =
∞∑
b k (x−c) k = b 0 +b 1 (x−c)+b 2 (x−c) 2 +b 3 (x−c) 3 +···
k=0
has a radius of convergence of r > 0, then
De-Yu Wang CSIE CYUT 129

7.7. POWER SERIES
(a) the derivative
f ′ (x) = d
dx (b 0 +b 1 (x−c)+b 2 (x−c) 2 +b 3 (x−c) 3 +···)
= b 1 +2b 2 (x−c)+3b 3 (x−c) 2 +···
∞∑
= b k k(x−c) k−1 ,
k=1
(b) and the antiderivative
∫ ∫
∑ ∞
f(x)dx = b k (x−c) k dx =
=
∞∑
k=0
k=0
b k
(x−c) k+1
k +1
+K
∞∑
∫
k=0
b k (x−c) k dx
have the same radii of convergence r.
Remark. The interval of convergence may differ at the endpoints.
Example 7.7.4. Usethepowerseries
∞∑
(−1) k x k tofindpowerseriesrepresentations
k=0
1
of
(1+x) 2. Also determine the radii of convergence and intervals of convergence.
∞∑ ∞∑
Solution: First, (−1) k x k = (−x) k 1
=
1−(−x) = 1 , for | − x| = |x| < 1.
1+x
k=0 k=0
Therefore, r = |x| = 1, and x ∈ (−1,1).
d 1
dx1+x = −1
(1+x) 2
1
(1+x) = − d 1
2 dx1+x = − d ∞∑
(−x) k
dx
k=0
∞∑
∞∑
= − k(−x) k−1 ·(−1) = (−1) k+1 kx k−1
k=1
For x = −1,
∞∑
(−1) k+1 kx k−1 =
k=1
For x = 1,
∞∑
(−1) k+1 kx k−1 =
k=1
∴ r = 1, and x ∈ (−1,1).
∞∑
(−1) 2k k =
k=1
k=1
∞∑
k, diverges (k-term)
k=1
∞∑
(−1) k+1 k, diverges (k-term)
k=1
De-Yu Wang CSIE CYUT 130

7.8. TAYLOR SERIES
Exercise 7.7.1. Find a power series representation of f(x) about c = 0. Also determine
the radius of convergence and interval of convergence.
(a) f(x) = 3
(c) f(x) = 3
4+x
x−1
(b) f(x) = 2 (d) f(x) = 3
1+x 2 6−x
Exercise 7.7.2. Determinetheradiusandintervalofconvergenceforthepowerseries
(a)
(b)
∞∑
(x+2) k
k=0
(c)
∞∑ ( x
) k
(−1) k (d)
2
k=0
∞∑
k=0
∞∑
k=1
2 k
k! (x−2)k
(−1) k+1k!
4 k
∞∑
Exercise 7.7.3. Use the power series (−1) k x k to find power series representations
k=0
of f(x). Also determine the radii of convergence and intervals of convergence.
(a) f(x) = ln(1+x 2 )
2
(b) f(x) =
(x−1) 2
(c) f(x) = 3tan −1 x
3x
(d) f(x) =
(1−x 2 ) 2
7.8 Taylor Series
Theorem 7.8.1. (a) Taylor series expansion of f(x) about x = c is given
f(x) =
∞∑
k=0
f (k) (c)
(x−c) k
k!
=f(c)+f ′ (c)(x−c)+ f′′ (c)
2!
(x−c) 2 +···+ f(k) (c)
(x−c) k +···
k!
(b) Taylor polynomial of degree n for f expanded about x = c
f(x) ≈P n (x)
n∑ f (k) (c)
= (x−c) k
k!
k=0
=f(c)+f ′ (c)(x−c)+ f′′ (c)
2!
(x−c) 2 +···+ f(n) (c)
(x−c) n
n!
Remark. A Taylor series expansion about x = 0 (i.e., take c = 0) is called a Maclaurin
series.
De-Yu Wang CSIE CYUT 131

7.8. TAYLOR SERIES
Proof:
f(x) =
∞∑
b k (x−c) k
k=0
= b 0 +b 1 (x−c)+b 2 (x−c) 2 +b 3 (x−c) 3 +··· f(c) = b 0 ,
f ′ (x) = b 1 +2b 2 (x−c)+3b 3 (x−c) 2 +4b 4 (x−c) 3 +··· f ′ (c) = b 1 ,
f ′′ (x) = 2b 2 +3·2b 3 (x−c)+4·3b 4 (x−c) 2 +··· f ′′ (c) = 2b 2 ,
f ′′′ (x) = 3·2b 3 +4·3·2b 4 (x−c)+··· f ′′′ (c) = 3!b 3 ,
··· f (k) (c) = k!b k ,
∴ b k = f(k) (c)
, for k = 0,1,2,···
k!
.
Example 7.8.1. Find the Taylor series about c = 0 and its interval of convergence
for e x
Solution: (a) Taylor series expansion
∞∑
f(x) = e x f (k) (0)
= x k , f(0) = 1
k!
k=0
f ′ (x) = e x , f ′ (0) = 1
f ′′ (x) = e x , f ′′ (0) = 1
f (3) (x) = e x , f (3) (0) = 1
.
f (k) (x) = e x , f (k) (0) = 1
∞∑
∴ e x 1
=
k! xk = 1+x+ 1 2! x2 + 1 3! x3 +···
k=0
(b) To test the convergence by ratio test,
∣ ∣∣∣∣∣∣∣ x ∣ k+1
lim
a k+1∣∣∣
(k +1)!
k→∞∣
= lim
a k k→∞ x k
∣
k!
∞∑
∴
k=0
1
= |x| lim
k→∞ k +1 = 0 < 1
1
k! xk converges for all x ∈ (−∞,∞) and r = ∞.
.
De-Yu Wang CSIE CYUT 132

7.8. TAYLOR SERIES
Example 7.8.2. For f(x) = e x , find the Taylor polynomial of degree n expanded
about x = 0 and use it to approximate the number e.
Solution: f (k) (x) = e x , for all k. So, the nth-degree Taylor polynomial is
n∑ f (k) (0)
n∑
f(x) ≈ P n (x) = (x−0) k 1
=
k! k! xk
k=0
k=0
= 1+x+ 1 2! x2 + 1 3! x3 +···+ 1 n! xn .
e = e 1 = 1+1+ 1 2! 12 + 1 3! 13 +···+ 1 n! 1n ≈ 2.718281828.
Example 7.8.3. Expand f(x) = sinx in a Taylor series about x = π 2
c = π ) and find its interval of convergence.
2
∞∑ f (k)( )
π (
2
Solution: (a) Taylor series expansion sinx = x− π ) k
k! 2
k=0
( π
f(x) = sinx, f = 1
( 2)
π
)
f ′ (x) = cosx, f ′ = 0
( 2
π
)
f ′′ (x) = −sinx, f ′′ = −1
( 2
π
)
f (3) (x) = −cosx, f (3) = 0
( 2
π
)
f (4) (x) = sinx, f (4) = 1
2
(i.e., take
.
( π
)
··· f (k) = (−1) k/2 for k = 0,2,4,···
2
∞∑ (−1) k/2 (
∴ sinx = x− π k,
for k = 0,2,4,···
k! 2)
k=0
∞∑ (−1) m (
= x− π ) 2m, k where m =
(2m)! 2 2
m=0
= 1− 1 (
x− π ) 2 1
(
+ x− π ) 4 1
(
− x− π 6
+···
2 2 4! 2 6! 2)
(b) To test the convergence by ratio test,
lim
k→∞
∣ a k+1∣∣∣
∣ = lim
a k
=
∣ ∣∣∣∣∣∣∣ (−1) k+1 ∣
( )
x−
π 2k+2∣∣∣∣∣∣∣
2
(2k +2)!
k→∞ (
x−
π 2k
2)
(−1) k
(2k)!
(
x− π 2) 2
lim
k→∞
1
(2k +2)(2k +1) = 0 < 1
De-Yu Wang CSIE CYUT 133

7.8. TAYLOR SERIES
∴
∞∑
k=0
(−1) k
(2k)!
(
x− π 2) 2k
converges absolutly for all x ∈ (−∞,∞) and r = ∞.
Example 7.8.4. Find Taylor series in power of x for e 2x , e x2 and e −2x .
Solution:
e x =
e 2x =
e x2 =
e −2x =
∞∑
k=0
∞∑
k=0
∞∑
k=0
∞∑
k=0
1
k! xk = 1+x+ 1 2! x2 + 1 3! x3 +··· for x ∈ (−∞,∞)
1
k! (2x)k =
1
k! (x2 ) k =
∞∑ 2 k
k! xk = 1+2x+ 22
2! x2 + 23
3! x3 +···
∞∑ 1
k! x2k = 1+x 2 + 1 2! x4 + 1 3! x6 +···
∞∑ (−1) k 2 k
x k = 1−2x+ 22
k! 2! x2 − 23
3! x3 +···
k=0
k=0
1
k! (−2x)k =
k=0
Summary. Common Taylor series
Taylor
Interval of
series
Convergence
∞∑
e x 1
=
k! xk = 1+x+ 1 2! x2 + 1 3! x3 +··· (−∞,∞)
k=0
∞∑ (−1) k
sinx =
(2k +1)! x2k+1 = x− 1 3! x3 + 1 5! x5 − 1 7! x7 +··· (−∞,∞)
k=0
∞∑ (−1) k (
= x− π ) 2k
= 1− 1 (
x− π ) 2 1
(
+ x− π 4
−··· (−∞,∞)
(2k)! 2 2 2 4! 2)
k=0
∞∑ (−1) k
cosx =
(2k)! x2k = 1− 1 2! x2 + 1 4! x4 − 1 6! x6 +··· (−∞,∞)
k=0
∞∑ (−1) k+1
lnx = (x−1) k = (x−1)− 1 k
2 (x−1)2 + 1 3 (x−1)3 −··· (0,2]
k=1
∞∑
tan −1 (−1) k
x =
(2k +1) x2k+1 = x− 1 3 x3 + 1 5 x5 − 1 7 x7 +··· [−1,1]
k=0
Exercise 7.8.1. Find the Maclaurin series (i.e. Taylor series about c = 0) and its
interval of convergence.
(a) f(x) = ln(1+x) (b) f(x) = 1
1−x
De-Yu Wang CSIE CYUT 134

7.8. TAYLOR SERIES
Exercise 7.8.2. Find the Taylor series about c and detremine the interval of convergence.
(a) f(x) = e x−1 , c = 1
(b) f(x) = 1 x , c = −1 (c) f(x) = cosx, c = − π 2
(d) f(x) = lnx, c = e
Exercise 7.8.3. Find Taylor series in the summary table and prove Theorem 7.8.1.
De-Yu Wang CSIE CYUT 135

Chapter 8
PARAMETRIC EQUATIONS
AND POLAR COORDINATES

8.1. PLANE CURVES AND PARAMETRIC EQUATIONS
8.1 Plane Curves and Parametric Equations
Definition 8.1.1.
Givenanypairoffunctionsx(t)andy(t)definedonthesamedomainD, theequations
x = x(t), y = y(t)
are called parametric equations.
The collection of all points (x(t),y(t)) is a plane curve.
Example 8.1.1. Sketch the plane curve defined by the parametric equations x =
t, y = √ t, for 0 ≤ t ≤ 4.
Solution:
t = 0, (x,y) = (0,0)
t = 1, (x,y) = (1,1)
(
t = 2, (x,y) = 2, √ )
2
(
t = 3, (x,y) = 3, √ )
3
(
t = 4, (x,y) = 4, √ )
4
y
3 ✻
2
1
t = 0
(0,0)
t = 4
(4,2)
✯ ✶ ✿
t = 2
✒ (2, √ 2)
✲x
1 2 3 4
Example 8.1.2. Find the parametric equation for the line segment from (0,1) to
(5,6).
Solution: For a line segment, the parametric equations can be chosen
{
x = a+bt
,t 1 ≤ t ≤ t 2 .
y = c+dt
For t = t 1 = 0, (x,y) = (x(0),y(0)) = (a,c) = (0,1),
∴ a = 0, c = 1.
For t = t 2 = 1, (x,y) = (x(1),y(1)) = (a+b,c+d) = (b,1+d) = (5,6),
∴ b = 5, d = 5.
{
x = 5t
We now have that
y = 1+5t , 0 ≤ t ≤ 1.
De-Yu Wang CSIE CYUT 137

8.2. CALCULUS AND PARAMETRIC EQUATIONS
Example 8.1.3. Find the parametric equation for the portion of the parabola y =
x 2 +1 from (1,2) to (2,5).
Solution:
{
{
x = t
y = t 2 +1 , 1 ≤ t ≤ 2, or x = 3t
y = 9t 2 +1 , 1
3 ≤ t ≤ 2 3 .
Exercise 8.1.1. Sketch the plane curve defined by the parametric equations
{
x = 2cost
{
x = −1+2t
(a)
y = 3t+1
{
x = −1+2t
(b)
y = t 2 −1
,1 ≤ t ≤ 2
,0 ≤ t ≤ 2
(c)
y = 3sint
{
x = −2t 2
(d)
y = 2−t
,0 ≤ t ≤ 2π
,−2 ≤ t ≤ 1
Exercise 8.1.2. Find the parametric equation for the line segment
(a) from (0,1) to (3,4)
(b) from (−3,1) to (1,3)
(c) from (2,−5) to (5,1)
(d) from (4,1) to (2,−3)
Exercise 8.1.3. Find the parametric equation for the portion of the parabola
(a) y = x 2 +1 from (1,2) to (3,10)
(b) y = 2x 2 −1 from (0,−1) to (2,7)
(c) y = 2−x 2 from (2,−2) to (0,2)
(d) y = x 2 +1 from (1,2) to (−1,2)
8.2 Calculus and Parametric Equations
Theorem 8.2.1. Parametric form of the derivative
For the parametric equations x(t) and y(t), then
( )
dy d dy
dy
dx = dt , and d2 y
dx dx = dt dx
, if dx
2 dx dt ≠ 0.
dt dt
Proof: By the chain rule
dy
dy
dx = dy dt
dt dx = dt , if dx
dx dt ≠ 0
dt
d 2 y
dx 2 = d
dx
( ) dy
=
dx
d
dt
( ) dy
dx
dx
dt
, if dx
dt ≠ 0.
De-Yu Wang CSIE CYUT 138

8.2. CALCULUS AND PARAMETRIC EQUATIONS
Example 8.2.1. Find the slope of the tangent lines to the curve x = 2cost +
sin2t, y = 2sint+cos2t at (a) t = 0, (b) t = π and (c) the point (0,−3).
4
Solution:
dy
dy
dx = dt
dx
dt
= 2cost−2sin2t
−2sint+2cos2t
(a) t = 0
dy
dx∣ = 2cost−2sin2t
t=0
−2sint+2cos2t∣ = 2cos0−2sin0
t=0
−2sin0+2cos0 = 1.
(b) t = π 4
dy
∣
dx
∣
t=
π
4
= 2cost−2sin2t
∣
−2sint+2cos2t
∣
t=
π
4
=
2cos π 4 −2sin π 2
−2sin π 4 +2cos π 2
=
√
2−2
− √ 2 = √ 2−1.
(c) The point (0,−3)
{
x = 2cost+sin2t = 0
y = 2sint+cos2t = −3
{
sin2t = −2cost
cos2t = −3−2sint
sin 2 2t+cos 2 2t = 1 = (−2cost) 2 +(−2sint−3) 2
1 = 4cos 2 t+9+12sint+4sin 2 t
−1 = sint, ⇒ t = 3π 2 .
dy
∣
dx
∣
t=
3π
2
= 2cost−2sin2t
∣
−2sint+2cos2t
= −2sint−4cos2t
∣
−2cost−4sin2t
∣
t=
3π
2
∣
t=
3π
2
( 0
0)
, By the L’Hôpital rule
=
−2sin 3π 2 −4cos3π
−2cos 3π 2 −2sin3π = ∞
does not exist.
De-Yu Wang CSIE CYUT 139

8.2. CALCULUS AND PARAMETRIC EQUATIONS
Theorem 8.2.2. Arc length in parametrical equations
For the curve defined parametrically by x = x(t), y = y(t), a ≤ t ≤ b, if x ′ and y ′
are continuous on [a,b] and the curve does not intersect itself, then the arc length of
the curve is given by
s =
∫ b
a
√
[x′ (t)] 2 +[y ′ (t)] 2 dt.
Proof: Divide the t-interval [a,b] into n subinterval of equal length, ∆t:
y
a = t 0 < t 1 < t 2 < ··· < t n = b, ✻
where t i −t i−1 = ∆t = b−a
n ,
for each i = 1,2,3,··· ,n.
The arc length of the subinterval [t i−1 ,t i ] is
s i ≈ √ [x(t i )−x(t i−1 )] 2 +[y(t i )−y(t i−1 )] 2
t = a = t
0
t = b = t n
✲x
√ [x(ti ] 2 [ ] 2
)−x(t i−1 ) y(ti )−y(t i−1 )
= + ∆t
∆t ∆t
By the mean value theorem,
s i ≈ √ [x ′ (c i )] 2 +[y ′ (d i )] 2 ∆t, where c i ,d i ∈ (t i−1 ,t i )
n∑ n∑ √
s = lim s i = lim [x′ (c i )] 2 +[y ′ (d i )] 2 ∆t
n→∞ n→∞
i=1 i=1
∫ b √
= [x′ (t)] 2 +[y ′ (t)] 2 dt.
a
Example 8.2.2. Find the arc length of the curve x = 2cost, y = 2sint for 0 ≤ t ≤
2π.
Solution:
s =
=
=
∫ b
a
∫ 2π
0
∫ 2π
√
[x′ (t)] 2 +[y ′ (t)] 2 dt
√
[−2sint]2 +[2cost] 2 dt
√
∫ 2π
4dt = 2dt = 4π.
0 0
y
✻
■
t = 0
✲
t = 2π x
De-Yu Wang CSIE CYUT 140

