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Mehdi Rahmani-Andebili
Practice Problems, Methods, and Solutions
Calculus II
Calculus II
Mehdi Rahmani-Andebili 
Calculus II 
Practice Problems, Methods, and Solutions
Mehdi Rahmani-Andebili 
Electrical Engineering Department 
Arkansas Tech University 
Russellville, AR, USA 
ISBN 978-3-031-45352-6 ISBN 978-3-031-45353-3 (eBook) 
https://doi.org/10.1007/978-3-031-45353-3 
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https://doi.org/10.1007/978-3-031-45353-3
Preface 
Calculus is one of the most important courses of many majors, including engineering and 
science, and even some non-engineering majors like economics and business, which is taught in 
three successive courses at universities and colleges worldwide. Moreover, in many universities 
and colleges, a precalculus course is mandatory for under-prepared students as the prerequisite 
course of Calculus 1. 
Unfortunately, some students do not have a solid background and knowledge in math and 
calculus when they start their education in universities or colleges. This issue prevents them 
from learning calculus-based courses such as physics and engineering. Sometimes, the problem 
escalates, so they give up and leave the university. Based on my real professorship experience, 
students do not have a serious issue comprehending physics and engineering courses. In fact, it 
is the lack of enough knowledge of calculus that hinder them from understanding those courses. 
Therefore, a series of calculus textbooks covering Precalculus, Calculus 1, Calculus 2, and 
Calculus 3 have been prepared to help students succeed in their major. The subjects of the 
calculus series books are as follows. 
Precalculus: Practice Problems, Methods, and Solution
• Real Number Systems, Exponents and Radicals, and Absolute Values and Inequalities
• Systems of Equations
• Quadratic Equations
• Functions, Algebra of Functions, and Inverse Functions
• Factorization of Polynomials
• Trigonometric and Inverse Trigonometric Functions
• Arithmetic and Geometric Sequences 
Calculus 1: Practice Problems, Methods, and Solution
• Characteristics of Functions
• Trigonometric Equations and Identities
• Limits and Continuities
• Derivatives and Their Applications
• Definite and Indefinite Integrals 
Calculus 2: Practice Problems, Methods, and Solution
• Applications of Integration
• Sequences and Series and Their Applications
• Polar Coordinate System
• Complex Numbers
v
vi Preface
Calculus 3: Practice Problems, Methods, and Solution
• Linear Algebra and Analytical Geometry
• Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System
• Multivariable Functions
• Double Integrals and their Applications
• Triple Integrals and their Applications
• Line Integrals and their Applications 
The textbooks include basic and advanced calculus problems with very detailed problem 
solutions. They can be used as practicing study guides by students and as supplementary 
teaching sources by instructors. Since the problems have very detailed solutions, the textbooks 
are helpful for under-prepared students. In addition, they are beneficial for knowledgeable 
students because they include advanced problems. 
In preparing the problems and solutions, care has been taken to use methods typically found 
in the primary instructor-recommended textbooks. By considering this key point, the textbooks 
are in the direction of instructors’ lectures, and the instructors will not see any untaught and 
unusual problem solutions in their students’ answer sheets. 
To help students study in the most efficient way, the problems have been categorized into 
nine different levels. In this regard, for each problem, a difficulty level (easy, normal, or hard) 
and a calculation amount (small, normal, or large) have been assigned. Moreover, problems 
have been ordered in each chapter from the easiest problem with the smallest calculations to the 
most difficult problems with the largest ones. Therefore, students are suggested to start studying 
the textbooks from the easiest problems and continue practicing until they reach the normal and 
then the hardest ones. This classification can also help instructors choose their desirable 
problems to conduct a quiz or a test. Moreover, the classification of computation amount can 
help students manage their time during future exams, and instructors assign appropriate 
problems based on the exam duration. 
Russellville, AR, USA Mehdi Rahmani-Andebili
The Other Works Published by the Author 
The author has already published the books and textbooks below with Springer Nature. 
Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2023. 
Calculus III – Practice Problems, Methods, and Solutions, Springer Nature, 2023. 
Calculus II – Practice Problems, Methods, and Solutions, Springer Nature, 2023. 
Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2023. 
Planning and Operation of Electric Vehicles in Smart Grid, Springer Nature, 2023. 
Applications of Artificial Intelligence in Planning and Operation of Smart Grid, Springer 
Nature, 2022. 
AC Electric Machines- Practice Problems, Methods, and Solutions, Springer Nature, 2022. 
DC Electric Machines, Electromechanical Energy Conversion Principles, and Magnetic Circuit 
Analysis- Practice Problems, Methods, and Solutions, Springer Nature, 2022. 
Differential Equations- Practice Problems, Methods, and Solutions, Springer Nature, 2022. 
Feedback Control Systems Analysis and Design- Practice Problems, Methods, and Solutions, 
Springer Nature, 2022. 
Power System Analysis – Practice Problems, Methods, and Solutions, Springer Nature, 2022. 
Advanced Electrical Circuit Analysis – Practice Problems, Methods, and Solutions, Springer 
Nature, 2022. 
Design, Control, and Operation of Microgrids in Smart Grids, Springer Nature, 2021. 
Applications of Fuzzy Logic in Planning and Operation of Smart Grids, Springer Nature, 2021. 
Operation of Smart Homes, Springer Nature, 2021. 
AC Electrical Circuit Analysis – Practice Problems, Methods, and Solutions, Springer Nature, 
2021. 
Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
vii
viii The Other Works Published by the Author
Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
DC Electrical Circuit Analysis – Practice Problems, Methods, and Solutions, Springer Nature, 
2020. 
Planning and Operationof Plug-in Electric Vehicles: Technical, Geographical, and Social 
Aspects, Springer Nature, 2019.
Contents 
1 Problems: Applications of integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 
1.1 Mean Value of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 
1.2 Surface Area Bounded by Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 
1.3 Volume Resulted from Rotation of an Enclosed Region . . . . . . . . . . . . . . . . 7 
1.4 Arc Length of a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 
1.5 Surface Area of a Solid of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 
1.6 Center of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 
2 Solutions of Problems: Applications of Integration . . . . . . . . . . . . . . . . . . . . . . 15 
2.1 Mean Value of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 
2.2 Surface Area Bounded by Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 
2.3 Volume Resulted from Rotation of an Enclosed Region . . . . . . . . . . . . . . . . 28 
2.4 Arc Length of a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 
2.5 Surface Area of a Solid of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 
2.6 Center of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 
3 Problems: Sequences and Series and Their Applications . . . . . . . . . . . . . . . . . . 53 
4 Solutions of Problems: Sequences and Series and Their Applications . . . . . . . . 59 
5 Problems: Polar Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 
6 Solutions of Problems: Polar Coordinate System . . . . . . . . . . . . . . . . . . . . . . . 77 
7 Problems: Complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 
8 Solutions of Problems: Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
ix
p
p
p
p
Problems: Applications of integration 1 
Abstract 
In this chapter, the basic and advanced problems related to the applications of integration are presented. The subjects 
include mean value of a function, surface area bounded by curves, volume resulted from rotation of an enclosed region, arc 
length of a curve, surface area of a solid of revolution, and center of gravity. In this chapter, the problems are categorized in 
different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). 
Additionally, the problems are ordered from the easiest problem with the smallest computations to the most difficult 
problems with the largest calculations. 
1.1 Mean Value of a Function 
1.1. For the range of 2 ≤ x ≤ 5, calculate the mean value of the following function [1–3]. 
y ¼ axþ b 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
5 
2 
aþ 3b 
2) 
7 
2 
aþ 3b 
3) 
5 
2 
aþ b 
4) 
7 
2 
aþ b 
1.2. Consider the functions of f (x)¼ 2x and g (x)¼ 3x2- 2x. Calculate the value of λ if the mean value of the functions in the 
range of [1, λ] is the same. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1þ 5 
2 
2) 
1þ 3 
2 
3) 
5 
2 
4) 
3 3 
2 
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 
M. Rahmani-Andebili, Calculus II, https://doi.org/10.1007/978-3-031-45353-3_1
1
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¼ ¼
p
p
p
¼ ¼ p
¼
¼ ¼
2 1 Problems: Applications of integration
1.2 Surface Area Bounded by Curves 
1.3. Calculate the surface area enclosed between the curves of y 2x2 - 2x and y x2 . 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1 
3 
2) 
2 
3 
3) 
4 
3 
4) 
7 
3 
1.4. Calculate the surface area bounded by the functions of y (x) ¼ sin x and y (x) ¼ cos x for x ¼ π 
4 
, 
5π 
4 
. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1 
2 
2) 2 
3) 2 2 
4) 3 2 
1.5. Calculate the surface area enclosed between the curves of y x2 and y x. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
2 
3 
2) 1 
3) 
1 
3 
4) 
1 
6 
1.6. Calculate the surface area enclosed between the curve of y x3 + 2x2 + x and x-axis. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1 
12 
2) 
1 
10 
3) 
1 
9 
4) 
1 
7 
1.7. Calculate the surface area enclosed between the curve of y x2 + 1 and the line of y 2. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1 
3 
2) 
2 
3
¼
¼
1
1.2 Surface Area Bounded by Curves 3
3) 1 
4) 
4 
3 
1.8. Calculate the surface area restricted by the curve of y2 (x) x2 - x4 . 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 1 
2) 2 
3) 
2 
3 
4) 
4 
3 
1.9. Calculate the surface area of the shaded region shown in Fig. 1.1. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 8 - 2π 
2) 8-
π 
2 
3) 4-
π 
4 
4) 2π - 4 
Figure 1.1 The graph of problem 1.9 
1.10. Calculate the surface area restricted by the following function, above the line of y(x) ¼ 0, and the right-hand side 
of x 1. 
y xð Þ ¼ 6 
2xþ 1ð Þ xþ 2ð Þ 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1 
2 
2) ln 2 
3) 2 ln 2 
4)
p
p
4 1 Problems: Applications of integration
1.11. Calculate the surface area enclosed between the curve with the function below and x-axis in the range of - 2 
p 
, 2 
p 
. 
y ¼ 1 
2þ x2 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
π 2 
4 
2) 
π 
4 
3) 
π 2 
2 
4) 
π 
2 
1.12. Calculate the surface area restricted by the function below and x-axis in the domain of [1, e2 ]. 
y xð Þ ¼ ln x 
x
p 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 5 
2) 4 
3) 3 
4) 2 
1.13. Calculate the surface area of the shaded region shown in Fig. 1.2. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
1 
3 
2) 
2 
3 
3) 
1 
ln 2
-
2 
3 
4) 
2 
3
- ln 2 
Figure 1.2 The graph of problem 1.13
¼ ¼ ¼
p
¼
1
1.2 Surface Area Bounded by Curves 5
1.14. Calculate the surface area of the shaded region shown in Fig. 1.3. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) π 
2) 
π 
2 
3) 
π 
2
-
1 
2 
4) 
π 
2
þ 1 
2 
Figure 1.3 The graph of problem 1.14 
1.15. Determine the value of parameter of c, where 0< c< 
π 
2 
, if the surface area restricted by the function of y (x) ¼ cos x, 
the function of y (x) ¼ cos (x - c), and the line of x ¼ 0 is equal to the surface area bounded by the function of 
y (x) cos (x - c), the line of x π, and the line of y (x) 0. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) 
π 
5 
2) 
π 
3 
3) 
π 
4 
4) 
π 
6 
1.16. Determine the surface area bounded by the function of y xð Þ ¼ x ln x, x-axis, and the lines with the equations of x ¼ 1 
and x 2. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 2 ln 2-
1 
2 
2) 2 ln 2 - 1 
3) ln 2 þ 3 
4 
4) ln 2-
3 
8 
1.17. Estimate the surface area of the shaded region shown in Fig. 1.4. 
Difficulty level○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
2) ln2
ð Þ ¼
p
¼
6 1 Problems: Applications of integration
3) cos2 
4) 2 ln 2 - cos 2 
Figure 1.4 The graph of problem 1.17 
1.18. Calculate the surface area of the shaded region shown in Fig. 1.5. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 12 
2) 14 
3) 16 
4) 18 
Figure 1.5 The graph of problem 1.18 
1.3 Volume Resulted from Rotation of an Enclosed Region 
1.19. Calculate the volume resulted from the rotation of the function of y x x2 - x3 around x-axis for x [-1, 1]. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
π 
3 
2) 
2π 
3
¼
1
1
¼ ¼
¼ ¼
¼ ¼ 1
¼ 1 ¼ 1
¼
4
1.3 Volume Resulted from Rotation of an Enclosed Region 7
3) π 
4) 
3π 
2 
1.20. Calculate the volume resulted from the rotation of the surface area around x-axis enclosed between one period of the 
curve of y sin (x) and x-axis. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) π2 
2) 2π2 
3) 
π2 
2 
4) 
π2 
4 
1.21. Calculate the volume resulted from the rotation of the function of y(x) ¼ cos x around x-axis for x ¼ 0, π 
2 
. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
π2 
4 
2) 
π2 
8 
3) 
π 
8 
4) 
π 
4 
1.22. The volume resulted from the rotation of the surface area restricted by the function of y(x) ¼ e-x , the coordinate axes, 
and x ¼ b, where b > 0, around x-axis is V. Calculate the value of lim 
b→ 
V . 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
π 
4 
2) 
π 
2 
3) π 
4) 
1.23. The surface area restricted by the function of y xð Þ ¼ 1 
x
p , the coordinate axes, and x ¼ 1 is called S. Moreover, the 
volume resulted from the rotation of the surface area around x-axis is called V. Which of the following options is 
correct? 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) S 2, V 8π 
2) S 4, V 8π 
3) S 2, V 
4) S , V 
1.24. Calculate the volume resulted from the rotation of the surface area around y-axis enclosed between the curve of 
y 1-
1 
x2 and x-axis. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large
¼
4
¼
2
¼ ¼
8 1 Problems: Applications of integration
1) π 
2) 2π 
3) 3π 
4) 4π 
1.25. Calculate the volume resulted from the rotation of the surface area around x-axis enclosed between the curve with the 
function below, x-axis, x 
π 
, and x 
π 
. 
y ¼ 1 
sin 2 xð Þ 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) π 
2) 
2π 
3 
3) 2π 
4) 
4π 
3 
1.26. The surface area confined by the function below, x-axis, and two lines with the equations of x ¼ π 
6 
and x ¼ π 
2 
are 
rotated around x-axis. Calculate the resultant volume. 
y xð Þ ¼ cos
3x 
p 
sin 2 x 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
7π 
3 
2) 
5π 
3 
3) 
4π 
3 
4) 
2π 
3 
1.27. Calculate the volume resulted from the rotation of the surface area bounded by the function of y (x) ¼ xe x and the lines 
with the equations of x 1 and y (x) 0 around x-axis. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
π 
2 
e2 - 1 
2) 
π 
2 
e2 - 2 
3) 
π 
4 
e2 - 1 
4) 
π 
4 
e2 þ 1 
1.28. Calculate the volume resulted from the rotation of the surface area restricted by the relation of y 
1 
2 xð Þ ¼ a 1 2 - x 1 2 and x- and 
y- axes around x-axis. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large
ð Þ ¼
p
þ
1.4 Arc Length of a Curve 9
1) 5πa3 
2) 
1 
2 
πa3 
3) 
1 
12 
πa3 
4) 
1 
15 
πa3 
1.29. Calculate the volume resulted from the rotation of the function of y x e- x sin x around x-axis for 0 ≤ x ≤ π. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) 
π 
5 
1 
1- e- 2π 
2) 
π 
5 
1 
1 e- 2π 
3) 
π 
5 
1þ e- 2π 
4) 
π 
5 
1- e- 2π 
1.30. Calculate the volume created by the rotation of the shaded region (see Fig. 1.6) around x-axis. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) 
4π 
15 
2) 
π 
2 
3) 
14π 
3 
4) 
3π 
5 
Figure 1.6 The graph of problem 1.30 
1.4 Arc Length of a Curve 
1.31. Calculate the arc length of the function of f xð Þ ¼ 
x 
1 
t4 - 1 
p 
dt for the interval of x ¼ [1, 3].
¼
p
p
þ
10 1 Problems: Applications of integration
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
19 
3 
2) 
25 
3 
3) 
26 
3 
4) 
28 
3 
1.32. Calculate the arc length of the parametric relation below for the interval of t [0, 4]. 
x tð Þ ¼ et cos t 
y tð Þ ¼ et sin t 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 2 e4 - 1 
2) 2 e4 1 
3) 2(e4 - 1) 
4) 2(e4 + 1) 
1.33. Calculate the arc length of the following parametric relation for the interval of 0 ≤ t ≤ 1. 
x tð Þ ¼ et cos t þ sin tð Þ 
y tð Þ ¼ et cos t- sin tð Þ 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) e - 1 
2) 
e- 1 
2 
3) 4(e - 1) 
4) 2(e - 1) 
1.34. Calculate the arc length of the function of y xð Þ ¼ 1 
2 
x2 -
1 
4 
ln x for the interval of x ¼ [1, 2]. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
3 
2
þ 1 
2 
ln 2 
2) 
3 
2
þ 1 
4 
ln 2 
3) 
3 
4
þ 1 
2 
ln 2 
4) 
3 
4
þ 1 
4 
ln 2 
1.35. Calculate the arc length of the function of y xð Þ ¼ x
4 
8 
þ 1 
4x2 
for the interval of x ¼ [1, 2]. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large
p
þ
p
1.4 Arc Length of a Curve 11
1) 15 
2) 21 
3) 
25 
3 
4) 
33 
16 
1.36. Calculate the arc length of the function of f (x) ¼ ln (sec x) for the interval of x ¼ 0, π 
4 
. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) ln 2 
2) ln 2- 2 
p 
3) ln 2 
p
- 1 
4) ln 1þ 2 
p 
1.37. Calculate the arc length of the function of f xð Þ ¼ 
x 
0 
cosh tð Þdt for the interval of x ¼ [0, 2]. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2 
p 
e-
1 
e 
2) 2 e-
1 
e 
3) 2 
p 
eþ 1 
e 
4) 2 eþ 1 
e 
1.38. Calculate the arc length of the function of f xð Þ ¼ 1 
2 
x x2 - 1 
p
- ln xþ x2 - 1 
p 
for the interval of x ¼ [1, 2]. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
1 
2 
2) 
3 
2 
3) 1 2 
4) 2 
1.39. Calculate the arc length of the function of f xð Þ ¼ ln e
x þ 1 
ex - 1 
for the interval of 0 ≤ x ≤ 2. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) ln e-
1 
e 
2) ln e þ 1 
e 
3) ln e2 -
1 
e2 
4) ln e2 þ 1 
e2
¼
12 1 Problems: Applications of integration
1.5 Surface Area of a Solid of Revolution 
1.40. The function of y (x) ¼ cosh x for the interval of 0 ≤ x ≤ 2 is rotated around x-axis. Calculate the surface area of the 
solid. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
π 
2 
2 þ cosh 2ð Þ 
2) 
π 
2 
2 þ sinh 2ð Þ 
3) 
π 
2 
4 þ cosh 4ð Þ 
4) 
π 
2 
4 þ sinh 4ð Þ 
1.41. The function of 3y(x) - x3 0 is rotated around x-axis. Calculate the surface area of the solid of revolution. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
π 
9 
2 2 
p
- 1 
2) 
π 
9 
2 
p
- 1 
3) 
π 
3 
2 
p
- 1 
4) 
π 
3 
2 2 
p
- 1 
1.42. The function of y (x) ¼ x2 for the interval of y ≤ 2 is rotated around y-axis. Calculate the surface area of the solid of 
revolution. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
8π 
3 
2) 
13π 
3 
3) 
11π 
3 
4) 
10π 
3 
1.43. The function of f xð Þ ¼ 6 cosh x 
6 
for the interval of 0 ≤ x ≤ 5 is rotatedaround x-axis. What is the ratio of the volume 
of the solid of revolution to its surface area. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 3 
2) 2 
3) 
3 
2 
4) 
2π 
3
¼
¼
4
1.6 Center of Gravity 13
1.44. The function of f xð Þ ¼ 
x 
0 
sinh t2 dt for the interval of 0≤ x≤ ln 1396 
p 
is rotated around y-axis. Calculate the surface 
area of the solid of revolution. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2π 1396-
1 
1396 
2) 
π 
4 
1396-
1 
1396 
3) π 1396-
1 
1396 
4) 
π 
2 
1396-
1 
1396 
1.6 Center of Gravity 
1.45. Determine the center of gravity of a surface from x-axis which is restricted by the function of f (x) 1 - x2 and x-axis. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
4 
15 
2) 
4 
5 
3) 
3 
5 
4) 
2 
5 
1.46. Determine the center of gravity of a surface from y-axis which is restricted by the function of f (x) 1 - x2 and x-axis. 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 0 
2) 
4 
5 
3) 
3 
5 
4) 
2 
5 
1.47. Determine the center of gravity of the parametric curve below from x-axis. 
x tð Þ ¼ t þ sin t 
y tð Þ ¼ 1- cos t , 0≤ t≤ π 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) 1 
2) 
1 
3 
3) 
2 
3 
4) 
3
14 1 Problems: Applications of integration
References 
1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
Solutions of Problems: Applications of Integration 2 
Abstract 
In this chapter, the problems of the first chapter are fully solved, in detail, step-by-step, and with different methods. 
2.1 Mean Value of a Function 
2.1. As we know, the average value of a function can be determined as follows [1–3]: 
f ave ¼ 1 b- a 
b 
a 
f xð Þdx 
In addition, from list of integral of functions, we know that: 
xn dx ¼ 1 
nþ 1 x
nþ1 þ c 
The problem can be solved as follows. 
f ave ¼ 1 5- 2 
5 
2 
axþ bð Þdx ¼ 1 
3 
a 
2 
x2 þ bx 5 
2 
¼ 1 
3 
25a 
2 
þ 5b- 4a 
2
- 2b 
⟹ f ave ¼ 7 2 aþ b 
Choice (4) is the answer. 
2.2. As we know, the average value of a function can be determined as follows: 
1 
b- a 
b 
a 
f xð Þdx 
Based on the problem, we know that: 
f ave ¼ gave 
Now, the problem can be solved as follows. 
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 
M. Rahmani-Andebili, Calculus II, https://doi.org/10.1007/978-3-031-45353-3_2
15
http://crossmark.crossref.org/dialog/?doi=10.1007/978-3-031-45353-3_2&domain=pdf
https://doi.org/10.1007/978-3-031-45353-3_2#DOI
p p
16 2 Solutions of Problems: Applications of Integration
1 
λ- 1 
λ 
1 
2x dx ¼ 1 
λ- 1 
λ 
1 
3x2 - 2x dx 
⟹ x2 
λ 
1 
¼ x3 - x2 λ 
1 
⟹ λ2 - 1 ¼ λ3 - λ2 - 0 ⟹ λ3 - 2λ2 þ 1 ¼ 0 
⟹ λ- 1ð Þ λ2 - λ- 1 ¼ 0 
⟹ λ ¼ 1- 5 
2 
, 
1 þ 5 
2 
, 1 
However, just 1þ 5
p 
2 is acceptable because the others are not within the range. 
⟹ λ ¼ 1þ 5
p 
2 
Choice (1) is the answer. 
In this problem, the rule below was used. 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
2.2 Surface Area Bounded by Curves 
2.3. First, we need to find the intersection points of the curves as follows: 
2x2 - 2x ¼ x2 ⟹ x2 - 2x ¼ 0 ⟹ x ¼ 0, 2 
Then: 
S ¼ 
x2 
x1 
y2 - y1ð Þdx 
⟹ S ¼ 
2 
0 
x2 - 2x2 þ 2x dx ¼ 
2 
0
- x2 þ 2x dx 
⟹ S ¼ - x
3 
3 
þ x2 2 
0 
¼ - 2
3 
3 
þ 22 - 0þ 0ð Þ 
⟹ S ¼ 4 
3 
Choice (3) is the answer.
p
p
p
p
p
2.2 Surface Area Bounded by Curves 17
In this problem, the rule below was used. 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
2.4. The problem can be solved as follows. 
S ¼ 
x2 
x1 
y2 - y1ð Þdx 
) S ¼ 
5π 
4 
π 
4 
sin x- cos xð Þdx 
) S ¼ - cos x- sin xð Þ 
5π 
4 
π 
4 
¼ - cos 5π 
4
- sin 
5π 
4
- - cos 
π 
4
- sin 
π 
4 
¼ 2
p 
2 
þ 2
p 
2
- -
2
p 
2
-
2
p 
2 
) S ¼ 2 2 
Choice (3) is the answer. 
In this problem, the rules below were used. 
sin xdx ¼ - cos xþ c 
cos xdx ¼ sin xþ c 
cos 
5π 
4 
¼ - 2 
2 
sin 
5π 
4 
¼ - 2 
2 
cos 
π 
4 
¼ 2 
2 
sin 
π 
4 
¼ 2 
2 
2.5. First, we need to find the intersection points of the curves as follows: 
y1 ¼ x2 
y2 ¼ x
p 
⟹ y2 ¼ y1 ⟹ x
p ¼ x2 ⟹ xp x xp - 1 ¼ 0 ⟹ x ¼ 0, 1
18 2 Solutions of Problems: Applications of Integration
S ¼ 
x2 
x1 
y2 - y1ð Þdx 
⟹ S ¼ 
1 
0 
x
p
- x2 dx 
⟹ S ¼ 2 
3 
x 
3 
2 -
x3 
3 
1 
0 
¼ 2 
3
-
1 
3 
⟹ S ¼ 1 
3 
Choice (3) is the answer. 
In this problem, the rule below was used. 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
2.6. First, we need to find the intersection points of the curves as follows: 
y2 ¼ y1 ⟹ x3 þ 2x2 þ x ¼ 0 ⟹ x x2 þ 2x þ 1 ¼ x xþ 1ð Þ2 ¼ 0 ⟹ x ¼ 0, - 1, - 1 
S ¼ 
x2 
x1 
y2 - y1ð Þdx 
⟹ S ¼ 
0
- 1 
x3 þ 2x2 þ x dx 
⟹ S ¼ x
4 
4 
þ 2 
3 
x3 þ x
2 
2 
0
- 1 
¼ 0- 1 
4
-
2 
3
þ 1 
2 
¼ - 3- 8 þ 6 
12 
¼ - 1 
12 
The surface area must be a positive quantity. Therefore, 
S ¼ 1 
12 
Choice (1) is the answer. 
In this problem, the rule below was used. 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
2.7. First, we need to find the intersection points of the curves as follows: 
y1 ¼ x2 þ 1 
y2 ¼ 2
¼
2.2 Surface Area Bounded by Curves 19
⟹ x2 þ 1 ¼ 2 ⟹ x2 ¼ 1 ⟹ x ¼ ± 1 
Then: 
S ¼ 
x2 
x1 
y2 - y1ð Þdx 
⟹ S ¼ 
1
- 1 
2- x2 þ 1 dx ¼ 2 
1 
0 
1- x2 dx 
⟹ S ¼ 2 x- x
3 
3 
1 
0 
¼ 2 1- 1 
3
- 0 
⟹ S ¼ 4 
3 
Choice (4) is the answer. 
In this problem, the rules below were used. 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
a
- a 
f xð Þdx ¼ 2 
a 
0 
f xð Þdx , if f xð Þ is an even function 
2.8. As can be noticed from the function of y2 ¼ x2 - x4 , the function is even with respect to both x and y. Therefore, we can 
calculate the surface area bounded in the first quadrant and then multiply its value by 4. 
To determine the range of x, we need to solve the equation of y ¼ 0 or x2 - x4
p 
¼ x 1- x2
p 
¼ 0 that for the first 
quadrant gives x 0, 1. 
S ¼ 
x2 
x1 
y2 - y1ð Þdx 
) S ¼ 4 
1 
0 
x 1- x2 
p 
dx 
By defining a new variable, we have: 
1- x2 ¼ u ) - 2xdx ¼ du ) xdx ¼ - 1 
2 
du 
) S ¼ 4 - 1 
2 
u 
1 
2du 
) S ¼ 4 - 1 
3 
u 
3 
2 ¼ - 4 
3 
1- x2 
3 
2 
1 
0 
¼ - 4 
3 
0- 1ð Þ 
) S ¼ 4 
3 
Choice (4) is the answer.
20 2 Solutions of Problems: Applications of Integration
2.9. The problem can be solved as follows. 
S ¼ y1 xð Þ- y2 xð Þð Þdx 
) S ¼ 2 
1 
0 
4-
4 
1þ x2 dx 
) S ¼ 2 4x- 4 arctan xð Þ 1 
0 
¼ 2 4- 4 π 
4
- 0 ¼ 2 4- πð Þ 
) S ¼ 8- 2π 
Choice (1) is the answer. 
Figure 2.1 The graph of problem 2.9 
In this problem, the rules below were used. 
adx ¼ axþ c 
1 
1þ x2 dx ¼ arctan xþ c 
a
- a 
f xð Þdx ¼ 2 
a 
0 
f xð Þdx , if f xð Þ is an even function 
arc tan 1 ¼ π 
4 
arc tan 0 ¼ 0 
2.10. The problem can be solved as follows. 
S ¼ y1 xð Þ- y2 xð Þð Þdx
p
p
2.2 Surface Area Bounded by Curves 21
) S ¼ 
1 
1 
6 
2x þ 1ð Þ xþ 2ð Þ - 0 dx ¼ 
1 
1 
4 
2xþ 1 -
2 
xþ 2 dx 
) S ¼ 2ln 2x þ 1ð Þ- 2 ln xþ 2ð Þð Þ 1 
1 
¼ 2 ln 2xþ 1 
xþ 2 
1 
1 
¼ 2 ln 2- ln 1½ �
) S ¼ 2 ln 2 
Choice (3) is the answer. 
In this problem, the rules below were used. 
1 
u 
du ¼ ln uþ c 
ln 1 ¼ 0 
2.11. The problem can be solved as follows. 
S ¼ y1 xð Þ- y2 xð Þð Þdx 
) S ¼ 
2
p
- 2
p 
1 
2þ x2 dx ¼ 2 
2
p 
0 
1 
2þ x2 dx ¼ 2 
2
p 
0 
1 
2 1þ x 
2
p 
2 
dx 
) S ¼ 2 
p 
arc tan 
x 
2
p 2 
0 
¼ 2 
p π 
4
- 0 
⟹ S ¼ π 2 
4 
Choice (1) is the answer. 
In this problem, the rules below were used. 
1 
1þ x a 
2 
dx ¼ a arc tan x 
a 
þ c 
a
- a 
f xð Þdx ¼ 2 
a 
0 
f xð Þdx , if f xð Þ is an even function 
arc tan 1 ¼ π 
4 
arc tan 0 ¼ 0
22 2 Solutions of Problems: Applications of Integration
2.12. From list of integral of functions or by using the method of integration by parts, weknow that: 
ln udu ¼ u ln u- uþ c 
The problem can be solved by defining a new variable as follows. 
x ¼ t2 ) xp ¼ t ) dx 
x
p ¼ 2dt 
S ¼ y1 xð Þ- y2 xð Þð Þdx 
) S ¼ 
e2 
1 
ln x 
x
p - 0 dx 
) S ¼ 
t2 
t1 
ln t2 2dtð Þ ¼ 4 
t2 
t1 
ln tdt ¼ 4 tln t- tð Þ t2 
t1 
) S ¼ 4 xp ln xp - xp e
2 
1 
) S ¼ 4 eln e- eð Þ- 1ln 1- 1ð Þ½ �
) S ¼ 4 
Choice (2) is the answer. 
In this problem, the rules below were used. 
ln ab ¼ b ln a 
ln e ¼ 1 
ln 1 ¼ 0 
2.13. From list of integral of functions, we know that: 
ax dx ¼ a
x 
ln a
þ c 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
The problem can be solved as follows. 
S ¼ y1 xð Þ- y2 xð Þð Þdx
2.2 Surface Area Bounded by Curves 23
) S ¼ 
1 
0 
2x - 1- x2 dx ¼ 
1 
0 
2x - 1þ x2 dx 
) S ¼ 2
x 
ln 2
- xþ x
3 
3 
1 
0 
¼ 2 
ln 2
- 1 þ 1 
3
-
1 
ln 2 
) S ¼ 1 
ln 2
-
2 
3 
Choice (3) is the answer. 
Figure 2.2 The graph of problem 2.13 
2.14. The problem can be solved as follows. 
S ¼ y1 xð Þ- y2 xð Þð Þdx 
) S ¼ 
1 
0 
2 
1þ x2 - x dx 
) S ¼ 2arctan x- x
2 
2 
1 
0 
¼ 2arctan 1- 1 
2
- 2arctan 0- 0ð Þ 
) S ¼ π 
2
-
1 
2 
Choice (3) is the answer. 
Figure 2.3 The graph of problem 2.14
¼
¼
¼ ¼
¼
24 2 Solutions of Problems: Applications of Integration
In this problem, the rules below were used. 
1 
1þ x2 dx ¼ arctan xþ c 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
arc tan 1 ¼ π 
4 
arc tan 0 ¼ 0 
2.15. First, we need to determine the points of intersection of two graphs related to the first surface area, that is, y1(x) ¼ cos x 
and y2(x) cos (x - c). 
y1 xð Þ ¼ y2 xð Þ ) cos x- cð Þ ¼ cos x 
) x- cð Þ ¼ ± x ) x- c ¼ - x ) x ¼ c 
2 
Also, from the problem, we have x 0. Thus, the problem can be solved as follows. 