8.2. CALCULUS AND PARAMETRIC EQUATIONS
Example 8.2.3. Findthearclengthofthecurvex = 2cost+sin2t, y = 2sint+cos2t
for 0 ≤ t ≤ 2π.
Solution:
s =
=
=
=
=
=
= 3
∫ b
a
∫ 2π
0
∫ 2π
0
∫ 2π
0
∫ 2π
0
∫ 2π
√
[x′ (t)] 2 +[y ′ (t)] 2 dt
√
[−2sint+2cos2t]2 +[2cost−2sin2t] 2 dt
√
8−8sintcos2t−8costsin2tdt
√
∫
√
2π
8(1−sin3t)dt = 8
√
0
∫ 5π
6
π
6
= 3 √ 8· 2
3
(
8 cos 3t ) 2
3t
−sin dt
2 2
∣
√ ∣∣∣
8 cos 3t
3t
−sin 2 2 ∣ dt
√
8
(sin 3t )
3t
−cos dt
2 2
(
−cos 3t )∣5π
3t ∣∣∣ 6
−sin
2 2
0
π
6
(
cos 3t
)
3t 3t 3t
2 +sin2 −2sin cos dt
2 2 2 2
= 16.
t =
5π 6
−3
y
✻
t = 3π 2
t = π
6
t = 0
✲x
Exercise 8.2.1. Sketch the graph and find the slope of the given curves at the
indicated points
{
x = t 2 −2
(a) at (i) t = −1, (ii) t = 1, (iii) (−2,0)
y = t 3 −t
(b)
(c)
(d)
{
x = t 3 −t
y = t 4 −5t 2 +4
{
x = 2cost
y = 3sint
{
x = 2cos2t
y = 3sin2t
at (i) t = −1, (ii) t = 1, (iii) (0,4)
at (i) t = π, (ii) t = π , (iii) (0,3)
4 2
at (i) t = π, (ii) t = π , (iii) (−2,0)
4 2
De-Yu Wang CSIE CYUT 141

8.3. POLAR COORDINATES
Exercise 8.2.2. Sketch the graph and find the arc length of the given curves.
(a)
(b)
{
x = t 3 −4t
y = t 2 −3
{
x = t 3 −4t
y = t 2 −3t
,−2 ≤ t ≤ 2
,−2 ≤ t ≤ 2
(c)
(d)
{
x = 2cost
y = 3sint
{
x = 2cos2t
y = 3sin7t
,0 ≤ t ≤ 2π
,0 ≤ t ≤ 2π
8.3 Polar Coordinates
Definition 8.3.1. Polar coordinates P = (r,θ)
r = directed distance from the pole (origin) O to P
θ = directed angle, counterclockwise from polar axis to segment OP
P = (r,θ)
r = OP
✧ ✧✧✧✧✧✧✧✧✧✧✧ ❑ θ ✲
O
Polar axis
Theorem 8.3.1. Coordinate conversion
y
✻
(r,θ)
r
y = rsinθ
❑θ
✲
O x = rcosθ
x
(a)
{
x = rcosθ
y = rsinθ
, (x,y) = (rcosθ,rsinθ)
(b)
{
r 2 = x 2 +y 2 , r = ± √ x 2 +y 2
tanθ = y x , θ = tan−1 y ,
x
θ can be any angle for which tanθ = y , while x −π < 2 tan−1 y < π , so that
x 2
De-Yu Wang CSIE CYUT 142

8.3. POLAR COORDINATES
(r,θ) = (+ √ x 2 +y 2 , tan −1 y +2nπ), for x > 0
x
= (− √ x 2 +y 2 , tan −1 y +π +2nπ),
x
(r,θ) = (+ √ x 2 +y 2 , tan −1 y +π +2nπ), for x < 0
x
where n is an integer.
= (− √ x 2 +y 2 , tan −1 y x +2nπ),
Example 8.3.1. Find the rectangular coordinate for the polar point (2,−π/3).
Solution:
(
(x,y) = (rcosθ,rsinθ) = 2cos −π )
−π
,2sin
3 3
( )
= 2· 1
2 , 2· −√ 3
= (1,− √ 3).
2
( Example ) 8.3.2. Find ( the ) polar coordinate representations for the points (r,θ) =
2,
π
6 and (r,θ) = 2,
7π
6 with r = −2.
Solution:
(
2, π )
=
(−2, π )
6 6 +π +2nπ =
(−2, 7π )
6 +2nπ (
2, 7π )
=
(−2, 7π )
6 6 +π +2nπ =
(−2, π )
6 +2nπ
y
✻
θ = π 6
✬✩
( )
2,
π
6
❑
❯ 1 2
( ✫✪ )
✧ ✧✧✧✧✧✧✧✧✧✧✧✧✧ 2,
7π
θ = 7π 6
6
✲x
Example 8.3.3. Find all polar coordinate representations for the rectangular point
(2,−2).
Solution:
{r 2 = x 2 +y 2 , r = ± √ 2 2 +(−2) 2 = ± √ 8
tanθ = y x , −2
θ = tan−1 = 2 tan−1 (−1) = − π .
4
(√ π
)
∵ x = 2 > 0 ∴ (r,θ) = 8, −
4 +2nπ =
(− √ 8, − π )
4 +π +2nπ .
De-Yu Wang CSIE CYUT 143

8.4. CALCULUS AND POLAR COORDINATES
Example 8.3.4. Find all polar coordinate representations for the rectangular point
(−2,2).
Solution:
⎧
⎨r 2 = x 2 +y 2 , r = ± √ (−2) 2 +2 2 = ± √ 8
⎩tanθ = y x , θ = tan−1 2 = .
−2 tan−1 (−1) = − π 4
(√ π
)
∵ x = −2 < 0 ∴ (r,θ) = 8, −
4 +π +2nπ =
(− √ 8, − π )
4 +2nπ .
Exercise 8.3.1. Find the rectangular coordinate for the polar point
(a) (−2,−π/3)
(b) (3,π/5)
(c) (0,3)
(d) (−3,1)
Exercise 8.3.2. Find all polar coordinate representations for the rectangular points
(a) (−2,−1)
(b) (0,3)
(c) (−2, √ 3)
(d) (3,−4)
8.4 Calculus and Polar Coordinates
Theorem 8.4.1. Slope in polar form
If f is a differentiable function of θ, then the slope of the tangent line to r = f(θ) at
the point (r,θ) is
dy
dy
dx = dθ
dx
dθ
= f(θ)cosθ +f′ (θ)sinθ
−f(θ)sinθ +f ′ (θ)cosθ
Proof: Using the product rule
provided that dx
dθ
≠ 0 at (r,θ).
dy d(rsinθ)
dy
dx = dθ = dθ =
dx d(rcosθ)
dθ dθ
= f(θ)cosθ +f′ (θ)sinθ
−f(θ)sinθ +f ′ (θ)cosθ
d[f(θ)sinθ]
dθ
d[f(θ)cosθ]
dθ
if dx
dθ ≠ 0.
De-Yu Wang CSIE CYUT 144

8.4. CALCULUS AND POLAR COORDINATES
Example 8.4.1. Find the slope of the tangent line to the polar curve at the point
r = cos2θ at θ = 0, θ = π and the point (−1, π). y
4 2
1
−1
1
x
Solution:
dy
dy
dx = dθ
dx
dθ
= f′ (θ)sinθ +f(θ)cosθ
f ′ (θ)cosθ −f(θ)sinθ
−1
=
−2sin(2θ)sinθ +cos(2θ)cosθ
−2sin(2θ)cosθ −cos(2θ)sinθ
(a) θ = 0
dy
dx∣ = 1
θ=0
0
does not exist.
(b) θ = π 4
dy
∣
dx
∣
θ=
π
4
= −2sin(2θ)sinθ+cos(2θ)cosθ
∣
−2sin(2θ)cosθ −cos(2θ)sinθ
= −2sin(π 2 )sin π 4 +cos(π 2 )cos π 4
−2sin( π 2 )cos π 4 −cos(π 2 )sin π 4
∣
θ=
π
4
=
√2 √2
−2·1· +0·
2
−2·1· √2 √2
−0·
2
2
2
= 1.
(c) The point (−1, π), θ = π 2 2
dy
dx∣ = −2sin(2θ)sinθ+cos(2θ)cosθ
∣
θ=
π −2sin(2θ)cosθ −cos(2θ)sinθ
2
= −2sin(π)sin( (
π
2)
+cos(π)cos
π
)
2
−2sin(π)cos ( (
π
2)
−cos(π)sin
π
)
2
= −2·0·1+(−1)·0
−2·0·0−(−1)·1 = 0.
∣
θ=
π
2
De-Yu Wang CSIE CYUT 145

8.4. CALCULUS AND POLAR COORDINATES
Note. Rose curves sinnθ, cosnθ :
{
n petals if n is odd
2n petals if n is even
cosθ
y
cos2θ
y
cos3θ
y
cos4θ
y
cos5θ
y
x
x
x
x
x
sinθ
y
sin2θ
y
sin3θ
y
sin4θ
y
sin5θ
y
x
x
x
x
x
Theorem 8.4.2. Area in polar coordinates
If f is continuous and nonnegative on the interval [a,b], 0 < b − a ≤ 2π, then the
area of the region bounded by the graph of r = f(θ) between the radial lines θ = a
and θ = b is given by
A =
∫ b
a
1
2 [f(θ)]2 dθ.
Proof: Divide the θ-interval [a,b] into n subinterval of equal length, ∆θ:
a = θ 0 < θ 1 < θ 2 < ··· < θ n = b,
where θ i − θ i−1 = ∆θ = b−a , for each i = 1,2,3,··· ,n. The area of the circular
n
sector of radius f(θ i ) and central angle ∆θ is
y
y
✻
θ = b
✁
✁
✁
✁
A θ = a
✁
✁
✁ r = f(θ)
✁✏ ✏✏✏✏✏✏✏✏✏✏✏ ✲
x
A i ≈ 1 2 r ·r∆θ = 1 2 [f(θ i)] 2 ∆θ
A = lim
n→∞
n∑
i=1
n∑ 1
A i = lim
n→∞ 2 [f(θ i)] 2 ∆θ =
i=1
∫ b
a
✻
θ = b
✁
θ = θ i
✁ θ = θ i−1
✁...
A
✁ i
θ = a
✁
✁
✁ r = f(θ)
✏✁
✏✏✏✏✏✏✏✏✏✏✏ ✲
x
1
2 [f(θ)]2 dθ.
De-Yu Wang CSIE CYUT 146

8.4. CALCULUS AND POLAR COORDINATES
Example 8.4.2. Find the area of one leaf of r = cos3θ.
Solution: If r = cos3θ = 0 then θ = π 6 , π 2 , 5π 6 ,··· . 1
y
A =
= 1 4
∫ π
2
1
2 (cos3θ)2 dθ = 1 4
(θ + 1 )∣π
∣∣∣
6 sin6θ 2
π
6
= π 12 .
π
6
∫ π
2
π
6
(1+cos6θ) dθ
−1
1
x
−1
Theorem 8.4.3. Arc length in polar coordinates
Let f be a function whose derivaive is continuous on [a,b]. The length of the graph
of r = f(θ) from θ = a to θ = b is
s =
∫ b
a
√
[f′ (θ)] 2 +[f(θ)] 2 dθ.
Proof: Divide the θ-interval [a,b] into n subinterval of equal length, ∆θ:
a = θ 0 < θ 1 < θ 2 < ··· < θ n = b,
whereθ i −θ i−1 = ∆θ = b−a
n , for each i = 1,2,3,··· ,n. Thearc lengthfrom θ = θ i−1
to θ = θ i is
s i ≈ √ [x(θ i )−x(θ i−1 )] 2 +[y(θ i )−y(θ i−1 )] 2
= √ [f(θ i )cosθ i −f(θ i−1 )cosθ i−1 ] 2 +[f(θ i )sinθ i −f(θ i−1 )sinθ i−1 ] 2
= √ f 2 (θ i )+f 2 (θ i−1 )−2f(θ i )f(θ i−1 )[cosθ i cosθ i−1 +sinθ i sinθ i−1 ]
= √ f 2 (θ i )+f 2 (θ i−1 )−2f(θ i )f(θ i−1 )cos∆θ
= √ [f(θ i )−f 2 (θ i−1 )] 2 +2f(θ i )f(θ i−1 )[1−cos∆θ]
[
(
= √ [f(θi )−f 2 (θ i−1 )] 2 +2f(θ i )f(θ i−1 ) 2 sin ∆θ ) ] 2
2
≈ √ [f(θ i )−f 2 (θ i−1 )] 2 +[f(θ i )] 2 (∆θ) 2
≈ √ [f ′ (θ i )] 2 +f 2 (θ i )∆θ
n∑ n∑ √
s = lim s i = lim [f′ (θ i )] 2 +[f(θ i )] 2 ∆θ
n→∞ n→∞
i=1 i=1
∫ b √
= [f(θ)]2 +[f ′ (θ)] 2 dθ.
a
De-Yu Wang CSIE CYUT 147

8.4. CALCULUS AND POLAR COORDINATES
Example 8.4.3. Sketch the graph and find the arc length of the curve r = 2−2cosθ.
Solution:
s =
=
=
= 4
∫ b
a
∫ 2π
0
∫ 2π
0
∫ 2π
0
√
[f(θ)]2 +[f ′ (θ)] 2 dθ
√
(2−2cosθ)2 +(2sinθ) 2 dθ
√
∫ 2π
√
8(1−cosθ)dθ = 4 sin 2 θ
0 2 dθ
∣ sin θ 2∣ dθ = −8cos θ 2π
2∣
= 16.
θ=0
−4
−3
−2
y
2
1
−1
−1
−2
1
x
Exercise 8.4.1. Sketch the graph and find the slope of the given curves at the
indicated points
(a) r = sin3θ at θ = π 3
(b) r = cos2θ at θ = 0
(c) r = 3sinθ at θ = 0
(d) r = cos2θ at θ = π 4
Exercise 8.4.2. Sketch the graph and find the area of the indicated region.
(a) One leaf of r = sin3θ
(b) One leaf of r = sin2θ
(c) Bounded by r = 2−2cosθ
(d) BOunded by r = 3−3sinθ
Exercise 8.4.3. Sketch the graph and find the arc length of the given curve.
(a) r = sin3θ
(b) r = 2−2sinθ
(c) r = 3−3cosθ
(d) r = 1+2sin2θ
De-Yu Wang CSIE CYUT 148

Chapter 9
VECTORS

9.1. VECTORS
9.1 Vectors
Definition 9.1.1. Vectors
(a) Notation
Vectors in the Plane: V 2 = {〈x,y〉|x,y ∈ R}
−→
PQ = 〈x 1 −x 0 , y 1 −y 0 〉 = 〈a 1 ,a 2 〉 = a
−→
PQ
Q(x 1 , y 1 )
terminal point
✟ ✟✟✟✟✟✟✟✟✟✟✟✯ initial point
P(x 0 , y 0 )
Vectors in the Space: V 3 = {〈x,y,z〉|x,y,z ∈ R} ]
−→
PQ = 〈x 1 −x 0 , y 1 −y 0 , z 1 −z 0 〉 = 〈a 1 ,a 2 ,a 3 〉 = a
−→
PQ
Q(x 1 , y 1 , z 1 )
terminal point
(b) Magnitude:
✟ ✟✟✟✟✟✟✟✟✟✟✟✯ initial point
P(x 0 , y 0 , z 0 )
√
‖a‖ = a 2 1 +a 2 2 for a ∈ V 2
√
‖a‖ = a 2 1 +a 2 2 +a 2 3 for a ∈ V 3 .
(c) Addition and subtraction:
a±b = 〈a 1 ,a 2 〉±〈b 1 ,b 2 〉 = 〈a 1 ±b 1 ,a 2 ±b 2 〉 for a ∈ V 2
a±b = 〈a 1 ,a 2 ,a 3 〉±〈b 1 ,b 2 ,b 3 〉
= 〈a 1 ±b 1 ,a 2 ±b 2 , a 3 ±b 3 〉 for a ∈ V 3
(d) Zero vector
0 = 〈0,0〉 for a ∈ V 2
0 = 〈0,0,0〉 for a ∈ V 3
De-Yu Wang CSIE CYUT 150

9.1. VECTORS
(e) The negative of a is
−a = 〈−a 1 ,−a 2 〉 for a ∈ V 2
−a = 〈−a 1 ,−a 2 ,−a 3 〉 for a ∈ V 3
(f) Parallel:
b = ca, for any real number c.
Example 9.1.1. For vector a = 〈5,−2〉 and b = 〈2,1〉, compute ‖5a−2b‖ .
Solution:
5a−2b = 5〈5,−2〉−2〈2,1〉
= 〈25,−10〉−〈4,2〉 = 〈21,−12〉
‖5a−2b‖ = ‖〈21,−12〉‖ = √ 21 2 +(−12) 2 = √ 585.
Example 9.1.2. Determine whether or not the given pair of vectors is parallel: (a)
a = 〈2,3〉 and b = 〈4,5〉, (b) a = 〈2,3〉 and b = 〈−4,−6〉.
Solution: (a) If b = ca, 〈4,5〉 = 〈2c,3c〉, then c = 2 and c = 5 . This is a
3
contradiction and so a and b are not parallel.
(b) If b = ca, 〈−4,−6〉 = 〈2c,3c〉, then c = −2. This says that b = −2a and so a
and b are parallel.
Example 9.1.3. Find the distance between the points (1,−3,5) and (5,2,−3).
Solution:
d = ‖〈5−1,2−(−3),−3−5〉‖ = ‖〈4,5,−8〉‖
= √ 4 2 +5 2 +(−8) 2 = √ 105.
De-Yu Wang CSIE CYUT 151

9.1. VECTORS
Theorem 9.1.1. Properties of vector operations
Let a,b and c be vectors, and let d and e be scales.
Proof:
1. a+b = b+a (commutativity)
2. a+(b+c) = (a+b)+c (associativity)
3. a+0 = a (zero vector)
4. a+(−a) = 0 (additive inverse)
5. d(a+b) = da+db (distributive law)
6. (d+e)a = da+ea (distributive law)
7. (1)a = a (multiplication by 1) and
8. (0)a = 0 (multiplication by 0).
1. a+b = 〈a 1 ,a 2 〉+〈b 1 ,b 2 〉 = 〈a 1 +b 1 ,a 2 +b 2 〉
= 〈b 1 +a 1 ,b 2 +a 2 〉 = b+a
2. a+(b+c) = 〈a 1 ,a 2 〉+(〈b 1 ,b 2 〉+〈c 1 ,c 2 〉)
= 〈a 1 ,a 2 〉+〈b 1 +c 1 ,b 2 +c 2 〉
= 〈a 1 +b 1 +c 1 ,a 2 +b 2 +c 2 〉
= 〈a 1 +b 1 ,a 2 +b 2 〉+〈c 1 ,c 2 〉 = (a+b)+c
Theorem 9.1.2. (a) Scalar multiplication:
ca = c〈a 1 ,a 2 〉 = 〈ca 1 ,ca 2 〉 for a ∈ V 2
ca = c〈a 1 ,a 2 ,a 3 〉 = 〈ca 1 ,ca 2 ,ca 3 〉 for a ∈ V 3
and ‖ca‖ = |c|‖a‖.
(b) For any nonzero position vector a, a unit vector having the same direction as a
is given by
u = 1
‖a‖ a.
Proof: (a) ‖ca‖ = √ (ca 1 ) 2 +(ca 2 ) 2 = |c| √ a 2 1 +a 2 2 = |c|‖a‖
∥ (b) ‖u‖ =
1 ∥∥∥
∥‖a‖ a = 1 ‖a‖ = 1.
‖a‖
De-Yu Wang CSIE CYUT 152