S1 ¼ y1 xð Þ- y2 xð Þð Þdx 
) S1 ¼ 
c 
2 
0 
cos x- cos x- cð Þ½ �dx 
) S1 ¼ sin x- sin x- cð Þ½ �
c 
2 
0 ¼ sin 
c 
2
- sin -
c 
2
- 0- sin - cð Þð Þ ¼ 2 sin c 
2
- sin c 
Likewise, we need to determine the points of intersection of two graphs related to the second surface area, that is, 
y3(x) 0 and y4 cos (x - c). 
y3 xð Þ ¼ y4 xð Þ ) cos x- cð Þ ¼ 0 
) x- c ¼ π 
2 
) x ¼ cþ π 
2 
Also, from the problem, we have x π. Therefore: 
S2 ¼ y3 xð Þ- y4 xð Þð Þdx 
S2 ¼ 
π 
cþπ 2 
0- cos x- cð Þ½ �dx 
) S2 ¼ - sin x- cð Þ½ �π cþπ 2 ¼ - sin π- cð Þ þ sin 
π 
2 
¼ - sin cþ 1 
Based on the problem, we have: 
S1 ¼ S2
2.2 Surface Area Bounded by Curves 25
) 2 sin c 
2
- sin c ¼ 1- sin c ) sin c 
2 
¼ 1 
2 
) c 
2
¼ π 
6 
) c ¼ π 
3 
Choice (2) is the answer. 
In this problem, the rules below were used. 
cos x ¼ sin x þ c 
cos xþ að Þ ¼ sin xþ að Þ þ c 
sin π- cð Þ ¼ sin cð Þ for 0< c< π 
2 
sin 
π 
2 
¼ 1 
2.15. The problem can be solved by using the method of integration by parts or using the list of integral of functions as 
follows. 
From list of integral of functions, we know that: 
xn ln xdx ¼ x
nþ1 
nþ 1 ln x-
1 
nþ 1 þ c , n≠ - 1 
S ¼ y1 xð Þ- y2 xð Þð Þdx 
) S ¼ 
2 
1 
xln x
p
- 0 dx ¼ 1 
2 
2 
1 
x ln xdx 
) S ¼ 1 
2 
x2 
2 
ln x-
1 
2 
2 
1 
¼ ln 2- 1 
2
-
1 
4 
0-
1 
2 
) S ¼ ln 2- 3 
8 
Choice (4) is the answer. 
In this problem, the rules below were used. 
ln ab ¼ b ln a 
ln 1 ¼ 0
� 1 ð Þ � 1
26 2 Solutions of Problems: Applications of Integration
2.17. From list of integral of functions, we know that: 
1 
u 
du ¼ ln u þ c 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
Also, from trigonometry, we know that: 
tanh x ¼ sinh x 
cosh x 
The problem can be solved as follows. 
S ¼ y1 xð Þ- y2 xð Þð Þdx 
S 
þ1 
1- tanh x dx 
þ1 
1-
sinh x 
cosh x 
dx) ¼ 
0 
ð Þ ¼ 
0 
) S ¼ x- ln cosh xð Þ½ �þ1 0 ¼ x- ln cosh xð Þð Þ x ¼ 1 - x- ln cosh xð Þð Þ x ¼ 0 
As we know, cosh x e
x 
2 when x → + . Therefore, x- ln cosh x x- ln 
ex 
2 , when x → + . Hence: 
) S ¼ x- ln e
x 
2 x ¼ 1 - x- ln cosh xð Þð Þ x ¼ 0 
) S ¼ x- ln ex - ln 2ð Þð Þ 
x ¼ 1 - x- ln cosh xð Þð Þ x ¼ 0 
) S ¼ ln 2ð Þ 
x ¼ 1 - x- ln cosh xð Þð Þ x ¼ 0 
) S ¼ ln 2- 0- ln 1ð Þ 
) S ¼ ln 2 
Choice (2) is the answer. 
In this problem, the rules below were used. 
tanh x dx ¼ ln cosh xð Þ 
ln 
a 
b 
¼ ln a- ln b 
ln ea ¼ a
¼
¼ ð Þ ¼
2.2 Surface Area Bounded by Curves 27
cosh 0 ¼ 1 
ln 1 ¼ 0 
Figure 2.4 The graph of problem 2.17 
2.18. Since the functions can be easily written in the form of x f( y), the formula below should be used. 
S ¼ x1 yð Þ- x2 yð Þð Þdy 
First, we need to determine the points of intersection of the two graphs of x1( y) y + 4 and x2 y 1 2 y
2. 
x1 xð Þ ¼ x2 xð Þ ) yþ 4 ¼ 1 2 y
2 
) y2 - 2y- 8 ¼ y- 4ð Þ yþ 2ð Þ ¼ 0 ) y ¼ - 2, 4 
S ¼ 
4
- 2 
y þ 4ð Þ- 1 
2 
y2 dy 
) S ¼ 1 
2 
y2 þ 4y- 1 
6 
y3 
4
- 2 
) S ¼ 1 
2
� 42 þ 4� 4- 1 
6
� 43 - 1 
2
� - 2ð Þ2 þ 4� - 2ð Þ- 1 
6
� - 2ð Þ3 
) S ¼ 18 
Choice (4) is the answer. 
In this problem, the rule below was used. 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c
28 2 Solutions of Problems: Applications of Integration
Figure 2.5 The graph of problem 2.18 
2.3 Volume Resulted from Rotation of an Enclosed Region 
2.19. The volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and x-axis is 
calculated as follows: 
V ¼ π 
x2 
x1 
f xð Þð Þ2 dx 
In addition, from list of integral of functions, we know that: 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
Therefore, based on the given information, we have: 
V ¼ π 
1
- 1 
x2 - x3 
p 2 
dx ¼ π 
1
- 1 
x2 - x3 dx 
V ¼ π x
3 
3
-
x4 
4 
1
- 1 
¼ π 1
3 
3
-
14 
4
- π
- 1ð Þ3 
3
-
- 1ð Þ4 
4 
⟹ V ¼ 2π 
3 
Choice (2) is the answer. 
2.20. The volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and x-axis is 
calculated as follows: 
V ¼ π 
x2 
x1 
f xð Þð Þ2 dx 
Therefore,
2.3 Volume Resulted from Rotation of an Enclosed Region 29
V ¼ π 
2π 
0 
sin 2 xð Þdx ¼ π 
2π 
0 
1 
2
-
cos 2xð Þ 
2 
dx 
⟹ V ¼ π x 
2
-
1 
4 
sin 2xð Þ 2π 
0 
¼ π π- 0ð Þ 
⟹ V ¼ π2 
Choice (1) is the answer. 
In this problem, the rules below were used. 
1- cos 2xð Þ ¼ 2 sin 2 xð Þ 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
cos axð Þdx ¼ 1 
a 
sin axð Þ þ c 
sin 4πð Þ ¼ 0 
sin 0ð Þ ¼ 0 
2.21. The volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and x-axis is 
calculated as follows: 
V ¼ π 
x2 
x1 
f xð Þð Þ2 dx 
In addition, from list of integral of functions, we know that: 
cos axð Þdx ¼ 1 
a 
sin axð Þ þ c 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
Also, from trigonometry, we know that: 
cos 2 x ¼ 1þ cos 2x 
2 
Therefore, based on the given information, we have: 
V ¼ π 
π 
2 
0 
cos 2 xdx ¼ π 
π 
2 
0 
1þ cos 2x 
2 
dx ¼ π 
2 
π 
2 
0 
1 þ cos 2xð Þdx 
⟹ V ¼ π 
2 
xþ 1 
2 
sin 2x 
π 
2 
0 
¼ π 
2 
π 
2
þ 1 
2 
sin 2� π 
2
- 0þ 0ð Þ
30 2 Solutions of Problems: Applications of Integration
⟹ V ¼ π
2 
4 
Choice (1) is the answer. 
2.22. The volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and x-axis is 
calculated as follows: 
V ¼ π 
x2 
x1 
f xð Þð Þ2 dx 
In addition, from list of integral of functions, we know that: 
eax dx ¼ e
ax 
a 
þ c 
Therefore: 
V ¼ π 
b 
0 
e- xð Þ2 dx ¼ π 
b 
0 
e- 2x dx 
) V ¼ - π 
2 
e- 2x 
b 
0 ¼ -
π 
2 
e- 2b - 1 
) lim 
b→1 
V ¼ - π 
2 
e-1 - 1ð Þ ¼ - π 
2 
0- 1ð Þ 
) lim 
b→1 
V ¼ π 
2 
Choice (2) is the answer. 
In this problem, the limit below was used. 
lim
x→1e
- x ¼ 0 
2.23. The surface area restricted by the function of y1(x) and y2(x) and the given boundaries can be calculated as follows. 
S ¼ 
x2 
x1 
y1 xð Þ- y2 xð Þð Þdx 
Moreover, the volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and 
x-axis is calculated as follows: 
V ¼ π 
x2 
x1 
f xð Þð Þ2 dx 
Therefore: 
S ¼ 
1 
0 
1 
x
p dx ¼ 
1 
0 
x-
1 
2dx
¼
2.3 Volume Resulted from Rotation of an Enclosed Region 31
) S ¼ 2x 1 2 
1 
0 
¼ 2- 0 
) S ¼ 2 
V ¼ π 
1 
0 
1 
x
p 
2 
dx ¼ π 
1 
0 
dx 
x 
) V ¼ π ln x½ �1 0 ¼ π 0- -1ð Þ½ �
) V ¼ 1 
Choice(3) is the answer. 
In this problem, the rules below were used. 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
1 
x 
dx ¼ ln xþ c 
ln 1 ¼ 0 
lim 
x→ 0þ 
ln x ¼ -1 
2.24. The volume resulted from the rotation of a surface area around y-axis, enclosed between the curve of f(x) and x-axis is 
calculated as follows: 
V ¼ π 
y2 
y1 
x2 dy 
Since x-axis is the boundary, y1 0. Another boundary for y can be determined as follows: 
x ¼ 0 ⟹ y2 ¼ 1- 1 4� 0 ¼ 1 
Therefore, 
V ¼ π 
1 
0 
x2 dy ¼ π 
1 
0 
4- 4yð Þdy 
⟹ V ¼ π 4y- 2y2 1 
0
¼ π 4- 2ð Þ 
⟹ V ¼ 2π 
Choice (2) is the answer. 
In this problem, the rule below was used.
32 2 Solutions of Problems: Applications of Integration
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
2.25. The volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and x-axis is 
calculated as follows: 
V ¼ π 
x2 
x1 
f xð Þð Þ2 dx 
Therefore, 
V ¼ π 
π 
2 
π 
4 
1 
sin 4 xð Þ dx ¼ π 
π 
2 
π 
4 
1 þ cot 2 xð Þ 1þ cot 2 xð Þ dx ð1Þ 
Now, we should change the variable of the integral as follows. 
cot xð Þ≜u ⟹ 1 þ cot 2 xð Þ dx ¼ - du ð2Þ 
Solving (1) and (2): 
V ¼ - π 
u2 
u1 
1þ u2 du 
⟹ V ¼ - π u þ 1 
3 
u3 
u2 
u1 
⟹ V ¼ - π cot xð Þ þ 1 
3 
cot 3 xð Þ 
π 
2 
π 
4 
V ¼ - π 0þ 0ð Þ- 1þ 1 
3 
⟹ V ¼ 4 
3 
π 
Choice (4) is the answer. 
In this problem, the rules below were used. 
1þ cot 2 xð Þ ¼ 1 
sin 2 xð Þ 
un du ¼ 1 
n þ 1 u
nþ1 þ c 
1þ cot 2 xð Þ dx ¼ - cot xð Þ þ c 
cot 
π 
2 
¼ 0
2.3 Volume Resulted from Rotation of an Enclosed Region 33
cot 
π 
4 
¼ 1 
2.26. The volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and x-axis is 
calculated as follows: 
V ¼ π 
x2 
x1 
f xð Þð Þ2 dx 
Therefore: 
V ¼ π 
π 
2 
π 
6 
cos 3x 
p 
sin 2 x 
2 
¼ π 
π 
2 
π 
6 
cos 3x 
sin 4 x 
¼ π 
π 
2 
π 
6 
cos x cos 2x 
sin 4 x 
⟹ V ¼ π 
π 
2 
π 
6 
cos x 1- sin 2 x 
sin 4 x 
dx ¼ π 
π 
2 
π 
6 
cos x 
sin 4 x
-
cos x 
sin 2 x 
dx 
⟹ V ¼ π - 1 
3 sin 3 x
þ 1 
sin x 
π 
2 
π 
6 
⟹ V ¼ π - 1 
3 sin 3 π 2 
þ 1 
sin π 2
-
- 1 
3 sin 3 π 6 
þ 1 
sin π 6 
¼ π - 1 
3 
þ 1 - - 8 
3
þ 2 
⟹ V ¼ 4π 
3 
Choice (3) is the answer. 
In this problem, the rules below were used. 
un du ¼ 1 
n þ 1 u
nþ1 þ c 
cos 2 x ¼ 1- sin 2 x 
sin 
π 
2 
¼ 1 
sin 
π 
6 
¼ 1 
2 
2.27. The volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and x-axis is 
calculated as follows: 
V ¼ π 
x2 
x1 
f xð Þð Þ2 dx 
In addition, from the method of integration by parts, we know that:
34 2 Solutions of Problems: Applications of Integration
udv ¼ uv- vdu 
Therefore: 
V ¼ π 
1 
0 
xexð Þ2 dx ¼ π 
1 
0 
x2 e2x dx 
By applying the method of integration by parts twice, we have: 
) V ¼ π 1 
2 
x2 e2x -
2 
4 
xe2x þ 2 
8 
e2x 
1 
0 
) V ¼ π 1 
2 
e2 -
1 
2 
e2 þ 1 
4 
e2 - 0- 0 þ 1 
4 
) V ¼ π 
4 
e2 - 1 
Choice (3) is the answer. 
2.28. The volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and x-axis is 
calculated as follows: 
V ¼ π 
x2 
x1 
f xð Þð Þ2 dx 
First, we need to arrange the relation based on y(x), as follows. 
y 
1 
2 xð Þ ¼ a 1 2 - x 1 2 ) y xð Þ ¼ aþ x- 2a 1 2x 1 2 
Moreover, the area is restricted by y-axis; therefore, x1 ¼ 0. The other boundary of the restricted area needs to be 
determined as follows. 
y xð Þ ¼ 0 ) a 1 2 - x 1 2 ¼ 0 ) x2 ¼ a 
Therefore: 
V ¼ π 
a 
0 
aþ x- 2a 1 2x 1 2 
2 
dx 
) V ¼ π 
a 
0 
a2 þ x2 þ 4axþ 2ax- 4a 3 2x 1 2 - 4a 1 2x 3 2 dx 
) V ¼ π 
a 
0 
a2 þ x2 þ 6ax- 4a 3 2x 1 2 - 4a 1 2x 3 2 dx 
) V ¼ π a2 xþ x
3 
3 
þ 3ax2 - 4a 3 2 � 2 
3 
x 
3 
2 - 4a 
1 
2 � 2 
5 
x 
5 
2 
a 
0
2.3 Volume Resulted from Rotation of an Enclosed Region 35
) V ¼ π a3 þ a
3 
3 
þ 3a3 - 4a 3 2 � 2 
3 
a 
3 
2 - 4a 
1 
2 � 2 
5 
a 
5 
2 - 0ð Þ ¼ π 1þ 1 
3
þ 3- 8 
3
-
8 
5 
a3 
) V ¼ 1 
15 
πa3 
Choice (4) is the answer. 
In this problem, the rule of factorization of polynomials was used as follows. 
a þ bþ cð Þ2 ¼ a2 þ b2 þ c2 þ 2abþ 2bcþ 2ac 
In addition, the integral below was applied. 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
2.29. The volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and x-axis is 
calculated as follows: 
V ¼ π 
x2 
x1 
f xð Þð Þ2 dx ð1Þ 
) V ¼ π 
π 
0 
e- x sin x 
p 2 
dx ¼ π 
π 
0 
e- 2x sin xdx ð2Þ 
Herein, let us first assume that: 
V ¼ πI ð3Þ 
where, 
I ¼ 
π 
0 
e- 2x sin xdx ð4Þ 
The integral can be solved by using the method of integration by parts twice as follows. 
udv ¼ uv- vdu ð5Þ 
u ¼ e- 2x ) du ¼ - 2e- 2x dx ð6Þ 
dv ¼ sin xdx ) v ¼ - cos x ð7Þ 
) I ¼ e- 2x sin xdx ¼ - e- 2x cos x- 2 e- 2x cos xdx ð8Þ 
u ¼ e- 2x ) du ¼ - 2e- 2x ð9Þ
36 2 Solutions of Problems: Applications of Integration
dv ¼ cos xdx ) v ¼ sin x ð10Þ 
) I ¼ - e- 2x cos x- 2 e- 2x sin xþ 2 e- 2x sin xdx ð11Þ 
Solving (4) and (11): 
) I ¼ - e- 2x cos x- 2e- 2x sin x- 4I ð12Þ 
) 5I ¼ - e- 2x cos x- 2e- 2x sin x ð13Þ 
) I ¼ - e
- 2x 
5 
cos x þ 2 sin xð Þ ð14Þ 
Solving (3) and (14): 
) V ¼ - π 
5 
e- 2x cos xþ 2 sin xð Þ π 0 ¼ -
π 
5
- e- 2π - 1 ð15Þ 
) V ¼ π 
5 
1 þ e- 2π 
Choice (3) is the answer. 
In this problem, the rules below were used. 
cos π ¼ - 1 
sin π ¼ 0 
cos 0 ¼ 1 
sin 0 ¼ 0 
2.30. The volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and x-axis is 
calculated as follows: 
V ¼ π 
x2 
x1 
f xð Þð Þ2 dx 
As can be noticed from Fig. 2.6, the volume of shaded region can be achieved by subtracting the volume of empty space 
from the volume of sphere as follows: 
V ¼ V2 -V1 ¼ π 
2 
0 
y2 1 xð Þ- y2 2 xð Þ dx 
Now, we need to determine the y1(x) and y2(x) as follows. 
x- 1ð Þ2 þ y2 ¼ 1 ) y2 ¼ 1- x- 1ð Þ2 ) y1 xð Þ ¼ 1- x- 1ð Þ2 ) y1 xð Þ ¼ 2x- x2 
p 
y2 xð Þ ¼ 2x- x2
2.4 Arc Length of a Curve 37
) V ¼ π 
2 
0 
2x- x2 
p 2
- 2x- x2 
2 
dx 
) V ¼ π 
2 
0 
2x- x2 - 4x2 þ 4x3 - x4 dx ¼ π 
2 
0
- x4 þ 4x3 - 5x2 þ 2x dx 
) V ¼ π - x
5 
5 
þ x4 - 5x
3 
3 
þ x2 2 
0 
) V ¼ π - 2ð Þ
5 
5 
þ 24 - 5 2ð Þ
3 
3 
þ 22 - 0 ¼ π - 96þ 240- 200 þ 60 
15 
) V ¼ 4π 
15 
Choice (1) is the answer. 
In this problem, the integral formula below was used. 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
Figure 2.6 The graph of problem 1.30 
2.4 Arc Length of a Curve 
2.31. The arc length of a curve can be calculated as follows. 
L ¼ 
b 
a 
1 þ f 0 xð Þð Þ2 dx 
Therefore, first we need to determine f ′ (x) as follows:
xp p
p
38 2 Solutions of Problems: Applications of Integration
f xð Þ ¼ 
1 
t4 - 1dt ) f 0 xð Þ ¼ x4 - 1 
Then: 
L ¼ 
3 
1 
1þ x4 - 1 
p 2 
dx ¼ 
3 
1 
1þ x4 - 1dx ¼ 
3 
1 
x2 dx 
) L ¼ 1 
3 
x3 
3 
1 
¼ 1 
3 
27- 1ð Þ 
) L ¼ 26 
3 
Choice (3) is the answer. 
In this problem, the rules below were used. 
F xð Þ ¼ 
u xð Þ 
v xð Þ 
f tð Þdt ) F0 xð Þ ¼ u0 xð ÞF u xð Þð Þ- v0 xð ÞF v xð Þð Þ 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
2.32. The arc length of a parametric curve can be calculated as follows. 
L ¼ 
b 
a 
x0t 
2 þ y0t 2 dt 
From the problem, we have: 
x tð Þ ¼ et cos t 
y tð Þ ¼ et sin t 
Thus: 
L ¼ 
4 
0 
etcos t- et sin tð Þ2 þ etsin t þ et cos tð Þ2 dt 
) L ¼ 
4 
0 
et cos 2t þ sin 2 t- 2 sin t cos t þ sin 2 t þ cos 2t þ 2 sin t cos tdt 
) L ¼ 
4 
0 
et 1þ 1p ¼ 2 
p 
et 
4 
0 
) L ¼ 2 e4 - 1 
Choice (1) is the answer. 
In this problem, the rules below were used.
Þ
2.4 Arc Length of a Curve 39
d 
dx 
u xð Þv xð Þð Þ ¼ u0 xð Þv xð Þ þ v0 xð Þu xð Þ 
d 
dx 
ex ¼ ex 
cos 2 xþ sin 2 x ¼ 1 
ex ¼ ex þ c 
2.33. The arc length of a parametric curve can be calculated as follows. 
L ¼ 
b 
a 
x0t 
2 þ y0t 2 dt 
From the problem, we have: 
x tð Þ ¼ et cos t þ sin tð Þ 
y tð Þ ¼ et cos t- sin tð Þ 
Thus: 
L ¼ 
1 
0 
et cos t þ sin tð Þ þ et - sin t þ cos tð Þð Þ2 þ et cos t- sin tð Þ þ et - sin t- cos tð Þð 2 dt 
) L ¼ 
1 
0 
2etcos tð Þ2 þ - 2et sin tð Þ2 dt ¼ 
1 
04e2t cos 2t þ 4e2tsin2 tdt ¼ 
) L ¼ 
1 
0 
4e2t cos 2t þ sin 2 t dt ¼ 
1 
0 
4e2t 
p 
dt ¼ 
1 
0 
2et dt 
) L ¼ 2et 1 
0 
) L ¼ 2 e- 1ð Þ 
Choice (4) is the answer. 
In this problem, the rules below were used. 
d 
dx 
u xð Þv xð Þð Þ ¼ u0 xð Þv xð Þ þ v0 xð Þu xð Þ 
d 
dx 
ex ¼ ex 
cos 2 xþ sin 2 x ¼ 1 
ex ¼ ex þ c
40 2 Solutions of Problems: Applications of Integration
2.34. The arc length of a curve can be calculated as follows. 
L ¼ 
b 
a 
1þ y0 xð Þð Þ2 dx 
Therefore, first we need to determine f ′ (x) as follows: 
y xð Þ ¼ 1 
2 
x2 -
1 
4 
ln x ) y0 xð Þ ¼ x- 1 
4x 
Then: 
L ¼ 
2 
1 
1þ x- 1 
4x 
2 
dx ¼ 
2 
1 
1 þ x2 - 1 
2
þ 1 
16x2
dx ¼ 
2 
1 
x2 þ 1 
2
þ 1 
16x2
dx 
) L ¼ 
2 
1 
x þ 1 
4x 
2 
dx ¼ 
2 
1 
xþ 1 
4x 
dx 
) L ¼ x
2 
2 
þ 1 
4 
ln x 
2 
1 
¼ 2
2 
2 
þ 1 
4 
ln 2 -
1 
2
þ 1 
4 
ln 1 
) L ¼ 3 
2
þ 1 
4 
ln 2 
Choice (2) is the answer. 
In this problem, the rules below were used. 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
a 
x 
dx ¼ a ln xþ c 
d 
dx 
xn ¼ nxn- 1 
d 
dx 
ln x ¼ 1 
x 
ln 1 ¼ 0 
2.35. The arc length of a curve can be calculated as follows. 
L ¼ 
b 
a 
1þ y0 xð Þð Þ2 dx 
Therefore, first we need to determine f ′ (x) as follows: 
y xð Þ ¼ x
4 
8 
þ 1 
4x2 
) y0 xð Þ ¼ x
3 
2
-
1 
2x3 
Then:
2.4 Arc Length of a Curve 41
L ¼ 
2 
1 
1þ x
3 
2
-
1 
2x3 
2 
dx ¼ 
2 
1 
1þ x
6 
2
-
1 
2
þ 1 
4x6
dx 
) L ¼ 
2 
1 
x6 
2 
þ 1 
2
þ 1 
4x6
dx ¼ 
2 
1 
x3 
2 
þ 1 
2x3 
2 
dx ¼ 
2 
1 
x3 
2 
þ 1 
2x3 
dx 
) L ¼ x
4 
8
-
1 
4x2 
2 
1 
¼ 2
4 
8
-
1 
4� 22 -
14 
8
-
1 
4� 12 ¼ 2-
1 
16
-
1 
8
þ 1 
4 
) L ¼ 33 
16 
Choice (4) is the answer. 
In this problem, the rules below were used. 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
d 
dx 
xn ¼ nxn- 1 
2.36. The arc length of a curve can be calculated as follows. 
L ¼ 
b 
a 
1 þ f 0 xð Þð Þ2 dx 
Therefore, first we need to determine f ′ (x) as follows: 
f xð Þ ¼ ln sec xð Þ ) f 0 xð Þ ¼ tan x 
Then, 
L ¼ 
π 
4 
0 
1þ tan 2xdx ¼ 
π 
4 
0 
1 
cos 2x 
dx ¼ 
π 
4 
0 
1 
cos x 
dx ¼ 
π 
4 
0 
sec xdx 
) L ¼ ln sec xþ tan xð Þ 
π 
4 
0 
¼ ln sec π 
4
þ tan π 
4
- ln sec 0þ tan 0ð Þ ¼ ln 2 
p 
þ 1 - ln 1þ 0ð Þ 
) L ¼ ln 1 þ 2 
p 
Choice (4) is the answer. 
In this problem, the rules below were used. 
d 
dx 
ln sec xð Þ ¼ d 
dx 
ln 
1 
cos x 
¼ 
sin x 
cos 2x 
1 
cos x 
¼ sin x cos x 
cos 2x 
¼ sin x 
cos x 
¼ tan x
42 2 Solutions of Problems: Applications of Integration
1þ tan 2 x ¼ 1 
cos 2x 
¼ sec 2 x 
sec xdx ¼ ln sec xþ tan xð Þ þ c 
sec 
π 
4 
¼ 2 
p 
tan 
π 
4 
¼ 1 
sec 0 ¼ 1 
tan 0 ¼ 0 
2.37. The arc length of a curve can be calculated as follows. 
L ¼ 
b 
a 
1 þ f 0 xð Þð Þ2 dx 
Therefore, first we need to determine f ′ (x) as follows: 
f xð Þ ¼ 
x 
1 
cosh tð Þdt ) f 0 xð Þ ¼ cosh xð Þ 
Then: 
L ¼ 
2 
0 
1þ cosh xð Þ 
2 
dx ¼ 
2 
0 
1þ cosh xð Þdx 
) L ¼ 
2 
0 
2 cosh 2 
x 
2 
dx ¼ 
2 
0 
2 
p 
cosh 
x 
2 
dx 
) L ¼ 2 2 
p 
sinh 
x 
2 
2 
0 
¼ 2 2 
p 
sinh 
2 
2
- sinh 0 ¼ 2 2 
p 
sinh 1 
) L ¼ 2 2 
p e1 - e- 1 
2 
) L ¼ 2 
p 
e-
1 
e 
Choice (1) is the answer. 
In this problem, the rules below were used. 
F xð Þ ¼ 
u xð Þ 
v xð Þ 
f tð Þdt ) F0 xð Þ ¼ u0 xð ÞF u xð Þð Þ- v0 xð ÞF v xð Þð Þ 
1 þ cosh xð Þ ¼ 2 cosh 2 x 
2
2.4 Arc Length of a Curve 43
cosh axð Þ ¼ 1 
a 
sinh axð Þ þ c 
sinh 1 ¼ e
1 - e- 1 
2 
sinh 0 ¼ 0 
2.38. The arc length of a curve can be calculated as follows. 
L ¼ 
b 
a 
1 þ f 0 xð Þð Þ2 dx 
Therefore, first we need to determine f ′ (x) as follows: 
f xð Þ ¼ 1 
2 
x x2 - 1 
p
- ln xþ x2 - 1 
p 
) f 0 xð Þ ¼ 1 
2 
x2 - 1 
p 
þ x
2 
x2 - 1 
p -
1þ x 
x2 - 1 
p 
xþ x2 - 1 
p 
) f 0 xð Þ ¼ 1 
2 
x2 - 1 
p 
þ x
2 
x2 - 1 
p - x
2 - 1 
p 
þ x 
xþ x2 - 1 
p 
x2 - 1 
p 
) f 0 xð Þ ¼ 1 
2 
x2 - 1 
p 
þ x
2 
x2 - 1 
p - 1 
x2 - 1 
p 
) f 0 xð Þ ¼ 1 
2 
x2 - 1þ x2 - 1 
x2 - 1 
p ¼ x
2 - 1 
x2 - 1 
p ¼ x2 - 1 
p 
Then: 
L ¼ 
2 
1 
1þ x2 - 1 
p 2 
dx ¼ 
2 
1 
1þ x2 - 1ð Þdx ¼ 
2 
1 
xdx 
) L ¼ x
2 
2 
2 
1 
¼ 2
2 
2
-
12 
2 
) L ¼ 3 
2 
Choice (2) is the answer. 
In this problem, the integral formula below was used. 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
d 
dx 
u xð Þv xð Þð Þ ¼ u0 xð Þv xð Þ þ v0 xð Þu xð Þ
44 2 Solutions of Problems: Applications of Integration
d 
dx 
u xð Þ ¼ u
0 xð Þ 
2 u xð Þ 
d 
dx 
ln u xð Þ ¼ u
0 xð Þ 
u xð Þ 
2.39. The arc length of a curve can be calculated as follows. 
L ¼ 
b 
a 
1 þ f 0 xð Þð Þ2 dx 
Therefore, first we need to determine f ′ (x) as follows: 
f xð Þ ¼ ln e
x þ 1 
ex - 1 
¼ ln ex þ 1ð Þ- ln ex - 1ð Þ 
) f 0 xð Þ ¼ e
x 
ex þ 1 -
ex 
ex - 1 
¼ e
x ex - 1ð Þ- ex ex þ 1ð Þ 
e2x - 1
¼ - 2e
x 
e2x - 1 
Then, 
L ¼ 
2 
0 
1 þ - 2e
x 
e2x - 1 
2 
dx ¼ 
2 
0 
1þ 4e
2x 
e2x - 1ð Þ2dx 
) L ¼ 
2 
0 
e2x - 1ð Þ2 þ 4e2x 
e2x - 1ð Þ2 dx ¼ 
2 
0 
e4x - 2e2x þ 1þ 4e2x 
e2x - 1ð Þ2 dx 
) L ¼ 
2 
0 
e4x þ 2e2x þ 1 
e2x - 1ð Þ2 dx ¼ 
2 
0 
e2x þ 1ð Þ2 
e2x - 1ð Þ2dx 
) L ¼ 
2 
0 
e2x þ 1 
e2x - 1 
dx ¼ 
2 
0 
ex þ e- x 
ex - e- x 
dx 
) L ¼ ln ex - e- xð Þ 2 
1 
¼ ln e2 - e- 2 - ln e1 - e- 1 
) L ¼ ln e
2 - e- 2 
e- e- 1 
¼ ln e- e
- 1ð Þ eþ e- 1ð Þ 
e- e- 1ð Þ ¼ ln eþ e
- 1 
) L ¼ ln eþ 1 
e 
Choice (2) is the answer. 
In this problem, the rules below were used. 
d 
dx 
ln u xð Þ ¼ u
0 xð Þ 
u xð Þ
þ ¼
þ ¼
þ
2.5 Surface Area of a Solid of Revolution 45
d 
dx 
ex ¼ ex 
1 
u xð Þ du ¼ ln u xð Þ þ c 
2.5 Surface Area of a Solid of Revolution 
2.40. If the function of f(x) is rotated around x-axis, the surface area of the solid of revolution can be calculated as follows. 
S ¼ 2π 
b 
a 
f xð Þ 1þ f 0 xð Þð Þ2 dx 
Moreover, if the function is rotated around y-axis, the surface area of the solid of revolution is calculated as follows. 
S ¼ 2π 
b 
a 
x 1þ f 0 xð Þð Þ2 dx 
For this problem, we have: 
S ¼ 2π 
2 
0 
cosh x 1þ sinh 2 x dx ¼ 2π 
2 
0 
cosh x cosh 2 x dx 
) S ¼ 2π 
2 
0 
cosh 2 x dx ¼ 2π 
2 
0 
1 
2 
1þ cosh 2xð Þdx 
) S ¼ π x þ sinh 2x 
2 
2 
0 
¼ π 2þ sinh 4 
2
- 0 
) S ¼ π 
2 
4 þ sinh 4ð Þ 
Choice (4) is the answer. 