9.1. VECTORS
Definition 9.1.2. The standard basis (unit) vectors
i = 〈1,0〉 and j = 〈0,1〉 for a ∈ V 2
i = 〈1,0,0〉, j = 〈0,1,0〉 and k = 〈0,0,1〉 for a ∈ V 3
For any vector a, we can write
a = 〈a 1 ,a 2 〉 = a 1 〈1,0〉+a 2 〈0,1〉
= a 1 i+a 2 j for a ∈ V 2
a = 〈a 1 ,a 2 ,a 3 〉 = a 1 〈1,0,0〉+a 2 〈0,1,0〉+a 3 〈0,0,1〉
= a 1 i+a 2 j+a 3 k for a ∈ V 3
Example 9.1.4. Find a vector a with the magnitude 3 and in the same direction as
the vector v = 〈4,1,−2〉.
Solution:
u = 1
‖v‖ v = 1 √
21
〈4,1,−2〉 =
a = 3u = 3
〈 4
√
21
,
〈 4
√
21
,
1
√ , −2 〉 〈 12
√ = √ , 21 21 21
1
√ , −2 〉
√
21 21
3
√ , −6 〉
√
21 21
Exercise 9.1.1. For a = 〈4,−1,−3〉 and b = 〈−1,3,1〉
(a) Compute 2a−3b.
(b) Compute ‖2a−3b‖
(c) Determine if the vectors a and b are parallel.
(d) Find a unit vector in the same direction as a.
Exercise 9.1.2. Repeat Exercise 9.1.1 with
(a) a = 〈2,1〉, b = 〈3,4〉
(b) a = 2i+j−3k, b = 6i+2j−k
(c) a = 2i−3k, b = 6i+2j
(d) a = 〈3,1,0〉, b = 〈−3,0,4〉
Exercise 9.1.3. Find a vector with the given magnitude and in the same direction
as the given vector v.
(a) Magnitude 6, vector v = 〈2,2,1〉 (c) Magnitude 3, vector v = 2i−j+3k
(b) Magnitude 10, vector v = 〈3,0,−4〉 (d) Magnitude 7, vector v = 3i−5j+k
Exercise 9.1.4. Prove Theorem 9.1.1
De-Yu Wang CSIE CYUT 153

9.2. THE DOT PRODUCT
9.2 The Dot Product
Definition 9.2.1. The dot product of two vectors
a·b = 〈a 1 ,a 2 〉·〈b 1 ,b 2 〉 = a 1 b 1 +a 2 b 2 for a ∈ V 2
a·b = 〈a 1 ,a 2 ,a 3 〉·〈b 1 ,b 2 ,b 3 〉 = a 1 b 1 +a 2 b 2 +a 3 b 3 for a ∈ V 3
Note. The dot product of two vectors is a scalar (i.e., a number, not a vector). For
this reason, the dot product is also called the scalar product.
Example 9.2.1. For vector a = 〈5,−2,1〉 and b = 〈2,1,−3〉, compute the dot
product a·b.
Solution:
a·b = 〈5,−2,1〉·〈2,1,−3〉 = 10−2−3 = 5.
Theorem 9.2.1. Let a,b and c be vectors, and let d be scale.
Proof:
1. a·b = b·a (commutativity)
2. a·(b+c) = a·b+a·c (distributive law)
3. (da)·b = d(a·b) = a·(db)
4. 0·a = 0
5. a·a = ‖a‖ 2 .
1. a·b = 〈a 1 ,a 2 〉·〈b 1 ,b 2 〉 = a 1 b 1 +a 2 b 2
= b 1 a 1 +b 2 a 2 = b·a
5. a·a = 〈a 1 ,a 2 〉·〈a 1 ,a 2 〉
= a 2 1 +a 2 2 = ‖a‖ 2
Theorem 9.2.2. Let θ be the angle between nonzero vectors a and b. Then
a·b = ‖a‖ ‖b‖cosθ
Proof: (a) If a and b are not parallel, then
‖a‖
‖a−b‖ 2 = (‖a‖sinθ) 2 +(‖b‖−‖a‖cosθ) 2
θ ■
‖a‖cosθ
= ‖a‖ 2 +‖b‖ 2 −2‖a‖‖b‖cosθ
✐
✒
‖a‖sinθ
‖a−b‖
‖b‖−‖a‖cosθ
✲
De-Yu Wang CSIE CYUT 154

9.2. THE DOT PRODUCT
(b) Using the properties of the dot product,
‖a−b‖ 2 = (a−b)·(a−b)
= a·a−a·b−b·a+b·b
= ‖a‖ 2 −2a·b+‖b‖ 2
= ‖a‖ 2 +‖b‖ 2 −2‖a‖‖b‖cosθ
∴ a·b = ‖a‖ ‖b‖cosθ
Definition 9.2.2. Two vector a and b are orthogonal if a·b = 0.
Example 9.2.2. For a = 〈4,−1,−3〉 and b = 〈−1,3,1〉
(a) Determine if the vectors a and b are orthogonal.
(b) Compute the angle between the vectors a and b.
Solution: (a) a·b = −4−3−3 = −10 ≠ 0, so that a and b are not orthogonal.
(b) From Theorem 9.2.2
cosθ = a·b
‖a‖‖b‖ = −10 √
26
√
11
0 ≤ θ = cos −1 ( −10
√
26
√
11
)
≤ π.
Exercise 9.2.1. For a = 〈3,−2,1〉 and b = 〈−1,4,2〉
(a) Compute the dot product a·b.
(b) Determine if the vectors a and b are orthogonal.
(c) Compute the angle between the vectors a and b.
Exercise 9.2.2. Repeat Exercise 9.2.1 with
(a) a = 〈2,1〉, b = 〈3,4〉
(b) a = 〈3,1,0〉, b = 〈−3,0,4〉
(c) a = 2i−3k, b = 6i+2j
(d) a = 2i+j−3k, b = 6i+2j−k
Exercise 9.2.3. Prove Theorem 9.2.1 and Theorem 9.2.2
De-Yu Wang CSIE CYUT 155

9.3. THE CROSS PRODUCT
9.3 The Cross Product
Definition 9.3.1. The determinate of
(a) 2×2
∣ a ∣
1 a 2∣∣∣
= a
b 1 b 1 b 2 −a 2 b 1 .
2
(b) 3×3
∣ a 1 a 2 a 3∣∣∣∣∣
∣ ∣ ∣ ∣ ∣ ∣ ∣∣∣ b b 1 b 2 b 3 = a 2 b 3∣∣∣
∣∣∣ b
1 −a 1 b 3∣∣∣
∣∣∣ b
∣ c
c 1 c 2 c 2 c 2 +a 1 b 2∣∣∣
3 c 1 c 3 .
3 c 1 c 2
3
Definition 9.3.2. For two vectors a = 〈a 1 ,a 2 ,a 3 〉 and b = 〈b 1 ,b 2 ,b 3 〉 in V 3 , the cross
(vector) product of a and b is
∣ i j k ∣∣∣∣∣ a×b =
a 1 a 2 a 3 =
∣ a ∣ 2 a 3∣∣∣
i−
∣ b
b 1 b 2 b 2 b 3
∣ a ∣ 1 a 3∣∣∣
j+
b 1 b 3
∣ a ∣
1 a 2∣∣∣
k
b 1 b 2
3
Example 9.3.1. For vectors a = 〈2,1,4〉, b = 〈−1,2,−1〉, compute the cross product
a×b.
Solution:
i j k
∣ ∣ ∣ ∣∣∣
a×b =
2 1 4
∣−1 2 −1∣ = 1 4
∣∣∣ 2 −1∣ i− 2 4
∣∣∣ −1 −1∣ j+ 2 1
−1 2∣ k = 〈−9,−2,5〉.
Theorem 9.3.1. Let a,b and c be vectors in V 3 , and let d be scale.
1. a×b = −(b×a) (anticommutativity)
2. a×(b+c) = a×b+a×c (distributive law)
(a+b)×c = a×c+b×c
3. (da)×b = d(a×b) = a×(db)
4. a·(b×c)=(a×b)·c
=(b×c)·a = b·(c×a)
=(c×a)·b
5. a×a = 0
6. a×0 = 0×a = 0
(scalar triple product)
De-Yu Wang CSIE CYUT 156

9.3. THE CROSS PRODUCT
Proof:
∣ i j k ∣∣∣∣∣ 1. a×b =
a 1 a 2 a 3 =
∣ a ∣ 2 a 3∣∣∣
i−
∣ b
b 1 b 2 b 2 b 3
∣ a ∣ 1 a 3∣∣∣
j+
b 1 b 3
∣ a ∣
1 a 2∣∣∣
k
b 1 b 2
3 = −
∣ b ∣ 2 b 3∣∣∣
i+
a 2 a 3
∣ b ∣ 1 b 3∣∣∣
j−
a 1 a 3
∣ b ∣
1 b 2∣∣∣
k = −(b×a)
a 1 a 2 ∣ ∣ ∣ a 1 a 2 a 3∣∣∣∣∣
a 1 a 2 a 3∣∣∣∣∣
c 1 c 2 c 3∣∣∣∣∣
4. a·(b×c) =
b 1 b 2 b 3 = −
c 1 c 2 c 3 =
a 1 a 2 a 3
∣c 1 c 2 c 3
∣b 1 b 2 b 3
∣b 1 b 2 b 3
= c·(a×b) = (a×b)·c
Theorem 9.3.2. Let θ be the angle between nonzero vectors a and b in V 3 . Then
(a) a×b is orthogonal to both a and b.
(b) ‖a×b‖ = ‖a‖ ‖b‖sinθ.
(c) a and b are parallel if and only if a×b = 0.
Proof for (a):
∣ ∣ a 1 a 2 a 3∣∣∣∣∣
0 0 0 ∣∣∣∣∣
a·(a×b) =
a 1 a 2 a 3 =
a 1 a 2 a 3 = 0
∣b 1 b 2 b 3
∣b 1 b 2 b 3 ∣ ∣ b 1 b 2 b 3∣∣∣∣∣
0 0 0 ∣∣∣∣∣
b·(a×b) =
a 1 a 2 a 3 =
a 1 a 2 a 3 = 0
∣b 1 b 2 b 3
∣b 1 b 2 b 3
Proof for (b):
‖a×b‖ 2 = (a×b)·(a×b)
= [a 2 b 3 −a 3 b 2 ] 2 +[a 1 b 3 −a 3 b 1 ] 2 +[a 1 b 2 −a 2 b 1 ] 2
= (a 2 1 +a 2 2 +a 2 3)(b 2 1 +b 2 2 +b 2 3)−(a 1 b 1 +a 2 b 2 +a 3 b 3 ) 2
= ‖a‖ 2 ‖b‖ 2 −(a·b) 2
= ‖a‖ 2 ‖b‖ 2 −‖a‖ 2 ‖b‖ 2 cos 2 θ
= ‖a‖ 2 ‖b‖ 2 (1−cos 2 θ)
= ‖a‖ 2 ‖b‖ 2 sin 2 θ
Since sinθ ≥ 0, for 0 ≤ θ ≤ π, so that ‖a×b‖ = ‖a‖ ‖b‖sinθ.
De-Yu Wang CSIE CYUT 157

9.3. THE CROSS PRODUCT
Proof for (c):
• If a and b are parallel (θ = 0 or π), then
‖a×b‖ = ‖a‖ ‖b‖sinθ = ‖a‖ ‖b‖·0 = 0.
• If ‖a×b‖ = 0 = ‖a‖ ‖b‖sinθ then sinθ = 0
⇒ θ = 0 or π, (a and b are parallel).
Theorem 9.3.3. The area of the parallelogram with two adjacent edges formed by
the vectors a and b.
A = ‖a×b‖
Proof:
A = ‖b‖
}{{}
base
‖a‖sinθ = ‖a×b‖
} {{ }
altitude
‖a‖
‖a‖sinθ
✑ ✑✑✑✑✑✸
■ θ
‖b‖
✲
Example 9.3.2. Find the area of the parallelogram with two adjacent edges formed
by the vectors a = 〈3,−1,1〉, and b = 〈−4,0,1〉.
Solution:
i j k
∣ ∣ ∣ ∣∣∣
a×b =
3 −1 1
∣−4 0 1∣ = −1 1
∣∣∣ 0 1∣ i− 3 1
∣∣∣ −4 1∣ j+ 3 −1
−4 0 ∣ k = 〈−1,−7,−4〉
A = ‖a×b‖ = ‖〈−1,−7,−4〉‖ = √ 66.
Theorem 9.3.4. The distance form the point Q, to the line through the points P
and R.
d = ‖−→ PQ× −→ PR‖
‖ −→ .
PR‖
Proof:
d = ‖ −→ PQ‖sinθ = ‖ −→ PQ‖sinθ · ‖−→ PR‖
= ‖−→ PQ× −→ PR‖
‖ −→ .
PR‖
‖ −→ PR‖
✒ Q
θ ■
P
d = ‖ −→ PQ‖sinθ
✲
R
De-Yu Wang CSIE CYUT 158

9.3. THE CROSS PRODUCT
Example 9.3.3. Find the distance form the point Q(1,2,−1), to the line through
the points P(0,1,3) and R(5,−2,1).
Solution:
−→
PQ = 〈1,1,−4〉 and −→ PR = 〈5,−3,−2〉
−→
PQ× −→ PR = 〈1,1,−4〉×〈5,−3,−2〉
i j k
=
1 1 −4
∣5 −3 −2∣ = 〈−14,−18,−8〉
d = ‖−→ PQ× −→ PR‖
‖ −→ PR‖
= ‖〈−14,−18,−8〉‖
‖〈5,−3,−2〉‖
=
√
584
√
38
Theorem 9.3.5. The volume of the parallelepiped with three adjacent edges formed
by the vectors a, b and c.
V = |c·(a×b)|.
Proof:
V = A·h
= ‖a×b‖
} {{ }
Area of base
·|‖c‖cosθ|
} {{ }
altitude
= ‖a×b‖· |c·(a×b)|
‖a×b‖
= |c·(a×b)|.
a×b
✻
✟ ✟✟✟ ✁
✁
c
✁ ✁
✁
✁ ✁✕
✁
✁ ✁
h
✁ ✁
✛θ
✁ b✁
✁
✟✁
✟✟✟✯ A
✟ ✟✟✟ ✁
✁
✁ ✁ ✁ ✁
✁ ✁
✁ ✁
✁ ✁
✁ ✁
✁
✲✁✟ ✟✟✟
a
Example 9.3.4. Find the volume of the parallelepiped with three adjacent edges
formed by the vectors a = 〈−4,−1,0〉, b = 〈2,2,−1〉, and c = 〈1,4,2〉.
Solution:
1 4 2
c·(a×b) =
−4 −1 0
∣ 2 2 −1∣
= 1
∣ −1 0
∣ ∣ ∣∣∣ 2 −1∣ −4 −4 0
∣∣∣ 2 −1∣ +2 −4 −1
2 2 ∣ = −27
V = |c·(a×b)| = |−27| = 27.
De-Yu Wang CSIE CYUT 159

9.4. VECTOR-VALUED FUNCTIONS
Exercise 9.3.1. For a = 〈3,−2,1〉 and b = 〈−1,4,2〉
(a) Compute the cross product a×b
(b) Determine if the vectors a and b are parallel.
(c) Findtheareaoftheparallelogramwithtwoadjacentedgesformedbythevectors
a and b.
Exercise 9.3.2. Repeat Exercise 9.3.1 with
(a) a = 〈2,1,−5〉, b = 〈3,4,1〉
(b) a = 〈3,1,0〉, b = 〈−3,0,4〉
(c) a = 2i−3k, b = 6i+2j
(d) a = 2i+j−3k, b = 6i+2j−k
Exercise 9.3.3. Find the distance form the point Q, to the line through the points
P and R.
(a) Q = (1,2,0), P = (6,−1,2) and R = (3,−4,1)
(b) Q = (1,0,−3), P = (−2,1,7) and R = (−3,1,1)
(c) Q = (3,2,0), P = (5,1,2) and R = (4,1,−1)
(d) Q = (1,3,1), P = (−2,1,3) and R = (1,0,−2)
Exercise 9.3.4. Find the volume of the parallelepiped with three adjacent edges
formed by the vectors a, b and c.
(a) a = 〈−4,−1,0〉, b = 〈2,2,−1〉, and c = 〈3,0,2〉.
(b) a = 〈1,−1,2〉, b = 〈1,2,3〉, and c = 〈0,4,2〉.
(c) a = 〈−3,2,1〉, b = 〈3,2,−5〉, and c = 〈1,5,−2〉.
(d) a = 〈−2,−1,5〉, b = 〈−3,6,−1〉, and c = 〈−1,4,2〉.
Exercise 9.3.5. Prove Theorem 9.3.1, Theorem 9.3.2, Theorem 9.3.3 and Theorem
9.3.4
9.4 Vector-Valued Functions
Definition 9.4.1. A function of the form
r(t) = f(t)i+g(t)j+h(t)k = 〈f(t), g(t), h(t)〉
is a vector-valued function, where f, g, and h are real-valued functions of the parameter
t.
De-Yu Wang CSIE CYUT 160