In this problem, the rules below were used. 
d 
dx 
cosh x ¼ sinh x 
sinh 2 x cosh 2 x 1 
1 cosh 2x 2 cosh 2 x 
xn dx ¼ 1 
n 1 
xnþ1 þ c 
cosh ax ¼ sinh ax 
a 
þ c
¼
2.42. If the function of f (x) is rotated around x-axis, the surface area of the solid of revolution is calculated as follows.
46 2 Solutions of Problems: Applications of Integration
sinh 0 0 
S ¼ 2π 
b 
a 
f xð Þ 1þ f 0 xð Þð Þ2 dx 
Moreover, if the function is rotated around y-axis, the surface area of the solid of revolution is calculated as follows. 
S ¼ 2π 
b 
a 
x 1þ f 0 xð Þð Þ2 dx 
For this problem, we have: 
3y xð Þ- x3 ¼ 0 ) y xð Þ ¼ 1 
3 
x3 
) S ¼ 2π 
1 
0 
1 
3 
x3 1 þ x2ð Þ2 ¼ 2π 
3 
1 
0 
x3 1þ x4dx 
The integral can be solved by defining a new variable as follows. 
1þ x4 ¼ u 
4x3 dx ¼ du ) x3 dx ¼ du 
4 
) S ¼ 2π 
3 
u 
1 
2 
du 
4 
¼ π 
6 
u 
1 
2du ¼ π 
6 
2 
3 
u 
3 
2 
) S ¼ π 
6 
2 
3 
1þ x4 
3 
2 
1 
0 
¼ π 
9 
1þ 1ð Þ3 2 - 1þ 0ð Þ3 2 
) S ¼ π 
9 
2 2 
p
- 1 
Choice (1) is the answer. 
In this problem, the rule below was used. 
un dx ¼ 1 
n þ 1 u
nþ1 þ c 
2.41. If the function of f(x) is rotated around x-axis, the surface area of the solid of revolution is calculated as follows. 
S ¼ 2π 
b 
a 
f xð Þ 1þ f 0 xð Þð Þ2 dx 
Moreover, if the function is rotated around y-axis, the surface area of the solid of revolution is calculated as follows.
2.5 Surface Area of a Solid of Revolution 47
S ¼ 2π 
b 
a 
x 1þ f 0 xð Þð Þ2 dx 
From y ¼ x2 and y ≤ 2, we have - 2p ≤ x≤ 2p ; however, the range of integration must be 0≤ x≤ 2p . Therefore: 
S ¼ 2π 
2
p 
0 
x 1þ 2xð Þ2 dx ¼ 2π 
2
p 
0 
x 1þ 4x2dx 
The problem can be solved by defining a new variable as follows. 
u ¼ 1þ 4x2 ) du ¼ 8xdx ) du 
8 
¼ xdx 
S ¼ 2π 
8 
2
p 
0 
u
p 
du ¼ π 
4 
2 
3 
u 
3 
2 
) S ¼ π4 
2 
3 
1þ 4x2 
3 
2 
2
p 
0 
¼ π 
6 
1þ 4� 2 
p 2 3 2
- 1 ¼ π 
6 
9 
3 
2 - 1 ¼ π 
6 
27- 1ð Þ ¼ 26π 
6 
S ¼ 13π 
3 
Choice (2) is the answer. 
In this problem, the rule below was used. 
un du ¼ 1 
n þ 1 u
nþ1 þ c 
2.42. The volume resulted from the rotation of a surface area around x-axis, enclosed between the curve of f(x) and x-axis is 
calculated as follows: 
V ¼ π 
x2 
x1 
f xð Þð Þ2 dx 
In addition, if the function of f(x) is rotated around x-axis, the surface area of the solid of revolution is calculated as 
follows. 
S ¼ 2π 
b 
a 
f xð Þ 1þ f 0 xð Þð Þ2 dx 
Moreover, if the function is rotated around y-axis, the surface area of the solid of revolution is calculated as follows. 
S ¼ 2π 
b 
a 
x 1þ f 0 xð Þð Þ2 dx 
For this problem, we have: 
V ¼ π 
5 
0 
6cosh 
x 
6 
2 
dx ¼ 36π 
5 
0 
cosh 2 
x 
6 
dx 
Moreover:
48 2 Solutions of Problems: Applications of Integration
S ¼ 2π 
b 
a 
6 cosh 
x 
6 
1þ sinh x 
6 
2 
dx ¼ 2π 
b 
a 
6 cosh 
x 
6 
cosh 2 
x 
6 
dx 
) S ¼ 12π 
b 
a 
6 cosh 2 
x 
6 
dx 
Therefore: 
V 
S 
¼ 
36π 
5 
0 
cosh 2 x 6 dx 
12π 
5 
0 
cosh 2 x 6 dx 
) V 
S 
¼ 3 
Choice (1) is the answer. 
In this problem, the rule below was used. 
cosh 2 x- sinh 2 x ¼ 1 
2.43. If the function of f(x) is rotated around x-axis, the surface area of the solid of revolution is calculated as follows. 
S ¼ 2π 
b 
a 
f xð Þ 1þ f 0 xð Þð Þ2 dx 
Moreover, if the function is rotated around y-axis, the surface area of the solid of revolution is calculated as follows. 
S ¼ 2π 
b 
a 
x 1þ f 0 xð Þð Þ2 dx 
Therefore: 
S ¼ 2π 
ln 1396ð Þ 
p 
0 
x 1þ sinh 2 x2ð Þdx ¼ 2π 
ln 1396ð Þ 
p 
0 
x cosh x2 dx 
) S ¼ 2π 1 
2 
sinh x2 
ln 1396ð Þ 
p 
0 
¼ π sinh ln 1396ð Þð Þ- 0 ¼ π sinh ln 1396ð Þð Þ 
) S ¼ π e
ln 1396 - e- ln 1396 
2 
) S ¼ π 
2 
1396-
1 
1396 
Choice (4) is the answer. 
In this problem, the rules below were used.
¼
¼
a
ð Þ ð Þð Þ
¼
2.6 Center of Gravity 49
F xð Þ ¼ 
u xð Þ 
v xð Þ 
f tð Þdt ) F0 xð Þ ¼ u0 xð ÞF u xð Þð Þ- v0 xð ÞF v xð Þð Þ 
cosh 2 x- sinh 2 x ¼ 1 
cosh u xð Þð Þdu ¼ sinh u xð Þð Þ þ c 
sinh 0ð Þ ¼ 0 
sinh u ¼ e
u - e- u 
2 
e ln a ¼ a 
e- ln a ¼ 1 
a 
2.6 Center of Gravity 
2.44. The center of gravity of a flat surface from y-axis which is restricted by the functions of y1(x) and y2(x) can be calculated 
as follows. 
x 
b 
a 
x y1 xð Þ- y2 xð Þð Þdx 
S 
where, S is the whole surface area. 
S 
b 
y1 x - y2 x dx 
In addition, the center of gravity of a flat surface from x-axis which is restricted by the functions of y1(x) and y2(x) can be 
calculated as follows. 
y 
1 
2 
b 
a 
y1 xð Þ- y2 xð Þð Þ2 dx 
S 
Therefore, for this problem, we have: 
y xð Þ ¼ 0 ) 1- x2 ¼ 0 ) x ¼ ± 1
50 2 Solutions of Problems: Applications of Integration
y ¼ 
1 
2 
1
- 1 
1- x2ð Þ2 dx 
1
- 1 
1- x2ð Þdx 
¼ 
1 
0 
1- 2x2 þ x4ð Þdx 
2 
1 
0 
1- x2ð Þdx 
) y ¼ x-
2 
3 x
3 þ 1 5 x5 
1 
0 
2 x- 1 3 x
3 1 
0 
¼ 1-
2 
3 þ 1 5 - 0 
2 1- 1 3 
¼ 
15- 10þ3 
15 
4 
3 
¼ 
8 
15 
4 
3 
) y ¼ 2 
5 
Choice (4) is the answer. 
In this problem, the rules below were used. 
a
- a 
f xð Þdx ¼ 2 
a 
0 
f xð Þdx , if f xð Þ is an even function 
xn dx ¼ 1 
n þ 1 x
nþ1 þ c 
2.46. The center of gravity of a flat surface from y-axis which is restricted by the functions of y1(x) and y2(x) can be calculated 
as follows. 
x ¼ 
b 
a 
x y1 xð Þ- y2 xð Þð Þdx 
S 
where, S is the whole surface area. 
S ¼ 
b 
a 
y1 xð Þ- y2 xð Þð Þdx 
In addition, the center of gravity of a flat surface from x-axis which is restricted by the functions of y1(x) and y2(x) can be 
calculated as follows. 
y ¼ 
1 
2 
b 
a 
y1 xð Þ- y2 xð Þð Þ2 dx 
S 
Therefore, for this problem, we have: 
y xð Þ ¼ 0 ) 1- x2 ¼ 0 ) x ¼ ± 1 
x ¼ 
1
- 1 
x 1- x2ð Þdx 
1
- 1 
1- x2ð Þdx 
The function of x(1 - x2 ) is an odd function; therefore, its integral over a symmetric boundary is zero. In other words:
Thus:
2.6 Center of Gravity 51
1 
2 
1
- 1 
x 1- x2 dx ¼ 0 
) x ¼ 0 
Choice (1) is the answer. 
In this problem, the rule below was used. 
a
- a 
f xð Þdx ¼ 0 , if f xð Þ is an odd function 
2.47. The center of gravity of a flat curve from y-axis can be calculated as follows. 
x ¼ 
b 
a 
x 1 þ y0 xð Þð Þ2 dx 
L 
where, L is the whole length of curve. 
L ¼ 
b 
a 
1þ y0 xð Þð Þ2 dx 
In addition, the center of gravity of a flat curve from y-axis can be calculated as follows. 
y ¼ 
b 
a 
y 1 þ y0 xð Þð Þ2 dx 
L 
Moreover, if the function is in the parametric form of x ¼ x(t) and y ¼ y(t), the center of gravity of a flat curve from 
y-axis and x-axis can be calculated as follows, respectively. 
x ¼ 
b 
a 
x xð Þ x0t 2 þ y0t 2 dt 
L 
y ¼ 
b 
a 
y xð Þ x0t 2 þ y0t 2 dt 
L 
where, L is the whole length of curve. 
L ¼ 
b 
a 
x0t 
2 þ y0t 2 dt 
Therefore, for this problem, we have:
52 2 Solutions of Problems: Applications of Integration
y ¼ 
π 
0 
y xð Þ x0t 2 þ y0t 2 dt 
π 
0 
x0t 
2 þ y0t 2 dt 
¼ 
π 
0 
1- cos tð Þ 1 þ cos tð Þ2 þ sin tð Þ2 dt 
π 
0 
1þ cos tð Þ2 þ sin tð Þ2 dt 
) y ¼ 
π 
0 
1- cos tð Þ 1þ 2 cos t þ cos tð Þ2 þ sin tð Þ2 dt 
π 
0 
1þ 2 cos t þ cos tð Þ2 þ sin tð Þ2 dt 
¼ 
π 
0 
1- cos tð Þ 2 1þ cos tð Þdt 
π 
0 
2 1þ cos tð Þdt 
) y ¼ 
π 
0 
2 sin 2 t 2 4 cos
2 t 
2dt 
π 
0 
4 cos 2 t 2dt 
¼ 
π 
0 
2sin 2 t 2 2cos 
t 
2 dt 
π 
0 
2 cos t 2 dt 
¼ 
π 
0 
4 sin 2 t 2 cos 
t 
2 dt 
π 
0 
2 cos t 2 dt 
) y ¼ 
8 
3 sin
3 t 
2 
π 
0 
4 sin t 2 
π 
0 
¼ 
8 
3 sin
3 π 
2 - sin
3 0 
4 sin π 2 - sin 0 
¼ 
8 
3 
4 
) y ¼ 2 
3 
Choice (3) is the answer. 
In this problem, the rules below were used. 
cos tð Þ2 þ sin tð Þ2 ¼ 1 
1þ cos t ¼ 2 cos 2 t 
2 
1- cos t ¼ 2 sin 2 t 
2 
cos axdx ¼ 1 
a 
sin ax þ c 
un du ¼ 1 
n þ 1 u
nþ1 þ c 
sin 
π 
2 
¼ 1 
sin 0 ¼ 0 
References 
1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
1
Problems: Sequences and Series 
and Their Applications 3 
Abstract 
In this chapter, the basic and advanced problems of sequences and series as well as their applications are presented. The 
subjects include limit of general term of a sequences and series, convergence or divergence status of a sequences and series, 
convergence range and radius of series, Maclaurin expansion, Taylor expansion, concept of growth rate, concept of 
equivalent functions, P-series, harmonic series, telescoping series, and Stirling’s approximation. In this chapter, the 
problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation 
amounts (small, normal, and large). Additionally, the problems are ordered from the easiest problem with the smallest 
computations to the most difficult problems with the largest calculations. 
3.1. Consider the sequence below and calculate its limit [1–3]. 
2n 
nþ 2ð Þ! 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 0 
2) 
3) 1 
4) Not available 
3.2. Consider the sequence below and calculate its limit. 
n!þ 2n8 þ ln n 
n!þ 5n þ 4n 
8 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 0 
2) 1 
3) 2 
4) Not available 
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 
M. Rahmani-Andebili, Calculus II, https://doi.org/10.1007/978-3-031-45353-3_3
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1
54 3 Problems: Sequences and Series and Their Applications
3.3. The general term of a sequence is as follows. Calculate the limit of the sequence. 
an ¼ n
p 
4n3 þ sin n þ 1p þ
n
p 
4n3 þ sin n þ 2p þ⋯þ
n
p4n3 þ sin n þ np 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
1 
2 
2) 1 
3) 
4) 0 
3.4. Determine the coefficient of x3 in the Maclaurin expansion of the function of f (x) for the following information. 
f 0ð Þ ¼ 1 
f 0 xð Þ ¼ 1þ f xð Þð Þ10 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
140 
3 
2) 
190 
3 
3) 280 
4) 380 
3.5. Some of the terms of a sequence is as follows. Calculate the limit of the sequence. 
1, -
2 
2 
, 
3 
4 
,
- 4 
8 
, 
5 
16 
,⋯ 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 0 
2) 
1 
2 
3) 
3 
4 
4) -
1 
2 
3.6. Determine the status of the series below. 
S ¼ 
þ1 
n¼1 
1þ cos n 
n2 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) Convergent 
2) Divergent
1
3 Problems: Sequences and Series and Their Applications 55
3) Can be convergent or divergent 
4) Impossible to determine 
3.7. Determine the status of the series below. 
S ¼ 
þ1 
n¼1 
10n2 þ 9nþ 8 
12n3 þ 11n2 þ 10nþ 9 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) Convergent 
2) Divergent 
3) Can be convergent or divergent 
4) Impossible to determine 
3.8. Calculate the final answer of the following term. 
S ¼ 
1 
n¼1 
nþ 1p - np 
n2 þ np 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1 
2 
2) 1 
3) 2 
4) 
3.9. Determine the convergence range of the series below. 
S ¼ 
1 
n¼1 
n1390 xn 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) [-1, 1] 
2) (-1, 1) 
3) (-1, 1] 
4) [-1, 1) 
3.10. Determine the convergence radius of the series below. 
S ¼ 
1 
n¼0 
n! xn 
3n
2 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 0 
2) 1
1
1
¼
p
1 1
1 1
1 1
1 1
56 3 Problems: Sequences and Series and Their Applications
3) 2 ln 3 
4) 
3.11. Determine the convergence radius of the series below. 
S ¼ 
1 
n¼1 
n!ð Þxn 
nþ 1ð Þn 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
1 
e 
2) e 
3) 
1 
e 
4) 
3.12. Determine the Taylor series of the function below around x 2. 
f xð Þ ¼ 1 
5- x 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) f xð Þ ¼ 1 
3
þ 1 
32 
x- 2ð Þ þ 1 
33 
x- 2ð Þ2 þ ⋯ 
2) f xð Þ ¼ 1 
5 
þ x- 2ð Þ þ x- 2ð Þ2 þ⋯ 
3) f xð Þ ¼ 1 
3
-
1 
32 
x- 2ð Þ þ 1 
33 
x- 2ð Þ3 -⋯ 
4) f xð Þ ¼ 1 
5 
1- x- 2ð Þ þ x- 2ð Þ2 þ x- 2ð Þ3 þ⋯ 
3.13. The general term of two sequence are as follows. What are the limits of the sequences? 
an ¼ n2 - n33
p 
þ n 
bn ¼ n
2 þ 1- np 
n3 þ n4p þ np 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
n 
lim
⟶ 
an ¼ 1, limn⟶ bn ¼ 0 
2) 
n 
lim
⟶ 
an ¼ 1, limn⟶ bn ¼ 1 
3) 
n 
lim
⟶ 
an ¼ 1 3 , limn⟶ bn ¼ 1 
4) 
n 
lim
⟶ 
an ¼ 1 3 , limn⟶ bn ¼ 0
1
p
þ þ þ
þ
3 Problems: Sequences and Series and Their Applications 57
3.14. The general term of a sequence is as follows. Calculate the limit of the sequence. 
an ¼ 
1þ 1 
2
þ 1 
3 
þ⋯þ 1 
n3 
ln n 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 0 
2) 
1 
3 
3) 3 
4) 
3.15. The general term of a sequence is as follows. Calculate the limit of the sequence. 
an ¼ 1 
n2 þ 1p þ 
1 
n2 þ 2p þ ⋯ 
1 
n2 þ np 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) ln 2 
2) 
1 
2 
3) 1 
4) Now available 
3.16. Calculate the sum of the series below. 
S ¼ 1 
1 
þ 1 
1 þ 2 þ
1 
1þ 2þ 3þ
1 
1þ 2þ 3þ 4þ ⋯ 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2 
2) 
5 
2 
3) 
9 
4 
4) 2 ln 2 
3.17. Which one of the sequences is divergent? 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) n!n
p 
2) 
n 
n2 1
þ n 
n2 2
þ⋯þ n 
n2 n 
3) 3n þ 2nn
p 
4) 
an - bn 
an bn 
, a> 0, b> 0
3.18.
58 3 Problems: Sequences and Series and Their Applications
Which one of the following series is convergent? 
S1 ¼ 
1 
n¼1 
n- 1 
n 
n 
S2 ¼ 
1 
n¼1 
nþ 1 
n 
n 
S3 ¼ 
1 
n¼1 
nþ 1 
n 
n2 
S4 ¼ 
1 
n¼1 
1 
n n
p 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) S1 
2) S2 
3) S3 
4) S4 
References 
1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
1
Solutions of Problems: Sequences and Series 
and Their Applications 4 
Abstract 
In this chapter, the problems of the third chapter are fully solved, in detail, step-by-step, and with different methods. 
4.1. From the concept of growth rate, for n ⟶ 1, a, b > 1 and k > 0, we know that the order of growth rate is as follows 
[1–3]. 
log a n< n
k < bn < n!< nn 
Thus, for this problem, we have: 
lim
n⟶1 
2n 
nþ 2ð Þ! � limn⟶1 
1 
nþ 2ð Þ! ¼ 
1 
1! ¼ 0 
) lim
n⟶1 
2n 
nþ 2ð Þ! ¼ 0 
Choice (1) is the answer. 
4.2. The problem can be solved as follows. 
lim
n⟶1 
n!þ 2n8 þ ln n 
n!þ 5n þ 4n 
8 � lim 
n→þ1 
n! 
n! 
8 ¼ lim 
n→þ11 ¼ 1 
) lim
n⟶1 
n!þ 2n8 þ ln n 
n! þ 5n þ 4n 
8 ¼ 1 
Choice (2) is the answer. 
In this problem, the rule below was used. 
Based on the concept of growth rate, for n ⟶ , a, b > 1 and k > 0, the order of growth rate is as follows. 
log a n< n
k < bn < n!< nn 
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 
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p p p
!
¼
60 4 Solutions of Problems: Sequences and Series and Their Applications
4.3. From the concept of equivalent functions, we know that for a > 0 and an even k: 
lim
n⟶1 an
k þ bnk- 1 þ . . .k ¼ lim
n⟶1 a
k
p 
n þ b 
ka 
And for an odd k, we have: 
lim
n⟶1 an
k þ bnk- 1 þ . . .k ¼ lim
n⟶1 a
k
p 
nþ b 
ka 
Moreover, based on growth rate, we know that: 
lim
n⟶1a1n
k1 þ a2nk2 � a1nk1 if k1 > k2 > 0 
Based on the information given in the problem, we have: 
an ¼ n
p 
4n3 þ sin n þ 1p þ
n
p 
4n3 þ sin n þ 2p þ⋯þ
n
p 
4n3 þ sin n þ np 
) lim
n→1an � limn→1 
n 
4n3
p þ n 
4n3
p þ ⋯ n 
4n3
p ¼ lim
n→1 
1 
2n
þ 1 
2n
þ ⋯þ 1 
2n 
) lim
n→1an ¼ limn→1n�
1 
2n 
) lim
n⟶1an ¼ 
1 
2 
Choice (1) is the answer. 
4.4. Based on the problem, we need to determine the coefficient of x3 in the Maclaurin expansion of the function below based 
on the following information. 
f 0ð Þ ¼ 1 
f 0 xð Þ ¼ 1þ f xð Þð Þ10 
From Maclaurin series or Maclaurin expansion, we know that f
000 0ð Þ 
3 is the coefficient of x
3 as can be seen in the following. 
f xð Þ ¼ f 0ð Þ þ f 0 0ð Þ x 
1!
þ f 00 0ð Þ x
2 
2! 
þþf 000 0ð Þ x
3 
3! 
þ . . .þ f nð Þ 0ð Þ x
n 
n! 
þ . . . 
Therefore: 
f 0 xð Þ ¼ 1þ f xð Þð Þ10 ) f 00 xð Þ ¼ 10f 0 xð Þ � f xð Þð Þ9 
) f 000 xð Þ ¼ 10f 00 xð Þ � f xð Þð Þ9 þ 90 f 0 xð Þð Þ2 � f xð Þð Þ8 
For x 0: 
f 0 0ð Þ ¼ 1þ f 0ð Þð Þ10 ¼ 1þ 1 ¼ 2
1
4 Solutions of Problems: Sequences and Series and Their Applications 61
f 00 0ð Þ ¼ 10f 0 0ð Þ � f 0ð Þð Þ9 ¼ 10� 2� 19 ¼ 20 
f 000 0ð Þ ¼ 10f 00 0ð Þ � f 0ð Þð Þ9 þ 90 f 0 0ð Þð Þ2 � f 0ð Þð Þ8 ¼ 10� 20� 19 þ 90 2ð Þ2 � 1ð Þ8 
) f 000 0ð Þ ¼ 380 
) coefficient of x3 ¼ f
000 0ð Þ 
3! 
¼ 380 
6 
¼ 190 
3 
Choice (2) is the answer. 
4.5. First, we need to find a general term for the sequence as follows. 
1, -
2 
2 
, 
3 
4 
,
- 4 
8 
, 
5 
16 
, . . . ¼ n - 1ð Þ
nþ1 
2n- 1 
From the concept of growth rate, for n ⟶ , a, b > 1 and k > 0, we knowthat the order of growth rate is as follows. 
log an< n
k < bn < n!< nn 
Hence: 
lim
n⟶1 
n - 1ð Þnþ1 
2n- 1 
¼ lim
n⟶1 
1 
2n- 1 
¼ 1 
21- 1 
¼ 1 1 
) lim
n⟶1 
n - 1ð Þnþ1 
2n- 1 
¼ 0 
Choice (1) is the answer. 
4.6. Based on the information given in the problem, we have: 
S ¼ 
þ1 
n¼1 
1þ cos n 
n2 
As we know, the series below, called P-series, is convergent if P > 1; otherwise, it is divergent. 
1 
n¼1 
1 
nP 
Based on the above-mentioned rule, the series below is convergent. 
S0 ¼ 
1 
n¼1 
2 
n2 
Since S0 is convergent and the relation below is held for every term, the series of S is convergent. 
1þ cos n 
n2 
≤ 2 
n2 
Choice (1) is the answer.
62 4 Solutions of Problems: Sequences and Series and Their Applications
In this problem, the theorems below were used. 
Theorem: The P-series, presented below, is convergent for P > 1 and it is divergent for P ≤ 1. 
1 
n¼1 
1 
nP 
Theorem: Suppose that for every term an ≤ bn. Then, if 1 n¼1an is divergent, 
1 
n¼1bn will be divergent as well. Also, if1 
n¼1bn is convergent, 
1 
n¼1an will be convergent. 
4.7. Based on growth rate, we know that: 
lim
n⟶1a1n
k1 þ a2nk2 � a1nk1 if k1 > k2 > 0 
Therefore: 
S ¼ 
þ1 
n¼1 
10n2 þ 9nþ 8 
12n3 þ 11n2 þ 10nþ 9 �
þ1 
n¼1 
10n2 
12n3 
¼ 5 
6 
þ1 
n¼1 
1 
n 
As we know, harmonic series, shown below, is divergent. 
þ1 
n¼1 
1 
n 
Hence, the series of S has a similar behavior, and consequently it is divergent as well. 
Choice (2) is the answer. 
In this problem, the theorem below was used. 
Theorem: Harmonic series is a divergent series presented in the following. 
þ1 
n¼1 
1 
n 
4.8. Based on the problem, we have: 
S ¼ 
1 
n¼1 
nþ 1p - np 
n2 þ np 
) S ¼ 
1 
n¼1 
nþ 1p - np 
n
p 
nþ 1p ¼ 
1 
n¼1 
1 
n
p - 1 
n þ 1p 
As can be noticed, the series is a telescoping series. Therefore: 
) S ¼ 1 
1
p - 1 1þ 1p ¼ 1- 0 
) S ¼ 1 
Choice (2) is the answer.
¼
4 Solutions of Problems: Sequences and Series and Their Applications 63
In this problem, the rule below was used. 
Telescoping series, presented below, is a series in which pairs of consecutive terms cancel each other so that only the 
initial and final terms are left. 
N 
n¼1 
an - anþ1ð Þ ¼ a1 - aNþ1 
4.9. Based on the problem, we have: 
S ¼ 
1 
n¼1 
n1390 xn 
To determine the convergence radius (R) and convergence range of a power series in the form of 1 n¼1an x- cð Þn , we can 
use two methods below. 
1 
R 
¼ lim
n→1 anj j
n 
1 
R 
¼ lim
n→1 
anþ1 
an 
The convergence range can be determined as follows. 
xj j≤R ) -R≤ x≤R 
For this problem, we can use the second method as follows. 
1 
R 
¼ lim
n→1 
anþ1 
an 
¼ lim
n→1 
nþ 1ð Þ1390 
n1390
� lim
n→1 
n1390 
n1390 
¼ 1 
) R ¼ 1 
) xj j≤ 1 ) - 1≤ x≤ 1 
It should be noted that for x ±1, the series do not have the necessary convergence criterion. Hence:
- 1< x< 1 
Choice (2) is the answer. 
4.10. Based on the problem, we have: 
S ¼ 
1 
n¼0 
n!xn 
3n
2
1
64 4 Solutions of Problems: Sequences and Series and Their Applications
To determine the convergence radius (R) and convergence range of a power series in the form of 
1 
n¼1 
an x- cð Þn , we can 
use two methods below. 
1 
R 
¼ lim
n→1 anj j
n 
1 
R 
¼ lim
n→1 
anþ1 
an 
The convergence range can be determined as follows. 
xj j≤R ) -R≤ x≤R 
For this problem, we can use the first method as follows. 
1 
R 
¼ lim
n→1 
n! 
3n
2 
n ¼ lim
n→1 
n!n
p 
3n
2n 
By using Stirling’s approximation, we have: 
) 1 
R 
¼ lim
n→1 
n 
e 
3n 
Using the concept of growth rate: 
) 1 
R 
¼ 1 
e 
lim
n→1 
n 
3n 
¼ 0 
Choice (4) is the answer. 
In this problem, the rules below were used. 
Stirling’s approximation: 
) R ¼ 1 
lim 
n→þ1 n!
n
p 
¼ lim 
n→þ1 
n 
e 
The concept of growth rate states that for n ⟶ , a, b > 1 and k > 0, the order of growth rate is as follows. 
log an< n
k < bn < n!< nn 
4.11. Based on the problem, we have: 
S ¼ 
1 
n¼1 
n!ð Þxn 
nþ 1ð Þn
¼
¼
¼
4 Solutions of Problems: Sequences and Series and Their Applications 65
To determine the convergence radius (R) and convergence range of a power series in the form of 
1 
n¼1 
an x- cð Þn , we can 
use two methods below. 
1 
R 
¼ lim
n→1 anj j
n 
1 
R 
¼ lim
n→1 
anþ1 
an 
The convergence range can be determined as follows. 
xj j≤R ) -R≤ x≤R 
For this problem, we can use the first method as follows. 
1 
R 
¼ lim 
n→þ1 
n! 
nþ 1ð Þn 
n ¼ lim 
n→þ1 
n!n
p 
nþ 1 
Based on Stirling’s approximation, we have: 
) 1 
R 
¼ lim 
n→þ1 
n 
e 
n þ 1 ¼ 
1 
e 
lim 
n→þ1 
n 
nþ 1 
) 1 
R 
¼ 1 
e 
) R ¼ e 
Choice (2) is the answer. 
In this problem, the rule below was used. 
Stirling’s approximation: 
lim 
n→þ1 n!
n
p 
¼ lim 
n→þ1 
n 
e 
4.12. Based on the problem, we need to determine the Taylor series of the function below around x 2. 
f xð Þ ¼ 1 
5- x 
The Taylor series or Taylor expansion of the function of f (x) around x a can be calculate as follows. 
f xð Þ ¼ f að Þ þ f 0 að Þ x- að Þ 
1! 
þ f 00 að Þ x- að Þ
2 
2! 
þ . . .þ f nð Þ að Þ x- að Þ
n 
n! 
þ . . . 
Moreover, if in the Taylor series a 0, the series is called Maclaurin series or Maclaurin expansion.
p
66 4 Solutions of Problems: Sequences and Series and Their Applications
f xð Þ ¼ f 0ð Þ þ f 0 0ð Þ x 
1!
þ f 00 0ð Þ x
2 
2! 
þ . . .þ f nð Þ 0ð Þ x
n 
n! 
þ . . . 
However, the problem can be easily solved without using the direct formula of Taylor series as follows. 
f xð Þ ¼ 1 
5- x 
¼ - 1 
x- 5 
¼ - 1 
x- 2ð Þ- 3 
) f xð Þ ¼ - 1
- 3 1- x- 2 3 
¼ 1 
3 
1 
1- x- 2 3 
As we know: 
1 
1- x 
¼ 1 þ xþ x2 þ ⋯ ¼ 
1 
n¼0 
xn 
Therefore: 
) f xð Þ ¼ 1 
3 
1 
n¼0 
x- 2 
3 
n 
¼ 1 
3
þ x- 2ð Þ 
32 
þ x- 2ð Þ
2 
33 
þ⋯ 
Choice (1) is the answer. 
4.13. From the concept of equivalent functions, we know that for a > 0 and even k: 
lim
n⟶1 an
k þ bnk- 1 þ . . .k ¼ lim
n⟶1 a
k
p 
n þ b 
ka 
And for odd k, we have: 
lim
n⟶1 an
k þ bnk- 1 þ . . .k ¼ lim
n⟶1 a
k
p 
nþ b 
ka 
Moreover, based on growth rate, we know that: 
lim
n⟶1a1n
k1 þ a2nk2 � a1nk1 if k1 > k2 > 0 
Therefore, for this problem, we have: 
lim
n⟶1an ¼ limn⟶1 n2 - n3
3
p 
þ n ¼ lim
n⟶1 - n
3 - n23
p 
þ n � lim
n⟶1 - nþ 
1 
3 - 1ð Þ þ n ¼ limn⟶1 
1 
3 
) lim
n⟶1an ¼ 
1 
3 
lim
n⟶1bn ¼ limn⟶1 
n2 þ 1- np 
n3 þ n4p þ np � limn→1 
n- n
p 
n34
p 
þ np � limn→1 
n 
n34
p ¼ lim
n→1 n
4
p ¼ 14p 
) lim
n⟶1bn ¼ 1 
Choice (3) is the answer.
1
þ þ
n→þ1 þ þ
4 Solutions of Problems: Sequences and Series and Their Applications 67
4.14. Based on the information given in the problem, we have: 
an ¼ 
1þ 1 2 þ 1 3 þ ⋯þ 1 n3 
ln n
ð1Þ 
lim 
n→þ1an ¼ lim n→þ1 
1 þ 1 2 þ 1 3 þ ⋯þ 1 n3 
ln n
¼ lim 
n→þ1 
1 
ln n
�
n3 
k¼1 
1 
k
ð2Þ 
From the concept of equivalent functions, we know that if n → + : 
n 
k¼1 
1 
k
� ln n ð3Þ 
Hence: 
n3 
k¼1 
1 
k
� ln n3 ð4Þ 
Solving (2) and (4): 
lim 
n→þ1an � limn→1 
ln n3 
ln n 
¼ lim
n→1 
3 ln n 
ln n 
¼ lim
n→13 
) lim
n→1an ¼ 3 
Choice (3) is the answer. 