9.4. VECTOR-VALUED FUNCTIONS
Example 9.4.1. Plot the values of the two-dimensional vector-valued function
r(t) = 〈t+1, t 2 −2〉 at t = −2, t = 0 and t = 2.
y
Solution:
3 ✻
〈−1,2〉
r(−2) = 〈f(−2), g(−2)〉 = 〈−1,2〉 2 〈3,2〉
❆❑ r(2)
r(0) = 〈f(0), g(0)〉 = 〈1,−2〉
r(−2)
❆ 1
❆
r(2) = 〈f(2), g(2)〉 = 〈3,2〉
❆✑ ✑✑✑✑✑✸ ✲x
−2 −1 ❆ 1 2 3 4
−1 ❆
r(0)
❆
−2 ❆❯
〈1,−2〉
Example 9.4.2. Sketch a graph of the curve traced out by the endpoint of the
vector-valued function r(t) = 〈t+1, t 2 −2〉 from t = −2 to t = 2.
Solution: The endpoint of all position vector r lie on the curve
y
{
3 ✻
x = f(t) = t+1
C : ,−2 ≤ t ≤ 2
〈−1,2〉
2 〈3,2〉
y = g(t) = t 2 −2
❆❑ r(2)
y = t 2 −2 = (x−1) 2 −2. r(−2)
❆ 1
◆
❆ ✗
❆✑ ✑✑✑✑✑✸ ✲x
−2 −1 ❆ 1 2 3 4
−1 ❆
❯ r(0)
❆ ✕
−2 ❆❯
〈1,−2〉
Note. If an object moves along the curve C traced out by the endpoint of r, then r ′
and r ′′ are the velocity and acceleration vectors, respectively.
Definition 9.4.2. Limit of Vector-Valued Functions
For a vector-valued function r(t) = 〈f(t), g(t), h(t)〉, the limit of r(t) as t approaches
a is given by
limr(t) = lim〈f(t), g(t), h(t)〉 =
t→a t→a
〈
〉
limf(t), limg(t), limh(t)
t→a t→a t→a
provided all of the indicated limits exist. If any one of the limits fails to exist, then
limr(t) does not exist.
t→a
〈
Example 9.4.3. Find the limit lim t 2 ,e 2t , √ 〉
t 2 +2t .
t→1
Solution:
〈
lim t 2 , e 2t , √ 〉
t 2 +2t
t→1
=
〈 √ 〉 〈
limt 2 , lime 2t , lim t2 +2t = 1, e 2 , √ 〉
3
t→1 t→1 t→1
De-Yu Wang CSIE CYUT 161

9.4. VECTOR-VALUED FUNCTIONS
Definition 9.4.3. Continuity of Vector-Valued Functions
The vector-valued function r(t) = 〈f(t), g(t), h(t)〉 is continuous at t = a whenever
limr(t) = r(a)
t→a
Theorem 9.4.1. A vector-valued function r(t) = 〈f(t), g(t), h(t)〉 is continuous at
t = a if and only if all of f, g and h are continuous at t = a.
Example 〈 9.4.4. 〉 Determine all values of t at which the vector-valued function r(t) =
t+1
t−1 , t2 , 2t is continuous.
Solution:
{t|t ≠ 1}
Example 9.4.5. Determine all values of t at which the vector-valued function
r(t) = 〈e 5t , ln(t+1), cost〉 is continuous.
Solution:
{t|t > −1}
Exercise 9.4.1. Sketch a graph of the curve traced out by the endpoint of the vectorvalued
function
(a) r(t) = 〈3t,t 2 〉, t = 0, t = 1
(b) r(t) = 〈t 2 ,2t−1〉, t = 0, t = 1
(c) r(t) = 〈cos2t,sint〉, t = 0, t = π 2
(d) r(t) = 〈cost,sin2t〉, t = 0, t = π 2
Exercise 9.4.2. Find the limit if it exists
(a) lim
t→0
〈
t 2 −1, e 2t , sint 〉
〈 sint
(b) lim , cost, t+1 〉
t→0 t t−1
Exercise 9.4.3. Determine all values of t at which the given vector-valued function
is continuous.
〈 〉 3
(a) r(t) =
t , cost, 2t (b) r(t) = 〈tant, cost 2 , 2t+3〉
De-Yu Wang CSIE CYUT 162

9.5. THE CALCULUS OF VECTOR-VALUED FUNCTIONS
9.5 The Calculus of Vector-Valued Functions
Definition 9.5.1. The derivative r ′ (t) of the vector-valued function r(t) is defined
as
r ′ (t) = lim
∆t→0
r(t+∆t)−r(t)
,
∆t
for any values of t for which the limit exists. When the limit exists for t = a, we say
that r is differentiable at t = a.
Theorem 9.5.1. Let r(t) = 〈f(t), g(t), h(t)〉 and suppose that the components f, g
and h are all differentiable for some value of t. Then
Proof:
r ′ (t) = 〈f ′ (t), g ′ (t), h ′ (t)〉
r(t+∆t)−r(t)
,
∆t
〈f(t+∆t), g(t+∆t), h(t+∆t)〉−〈f(t), g(t), h(t)〉
= lim
∆t→0
∆t
〈f(t+∆t)−f(t), g(t+∆t)−g(t), h(t+∆t)−h(t)〉
= lim
∆t→0
〈
∆t
f(t+∆t)−f(t)
= lim , g(t+∆t)−g(t) , h(t+∆t)−h(t) 〉
∆t→0 ∆t ∆t ∆t
〈
f(t+∆t)−f(t) g(t+∆t)−g(t)
= lim , lim , lim
∆t→0 ∆t ∆t→0 ∆t ∆t→0
r ′ (t) = lim
∆t→0
= 〈f ′ (t), g ′ (t), h ′ (t)〉
〉
h(t+∆t)−h(t)
∆t
Example 9.5.1. Find the derivative of the vector-valued function r(t) =
〈3t 3 +2t, sin(t 2 +3t), e 3 t 3 〉 and s(t) = 〈7te t , cost 3 , te 2 〉.
Solution:
r ′ (t) = 〈 9t 2 +2, (2t+3)cos(t 2 +3t), 3e 3 t 2〉
s ′ (t) = 〈 7e t +7te t , −3t 2 sint 3 , e 2〉
De-Yu Wang CSIE CYUT 163

9.5. THE CALCULUS OF VECTOR-VALUED FUNCTIONS
Theorem 9.5.2. Supposethatr(t)ands(t)aredifferentiablevector-valuedfunctions,
f(t) is a differentiable scalar function and c is any scalar constant. Then
(a) d dt [r(t)+s(t)] = r′ (t)+s ′ (t)
(b) d dt [cr(t)] = cr′ (t)
(c) d dt [f(t)r(t)] = f′ (t)r(t)+f(t)r ′ (t)
(d) d dt [r(t)·s(t)] = r′ (t)·s(t)+r(t)·s ′ (t)
(e) d dt [r(t)×s(t)] = r′ (t)×s(t)+r(t)×s ′ (t)
Proof: Let r(t) = 〈f 1 (t), g 1 (t), h 1 (t)〉 and s(t) = 〈f 2 (t), g 2 (t), h 2 (t)〉
(a) d dt [r+s] = d dt [〈f 1, g 1 , h 1 〉+〈f 2 , g 2 , h 2 〉]
= d dt [〈f 1 +f 2 , g 1 +g 2 , h 1 +h 2 〉]
= 〈f 1 ′ +f 2, ′ g 1 ′ +g 2, ′ h ′ 1 +h ′ 2〉
= 〈f 1, ′ g 1, ′ h ′ 1〉+〈f 2, ′ g 2, ′ h ′ 2〉 = r ′ +s ′
(d) d dt [r·s] = d dt [〈f 1, g 1 , h 1 〉·〈f 2 , g 2 , h 2 〉]
= d dt [f 1f 2 +g 1 g 2 +h 1 h 2 ]
= f ′ 1f 2 +f 1 f ′ 2 +g ′ 1g 2 +g 1 g ′ 2 +h ′ 1h 2 +h 1 h ′ 2
= [f ′ 1f 2 +g ′ 1g 2 +h ′ 1h 2 ]+[f 1 f ′ 2 +g 1 g ′ 2 +h 1 h ′ 2]
= r ′ ·s+r·s ′
Example 9.5.2. For f(t) = 3t 3 +2t, and r(t) = 〈7te t , cost 3 , te 2 〉, find d dt [f(t)r(t)].
Solution:
d
dt [f(t)r(t)] = d dt [(3t3 +2t) 〈 7te t , cost 3 , te 2〉 ]
=(9t 2 +2) 〈 7te t , cost 3 , te 2〉 +
(3t 3 +2t) 〈 7e t +7te t , −3t 2 sint 3 , e 2〉
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9.5. THE CALCULUS OF VECTOR-VALUED FUNCTIONS
Example 9.5.3. For r(t) = 〈lnt, t 2 , 3t+1〉, and s(t) = 〈3 t , t 2 −1, te 2 〉, find
d
dt [r(t)·s(t)].
Solution:
d
dt [r(t)·s(t)] =r′ (t)·s(t)+r(t)·s ′ (t)
= d 〈
lnt, t 2 , 3t+1 〉·〈3t , t 2 −1, te 2〉
dt
+ 〈 lnt, t 2 , 3t+1 〉· d 〈
3 t , t 2 −1, te 2〉
〈 〉 dt
1
=
t , 2t, 3 ·〈3 t , t 2 −1, te 2〉 + 〈 lnt, t 2 , 3t+1 〉·〈3t ln3, 2t, e 2〉
= 3t
t +2t3 −2t+e 2 3t+3 t lntln3+2t 3 +e 2 (3t+1)
Example 9.5.4. For r(t) =
Solution:
〈 〉
5 t2 , 3t , and s(t) = 〈sect, sint〉, find d dt [r(t)×s(t)].
d
dt [r(t)×s(t)] =r′ (t)×s(t)+r(t)×s ′ (t)
= d 〈 〉 〈 〉
5 t2 , 3t ×〈sect, sint〉+ 5 t2 , 3t × d 〈sect, sint〉
dt 〈 〉 〈 〉 dt
= 5 t2 2t ln5, 3 ×〈sect, sint〉+ 5 t2 , 3t ×〈secttant, cost〉
Definition 9.5.2. The vector-valued function R(t) is an antiderivative of the vectorvalued
function r(t) whenever
R ′ (t) = r(t)
Definition 9.5.3. If R(t) is any antiderivative of r(t) = 〈f(t), g(t), h(t)〉, then
(a) the indefinite integral of r(t) is defined to be
∫ ∫
r(t)dt = 〈f(t), g(t), h(t)〉 dt
〈∫ ∫ ∫ 〉
= f(t)dt, g(t)dt, h(t)dt = R(t)+c.
where c is an arbitrary constant vector.
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9.5. THE CALCULUS OF VECTOR-VALUED FUNCTIONS
(b) the definite integral of r(t) is defined to be
∫ b
a
∫ b
r(t)dt = 〈f(t), g(t), h(t)〉 dt
a
〈∫ b ∫ b
= f(t)dt, g(t)dt,
a
= R(t)| b t=a = R(b)−R(a)
a
∫ b
a
〉
h(t)dt
∫ 〈cost,
Example 9.5.5. Evaluate the indefinite integral 2t 2 +3, e 4t〉 dt.
Solution:
∫ 〈cost,
2t 2 +3, e 4t〉 〈
dt = sint+c 1 ,
=
〈
sint,
2
3 t3 +3t+c 2 ,
〉
e 4t
+c
4
2
3 t3 +3t,
〉
e 4t
4 +c 3
Example 9.5.6. Evaluate the definite integral
∫ 4
1
〈
sint,
t 2 3
〉
dt.
Solution:
∫ 4
1
〈
sint, t 2 /3 〉 dt =
〈
−cost,
〉∣
t 3 ∣∣∣
4
= 〈−cos4+cos1, 7〉
9
t=1
Exercise 9.5.1. Find the derivative of the vector-valued function
(a) r(t) = 〈3t 3 , sint 2 , te 3t 〉 (b) r(t) = 〈3t 3 +2t, sin(t 2 +3t), e 3 t 3 〉
Exercise 9.5.2. Evaluate the indefinite or definite integral
(a)
(b)
∫ 〈cost,
2t 2 +3, e 4t〉 dt
∫ 〈 〉
3t
te t2 , 3tsint, dt
t 2 +1
(c)
(d)
∫ 4
1
∫ 2
0
〈
sint, t 2 /3 〉 dt
〈 〉 4
t+1 , et−2 , te t , dt
De-Yu Wang CSIE CYUT 166

9.6. ARC LENGTH AND CURVATURE
9.6 Arc Length and Curvature
Theorem 9.6.1. Arc length in the vector-valued function
The arc length of the curve traced out by the endpoint of the vector-valued function
r(t) = 〈f(t), g(t), h(t)〉, a ≤ t ≤ b is given by
s =
∫ b
a
‖r ′ (t)‖dt.
Proof: The curve described parametrically by
C : x = f(t), y = g(t), z = h(t), a ≤ t ≤ b.
The arc length (see page 140) is
s =
∫ b
a
√
[f′ (t)] 2 +[g ′ (t)] 2 +[h ′ (t)] 2 dt =
∫ b
a
‖r ′ (t)‖dt.
Example 9.6.1. Find the arc length of the curve traced out by the endpoint of the
vector-valued function r = 〈2t, lnt, t 2 〉, for 1 ≤ t ≤ e.
Solution:
s =
=
=
=
∫ b
a
∫ e
1
∫ e
1
∫ e
1
√
[f′ (t)] 2 +[g ′ (t)] 2 +[h ′ (t)] 2 dt
√
( ) 2 1
2 2 + +(2t)
t
2 dt
√ √
4+ 1 ∫ e (1 ) 2
t 2 +4t2 dt =
t +2t dt
( ) 1
t +2t dt =
1
(
ln|t|+2 t2 2
= (lne+e 2 )−(ln1+1) = e 2
)∣ ∣∣∣
e
1
z
✻
✍
✘ ✘✘✘ ✘ ✘✘✘ ✘ ✘✿ y
x
Definition 9.6.1. The curvature of a curve C is
κ =
dT
∥ ds ∥ ,
where T is the unit tangent vector.
Note. The curvature of a straight line is zero and
the curvature for a circle of radius a > 0 is a constant 1 a .
De-Yu Wang CSIE CYUT 167

9.6. ARC LENGTH AND CURVATURE
z
✻
C
y
✘✘ ✘✘✘ ✘ ✘✘✘ ✁ ✁✕ T i
✘✿
✁
✁
✏
✏✏✏✶ T i−1
x
Theorem 9.6.2. The curvature of the curve given by the vector-valued function r(t)
is
κ(t) = ‖T′ (t)‖
‖r ′ (t)‖ = ‖r′ (t)×r ′′ (t)‖
‖r ′ (t)‖ 3 .
where T(t) = r′ (t)
‖r ′ (t)‖ .
Proof: Since ds = √ [f
dt ′ (t)] 2 +[g ′ (t)] 2 +[h ′ (t)] 2 = ‖r ′ (t)‖, so that
∥ κ(t) =
dT
∥∥∥∥ dT
∥ ds ∥ = dt
∥ = ‖T′ (t)‖
‖r ′ (t)‖ .
ds
dt
Example 9.6.2. Using 〈 two different 〉 methods in Theorem 9.6.2 to find the curvature
of the curve r(t) = 2t,t 2 ,− t3 at the point t = 0.
3
Solution with Method 1: κ(t) = ‖T′ (t)‖
‖r ′ (t)‖
r ′ (t) = 〈 2,2t,−t 2〉
‖r ′ (t)‖ = √ 4+4t 2 +t 4 = t 2 +2
T(t) = r′ (t)
‖r ′ (t)‖ = 〈2,2t,−t2 〉
t 2 +2
T ′ (t) = 〈0,2,−2t〉·(t2 +2)−〈2,2t,−t 2 〉·2t
(t 2 +2) 2
= 〈−4t,4−2t2 ,−4t〉
(t 2 +2)
√ 2
‖T ′ 16t2 +16−16t
(t)‖ =
2 +4t 4 +16t 2
= 2
(t 2 +2) 2 t 2 +2
κ(t) = ‖T′ (t)‖
‖r ′ (t)‖ = 2
(t 2 +2) 2, κ(0) = 1 2 .
De-Yu Wang CSIE CYUT 168

9.6. ARC LENGTH AND CURVATURE
Solution with Method 2: κ(t) = ‖r′ (t)×r ′′ (t)‖
‖r ′ (t)‖ 3
r ′ (t) = 〈 2,2t,−t 2〉
‖r ′ (t)‖ = √ 4+4t 2 +t 4 = t 2 +2
r ′′ (t) = 〈0,2,−2t〉
i j k
r ′ (t)×r ′′ (t) =
2 2t −t 2
∣0 2 −2t∣ = 〈 −4t 2 +2t 2 ,4t,4 〉 = 〈 −2t 2 ,4t,4 〉
‖r ′ (t)×r ′′ (t)‖ = √ 4t 4 +16t 2 +16 = 2(t 2 +2)
κ(t) = ‖r′ (t)×r ′′ (t)‖
‖r ′ (t)‖ 3 = 2(t2 +2)
(t 2 +2) 3 = 2
(t 2 +2) 2, κ(0) = 1 2 .
Exercise 9.6.1. Find the arc length of the curve traced out by the of the vectorvalued
function
(a) r(t) = 〈cost,sint,2t〉, 0 ≤ t ≤ 2π
(b) r(t) = 〈√ 8t, t 2 , ln(t 2 ) 〉 , 1 ≤ t ≤ 2
(c) r(t) = 〈t 2 +1,2t,t 2 −1〉, 0 ≤ t ≤ 2
(d) r(t) = 〈t,t 2 −1,t 3 〉, 0 ≤ t ≤ 2
Exercise 9.6.2. Using two different methods to find the curvature of the curve at
the given point.
(a) r(t) = 〈e −2t ,2t,4〉, t = 0
(b) r(t) = 〈2,sinπt,lnt〉, t = 1
(c) r(t) = 〈t,t,t 2 −1〉, t = 2
(d) r(t) = 〈t,t 2 +t−1,t〉, t = 0
De-Yu Wang CSIE CYUT 169