In this problem, the rule below was used. 
ln ab ¼ b ln a 
4.15. Based on the information given in the problem, we have: 
an ¼ 1 
n2 þ 1p þ 
1 
n2 þ 2p þ ⋯ 
1 
n2 þ np 
Each term of the sequence is equal or larger than 1 
n2 n
p and equal or smaller than 1 
n2 1
p . In other words: 
1 
n2 þ np ≤ an ≤
1 
n2 þ 1p ,8n E ℕ
Therefore, the value of lim an is larger than n n2 n
p and smaller than n 
n2 1
p . In other words: 
lim 
n→þ1 
n 
n2 þ np ≤ lim n→þ1an ≤ lim n→þ1 
n 
n2 þ 1p 
Based on the concept of equivalent functions, we have: 
lim 
n→þ1 
n 
n þ 1 2 
≤ lim 
n→þ1an ≤ lim n→þ1 
n 
nj j
68 4 Solutions of Problems: Sequences and Series and Their Applications
) 1≤ lim 
n→þ1an ≤ 1 
Thus, based on Sandwich theorem, we have: 
Choice (3) is the answer. 
In this problem, the rules below were used. 
lim 
n→þ1an ¼ 1 
From the concept of equivalent functions, we know that for a > 0 and an even k: 
lim
n⟶1 an
k þ bnk- 1 þ .. .k ¼ lim
n⟶1 a
k
p 
n þ b 
ka 
And for an odd k, we have: 
lim
n⟶1 an
k þ bnk- 1 þ . . .k ¼ lim
n⟶1 a
k
p 
nþ b 
ka 
Theorem: Sandwich theorem states that if we have: 
bn ≤ an ≤ cn 
lim 
n→þ1bn ¼ L 
lim 
n→þ1cn ¼ L 
Then: 
lim 
n→þ1an ¼ L 
4.16. Based on the problem, we have: 
S ¼ 1 
1
þ 1 
1þ 2 þ
1 
1 þ 2þ 3 þ
1 
1þ 2 þ 3þ 4þ ⋯ 
) S ¼ 
1 
n¼1 
1 
n nþ1ð Þ 
2 
¼ 
1 
n¼1 
2 
n n þ 1ð Þ 
) S ¼ 2 
1 
n¼1 
1 
n
-
1 
nþ 1 
As can be noticed, the series is a telescoping series. Therefore: 
S ¼ 2 1- 1 1þ 1 ¼ 2 1- 0ð Þ 
) S ¼ 2 
Choice (1) is the answer.
p
þ þ
n→þ1 þ þ
4 Solutions of Problems: Sequences and Series and Their Applications 69
In this problem, the rules below were used. 
1þ 2þ 3þ . . .þ n ¼ n nþ 1ð Þ 
2 
Telescoping series, presented below, is a series in which pairs of consecutive terms cancel each other so that only the 
initial and final terms are left. 
N 
n¼1 
an - anþ1ð Þ ¼ a1 - aNþ1 
4.17. A sequence in the form of an¼ {a1, a2, . . .} is convergent if its limit when n →1 is a unique finite number ( lim 
n→þ1an); 
otherwise, the sequence is divergent. 
Choice (1): Based on Stirling’s approximation, we have: 
lim 
n→þ1 n!
n
p 
¼ lim 
n→þ1 
n 
e 
) lim 
n→þ1 n!
n ¼ 1 
Hence, the sequence is divergent. 
Choice (2): Each term of the sequence is equal or larger than n n2 n and equal or smaller than 
n 
n2 1. In other words: 
n 
n2 þ n ≤ 
n 
n2 þ 1þ 
n 
n2 þ 2þ⋯þ 
n 
n2 þ n ≤ 
n 
n2 þ 1 , 8nEℕ
Therefore, the value of lim an is larger than n
2 
n2 n and smaller than 
n2 
n2 1. In other words: 
lim 
n→þ1 
n2 
n2 þ n ≤ lim n→þ1an ≤ lim n→þ1 
n2 
n2 þ 1 
) 1≤ lim 
n→þ1an ≤ 1 
Therefore, based on Sandwich theorem: 
lim 
n→þ1an ¼ 1 
Hence, the sequence is convergent. 
Choice (3): Based on growth rate, we know that: 
lim
n⟶1a1 
n þ a2 n � a1 n if a1 > a2 > 0 
Thus: 
lim 
n→þ1 3
n þ 2nn
p
� lim 
n→þ1 3
nn
p 
¼ lim 
n→þ13 ¼ 3
70 4 Solutions of Problems: Sequences and Series and Their Applications
) lim 
n→þ1an ¼ 3 
Thus, the sequence is convergent. 
Choice (4): Based on growth rate, we know that: 
lim
n⟶1a1 
n þ a2 n � a1 n if a1 > a2 > 0 
Now, if a > b: 
lim 
n→þ1 
an - bn 
an þ bn � lim n→þ1 
an 
an 
¼ 1 
) lim 
n→þ1an ¼ 1 
If b > a: 
lim 
n→þ1 
an - bn 
an þ bn � lim n→þ1
- bn 
bn 
¼ - 1 
) lim 
n→þ1an ¼ - 1 
Therefore, in any condition, the sequence is convergent. 
Choice (1) is the answer. 
In this problem, the rule below was used. 
Theorem: Sandwich theorem states that if we have: 
bn ≤ an ≤ cn 
lim 
n→þ1bn ¼ L 
lim 
n→þ1cn ¼ L 
Then: 
lim 
n→þ1an ¼ L 
4.18. The problem can be solved by using the following theorem. 
Theorem: It states that the necessary, but not enough, criterion for a series in the form of Sn ¼ 
þ1 
n¼1 
an to be convergent is 
that the limit of its general term must be zero when n → 1. In other words: 
lim 
n→þ1an ¼ 0 
Choice (1): 
lim
n→1 
n- 1 
n 
n 
¼ lim
n→1 1-
1 
n 
n 
¼ e- 1 ≠ 0
¼
References 71
Choice (2): 
lim
n→1 
nþ 1 
n 
n 
¼ lim
n→1 1þ
1 
n 
n 
¼ e≠ 0 
Choice (3): 
lim
n→1 
nþ 1 
n 
n2 
¼ lim
n→1 
nþ 1 
n 
n n 
¼ lim
n→1 1þ 
1 
n 
n n 
¼ e1 ¼ 1≠ 0 
Choice (4): 
lim
n→1 
1 
n n
p ¼ lim
n→1 
1 
n 
3 
2 
¼ 1 1 ¼ 0 
Thus, the series of S4 has the necessary criterion of convergence. 
On the other hand, we know that P-series, shown below, is convergent for P > 1. 
1 
n¼1 
1 
nP 
For Choice (4), P 1.5 > 1; therefore, the series is convergent. Choice (4) is the answer. 
In this problem, the rule below was used. 
lim
n→1 1þ 
a 
bn 
cn 
¼ e ac b 
References 
1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
p
¼ ¼
Problems: Polar Coordinate System 5 
Abstract 
In this chapter, the basic and advanced problems concerned with polar coordinate system are presented. The subjects 
include tangent line on a curve, radius of curve, polar equation, spiral, transferring from cartesian coordinate to polar 
coordinate and vice versa, and curve types such as ellipse, straight line, parabola, hyperbola. In this chapter, the problems 
are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, 
normal, and large). Additionally, the problems are ordered from the easiest problem with the smallest computations to the 
most difficult problems with the largest calculations. 
5.1. Express the cartesian position of the point of 
1 
2 
, -
3 
2 
in polar coordinate system (r, θ) [1–3]. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 2, 
π 
3 
2) 1, 
π 
6 
3) 1, -
π 
6 
4) 1, -
π 
3 
5.2. Express the cartesian position of the point of - 3, 3 
p 
in polar coordinate system (r, θ). 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 2 3 
p 
, 
5π 
6 
2) 3 3 
p 
, 
5π 
6 
3) 2 3 
p 
, 
π 
6 
4) 2 3 
p 
, 
π 
3 
5.3. Express the polar position of the point of r 1, θ 0 in cartesian coordinate system (x, y). 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) (1, 1) 
2) (1, 0) 
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 
M. Rahmani-Andebili, Calculus II, https://doi.org/10.1007/978-3-031-45353-3_5
73
http://crossmark.crossref.org/dialog/?doi=10.1007/978-3-031-45353-3_5&domain=pdf
https://doi.org/10.1007/978-3-031-45353-3_5#DOI
¼ ¼
4
p p
¼ ¼
¼
6
¼
¼
¼
¼
¼ ¼ ¼
p
p
74 5 Problems: Polar Coordinate System
3) (0, 1) 
4) (0, 0) 
5.4. Express the polar position of the point of r 2 
p 
, θ -
π 
in cartesian coordinate system (x, y). 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 2 
p 
, 2 
p 
2) 
2 
2 
, 
2 
2 
3) (1, -1) 
4) (1, 1) 
5.5. Calculate the amount of angle between the tangent line and the radius of the curve of r θ2 + 1 at θ 1. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
π 
6 
2) 
π 
4 
3) 
π 
3 
4) 
π 
2 
5.6. Calculate the amount of angle between the tangent line and the radius of the curve of r 2 + 2 sin θ at the point of (3, 
π 
). 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 
π 
6 
2) 
π 
4 
3) 
π 
3 
4) 
π 
2 
5.7. Determine the polar equation of a circle with the center on positive side of x-axis and radius of two while passing from 
the origin. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) r 4 cos θ 
2) r 2 cos θ 
3) r 4 sin θ 
4) r 2 sin θ 
5.8. Calculate the surface area of the spiral of r e 
θ 
4π from θ 0 to θ 2π. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) π(e - 1) 
2) 2π e- 1 
3) 2π(e - 1) 
4) 4π e- 1
¼ ð Þ
¼
¼
¼
¼
¼
¼
5 Problems: Polar Coordinate System 75
5.9. Calculate the surface area that the curve of r 4- 4 sin θ cos θ 
1 
2 separates from the first quadrant. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) π - 1 
2) π 
3) π + 1 
4) π - 2 
5.10. What is the curve type of the polar function of r cot θ csc θ. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) Ellipse 
2) Straight line 
3) Parabola 
4) Hyperbola 
5.11. Determine the curve type of the polar function below. 
r ¼ 5 
3 cos θ þ 2 sin θ 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) Straight line 
2) Ellipse 
3) Parabola 
4) Hyperbola 
5.12. Express the function of cos 2θ 1 in cartesian coordinate system. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) x - y 0 
2) x + y 03) x 0 
4) y 0 
5.13. Express the function below in cartesian coordinate system. 
x ¼ cos 2θ 
y ¼ sin θ cos θ 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) x-
1 
2 
2 
þ y2 ¼ 1 
4 
2) x2 þ y- 1 
2 
2 
¼ 1 
4 
3) xþ 1 
2 
2 
þ y2 ¼ 1 
4 
4) x-
1 
2 
2 
þ y- 1 
2 
2 
¼ 1 
4
¼ ¼
1
76 5 Problems: Polar Coordinate System
5.14. Calculate the amount of angle between the curves of r 2(1 + sin θ) and r 3(1 - sin θ) at the intersection point. 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
π 
2 
2) 
π 
4 
3) arctan 
2 
3 
4) arctan 
1 
7 
5.15. Determine the curve type of the polar function below. 
r ¼ 4 
2- cos θ 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) Ellipse 
2) Parabola 
3) Hyperbola 
4) Two crossing straight lines 
5.16. Determine the number of intersection points of the following polar functions. 
r ¼ cos θ 
sin 2 θ 
r ¼ - 3 
cos θ þ 4 sin θ 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) 1 
2) 2 
3) 
4) 0 
References 
1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
6Solutions of Problems: Polar Coordinate System 
Abstract 
In this chapter, the problems of the fifth chapter are fully solved, in detail, step-by-step, and with different methods. 
6.1. Based on the information given in the problem, we have [1–3]: 
x, yð Þ ¼ 1 
2 
, -
3
p 
2 
The relations below are used to transfer from cartesian coordinate to polar coordinate. 
r ¼ x2 þ y2 
θ ¼ tan - 1 y 
x 
if x> 0, y> 0 
θ ¼ π- tan - 1 y 
x 
if x< 0, y> 0 
θ ¼ π þ tan - 1 y 
x 
if x< 0, y< 0 
θ ¼ - tan - 1 y 
x 
if x> 0, y< 0 
Therefore, for this problem, we have: 
r ¼ 1 
2 
2 
þ - 3
p 
2 
2 
¼ 1 
4
þ 3 
4 
¼ 1 
θ ¼ - tan - 1 
3
p 
2 
1 
2 
¼ - tan - 1 3 
p 
¼ - π 
3 
) r, θð Þ ¼ 1, - π 
3 
Choice (4) is the answer. 
In this problem, the rule below was used. 
# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 
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p π
p
78 6 Solutions of Problems: Polar Coordinate System
tan - 1 3 ¼
3 
6.2. Based on the information given in the problem, we have: 
x, yð Þ ¼ - 3, 3 
p 
The relations below are used to transfer from cartesian coordinate to polar coordinate. 
r ¼ x2 þ y2 
θ ¼ tan - 1 y 
x 
if x> 0, y> 0 
θ ¼ π- tan - 1 y 
x 
if x< 0, y> 0 
θ ¼ π þ tan - 1 y 
x 
if x< 0, y< 0 
θ ¼ - tan - 1 y 
x 
if x> 0, y< 0 
Therefore, for this problem, we have: 
r ¼ - 3ð Þ2 þ 3 
p 2 
¼ 9þ 3p ¼ 2 3 
p 
θ ¼ π- tan - 1 3 
3 
¼ π- π 
6 
¼ 5π 
6 
) r, θð Þ ¼ 2 3 
p 
, 
5π 
6 
Choice (1) is the answer. 
In this problem, the rule below was used. 
tan - 1 
3
p 
3 
¼ π 
6 
6.3. Based on the information given in the problem, we have: 
r, θð Þ ¼ 1, 0ð Þ 
The relations below are used to transfer from polar coordinate to cartesian coordinate. 
x ¼ r cos θ 
y ¼ r sin θ 
Therefore, for this problem, we have:
p
6 Solutions of Problems: Polar Coordinate System 79
x ¼ 1 cos 0 ¼ 1 
y ¼ 1 sin 0 ¼ 0 
) x, yð Þ ¼ 1, 0ð Þ 
Choice (2) is the answer. 
In this problem, the rules below were used. 
cos 0 ¼ 1 
sin 0 ¼ 0 
6.4. Based on the information given in the problem, we have: 
r, θð Þ ¼ 2 
p 
, -
π 
4 
The relations below are used to transfer from polar coordinate to cartesian coordinate. 
x ¼ r cos θ 
y ¼ r sin θ 
Therefore, for this problem, we have: 
x ¼ 2 
p 
cos -
π 
4 
¼ 1 
y ¼ 2 
p 
sin -
π 
4 
¼ - 1 
) x, yð Þ ¼ 1, - 1ð Þ 
Choice (3) is the answer. 
In this problem, the rules below were used. 
cos -
π 
4 
¼ 2
p 
2 
sin -
π 
4 
¼ - 2 
2 
6.5. The amount of angle between the tangent line on a polar curve and the radius of the curve can be calculated as follows. 
tan α ¼ f θð Þ 
f 0 θð Þ ¼ 
r θð Þ 
dr θð Þ 
dθ 
ð1Þ 
Based on the information given in the problem, we have: 
r ¼ θ2 þ 1 ð2Þ
p
80 6 Solutions of Problems: Polar Coordinate System
θ ¼ 1 ð3Þ 
Solving (1)–(3): 
tan α ¼ θ
2 þ 1 
2θ θ ¼ 1 ¼ 1 
) α ¼ tan - 1 1ð Þ ¼ π 
4 
Choice (2) is the answer. 
In this problem, the rule below was used. 
tan - 1 1ð Þ ¼ π 
4 
6.6. Based on the information given in the problem, we have: 
r ¼ 2þ 2 sin θ ð1Þ 
r, θð Þ ¼ 3, π 
6
ð2Þ 
The amount of angle between the tangent line on a polar curve and the radius of the curve can be calculated as follows. 
tan α ¼ f θð Þ 
f 0 θð Þ ¼ 
r θð Þ 
dr θð Þ 
dθ 
ð3Þ 
Solving (1)–(3): 
tan α ¼ 2 þ 2 sin θ 
2 cos θ 
r, θð Þ ¼ 3, π 
6 
¼ 2þ 2 sin 
π 
6 
2 cos π 6 
¼ 3 
3
p ¼ 3 
p 
) α ¼ tan - 1 3 
p 
¼ π 
3 
Choice (3) is the answer. 