Chapter 10
PARTIAL DIFFERENTIATION

10.1. FUNCTIONS OF SEVERAL VARIABLES
10.1 Functions of Several Variables
Definition 10.1.1. A function f is a rule that assign a unique real number
z = f(x 1 ,x 2 ,··· ,x n ),
to each n-tuples of real numbers (x 1 ,x 2 ,··· ,x n ) in D. The set D = {(x 1 ,x 2 ,··· ,x n )}
is the domain of f, and the set R = {z} is the range of f.
Note. x 1 ,x 2 ,··· and x n are the independent variables and z is the dependent variable.
Example 10.1.1. Find the domain and range of the function z = f(x,y) =
√ x+3y −2 and evaluate the value of f at the point (2,1).
Solution: (a) The domain
D = {(x,y)|x+3y −2 ≥ 0}.
(b) The range
R = {z|z ≥ 0}
(c) f(2,1) = √ x+3y −2 = √ 2+3·1−2 = √ 3.
Example 10.1.2. Find the domain and range of the function z = f(x,y) = cos(x 2 +
y 2 ) and evaluate the value of f at the point (−2,1).
Solution: (a) The domain
D = {(x,y)|all x,y ∈ R}.
(b) The range
R = {z|−1 ≤ z ≤ 1}
(c) f(−2,1) = cos(x 2 +y 2 ) = cos[(−2) 2 +1 2 ] = cos(5).
Exercise 10.1.1. Find the domain and range of the function f and evaluate the
value of f at the given point.
(a) f(x,y) = 1
x+y , (3,1)
(b) f(x,y) = ln(2+x+y), (−3,7)
(c) f(x,y) = sin 2 x+cos 2 y, (π,0)
(d) f(x,y,z) = eyz
z−x 2 −y 2 , (2,−3,1)
De-Yu Wang CSIE CYUT 171

10.2. LIMITS AND CONTINUITY
10.2 Limits and Continuity
Definition 10.2.1. Formal definition of limit
Let f be defined on an open disk centered at (a,b), except possibly at (a,b). We say
that
lim f(x,y) = L,
(x,y)→(a,b)
if for every ε > 0 there exists a δ > 0 such that
|f(x,y)−L| < ε whenever 0 < √ (x−a) 2 +(y −b) 2 < δ.
Remark. If f(x,y) approaches L 1 as (x,y) approaches (a,b) along a path P 1 and
f(x,y) approaches L 2 ≠ L 1 as (x,y) approaches (a,b) along a path P 2 , then
f(x,y) does not exist.
lim
(x,y)→(a,b)
y
✻
✬✩f
❈
δ❈
(x,y)
(a,b)
✫✪
✲x
z
✻
L+ε
❘
f(x,y)
L
L−ε
Example 10.2.1. Compute the limit
Solution:
e x+y−z
lim
(x,y,z)→(1,1,2) x−z = e0
−1 = −1.
e x+y−z
lim
(x,y,z)→(1,1,2) x−z .
y
✻ P 1
P 5 ✬✩P 7
❅
✲ ❅❘ ❄
✠
P ✛ 3 P 4
✻
✫✪ ✒ ❅■ P ❅
8 P 6
P 2
✲x
y
Example 10.2.2. Compute the limit lim
(x,y)→(1,0) x+y −1 .
Solution:
P 1 : x = 1
P 2 : y = 0
∴
lim
(1,y)→(1,0)
lim
(x,0)→(1,0)
lim
(x,y)→(1,0)
y
1+y −1 = lim1
= 1
y→0
0
x+0−1 = lim0
= 0 ≠ 1
x→1
y
does not exist.
x+y −1
De-Yu Wang CSIE CYUT 172

10.2. LIMITS AND CONTINUITY
xy 2
Example 10.2.3. Evaluate lim
(x,y)→(0,0) x 2 +y 4.
Solution:
P 1 : x = 0
P 2 : y = 0
P 3 : y = x
P 4 : x = y 2
∴
lim
(x,y)→(0,0)
lim
(0,y)→(0,0)
lim
(x,0)→(0,0)
lim
(x,x)→(0,0)
lim
(y 2 ,y)→(0,0)
0
0+y = lim0
= 0 4 y→0
0
x 2 +0 = lim0
= 0
x→0
x 3
x 2 +x 4 = lim
x→0
y 2 (y 2 )
(y 2 ) 2 +y 4 = lim
y→0
xy 2
x 2 +y 4 does not exist.
x
1+x 2 = 0
y 4
2y 4 = 1 2 ≠ 0
Theorem 10.2.1. Suppose that |f(x,y)−L| ≤ g(x,y) for all (x,y) in the interior of
some circle centered at (a,b), except possibly at (a,b). If lim g(x,y) = 0, then
(x,y)→(a,b)
lim f(x,y) = L.
(x,y)→(a,b)
Proof: If lim g(x,y) = 0, given any ε > 0 there exists a δ > 0 such that
(x,y)→(a,b)
|g(x,y)−0| = |g(x,y)| < ε whenever 0 < √ (x−a) 2 +(y −b) 2 < δ
|f(x,y)−L| ≤ g(x,y) ≤ |g(x,y)| < ε
|f(x,y)−L| < ε whenever 0 < √ (x−a) 2 +(y −b) 2 < δ
∴ lim f(x,y) = L.
(x,y)→(a,b)
xy 2
Example 10.2.4. Show that the limit lim
(x,y)→(0,0) x 2 +y exist. 2
Proof:
xy 2
lim
(x,0)→(0,0) x 2 +y = lim 0
2 (x,0)→(0,0) x = 0, 2
xy 2
lim
(0,y)→(0,0) x 2 +y = lim 0
2 (0,y)→(0,0) y = 0
∣ 2 ∣ ∵ |f(x,y)−L| =
xy 2 ∣∣∣
∣x 2 +y −0 ≤
xy 2 ∣∣∣ 2 ∣ = |x| and
y 2
∴
lim
(x,y)→(0,0)
xy 2
x 2 +y 2 = 0
lim |x| = 0
(x,y)→(0,0)
De-Yu Wang CSIE CYUT 173

10.2. LIMITS AND CONTINUITY
Definition 10.2.2. Suppose that f(x,y) is defined in the interior of a circle centered
at (a,b). We say that f is continuous at (a,b) if lim f(x,y) = f(a,b).
(x,y)→(a,b)
Example 10.2.5. Determine all points at which the function f(x,y) = ln(x 2 +y 2 −5)
is continuous.
Solution:
{(x,y)|x 2 +y 2 > 5}
Exercise 10.2.1. Compute the indicated limit
(a)
(b)
e x+y−z
lim
(x,y,z)→(1,1,2) x−z .
lim
(x,y)→(π,2)
cosxy
y 2 +1
(c)
(d)
lim
(x,y,z)→(1,0,2)
lim
(x,y)→(−3,0)
4xz
y 2 +z 2.
e xy
x 2 +y 2
Exercise 10.2.2. Show that the indicated limit exist.
(a)
(b)
lim
(x,y)→(0,0)
lim
(x,y)→(0,0)
xy 2
x 2 +y 2
(x,y,z)→(0,0,0)
x 2 y
x 2 +y 2 (d)
(c) lim
lim
(x,y,z)→(0,0,0)
3x 3
x 2 +y 2 +z 2
x 2 y 2 z 2
x 2 +y 2 +z 2
Exercise 10.2.3. Show that the indicated limit does not exist.
(a)
(b)
lim
(x,y)→(0,0)
lim
(x,y)→(0,0)
3x 2
x 2 +y 2
(x,y,z)→(0,0,0)
(c) lim
2x 2 y
x +y
(d)
lim
(x,y,z)→(0,0,0)
x 2 yz
x 4 +y 4 +z 4
xyz
x 3 +y 3 +z 3
Exercise 10.2.4. Determine all points at which the function is continuous.
(a) f(x,y) = √ x 2 −y 2 −5
(b) f(x,y) = ln(x 2 +y 2 −5)
(c) f(x,y) = 4xy +sin3x 2 y
(d) f(x,y) = ln(5−x 2 +y)
(e) f(x,y) = √ 9−x 2 −y 2
(f) f(x,y) = tan(x+y)
De-Yu Wang CSIE CYUT 174

10.3. PARTIAL DERIVATIVES
10.3 Partial Derivatives
Definition 10.3.1. The partial derivative of f(x,y) with respect to x and y are
defined as
f x (x,y) = ∂f f(x+h,y)−f(x,y)
(x,y) = lim ,
∂x h→0 h
f y (x,y) = ∂f f(x,y +h)−f(x,y)
(x,y) = lim ,
∂y h→0 h
provided the limits exist.
Example 10.3.1. For f(x,y) = 3x 2 +x 3 y+4y 2 , compute ∂f
∂x (x,y), ∂f
∂y (x,y), f x(1,0)
and f y (2,−1).
Solution:
∂f
∂x = 6x+3x2 y +0 = 6x+3x 2 y
∂f
∂y = 0+x3 ·1+8y = x 3 +8y
f x (1,0) = 6+0 = 6
f y (2,−1) = 8−8 = 0
Guidelines 10.3.1. Higher order partial derivatives
( )
∂ ∂f
= ∂2 f
∂x ∂x ∂x =f 2 xx
( )
∂ ∂f
= ∂2 f
∂y ∂y ∂y =f 2 yy
( )
∂ ∂f
= ∂2 f
∂y ∂x ∂x∂y =f xy
( )
∂ ∂f
= ∂2 f
∂x ∂y ∂y∂x =f yx = f xy
Example 10.3.2. Findallsecond-orderpartialderivativesoff(x,y) = x 2 y−y 3 +lnx.
Solution:
f x = 2xy + 1 x
f xx = 2y − 1 x 2
f xy = 2x
f y = x 2 −3y 2
f yx = 2x = f xy
f yy = −6y
De-Yu Wang CSIE CYUT 175

10.4. THE CHAIN RULE
Example 10.3.3. Find the partial derivative f yxy of f(x,y) = 2sec −1 (x 3 )−4xy 2 +
xln3y.
Solution:
f y = −8xy + x y
f yx = −8y + 1 y
f yxy = −8− 1 y 2
Exercise 10.3.1. Evaluate f x and f y at the indicated point.
(a) f(x,y) = cos −1 xy, (1,0)
(b) f(x,y) = 6xy
4x 2 +5y 2 , (1,1)
(c) f(x,y) = tan −1 y x , (1,−2)
(d) f(x,y) = xy
x−y , (2,−2)
Exercise 10.3.2. Find all second-order partial derivatives
(a) f(x,y) = x 3 −4xy 2 +y 4
(c) f(x,y,z) = √
2
(b) f(x,y) = x 3 sinxy 2 −3y 2
x 2 +y 2 +z 2
Exercise 10.3.3. Find the partial indicated derivative
(a) f yxy ; f(x,y) = 2sec −1 (x 3 )−4xy 2 +xln3y
(b) f yxy ; f(x,y) = 2cos(y 3 )−3x 3 y 2 +ln3xy
(c) f xx ,f yy ,f yyzz ; f(x,y,z) = e 2xy − z2
y +xzsiny
(d) f xx ,f yy ,f wxyz ; f(w,x,y,z) = √ wyz −x 3 sinw
10.4 The Chain Rule
Definition 10.4.1. For z = f(x,y), the increment of z is given by
∆z = f(x+∆x,y +∆y)−f(x,y)
where ∆x and ∆y are the increments of x and y.
De-Yu Wang CSIE CYUT 176

10.4. THE CHAIN RULE
Theorem 10.4.1. The differential of z is
Proof:
dz = f x (x,y)dx+f y (x,y)dy.
∆z =f(x+∆x,y +∆y)−f(x,y)
=f(x+∆x,y +∆y)−f(x,y +∆y)+f(x,y +∆y)−f(x,y)
f(x+∆x,y +∆y)−f(x,y +∆y)
= ∆x+
∆x
dz = lim ∆z
∆x→0
∆y→0
= lim
∆x→0
∆y→0
f(x+∆x,y +∆y)−f(x,y +∆y)
∆x
=f x (x,y)dx+f y (x,y)dy.
f(x,y +∆y)−f(x,y)
∆y
∆y
∆x+ lim
∆x→0
∆y→0
f(x,y +∆y)−f(x,y)
∆y
∆y
Example 10.4.1. Find the total differential of f(x,y) = ye x +2sinx
Solution:
dz = f x (x,y)dx+f y (x,y)dy
= (ye x +2cosx)dx+(e x )dy
Theorem 10.4.2. Chain Rule
Ifz = f(x(t),y(t)), wherex(t)andy(t)aredifferentiableandf(x,y)isadifferentiable
function of x and y, then
dz
dt = ∂f dx
∂x dt + ∂f dy
∂y
z = f(x,y)
✡
❏❏
✡ ❏
∂f
✡ ❏
∂y
✡ ❏
✡ ❏
x y
∂f
∂x
dt . t t
dx
dt
dy
dt
Proof:
dz =f x (x,y)dx+f y (x,y)dy
dz
dt =f x(x,y) dx
dt +f y(x,y) dy
dt .
De-Yu Wang CSIE CYUT 177

10.4. THE CHAIN RULE
Example 10.4.2. Use the chain rule to find the derivative g ′ (t) where
g(t) = f(x(t),y(t)), f(x,y) = x 2 y +siny, x(t) = t 3 +1, y(t) = e 2t .
Solution:
g ′ (t) = ∂f dx
∂x dt + ∂f dy
∂y dt
= 2xy ·3t 2 +(x 2 +cosy)·2e 2t
= 2(t 3 +1)e 2t ·3t 2 + [ (t 3 +1) 2 +cos(e 2t 2t
)
]·2e
Theorem 10.4.3. Chain Rule
If z = f(x(s,t),y(s,t)), where f(x,y) is a differentiable function of x and y and
where x = x(s,t) and y = y(s,t) both have first-order partial derivatives. Then
∂z
∂s = ∂f ∂x
∂x ∂s + ∂f ∂y
∂y ∂s
∂z
∂t = ∂f ∂x
∂x ∂t + ∂f ∂y
∂y ∂t
Comments:
dz =f x (x,y)dx+f y (x,y)dy
dz
dt =f x(x,y) dx
dt +f y(x,y) dy
dt
∂z
∂t =f x(x,y) ∂x
∂t +f y(x,y) ∂y
∂t
∂z
∂s =f x(x,y) ∂x
∂s +f y(x,y) ∂y
∂s .
z = f(x,y)
✡
❏❏
✡ ❏
∂f ∂f
✡ ❏
∂x ∂y
✡ ❏
✡ ❏
x
y
✁❆
✁❆
∂x ✁ ❆ ∂x ∂y ✁ ❆ ∂y
∂s ✁ ❆ ∂t ∂s ✁ ❆ ∂t
✁ ❆ ✁ ❆
✁ ❆ ✁ ❆
s t s t
Example 10.4.3. Use the chain rule to find the derivative ∂g ∂g
and where
∂u ∂v
g(u,v) = f(x(u,v),y(u,v)), f(x,y) = 4x 2 y 3 , x(u,v) = u 3 −vsinu, y(u,v) = 4u 2 .
Solution:
∂g
∂u = ∂f ∂x
∂x∂u + ∂f ∂y
∂y ∂u
= (8xy 3 )(3u 2 −vcosu)+(12x 2 y 2 )(8u)
= [8(u 3 −vsinu)(4u 2 ) 3 ](3u 2 −vcosu)+[12(u 3 −vsinu) 2 (4u 2 ) 2 ](8u)
∂g
∂v = ∂f ∂x
∂x∂v + ∂f ∂y
∂y ∂v
= (8xy 3 )(−sinu)+(12x 2 y 2 )(0)
= −sinu[8(u 3 −vsinu)(4u 2 ) 3 ]
De-Yu Wang CSIE CYUT 178

10.4. THE CHAIN RULE
Theorem 10.4.4. Implicit differentiation
Suppose that the equation F(x,y,z) = 0 implicitly defines a function of z = f(x,y),
then
∂z
∂x = −F x
F z
,
∂z
∂y = −F y
F z
.
Proof: Let w = F(x,y,z) = 0 and by the chain rule,
∂w
∂x = F ∂x
x
∂x +F ∂y
y
∂x +F ∂z
z
∂x
∂z
0 = F x +F z
∂x
∂z
∂x = −F x
for F z ≠ 0;
F z
∂w
∂y = F ∂x
x
∂y +F ∂y
y
∂y +F ∂z
z
∂y
∂z
0 = F y +F z
∂y
∂z
∂y = −F y
for F z ≠ 0;
F z
Example 10.4.4. ForF(x,y,z) = xy 2 +z 3 +sin(xyz) = 0, useimplicitdifferentiation
to find ∂z ∂z
and . ∂x ∂y
Solution:
∂z
∂x = −F x
= − y2 +yzcos(xyz)
F z 3z 2 +xycos(xyz) ;
∂z
∂y = −F y 2xy +xzcos(xyz)
= −
F z 3z 2 +xycos(xyz) .
Exercise 10.4.1. Find the total differential of f(x,y)
(a) f(x,y) = ye x +sinx
(b) f(x,y) = √ x+y
(c) f(x,y) = x 2 y 2 +3y −4x,
(d) f(x,y) = x+3
y
Exercise 10.4.2. Use the chain rule to find the indicated derivative(s)
(a) g ′ (t) where g(t) = f(x(t),y(t)), f(x,y) = e x2 y +siny, x(t) = t 3 +1, y(t) = e 2t
(b) g ′ (t)whereg(t) = f(x(t),y(t)), f(x,y) = lnxy 2 +sin(xy), x(t) = 2t 3 +t, y(t) =
e 2t
(c) ∂g ∂g
and where g(u,v) = f(x(u,v),y(u,v)), f(x,y) = ∂u ∂v 4x2 y 3 , x(u,v) = u 3 −
vsinu, y(u,v) = 4u 2 .
(d) ∂g
∂u
∂g
and where g(u,v) = f(x(u,v),y(u,v)), f(x,y) = ∂v xy3 − 4x 2 , x(u,v) =
e u2 , y(u,v) = √ v 2 +1sinu
Exercise 10.4.3. Use implicit differentiation to find ∂z
∂x
(a) F(x,y,z) = xyz −4y 2 z 2 +cos(xy) = 0
(b) F(x,y,z) = 3y 2 z −e 4x cos(3z)−3y 2 = 0.
and
∂z
∂y .
De-Yu Wang CSIE CYUT 179

10.5. THE GRADIENT AND DIRECTIONAL DERIVATIVES
10.5 The Gradient and Directional Derivatives
Definition 10.5.1. The directional derivative of f(x,y) at the point (a,b) and in the
direction of the unit vector u = 〈u 1 ,u 2 〉 is given by
f(a+hu 1 ,b+hu 2 )−f(a,b)
D u f(a,b) = lim
,
h→0 h
provided the limit exists.
Theorem 10.5.1. Suppose that f is differentiable at (a,b) and u = 〈u 1 ,u 2 〉 is any
unit vector. Then,
D u f(a,b) = f x (a,b)u 1 +f y (a,b)u 2 .
Proof: Let g(h) = f(a+hu 1 ,b+hu 2 ) = f(x,y), where x = a+hu 1 and y = b+hu 2 .
Then g(0) = f(a,b) and so
g ′ (h) = ∂g
∂h = ∂f dx
∂xdh + ∂f dy
∂y dh = f x(x,y)u 1 +f y (x,y)u 2
g ′ (0) = f x (a,b)u 1 +f y (a,b)u 2
f(a+hu 1 ,b+hu 2 )−f(a,b) g(h)−g(0)
D u f(a,b) = lim
= lim
h→0 h
h→0 h
= g ′ (0) = f x (a,b)u 1 +f y (a,b)u 2 .
Example 〈 10.5.1. For f(x,y) = x 2 y −4y 3 , compute D u f(2,1) for the directions (a)
1
u = , √ 〉
3
and (b) u in the direction from (2,1) to (4,0).
2 2
Solution:
f x (x,y) = 2xy,
f y (x,y) = x 2 −12y 2
〈
1
(a) For u = , √ 〉
3
2 2
D u f(a,b) = f x (2,1)u 1 +f y (2,1)u 2
= 4· 1
2 −8· √
3
2 = 2−4√ 3.
(b) For u in the direction from (2,1) to (4,0).
u =
〈4−2, 0−1〉
‖〈4−2, 0−1〉‖ = 〈2,−1〉 〈 2
‖〈2,−1〉‖ = √5 ,−√ 1 〉
5
D u f(a,b) = f x (2,1)u 1 +f y (2,1)u 2
2
= 4· √ −8·
(− 1 )
√ = √ 16 . 5 5 5
De-Yu Wang CSIE CYUT 180

10.5. THE GRADIENT AND DIRECTIONAL DERIVATIVES
Definition 10.5.2. The gradient of f(x,y) is the vector-valued function
∇f(x,y) =
〈 ∂f
∂x , ∂f 〉
∂y
= ∂f ∂f
i+
∂x ∂y j
provided both partial derivatives exist.
Example 10.5.2. Find the gradient of f(x,y) = x 2 +4xy 2 −y 5 .
Solution:
▽f(x,y) = 〈 2x+4y 2 , 8xy −5y 4〉
Theorem 10.5.2. If f is differentiable function of x and y and u is any unit vector,
then
Proof:
D u f(x,y) = ∇f(x,y)·u.
D u f(x,y) =f x (x,y)u 1 +f y (x,y)u 2
〈 ∂f
=
∂x , ∂f 〉
·〈u 1 ,u 2 〉
∂y
=∇f(x,y)·u.
Example 10.5.3. Usingthegradientoff(x,y) = x 2 y−4y 2 tocomputethedirectional
derivative of f(x,y) at the point (2,−1) in the direction of the vector 〈1, 2〉.
Solution:
∇f(x,y) = 〈 2xy, x 2 −8y 〉
∇f(2,−1) = 〈−4, 12〉
u =
〈1, 2〉
√
5
=
〈 1√5
,
D u f(2,−1) = 〈−4, 12〉·〈 1√5
,
〉
2
√
5
〉
2
√ = √ 20
5 5
De-Yu Wang CSIE CYUT 181

10.6. EXTREMA OF FUNCTIONS OF SEVERAL VARIABLES
Exercise 10.5.1. Compute the directional derivative of f at the given point in the
direction of the indicated vector.
〈
(a) f(x,y) = x 2 y +4y 2 1
, (2,1), u = , √ 〉
3
2 2
〈 〉
(b) f(x,y) = x 3 y −4y 2 1
, (2,−1), u = √ 1
2
, √2
(c) f(x,y) = √ x 2 +y 2 , (3,−4), u in the direction of 〈3,−2〉
(d) f(x,y) = cos(2x−y), (π,0), u in the direction from (π,0) to (2π,π)
Exercise 10.5.2. Find the gradient of the given function at the indicated point.
(a) f(x,y) = x 3 e 3y −y 4 , (2,−1)
(b) f(x,y) = sin3xy +y 2 , (π,1)
(c) f(x,y) = e 3y/x −x 2 y 4 , (1,2)
(d) f(x,y) = xe xy2 +cosy 2 , (1,−1)
Exercise 10.5.3. Repeat Exercise 10.5.1 using Theorem 10.5.2.
10.6 Extrema of Functions of Several Variables
Definition 10.6.1. Absolute extremum (maximum or minimum)
Let f be a function with domain D. Then
(a) f(a,b) is an absolute maximum of f if
f(x,y) ≤ f(a,b)
for all (x,y) in D
(b) f(a,b) is an absolute minimum of f if
f(x,y) ≥ f(a,b)
Definition 10.6.2. Local extremum
for all (x,y) in D
(a) f(a,b) is a local maximum of f if there is an open disk R centered at (a,b), for
which f(a,b) ≥ f(x,y) for all (x,y) ∈ R.
(b) f(c) is a local minimum of f if there is an open disk R centered at (a,b), for
which f(a,b) ≤ f(x,y) for all (x,y) ∈ R.
z
✻ absolute maximum
✘✾ local maximum
✘✘
O
✑
❳ ❳❳❳
✑ absolute ❳ ❳❳❳
minimum
✻ ✑
✑ ✘✾✘ ✘✘ ✘
❳ ❳3 y
✑
local ✑
x
✑✰ ✑ minimum
De-Yu Wang CSIE CYUT 182

10.6. EXTREMA OF FUNCTIONS OF SEVERAL VARIABLES
Definition 10.6.3. The point (a,b) is a critical number of the function f(x,y) if
(a,b) is in the domain of f and one of the following is true.
(a) f x (a,b) = f y (a,b) = 0
(b) f x (a,b) or f y (a,b) does not exist.
Theorem 10.6.1. If f(x,y) has a local extremum at (a,b), then (a,b) must be a
critical number of f.
Proof: Suppose that f has a local extremum at (a,b).
(a) Holding y = b, then g(x) = f(x,b) has a local extremum at x = a. By the
Fermat’s theorem (page 44) either g ′ (a) = f x (a,b) = 0 or g ′ (a) does not exist.
(b) Holding x = a, then h(y) = f(a,y) has a local extremum at y = b. It follows
that h ′ (b) = f y (a,b) = 0 or h ′ (b) does not exist.
}
f x (a,b) = g ′ (a) = 0 or does not exist
f y (a,b) = h ′ ⇒ (a,b) must be a critical number of f.
(b) = 0 or does not exist
Example 10.6.1. Find all critical numbers of f(x,y) = xe −x2 2 −y3 3 +y .
Solution:
f x (x,y) = e −x2 2 −y3 3 +y +x(−x)e −x2 2 −y3 3 +y = (1−x 2 )e −x2 2 −y3 3 +y = 0
⇒ x = ±1
f y (x,y) = x(−y 2 +1)e −x2 2 −y3 3 +y = 0
⇒ x = 0 or y = ±1.
The critical numbers are (1,1), (−1,1), (1,−1) and (−1,−1).
Theorem 10.6.2. Second Derivatives Test
Supposethatf(x,y)hascontinuoussecond-orderpartialderivativesinsomeopendisk
containing the point (a,b) and that f x (a,b) = f y (a,b) = 0. Define the discriminant
D(a,b) = f xx (a,b)f yy (a,b)−[f xy (a,b)] 2 f xx (a,b) f xy (a,b)
=
∣f yx (a,b) f yy (a,b) ∣ .
(a) If D(a,b) > 0 and f xx (a,b) > 0, then f has a local minimum at (a,b).
(b) If D(a,b) > 0 and f xx (a,b) < 0, then f has a local maximum at (a,b).
(c) If D(a,b) < 0, then f has a saddle point at (a,b).
(d) If D(a,b) = 0, then no conclusion.
De-Yu Wang CSIE CYUT 183

10.6. EXTREMA OF FUNCTIONS OF SEVERAL VARIABLES
Comments:
{
f xx (a,b) > 0 and f yy (a,b) > 0 a local minimum at (a,b)
D(a,b) > 0
f xx (a,b) < 0 and f yy (a,b) < 0 a local maximum at (a,b)
{
f xx (a,b) > 0 and f yy (a,b) < 0 a saddle point at (a,b)
D(a,b) < 0
f xx (a,b) < 0 and f yy (a,b) > 0 a saddle point at (a,b)
z
✻
4
saddle point
3 ✡
✡
2 ✡
✡
✡✢
1
O
1 ✑
❳ ❳ 1
❳
✑ ❳❳❳ 2 3
2✑
❳ 4 ❳❳ 5
3 ✑
❳3 y
4 ✑
✑
x
✑✰ ✑
Example 10.6.2. Locateandclassifyallcriticalnumbersforf(x,y) = 2x 2 −y 3 −2xy.
Solution:
(a) The critical numbers
f x = 4x−2y = 0, ⇒ y = 2x
f y = −3y 2 −2x = 0
= −12x 2 −2x = −2x(6x+1) = 0
⇒ x = 0 or x = − 1 6
f have two critical numbers(0,0) and
(
− 1 )
6 ,−1 .
3
De-Yu Wang CSIE CYUT 184

10.6. EXTREMA OF FUNCTIONS OF SEVERAL VARIABLES
(b) Second derivatives test:
(
D − 1 6 ,−1 3
f xx = 4, f yy = −6y, f xy = −2
D(0,0) = f xx (0,0)f yy (0,0)−[f xy (0,0)] 2
= 4·0−(−2) 2 = −4 < 0,
∴ (0,0) is a saddle point.
) (
= f xx − 1 6 ,−1 3
)f yy
(
− 1 6 ,−1 3
= 4·2−(−2) 2 = 4 > 0
(
and f xx − 1 )
6 ,−1 = 4 > 0,
3
(
∴ − 1 )
6 ,−1 is a local minimum.
3
)
−
( [f xy − 1 )] 2
6 ,−1 3
Exercise 10.6.1. Locateandclassify(usesecondderivativestest)allcriticalnumbers
(a) f(x,y) = e −x2 −y 2
(b) f(x,y) = e −x2 (y 2 +1)
(c) f(x,y) = x 3 −3xy +y 3
(d) f(x,y) = y 2 +x 2 y +x 2 −2y
De-Yu Wang CSIE CYUT 185

Chapter 11
MULTIPLE INTEGRALS

11.1. DOUBLE INTEGRALS
11.1 Double Integrals
Definition 11.1.1. Double integrals over rectangles
For any function f(x,y) defined on a rectangle R = {(x,y)|a ≤ x ≤ b and c ≤ y ≤ d},
we define the double integral of f over R by
∫∫
n∑
f(x,y)dA = lim f(u i ,v i )∆A i ,
‖∆‖→0
R
i=1
provided the limit exists and is the same for every choice of the evaluation points
(u i ,v i ) in R i , for i = 1,2,··· ,n. When this happens, we say that f is integrable over
R. z
✻
y
z = f(x,y) ✻
d
R i
O
✑
❳❳ ❳❳❳ c
a ✑ ❳
✑ ❳❳❳ d c
✑
❳3
b ✑ ✑ ✑❳ ❳❳❳ y
✑
✑ R
✑
✑✰
✑❳❳❳❳❳ ✑ ✑✑✑❳ O a
x ∫∫
∫∫
Note. Area of R: A = 1dA and Volume V = f(x,y)dA.
R
R
✲x
b
Theorem 11.1.1. Fubini’s Theorem (First Form)
Supposethatf isintegrableover therectangleR = {(x,y)|a ≤ x ≤ b and c ≤ y ≤ d}.
Then,
∫∫ ∫ b ∫ d ∫ d ∫ b
f(x,y)dA = f(x,y)dydx = f(x,y)dxdy
R
a
c
Proof for dA = dydx: For each fixed value of x, the area of the cross section is the
area under the curve z = f(x,y) for c ≤ y ≤ d, which is given by
c
a
∫ d
A(x) = f(x,y)dy
c
∫∫
V = f(x,y)dA
=
=
R
∫ b
a
∫ b ∫ d
a
A(x)dx
f(x,y)dy dx.
c
} {{ }
A(x)
z
✻
z = f(x,y)
O
✑
❳❳ ❳❳❳ c
a ✑ A(x)
❳
✑ ❳❳❳ d
✑
❳3
b ✑ ✑ ✑❳ ❳❳❳ y
✑
✑ R
✑
✑✰
✑❳❳❳❳❳ ✑ ✑✑✑❳
x
De-Yu Wang CSIE CYUT 187

11.1. DOUBLE INTEGRALS
Proof for dA = dxdy: For each fixed value of y, the area of the cross section is the
area under the curve z = f(x,y) for a ≤ x ≤ b, which is given by
∫ b
A(y) = f(x,y)dx
a
∫∫
V = f(x,y)dA
=
=
R
∫ d
c
∫ d ∫ b
c
A(y)dy
f(x,y)dx dy.
a
} {{ }
A(y)
z
✻
z = f(x,y)
O
✑
❳❳ ❳❳❳ c
a ✑ A(y) ❳
✑ ❳❳ d
✑
❳ ❳3
b ✑ ✑ ✑❳ ❳❳❳ y
✑
✑ R
✑
✑✰
✑❳❳❳❳❳ ✑ ✑✑✑❳
x
Example 11.1.1. Evaluate
y ≤ 4}.
∫∫
R
(6x 2 +4xy 3 )dA, where R = {(x,y)|0 ≤ x ≤ 2, 1 ≤
Solution:
∫∫
(6x 2 +4xy 3 )dA =
R
=
=
∫ 4 ∫ 2
1
∫ 4
1
∫ 4
1
0
(6x 2 +4xy 3 )dxdy
(
2x 3 +2x 2 y 3)∣ ∣ 2 x=0 dy
(
16+8y
3 ) dy
= (16y +2y 4 ) ∣ ∣ 4 y=1 = 558.
Definition 11.1.2. Double integrals over general regions
For any function f(x,y) defined on a bounded region R ⊂ R 2 , we define the double
integral of f over R by
∫∫
R
f(x,y)dA = lim
‖∆‖→0
n∑
f(u i ,v i )∆A i ,
i=1
provided the limit exists and is the same for every choice of the evaluation points
(u i ,v i ) in R i , for i = 1,2,··· ,n. When this happens, we say that f is integrable over
De-Yu Wang CSIE CYUT 188

11.1. DOUBLE INTEGRALS
R.
z
✻
z = f(x,y)
y
✻
d
O
✑
❳ ❳❳
✑ ❳
c
a ❳❳❳
✑ ❳
d
✑
❳❳3
b
y
✑
✑ R
✑
✑✰
x
c
O
a
R
b
✲x
Theorem 11.1.2. Fubini’s Theorem (Stronger Form)
Suppose that f(x,y) is continuous on a region R ⊂ R 2 .
(a) If R is defined by R = {(x,y)|a ≤ x ≤ b and g 1 (x) ≤ y ≤ g 2 (x)}, with g 1 and
g 2 continuous on [a,b], then
∫∫ ∫ b ∫ g2 (x)
f(x,y)dA = f(x,y)dydx.
R
a
g 1 (x)
(b) If R is defined by R = {(x,y)|c ≤ y ≤ d and h 1 (y) ≤ x ≤ h 2 (y)}, with h 1 and
h 2 continuous on [c,d], then
∫∫ ∫ d ∫ h2 (y)
f(x,y)dA = f(x,y)dxdy.
R
c
h 1 (y)
Proof for (a): For each fixed value of x, the area of the cross section is the area under
the curve z = f(x,y) for g 1 (x) ≤ y ≤ g 2 (x), which is given by
A(x) =
∫ g2 (x)
g 1 (x)
∫∫
V =
R
f(x,y)dy
f(x,y)dA =
z
✻
∫ b
a
z = f(x,y)
A(x)dx =
∫ b ∫ g2 (x)
a
g 1 (x)
f(x,y)dy dx.
} {{ }
A(x)
y
✻
y = g 2 (x)
O
✑
❳❳ ❳❳❳ c
a ✑ A(x)
❳
✑ ❳❳❳ d
✑
❳3
b
y
✑
✑ R
✑
✑✰
x
O
a
y = g 1 (x)
b
✲x
De-Yu Wang CSIE CYUT 189

11.1. DOUBLE INTEGRALS
Proof for (b): For each fixed value of y, the area of the cross section is the area under
the curve z = f(x,y) for h 1 (y) ≤ x ≤ h 2 (y), which is given by
A(y) =
∫ h2 (y)
∫∫
V =
h 1 (y)
R
f(x,y)dx
f(x,y)dA =
z
✻
∫ h2 (y)
h 1 (y)
z = f(x,y)
A(y)dy =
∫ d ∫ h2 (y)
c
y
✻
d
h 1 (y)
f(x,y)dx dy.
} {{ }
A(y)
O
✑
❳ ❳❳❳ c
a ✑ A(y) ❳ ❳❳❳ d
✑
✑
❳ ❳3
b
y
✑ R
✑
✑
✑✰
x
R
x = h 1 (y) x = h 2 (y)
c
O
✲x
Example ∫∫ 11.1.2. LetRbetheregionboundedbythegraphsofx = y 2 andx = 2−y.
Write f(x,y)dA as the iterated integrals with dA = dxdy and dA = dydx.
Solution:
∫∫
R
R
y 2 +y −2 = 0
f(x,y)dA =
x = y 2 = 2−y
y = 2, −1
=
∫ 1 ∫ 2−y
−2 y
∫ 2
1 ∫ √ x
0
− √ x
f(x,y)dxdy
f(x,y)dydx+
y
3 ✻
2
❅
1 ❅
❅
❅
−2 −1 1 2❅
3 4
−1 R
−2
∫ 4 ∫ 2−x
1
− √ x
❅
❅
❅
❅
f(x,y)dydx
x = y 2
✲x
x = 2−y
Example 11.1.3. Let R be the region bounded by the graphs of y = √ ∫∫
x, x = 0 and
y = 3. Evaluate (2xy 2 +2ysinx)dA.
R
De-Yu Wang CSIE CYUT 190

11.1. DOUBLE INTEGRALS
Solution:
y
4 ✻
3
2
1
R
y = 3
y = √ x
✲x
−3 3 6 9 12 15
∫∫
−1
∫ 3 ∫ y 2
(2xy 2 +2ysinx)dA = (2xy 2 +2ysinx)dxdy
R
0 0
∫ 3
= (x 2 y 2 −2ycosx) ∣ y2
dy x=0
0
∫ 3
= [(y 6 −2ycosy 2 )−(0−2ycos0)]dy
0
∫ 3
= (y 6 −2ycosy 2 +2y)dy
0
( )∣ y
7 ∣∣∣
3
=
7 −siny2 +y 2 = 37
y=0
7 −sin9+9
∫∫
∫ 9 ∫ 3
(2xy 2 +2ysinx)dA = (2xy 2 +2ysinx)dydx
√
R
0 x
Exercise 11.1.1. Evaluate the double integral
(a)
(b)
(c)
(d)
∫ 2 ∫ 1
−2 0
∫ 2 ∫ 1
−1
∫∫
∫∫
R
R
0
(6−2x−3y)dydx
(6−2x 2 +y 2 )dxdy
(2xy +y 2 )dA, where R = {(x,y)|0 ≤ x ≤ 2, 0 ≤ y ≤ 1}
4xe 2y dA, where R = {(x,y)|1 ≤ x ≤ 4, 0 ≤ y ≤ 1}
Exercise 11.1.2. Evaluate the double integral
(a)
(b)
∫ 2 ∫ x 2
0 0
∫ 2 ∫ y
0 0
(2x+1)dydx
3e y dxdy
De-Yu Wang CSIE CYUT 191

11.2. DOUBLE INTEGRALS IN POLAR COORDINATES
(c)
(d)
∫∫
(2x−y)dA, where R is bound by y = −3 and y = 1−x 2
∫∫
R
R
e x2 dA, where R is bound by y = x 2 and y = 1
11.2 Double Integrals in Polar Coordinates
Theorem 11.2.1. Fubini’s Theorem
Suppose that f(r,θ) is continuous on a region R = {(r,θ)|α ≤ θ ≤ β and g 1 (θ) ≤ r ≤
g 2 (θ)}. Then
Proof:
∫∫
R
f(r,θ)dA =
∫ β ∫ g2 (θ)
α
g 1 (θ)
f(r,θ)rdrdθ.
y
θ = β
✻
✂ ✁ ✓ ✂ ✁ ✓ ✂ ✁ ✓
✂ ✁ ✓ R
✟
✂ ✁ ✓
θ = α
✂✁✓
✂✁✓
r = g 2 (θ)
✂✁✓
✚✓
✟ ✁✂
✏ ✚✚✚✚✚✚✚✚✚✚ ✟✟✟✟✟✟✟✟✟✟ ✏✏✏✏✏✏✏✏✏✏✏ r = g 1 (θ)
✲
x
∆A = πr 2 2 · ∆θ
2π −πr2 1 · ∆θ
2π = 1 2 (r2 2 −r 2 1)∆θ
= 1 2 (r 2 +r 1 )(r 2 −r 1 )∆θ
= r∆r∆θ
where r = 1 2 (r 2 +r 1 ), ∆r = r 2 −r 1 .
V = lim
n→∞
n∑
= lim
=
i=1
n∑
n→∞
i=1
∫ β ∫ g2 (θ)
α
g 1 (θ)
V i = lim
n→∞
n∑
i=1
f(r i ,θ i )r i ∆r i ∆θ i
f(r,θ)rdrdθ
f(r i ,θ i )
} {{ }
height
y
✻
∆r
θ = θ 2
r = r 2
∆A
θ = θ 1
r = r 1
∆θ
❦
✧✔ ✔✔✔✔✔✔✔✔✔✔✔ ✧✧✧✧✧✧✧✧✧✧✧
∆A i }{{}
area of base
✲
x
De-Yu Wang CSIE CYUT 192

11.2. DOUBLE INTEGRALS IN POLAR COORDINATES
Example 11.2.1. Find the area of the region bounded by the curves r = 2−2sinθ.
Solution:
∫∫
A = dA =
=
R
∫ 2π
r 2
= 1 2
0
∫ 2π
0
∫ 2π
2 ∣
∫ 2π ∫ 2−2sinθ
0
2−2sinθ
r=0
0
dθ
[(2−2sinθ) 2 −0]dθ
rdrdθ
= (2−4sinθ +2sin 2 θ)dθ
0
∫ 2π
= [2−4sinθ+(1−cos2θ)]dθ
0
=
(3θ +4cosθ− 1 )∣ ∣∣∣
2π
2 sin2θ = 6π
0
−3
−2
y
1
−1
−1
−2
−3
−4
1 2 3
R
x
Example 11.2.2. Evaluate the iterated integral by converting to polar coordinates
∫ 2 ∫ √ 4−x 2
−2
− √ 4−x 2 √
x2 +y 2 dydx.
Solution: R = {(x,y)| − 2 ≤ x ≤ 2 and − √ 4−x 2 ≤ y ≤ √ 4−x 2 } is a circle of
radius centered at the origin.
∫ 2 ∫ √ 4−x 2
−2
− √ 4−x 2 √
x2 +y 2 dydx =
=
=
∫ 2π ∫ 2
0
∫ 2π
0
∫ 2π
0
0
r 3
3
∣
√
r2 rdrdθ
2
r=0
dθ
8 16π
dθ =
3 3
R
y
✻
r = 2
✟ ✟✟✟ ▼θ
✲
2 x
Exercise 11.2.1. Find the area of the region bounded by the curves
(a) r = 3+2sinθ
(b) r = 2−2cosθ
(c) r = 2sinθ
(d) r = 3cosθ
De-Yu Wang CSIE CYUT 193

11.3. SURFACE AREA
Exercise 11.2.2. Evaluate the iterated integral by converting to polar coordinates
(a)
(b)
∫ 2 ∫ √ 4−x 2
−2 0
∫ 2 ∫ √ 4−x 2
−2
√
x2 +y 2 dydx
− √ 4−x 2 e −x2 −y 2 dydx
(c)
(d)
∫ 2 ∫ 0
− √ 4−x 2 ydydx
0
∫ 2 ∫ √ 4−x 2
0
− √ 4−x 2 e −x2 −y 2 dydx
11.3 Surface Area
Theorem 11.3.1. Surface Area
If f and its first partial derivatives are continuous on the closed region R in the
xy-plane, then the area of the surface S given by z = f(x,y) over R is given by
∫∫ √
S = [f x (x,y)] 2 +[f y (x,y)] 2 +1dA.
Note.
R
Length on x-axis:
Arc length in xy-plane:
Area in xy-plane:
Surface area in space:
∫ b
a
∫ b
∫∫
a
∫∫
R
R
dx = b−a
ds =
∫ b
a
dA = A
∫∫
dS =
√
1+[f′ (x)] 2 dx
R
√
[f x (x,y)] 2 +[f y (x,y)] 2 +1dA
Proof: Let T i be the portion of the tangent plane lying above R i . To find the area of
the parallelogram ∆T i note that its sides are given by the vectors
a i =< ∆x i ,0,f x (x i ,y i )∆x i >,
b i =< 0,∆y i ,f y (x i ,y i )∆y i >,
∣ i j k ∣∣∣∣∣
a i ×b i =
∆x i 0 f x (x i ,y i )∆x i
∣ 0 ∆y i f y (x i ,y i )∆y i
=< −f x (x i ,y i )∆x i ∆y i ,−f y (x i ,y i )∆x i ∆y i ,∆x i ∆y i >,
√
[f x (x i ,y i )] 2 +[f y (x i ,y i )] 2 +1 ∆x i ∆y } {{ } i
∆A i
n∑
∆T i = ‖a i ×b i ‖ =
S = lim
‖∆|→0
∫∫
=
R
n∑
i=1
∆T i = lim
‖∆‖→0
i=1
√
[f x (x,y)] 2 +[f y (x,y)] 2 +1dA.
√
[f x (x i ,y i )] 2 +[f y (x i ,y i )] 2 +1∆A i
De-Yu Wang CSIE CYUT 194

11.3. SURFACE AREA
Example 11.3.1. Find the surface area of that portion of the surface z = y 2 + 4x
lying above the triangular region R in the xy-plane with vertices at (0,0), (0,2) and
(2,2).
y
z
✻
✻ 3
4 ❤❤❤❤❤❤❤
3
z = y 2 +4x 2 (0,2)
(2,2)
2
✭✭✭✭✭✭✭
R
1
1
y = x
O
✑
❳
1 ❳❳❳ 1
2
✑ ❳ 3
2 ❳❳❳ 4
✑ R
✑
❳ 5
3 ✑ ❳3
✑
✑✑✑
y
✲x
4✑
(0,0)
✑✰
x
✑
1 2 3
Solution:
∫∫
S =
∫∫
=
R
R
∫ 2
√
[f x (x,y)] 2 +[f y (x,y)] 2 +1 2 dA
√
42 +4y 2 +1dA =
√ ∣
= 4y2 +17x
0
= 1 8 (4y2 +17) 3 2
2 · 3∣
∣ y x=0
2
y=0
dy =
= 1
12
∫ 2 ∫ y
0 0
∫ 2
(
0
√
4y2 +17dxdy
√
4y2 +17ydy
)
33 3 3
2 −17 2
Example 11.3.2. Find the surface area of that portion of the paraboloid z = 1 +
x 2 +y 2 that lies below the plane z = 5.
✘✘✘✘✘
z
✻
8
6
4
z = 1+x 2 +y 2
✘✘✘✘✘✘✘
z = 5
2
O
1 ✑
❳❳ 1 ❳❳❳ 2
✑ 3
2 ❳
✑ ❳❳❳ 4 5
3 ✑
❳3 y
4 ✑
✑
x
✑✰ ✑
R
y
✻
r = 2
✟ ✟✟✟ ▼ θ
✲
2 x
De-Yu Wang CSIE CYUT 195

11.4. TRIPLE INTEGRALS
Solution:
z = 5 = 1+x 2 +y 2 , ⇒ x 2 +y 2 = 4
∫∫ √
S = [f x (x,y)] 2 +[f y (x,y)] 2 +1dA
R
∫∫
√
= 4x2 +4y 2 +1dA
=
= 1 8
R
∫ 2π ∫ 2
= 1
12
0 0
∫ 2π
0
∫ 2π
0
= π 6 (173 2 −1)
√
4r2 +1rdrdθ
2
3 (4r2 +1) 3 2
∣
(17 3 2 −1
3
2 )dθ
r=2
r=0
dθ
Exercise 11.3.1. Evaluate or estimate the surface area.
(a) The portion of z = √ x 2 +y 2 between y = x 2 and y = 4.
(b) The portion of x+3y +z = 6 in the first octant.
(c) The portion of z = 2y +3x between y = 2x, y = 0 and x = 3.
(d) The portion of the surface z = 3x + y lying above the triangular region R in
the xy-plane with vertices at (0,0), (2,2) and (2,0).
11.4 Triple Integrals
Definition 11.4.1. Triple integrals over rectangles
For any function f(x,y,z) defined on a rectangle box Q = {(x,y,z)|a ≤ x ≤ b, c ≤
y ≤ d and r ≤ z ≤ s}, we define the triple integral of f over Q by
∫∫∫
Q
f(x,y,z)dV = lim
‖∆‖→0
n∑
f(u i ,v i ,w i )∆V i ,
i=1
provided the limit exists and is the same for every choice of the evaluation points
(u i ,v i ,w i ) in Q i , for i = 1,2,··· ,n. When this happens, we say that f is integrable
De-Yu Wang CSIE CYUT 196

11.4. TRIPLE INTEGRALS
over Q.
z
✻
4 Q
3
❳❳❳❳❳❳❳❳
❳❳❳❳❳❳❳❳
❳❳❳❳❳❳❳❳
2
✑ ✑✑✑✑✑
✑
❳❳❳❳❳❳❳❳
❳❳❳❳❳❳❳❳
✑✑✑✑✑
✑
❳❳❳❳❳❳❳❳
✑✑✑✑✑
✑ ❳❳❳❳❳❳❳❳✑ ✑✑✑✑✑
✑
❳❳❳❳❳❳❳❳✑ ✑✑✑✑✑
✑ ✑✑✑✑✑
1
✑✑✑✑✑
❳❳❳❳❳❳❳❳✑ ✑✑✑✑✑
❳❳❳❳❳❳❳❳✑ O
1 ✑
❳ 1 ✑✑✑✑✑
2 ❳❳❳❳❳❳❳❳✑
✑ ❳ 3
❳❳❳❳❳❳❳❳✑ ✑✑✑✑✑
2 ❳❳❳ 4
✑
✑✑✑✑✑
✑
❳ 5
3
❳3 y
4 ✑
✑
❳❳❳❳❳❳❳❳✑ ✑✑✑✑✑
✑✑✑✑✑
✑✰
x
✑
x
∫∫∫
Note. The volume of box Q: V = 1dV.
Q
z
✻
4 Q
✑
✑
✑3
❳❳❳❳❳❳❳ ❳
✑
✑
2 ✑❳❳❳❳❳❳❳❳✑
1
✑✑✑✑✑
O
1 ✑
❳❳❳ 1
❳ 2
✑ ❳ 3
2 ❳ ❳❳❳ 4
✑
✑
❳ 5
3 ❳ ❳3 y
4 ✑
✑
❳❳❳❳❳❳❳❳✑ ✑✑✑✑✑
✑✑✰ ∆V i = ∆z i ∆y i ∆x i
Theorem 11.4.1. Fubini’s Theorem (First Form)
Suppose that f(x,y,z) is continuous on the box Q = {(x,y,z)|a ≤ x ≤ b, c ≤ y ≤
d and r ≤ z ≤ s}. Then,
∫∫∫
Q
f(x,y,z)dV =
∫ s ∫ d ∫ b
or in any of the four remaining orders.
∫∫∫
Example 11.4.1. Evaluate
y ≤ 1 and 0 ≤ z ≤ π}.
Solution:
∫∫∫
Q
2xe y sinzdV =
r
=
=
= 3
c
a
Q
∫ π ∫ 1 ∫ 2
0 0
∫ π ∫ 1
0 0
∫ π ∫ 1
0 0
∫ π
0
f(x,y,z)dxdydz =
∫ b ∫ d ∫ s
a
c
r
f(x,y,z)dzdydx,
2xe y sinzdV, where Q = {(x,y,z)|1 ≤ x ≤ 2, 0 ≤
1
2xe y sinzdxdydz
e y sinz 2x2
2
∣
2
x=1
3e y sinzdydz
e y sinz| 1 y=0 dz
dydz
= 3(e 1 −1)(−cosz) ∣ ∣ π z=0 = 6(e−1).
De-Yu Wang CSIE CYUT 197

11.4. TRIPLE INTEGRALS
Definition 11.4.2. Triple integrals over general regions
For any function f(x,y,z) defined on a bounded solid Q, we define the triple integral
of f over Q by
∫∫∫
Q
f(x,y,z)dV = lim
‖∆‖→0
n∑
f(u i ,v i ,w i )∆V i ,
i=1
provided the limit exists and is the same for every choice of the evaluation points
(u i ,v i ,w i ) in Q i , for i = 1,2,··· ,n. When this happens, we say that f is integrable
over Q.
z
z
✻
✻
4
4
3
❳❳❳❳❳❳❳❳
❳❳❳❳❳❳❳❳
3
❳❳❳❳❳❳❳❳
2
✑ ✑✑✑✑✑
✑
❳❳❳❳❳❳❳❳
❳❳❳❳❳❳❳❳
✑✑✑✑✑
✑ ✑✑✑✑✑
✑ ❳❳❳❳❳❳❳❳✑ ✑✑✑✑✑
✑ ✑✑✑✑✑
✑ ✑✑✑✑✑
2
❳❳❳❳❳❳❳❳
❳❳❳❳❳❳❳❳✑
1
✑✑✑✑✑
Q
❳❳❳❳❳❳❳❳✑ ✑✑✑✑✑ 1
❳❳❳❳❳❳❳❳✑ O
1 ✑
❳ ❳ 1 ✑✑✑✑✑
Q
❳
✑ ❳❳❳ 2 ❳❳❳❳❳❳❳❳✑ 3
❳❳❳❳❳❳❳❳✑ ✑✑✑✑✑ O
2✑
✑✑✑✑✑ ❳ 4
1 ✑
❳ 1
❳ ❳
❳❳ 5
3 ✑
❳3 y
4 ✑
✑
❳❳❳❳❳❳❳❳✑ ✑✑✑✑✑
✑ ❳❳❳ 2 3
2 ❳
✑ ❳ 4 ❳❳ 5
✑✑✑✑✑ 3 ✑ ❳ ❳3 y
4 ✑
x
✑✰ ✑
✑
x
✑✑✰ ∆V i = ∆z i ∆y i ∆x i
Theorem 11.4.2. Fubini’s Theorem (Stronger Form)
Suppose that f(x,y,z) is continuous on a region Q = {(x,y,z)|(x,y) ∈
R and g 1 (x,y) ≤ z ≤ g 2 (x,y)}, with R is some region in the xy-plane, then
∫∫∫
Q
∫∫
f(x,y,z)dV =
∫∫∫
Example 11.4.2. Evaluate
0 and 2x+y +z = 4.
R
z
✻
4
✂❏
✂
3✂
✂
✂2
∫ g2 (x,y)
g 1 (x,y)
Q
❏
❏
❏
f(x,y,z)dzdA.
✂ ❏
✂ ❏
✂ 1 ❏
✂
✑
❏
✂ O ❳ ❳
✑
❏
✂
1 ❳ 1 2
✑
✂✑
❳❳❳❳❳❳
3
2 ✑✑✑❳ R ❏❳ 4 ❳❳❳3 5
3 ✑
y
4 ✑
✑
✑✰
x
✑
6xydV, where Q is bounded by x = 0, y = 0, z =
2x+y +z = 4
y
✻
4
❆
❆
❆
2 ❆
R
❆
2x+y = 4
❆
❆
❆
2 4
✲x
De-Yu Wang CSIE CYUT 198

11.5. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL
COORDINATES
Solution:
∫∫∫
Q
∫∫
f(x,y,z)dV =
∫∫
=
=
=
=
=
∫ g2 (x,y)
R g 1 (x,y)
∫ 4−2x−y
R 0
∫ 2 ∫ 4−2x
0
∫ 2
0
∫ 2
0
∫ 2
0
= 64
5 .
6
0
f(x,y,z)dzdA
6xydzdA
6xy(4−2x−y)dydx
)∣ ∣∣∣
(4x y2
y=4−2x
2 −2x2y2 2 −xy3 3
8x(2−x) 3 dx = −
∫ 0
8(2−u)u 3 du = 4u 4 − 8u5
5
2
y=0
dx
8(2−u)u 3 du
2
∣
u=0
(u = 2−x)
Exercise 11.4.1. Evaluate the triple integrals
∫∫∫
(a) (2x+y 2 )dV, where Q = {(x,y,z)|0 ≤ x ≤ 3,−2 ≤ y ≤ 1,1 ≤ z ≤ 2}.
Q
∫∫∫
(b) (2x−y)dV, where Q = {(x,y,z)|1 ≤ x ≤ 3,0 ≤ y ≤ 1,1 ≤ z ≤ 2}.
Q
∫∫∫
(c) (2x + 3y + z)dV, where Q is bounded by x = 0, y = 0, z = 0 and
Q
3x−y −z = 4
∫∫∫
(d) (x − 3y + z)dV, where Q is bounded by x = 0, y = 0, z = 0 and
Q
−3x+y −z = 3
11.5 Triple Integrals in Cylindrical and Spherical
Coordinates
Definition 11.5.1. Cylindrical Coordinates
Cylindrical coordinates represent a point P in space by (r,θ,z) in which
(a) (r,θ) is the polar coordinate for the vertical projection of P on the xy-plane.
(b) z is the rectangular vertical coordinate.
De-Yu Wang CSIE CYUT 199

11.5. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL
COORDINATES
The rectangular conversion equations for cylindrical coordinates:
x = rcosθ
y = rsinθ
z = z
z
✻
P(r,θ,z)
z
O
✑
❳ ❳❳❳
✑ ✿ ❳ ❳❳❳3
✑❳ ✑
✑ ✑ θ r ❳
✑✰ ❙
x y
x
y
Example 11.5.1. Write the equations (a) x 2 + y 2 = 16 and (b) z 2 = x 2 + y 2 in
cylindrical coordinates.
z
✻
z
✻
8
8
6
6
4 x 2 +y 2 = 16 4 z 2 = x 2 +y 2
2
❅2
O
1 ✑
❳ ❳❳ 1 ❅❅❅❅❅
✑ ❳ 2
O
❳❳ 3 1 ✑
❳ ❳❳ 1
2✑
❳ 4 ❳❳❳3 5 ✑ ❳ 2 ❳❳ 3
2✑
❳ 4 ❳❳❳3 5
3 ✑
y 3 ✑
y
4 ✑
✑
✑
4
✑✰
x
✑
✑
✑✰
x
✑
Solution: (a) x 2 +y 2 = 16, r 2 = 16, r = ±4.
(b) z 2 = x 2 +y 2 = r 2 ,
z = ±r.
Theorem 11.5.1. Triple integrals in cylindrical coordinates
Suppose that the solid Q as
Q = {(r,θ,z)|(r,θ) ∈ R and k 1 (r,θ) ≤ z ≤ k 2 (r,θ)},
where k 1 (r,θ) ≤ k 2 (r,θ), for all (r,θ) in the region R of the xy-plane defined by
R = {(r,θ)|α ≤ β and g 1 (θ) ≤ r ≤ g 2 (θ)}.
The triple integrals in cylindrical coordinates:
∫∫∫
Q
f(r,θ,z)dV =
∫ β ∫ g2 (θ) ∫ k2 (r,θ)
α
g 1 (θ)
k 1 (r,θ)
f(r,θ,z) dzrdrdθ } {{ }.
dV
De-Yu Wang CSIE CYUT 200

11.5. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL
COORDINATES
z
✻
O
✑
❳❳ ❍
✑ ❡ ❍❍❍❍❍❍ ❳❳❳
❳
✑ ❡❡❡❡❡ ✣ ❳❳3
✑ ∆θ y
✑✰
x
r
r∆θ
Example 11.5.2. Evaluate the triple integral
y 2 ) 3 2 dzdydx in cylindrical coordinates.
∫ 1 ∫ √ 1−x 2
−1
− √ 1−x 2 ∫ 2−x 2 −y 2
x 2 +y 2 (x 2 +
Solution:
∫ 1 ∫ √ 1−x 2
−1
− √ 1−x 2 ∫ 2−x 2 −y 2
x 2 +y 2
(x 2 +y 2 ) 3 2 dzdydx =
∫ 2π
=
=
= 2
∫ 1 ∫ 2−r 2
0 0 r
∫ 2
2π ∫ 1 ∫ 2−r 2
0 0
∫ 2π ∫ 1
0 0
∫ 2π
0
= 8π
35 .
r 2
(r 2 ) 3 2 dzrdrdθ
r 4 dzdrdθ
r 4 (2−r 2 −r 2 )drdθ
( r
5
5 − r7
7
)∣ ∣∣∣
1
r=0
dθ
Definition 11.5.2. Spherical Coordinates
Spherical coordinates represent a point P in space by (ρ,φ,θ) in which
(a) ρ = √ x 2 +y 2 +z 2 ≥ 0 is the distance from the origin O;
(b) 0 ≤ φ ≤ π is angle from the positive z-axis to the vector −→ OP;
(c) 0 ≤ θ ≤ 2π the same as the θ in the polar coordinates.
The rectangular conversion equations for spherical coordinates:
De-Yu Wang CSIE CYUT 201

11.5. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL
COORDINATES
z
✻
x = ρsinφcosθ
y = ρsinφsinθ
z = ρcosφ
P(ρ,φ,θ)
φ
✓ ✓✼
❥
✓
ρ
z
O✓
✑
❳ ❳❳
✑ ❳ ❳❳❳
✑❳ ✑ ❳3
✑ ✑ θ ✿ r ❳
✑✰ ❙
x y
x
y
Example 11.5.3. Write the equations (a) x 2 +y 2 +z 2 = 4 and (b) z 2 = x 2 +y 2 in
spherical coordinates.
Solution: (a) x 2 +y 2 +z 2 = ρ 2 = 4, ρ = 2.
(b) z 2 = (ρcosφ) 2 = x 2 +y 2 = (ρsinφcosθ) 2 +(ρsinφsinθ) 2
ρ 2 cos 2 φ = ρ 2 sin 2 φ(cos 2 θ +sin 2 θ)
ρ 2 cos 2 φ = ρ 2 sin 2 φ,
∴ ρ = 0 or cos 2 φ = sin 2 φ, φ = π 4 , 3π 4 .
Theorem 11.5.2. Triple integrals in spherical coordinates
Suppose that the solid Q as
Q = {(ρ,φ,θ)|α ≤ θ ≤ β,c ≤ φ ≤ d and g 1 (θ,φ) ≤ ρ ≤ g 2 (θ,φ)}.
The triple integrals in cylindrical coordinates:
∫∫∫ ∫ β ∫ d ∫ g2 (θ,φ)
f(ρ,φ,θ)dV = f(ρ,φ,θ) ρ 2 sinφdρdφdθ.
Q
α c g 1
} {{ }
(θ,φ)
dV
z
✻
ρ∆φ
Proof:
ρ
∆φ
❫
O
✑
❳ ✑✡ ✡✡✡✡✡✡ ✑✑✑✑✑✑
❍ ❳❳❳
✑ ❡ ❍❍❍❍❍❍ ❳
✑ ❡❡❡❡❡ ✣ ❳❳
❳3
✑ ∆θ y
✑✰
x
r
r∆θ
∆V k ≈ ∆ρ k (ρ k ∆φ k )(ρ k sinφ k ∆θ k ) = ρ 2 ksinφ k ∆ρ k ∆φ k ∆θ k
dV = ρ 2 sinφdρdφdθ
De-Yu Wang CSIE CYUT 202

11.5. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL
COORDINATES
Example 11.5.4. Evaluate the triple integral
z 2 )dzdydx in spherical coordinates.
∫ 2 ∫ √ 4−x 2
−2 0 0
∫ √ 4−x 2 −y 2
(x 2 + y 2 +
Solution:
∫ 2 ∫ √ 4−x 2
−2
0
∫ √ 4−x 2 −y 2
(x 2 +y 2 +z 2 )dzdydx =
0
=
=
=
∫ π
0
∫ π
0
∫ π
0
∫ π
0
= 32π
5 .
∫ π
2
0
∫ π
2
0
∫ π
2
0
∫ 2
0
ρ 2 ρ 2 sinφdρdφdθ
ρ 5
5 sinφ ∣ ∣∣∣
2
ρ=0
32
5 sinφdφdθ
− 32 ∣π
∣∣∣
5 cosφ 2
φ=0
dθ
dφdθ
Exercise 11.5.1. Write the given equation in cylindrical coordinates.
(a) (x−2) 2 +y 2 = 9
(b) z = √ (c) z = e −x2 −y 2
x 2 +y 2 (d) y = 2x
Exercise 11.5.2. Evaluate the triple integral in cylindrical coordinates.
(a)
(b)
(c)
∫ 1 ∫ √ 1−x 2
−1
− √ 1−x 2 ∫ 2−x 2 −y 2
∫ 1 ∫ √ 1−x 2
−1
∫ 2
0
0
∫ √ x 2 +y 2
− √ 1−x 2 0
√
x2 +y 2 dzdydx
3zdzdydx
∫ √ 4−y ∫ √ 2 8−x 2 −y 2
√ 2dzdxdy
− 4−y
√x 2 2 +y 2
Exercise 11.5.3. Convert the equation into spherical coordinates.
(a) x 2 +y 2 +z 2 = 6 (c) z = 2
(b) y = x 2 (d) z = − √ x 2 +y 2
De-Yu Wang CSIE CYUT 203

11.6. CHANGE OF VARIABLES: JACOBIANS
Exercise 11.5.4. Evaluate the triple integral in the spherical coordinate system.
(a)
(b)
(c)
∫ 2 ∫ √ 4−x 2
−2
0
∫ 1 ∫ √ 1−x 2
0
∫ √ 4−x 2 −y 2
√
x2 +y
−
√4−x 2 +z 2 dzdydx
2 −y 2
∫ √ 1−x 2 −y 2
(x
− √ 1−x 2 −
√1−x 2 +y 2 +z 2 ) 3 2 dzdydx
2 −y 2
∫ 4 ∫ √ 16−x ∫ 2 0
0
0
√
x2 +y
−
√16−x 2 +z 2 dzdydx
2 −y 2
11.6 Change of Variables: Jacobians
Definition 11.6.1. Jacobian
(a) For x = g(u,v) and y = h(u,v), the Jacobian is
∣ ∣∣∣∣∣∣ ∂x
∂(x,y)
∂(u,v) = ∂u
∂y
∂u
∂x
∂v
.
∂y
∣
∂v
(b) For x = g(u,v,w), y = (u,v,w) and z = k(u,v,w), the Jacobian is
∣ ∣∣∣∣∣∣∣∣∣∣∣ ∂x
∂u
∂(x,y,z)
∂(u,v,w) = ∂y
∂u
∂z
∂u
∂x
∂v
∂y
∂v
∂z
∂v
∂x
∂w
∂y
∂w
.
∂z
∣
∂w
Example 11.6.1. Find the Jacobian of the transformation x = 2u v , y = 3uv +v2 .
Solution:
∣ ∣∣∣∣∣∣ ∂x
∂(x,y)
∂(u,v) = ∂u
∂y
∂u
∂x
2
∂v
− 2u
=
∂y
v v 2
∣ ∣3v 3u+2v∣ = 12u
v +4.
∂v
De-Yu Wang CSIE CYUT 204

11.6. CHANGE OF VARIABLES: JACOBIANS
Example 11.6.2. Compute the Jacobian for the spherical-like transformation x =
ρsinφcosθ, y = ρsinφsinθ and z = ρcosφ.
Solution:
∣ ∣∣∣∣∣∣∣∣∣∣∣∣ ∂x ∂x ∂x
∂ρ ∂φ ∂θ
∂(x,y,z)
∂(ρ,φ,θ) = ∂y ∂y ∂y
sinφcosθ ρcosφcosθ −ρsinφsinθ
=
∂ρ ∂φ ∂θ
sinφsinθ ρcosφsinθ ρsinφcosθ
∣ cosφ −ρsinφ 0 ∣
∂z ∂z ∂z
∂ρ ∂φ ∂θ∣
=cosφ
ρcosφcosθ −ρsinφsinθ
∣ρcosφsinθ ρsinφcosθ ∣
+ρsinφ
sinφcosθ −ρsinφsinθ
∣sinφsinθ ρsinφcosθ ∣
=cosφ(ρ 2 sinφcosφcos 2 θ +ρ 2 sinφcosφsin 2 θ)
+ρsinφ(ρsin 2 φcos 2 θ +ρsin 2 φsin 2 θ)
=ρ 2 sinφ(cos 2 φ+sin 2 φ) = ρ 2 sinφ.
Theorem 11.6.1. Change of Variables
(a) For x = g(u,v) and y = h(u,v),
∫∫ ∫∫
f(x,y)dA =
R
S
f(g(u,v),h(u,v))
∂(x,y)
∣∂(u,v)
∣ dudv.
(b) For x = g(u,v,w), y = (u,v,w) and z = k(u,v,w),
∫∫∫ ∫∫∫
f(x,y,z)dV =
Q
S
f(g(u,v,w),h(u,v,w),k(u,v,w))
∂(x,y,z)
∣∂(u,v,w)
∣ dudvdw.
Proof: Consider the region ∆A with vertices M(g(u,v),h(u,v)), N(g(u +
∆u,v),h(u + ∆u,v)), P(g(u + ∆u,v + ∆v),h(u + ∆u,v + ∆v)), and Q(g(u,v +
∆v),h(u,v +∆v)),
De-Yu Wang CSIE CYUT 205

11.6. CHANGE OF VARIABLES: JACOBIANS
Q
P
✑ ✑✑✑✑✑
M
✑ ✑✑✑✑✑
N
∆x∆y = ∆A ≈ ∥ −−→ MN × −−→
MQ∥
−−→
MN × −−→
i j k
MQ =
g(u+∆u,v)−g(u,v) h(u+∆u,v)−h(u,v) 0
∣g(u,v +∆v)−g(u,v) h(u,v +∆v)−h(u,v) 0∣
∣ i j k ∣∣∣∣∣∣∣∣∣ ∂x ∂y
∂x ∂y
≈
∆u ∂u ∂u ∆u 0
=
∂u ∂u
∆u∆vk
∂x ∂y
∂x
∣
∂v ∆v ∂y
∂v ∆v 0 ∣ ∣
∂v ∂v
= ∂(x,y)
∂(u,v) ∆u∆vk
∆A ≈ ∥ −−→ MN × −−→
MQ∥ ≈
∂(x,y)
∣∂(u,v)
∣ ∆u∆v
Example 11.6.3. Use Theorem 11.6.1 to derive the evaluation formula for polar
coordinates:
∫∫ ∫∫
f(x,y)dA = f(rcosθ,rsinθ)rdrdθ.
R
S
Proof:
∣ ∣∣∣∣∣∣ ∂x ∂x
∂(x,y)
∂(r,θ) = ∂r ∂θ
cosθ −rsinθ
=
∂y ∂y
∣
∣
sinθ rcosθ ∣ = rcos2 θ+rsin 2 θ = r
∫∫ ∫∫
∂r ∂θ
f(x,y)dA = f(rcosθ,rsinθ)
∂(x,y)
∣
R
S ∂(r,θ) ∣ drdθ
∫∫
= f(rcosθ,rsinθ)rdrdθ.
S
∫∫
Example 11.6.4. Evaluate the integral (x + 2y)dA, where R is the region
bounded by the lines y = −1−2x, y = 3−2x, y = x−1 and y = x−3.
De-Yu Wang CSIE CYUT 206
R

11.6. CHANGE OF VARIABLES: JACOBIANS
Solution: Let u = y +2x and v = x−y, then x = u+v and y = u−2v
3 3
∣ ∂(x,y) ∣∣∣∣ 1 1
∂(u,v) = 3 3
1 −2∣ = −1 3
3 3
∫∫ ∫ 3 ∫ 3
( u+v
(x+2y)dA = + 2(u−2v) )∣ ∣∣∣ ∂(x,y)
R
1 −1 3 3 ∂(u,v) ∣ dudv
∫ 3 ∫ 3
= (u−v)
∣ −1 1 −1 3∣ dudv y
✻ y = x−1
= 1 ∫ 3
( )∣ u
2 ∣∣∣
3
❅
3
1 ❅
v = 1
1 2 −uv dv
y = x−3
u=−1
❅ v = 3
❅ ✲
= 1 ∫ 3
(4−4v) dv
x
❅ 1 ❅
3
2 3
1
−1❅
✟ ✟✟✟✟✟✟✟ R ❅
❅
❅
= 1 ( )∣ 4v −2v
2 ∣ 3 y = 3−2x
❅
❅
3
v=1
❅
−3 ✟ ✟✟✟✟✟✟✟✟✟✟✟✟ u = 3
= − 8 ❅ y = −1−2x
3 .
❅ u = −1
Exercise 11.6.1. Find the Jacobian of the transformation.
(a) x = ue v , y = ue −v
x = 2uv, y = 3u−v
(b) x = 2u v
, y = 3uv +v2
(c)
(d) x = u v , y = v2
Exercise 11.6.2. Evaluate the integral
∫∫
(a) (x−2y)dA,whereRistheregionboundedbyy = 3x, y = 3x−4, y = 1−2x
R
and y = 2−2x.
∫∫
(b) (x+2y)dA, whereRis theregion boundedby y = e x , y = e x +2, y = 3−e x
R
and y = 5−e x .
∫∫
(c) (x+2y)dA, where R is the region bounded by y = 1−2x, y = 3−2x, y =
R
x−1 and y = x−3.
Exercise 11.6.3. Prove that
∫∫ ∫∫
(a) f(x,y)dA = f(rcosθ,rsinθ)rdrdθ.
R
∫∫∫ ∫∫∫
(b) f(x,y,z)dV =
Q
S
S
f(ρsinφcosθ,ρsinφsinθ,ρcosφ)ρ 2 sinφdρdφdθ.
De-Yu Wang CSIE CYUT 207