In this problem, the rules below were used. 
d 
dx 
sin x ¼ cos x 
cos 
π 
6 
¼ 3 
2 
sin 
π 
6 
¼ 1 
2 
tan - 1 3 
p 
¼ π 
3
¼
6 Solutions of Problems: Polar Coordinate System 81
6.7. The polar equations of a circle with the center on positive and negative sides of x-axis and radius of “a” while passing 
from the origin are as follows, respectively. 
r ¼ 2a cos θ 
r ¼ - 2a cos θ 
Moreover, the polar equations of a circle with the center on positive and negative sides of y-axis and radius of “a” while 
passing from the origin are as follows, respectively. 
r ¼ 2a sin θ 
r ¼ - 2a sin θ 
Thus, for this problem, we have: 
r ¼ 4 cos θ 
Choice (1) is the answer. 
6.8. The surface area of a spiral in the form of r f(θ) from θ1 to θ2 can be calculated as follows. 
S ¼ 1 
2 
θ2 
θ1 
r2 dθ 
Based on the information given in the problem, we have: 
r ¼ e θ 4π ð1Þ 
0≤ θ≤ 2π ð2Þ 
For this problem, we have: 
S ¼ 1 
2 
2π 
0 
e 
θ 
2πdθ 
) S ¼ πe θ 2π 2π 
0 
¼ π e1 - e0 
) S ¼ π e- 1ð Þ 
Choice (1) is the answer. 
In this problem, the rule below was used. 
eax dx ¼ 1 
a 
eax þ c
¼
82 6 Solutions of Problems: Polar Coordinate System
6.9. The surface area of a spiral in the form of r f(θ) from θ1 to θ2 can be calculated as follows. 
S ¼ 1 
2 
θ2 
θ1 
r2 dθ 
Based on the information given in the problem, we have: 
r ¼ 4- 4 sin θ cos θð Þ1 2 ð1Þ 
0≤ θ≤ π 
2
ð2Þ 
Thus, for this problem, we have: 
) S ¼ 1 
2 
θ2 
θ1 
r2 dθ ¼ 1 
2 
π 
2 
0 
4- 4 sin θ cos θð Þdθ 
) S ¼ 1 
2 
4θ- 2 sin 2 θ 
π 
2 
0 
¼ 1 
2 
4� π 
2
- 2 sin 2 
π 
2
- 0- 0ð Þ 
) S ¼ π- 1 
Choice (1) is the answer. 
In this problem, the rule below was used. 
un du ¼ 1 
n þ 1 u
nþ1 þ c 
6.10. Based on the information given in the problem, we have: 
r ¼ cot θ csc θ ð1Þ 
It can be simplified as follows. 
r ¼ cos θ 
sin θ 
1 
sin θ 
¼ cos θ 
sin 2 θ
ð2Þ 
By transferring from polar coordinate to cartesian coordinate, we have: 
y ¼ r sin θ ð3Þ 
x ¼ r cos θ ð4Þ 
x2 þ y2 ¼ r2 ð5Þ 
Solving (2)–(5): 
) x2 þ y2 ¼ 
x 
x2 þ y2 
y 
x2 þ y2 
2 )
y2 
x2 þ y2 ¼
x 
x2 þ y2
6 Solutions of Problems: Polar Coordinate System 83
) y2 ¼ x 
It is the equation of a parabola. Choice (3) is the answer. 
In general, the equation of a parabola is as follows. 
x- x0ð Þ2 ¼ 4a y- y0ð Þ 
y- y0ð Þ2 ¼ 4a x- x0ð Þ 
In this problem, the rules below were used. 
cot θ ¼ cos θ 
sin θ 
cscθ ¼ 1 
sin θ 
6.11. Based on the information given in the problem, we have: 
r ¼ 5 
3 cos θ þ 2 sin θ ð1Þ 
) 3r cos θ þ 2r sin θ ¼ 5 ð2Þ 
By transferring from polar coordinate to cartesian coordinate, we have: 
y ¼ r sin θ ð3Þ 
x ¼ r cos θ ð4Þ 
x2 þ y2 ¼ r2 ð5Þ 
Solving (2)–(5): 
3x þ 2y ¼ 5 
It is the equation of a straight line. Choice (1) is the answer. 
In general, the equation of a straight line is as follows. 
axþ by ¼ c 
6.12. Based on the informationgiven in the problem, we have: 
cos 2θ ¼ 1 ð1Þ 
) cos 2 θ- sin 2 θ ¼ 1 ð2Þ 
By transferring from polar coordinate to cartesian coordinate, we have:
Þ
Þ
84 6 Solutions of Problems: Polar Coordinate System
y ¼ r sin θ ð3Þ 
x ¼ r cos θ ð4Þ 
x2 þ y2 ¼ r2 ð5Þ 
Solving (2)–(5): 
x 
x2 þ y2 
2
-
y 
x2 þ y2 
2 
¼ 1 ) x2 - y2 ¼ x2 þ y2 ) 2y2 ¼ 0 
) y ¼ 0 
Choice (4) is the answer. 
In this problem, the rule below was used. 
cos 2θ ¼ cos 2 θ- sin 2 θ 
6.13. Based on the information given in the problem, we have: 
x ¼ cos 2θ 
y ¼ sin θ cos θ ð1Þ 
By transferring from polar coordinate to cartesian coordinate, we have: 
y ¼ r sin θ ð2Þ 
x ¼ r cos θ ð3Þ 
x2 þ y2 ¼ r2 ð4Þ 
Solving (1)–(4): 
x ¼ x 
x2 þ y2 
2 
y ¼ y 
x2 þ y2 
x 
x2 þ y2 
) 
x ¼ x
2 
x2 þ y2 5ð 
y ¼ xy 
x2 þ y2 6ð 
From (5) or (6), we have: 
x2 þ y2 ¼ x ) x2 - xþ y2 ¼ 0 
) x- 1 
2 
2
-
1 
4 
þ y2 ¼ 0 ) x- 1 
2 
2 
þ y2 ¼ 1 
4 
Choice (1) is the answer.
¼
6 Solutions of Problems: Polar Coordinate System 85
6.14. Suppose α1 is the angle between the tangent line and the radius of the curve of r¼ f1(θ) at θ0 and α2 is the angle between 
the tangent line and the radius of the curve of r¼ f2(θ) at θ0. The acute or straight angle between the two polar curves at 
the intersection point (θ θ0) can be calculated as follows. 
tanψ ¼ tan α1 - tan α2 
1þ tan α1 tan α2 ð1Þ 
Moreover, the amount of angle between the tangent line on a polar curve and the radius of the curve can be calculated as 
follows. 
tan α ¼ f θð Þ 
f 0 θð Þ ¼ 
r θð Þ 
dr θð Þ 
dθ 
ð2Þ 
Based on the information given in the problem, we have: 
r ¼ 2 1þ sin θð Þ ð1Þ 
r ¼ 3 1- sin θð Þ ð2Þ 
First, we need to find the intersection point of the polar curves as follows. 
2 1þ sin θð Þ ¼ 3 1- sin θð Þ ) 5 sin θ ¼ 1 ) sin θ ¼ 1 
5 
) cos θ ¼ 24
p 
5 
Then: 
tan α1 ¼ f 2 θð Þ f 0 2 θð Þ
¼ 2 1þ sin θð Þ 
2 cos θ 
¼ 1 þ sin θ 
cos θ 
¼ 1þ
1 
5 
24
p 
5 
¼ 6 
24
p ¼ 6
p 
2 
tan α2 ¼ f 2 θð Þ f 0 2 θð Þ
¼ 3 1- sin θð Þ
- 3 cos θ 
¼ 1- sin θ
- cos θ 
¼ 1-
1 
5
- 24
p 
5 
¼ 4
- 24
p ¼ - 6
p 
3 
Therefore: 
) tanψ ¼ 
6
p 
2 -
6
p 
3 
1þ 6
p 
2
- 6
p 
3 
¼ 
6
p 
6 
1- 1 
¼ 1 
) ψ ¼ tan - 1 1ð Þ ¼ π 
2 
Choice (1) is the answer. 
In this problem, the rules below were used. 
sin 2 x þ cos 2 x ¼ 1 
d 
dx 
sin x ¼ cos x 
tan - 1 1ð Þ ¼ π 
2
86 6 Solutions of Problems: Polar Coordinate System
6.15. Based on the information given in the problem, we have: 
r ¼ 4 
2- cos θ
ð1Þ 
) 2r- r cos θ ¼ 4 ð2Þ 
By transferring from polar coordinate to cartesian coordinate, we have: 
y ¼ r sin θ ð3Þ 
x ¼ r cos θ ð4Þ 
x2 þ y2 ¼ r2 ð5Þ 
Solving (2)–(5): 
2 x2 þ y2 - x ¼ 4 ) 2 x2 þ y2 ¼ xþ 4 
) 4 x2 þ y2 ¼ x2 þ 8xþ 16 ) 3x2 - 8xþ 4y2 ¼ 16 
) 3 x2 - 8 
3 
xþ 16 
9 
þ 4y2 ¼ 16þ 16 
3 
) 3 x- 4 
3 
2 
þ 4y2 ¼ 64 
3 
) x-
4 
3 
2 
8 
3 
2 þ 
y- 0ð Þ2 
8 
2 3
p 
2 ¼ 1 
It is the equation of an ellipse. Choice (1) is the answer. 
In general, the equation of an ellipse is as follows. 
x- x0ð Þ2 
a2 
þ y- y0ð Þ
2 
b2 
¼ 1 
6.16. Based on the information given in the problem, we have: 
r ¼ cos θ 
sin 2 θ
ð1Þ 
r ¼ - 3 
cos θ þ 4 sin θ ð2Þ 
The problem can be easily solved by transferring from polar coordinate to cartesian coordinate as follows: 
y ¼ r sin θ ð3Þ 
x ¼ r cos θ ð4Þ 
x2 þ y2 ¼ r2 ð5Þ 
Solving (1) and (3)–(5):
xp
p
p
References 87
x2 þ y2 ¼ x2þy2 
y 
x2þy2 
p 
2 )
y2 
x2 þ y2 ¼ 
x 
x2 þ y2 ) y
2 ¼ x ð6Þ 
Solving (2) and (3)–(5): 
x2 þ y2 ¼ - 3 x 
x2þy2 
p þ 4 y 
x2þy2 
p ) xþ 4y ¼ - 3 ð7Þ 
Solving (6) and (7): 
y2 þ 4y ¼ - 3 ) y2 þ 4y þ 3 ¼ 0 
) y ¼ - 4± 4
2 - 4� 1� 3 
2
¼ - 4± 2 
2 
) y ¼ - 1, - 3 
Choice (2) is the answer. 
In this problem, the rule below was used. 
ax2 þ bx þ c ¼ 0 
) x ¼ - b± b
2 - 4ac 
2a 
References 
1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
7
1
1
ð Þ ¼ þ
2)
Problems: Complex numbers 
Abstract 
In this chapter, the basic and advanced problems of complex numbers are presented. The subjects include the operations on 
complex numbers as well as on functions in complex form. In this chapter, the problems are categorized in different levels 
based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Additionally, 
the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the 
largest calculations. 
7.1. Calculate the value of e0i . 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 0 
2) 1 
3) -1 
4) 
7.2. Calculate the value of eπi [1–3]. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 0 
2) 1 
3) -1 
4) 
7.3. Calculate the complex conjugate of i. 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) 1 - i 
2) 0 
3) 1 + i 
4) -i 
7.4. Which one of the relations below is wrong? 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) a- ib a ib 
z1 
z2 
¼ z1 
z2 
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¼
¼
p
þj j ¼ þ
¼
¼
¼
1) j j
2) j j
¼ j jj j
þ ¼ j j þ j jð Þ
1)
2)
3)
4)
90 7 Problems: Complex numbers
3) (a + ib)(a - ib) a2 - b2 
4) i - 1 
7.5. Which one of the relations below is wrong? 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) a ib a2 b2 
2) Re(z1z2) x1x2 - y1y2 
3) Im(z1z2) x1x2 + y1y2 
4) eix cos x + i sin x 
7.6. Which one of the relations below is wrong? 
Difficulty level ● Easy ○ Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
z1 
z2 
¼ z1j j 
z2 
ei θ1 - θ2ð Þ 
1 
z2 
¼ 1 
z2 
e- iθ2 
3) z1z2 z1 z2 e
i θ1þθ2ð Þ 
4) z1 z2 z1 z2 e
i θ1þθ2ð Þ 
7.7. Calculate the value of zz if: 
z ¼ 1þ 2ið Þ 1þ 3ið Þ 1þ 4ið Þ 
2- 3ið Þ 2- 4ið Þ 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
86 
25 
96 
25 
85 
26 
95 
26 
7.8. Calculate the value of ii . 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ● Small ○ Normal ○ Large 
1) e 
2 
π 
2) e-
π 
2 
3) Ln i 
4) Ln π 
7.9. Present the equation below in complex form. 
x2 - y2 ¼ 1 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large
ð Þ ¼
þ ð Þ ¼
ð Þ ¼
þ ð Þ ¼
p
p
3)
4)
7 Problems: Complex numbers 91
1) z 2 - z2 2 
2) z2 z 2 2 
3) z2 - z 2 1 
4) z2 z 2 1 
7.10. Calculate the value of following complex number if n E ℕ . 
1 þ ið Þn 
1- ið Þn- 2 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2i n - 1 
2) 2i2 
3) 2i 
4) i-2 
7.11. Calculate the value of θ if the complex number below does not have a real part. 
3þ 2i sin θ 
1- 2i sin θ 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 
π 
6 
2) 
π 
4 
3) 
π 
3 
4) 
π 
2 
7.12. Calculate the value of following relation. 
z ¼ 1- ið Þð1þ i 3 
p 
Þ 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2 2 
p 
cos 
π 
12 
þ i sin π 
12 
2) 2 2 
p 
cos 
π 
12
- i sin 
π 
12 
2 2 
p 
cos 
7π 
12 
þ i sin 7π 
12 
2 2 
p 
cos 
7π 
12
- i sin 
7π 
12
1)
p
2)
p
3)
p
4)
p
p
p
92 7 Problems: Complex numbers
7.13. Calculate the value of following relation. 
1 þ ið Þ15 
1- ið Þ13 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) 2 
2) -2 
3) -3 
4)3 
7.14. Calculate the value of relation below. 
3
p þ i 
3
p
- i 
10 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1þ i 3 
2
- 1þ i 3 
2 
1- i 3 
2
- 1- i 3 
2 
7.15. Calculate the value of following relation. 
1þ cos 2π 
3 
þ i sin 2π 
3 
120 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -1 
2) 1 
3) -i 
4) i 
7.16. In the equation below, calculate the value of a. 
eaþib ¼ 1- i 3 
p 
Difficulty level ○ Easy ● Normal ○ Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) ln 2 
2) ln 2 
3) 2 
4) 2
¼
2
þ
2
¼
1)
2)
3)
4)
4)
7 Problems: Complex numbers 93
7.17. Calculate the value of relation below if zm cos 
π 
m i sin 
π 
m. 
∏ 
1 
m¼1 
zm 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) -πi 
2) πi 
3) 1 
4) -1 
7.18. Which one of the following choices is correct if z1 and z2 are two non-zero complex numbers where: 
z1 - z2 
z1 þ z2 ¼ 1 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1) Re(z1z2) > 0 
2) Re(z1z2) < 0 
3) Re(z1z2) 0 
4) Im(z1z2) > 0 
7.19. Calculate the maximum value of z if the relation below is held. 
6z- i 
2þ 3iz ≤ 1 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ● Normal ○ Large 
1 
5 
1 
4 
1 
3 
1 
2 
7.20. Calculate the value of the relation below in complex form. 
1þ sin θ þ i cos θ 
1 þ sin θ þ i cos θ 
Difficulty level ○ Easy ○ Normal ● Hard 
Calculation amount ○ Small ○ Normal ● Large 
1) sin θ + i cos θ 
2) sin θ - i cos θ 
3) 1 - i 
sin 
θ 
2
- i cos 
θ 
2
94 7 Problems: Complex numbers
References 
1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
8
¼ þ
¼
¼ þ
Solutions of Problems: Complex Numbers 
Abstract 
In this chapter, the problems of the seventh chapter are fully solved, in detail, step-by-step, and with different methods. 
8.1. As we know from Euler’ formula [1–3]: 
eθi cos θ i sin θ 
Therefore: 
e0i ¼ cos 0 þ i sin 0 ¼ 1þ 0i 
e0i 1 
Choice (2) is the answer. 
In this problem, the rules below were used. 
cos 0 ¼ 1 
sin 0 ¼ 0 
8.2. As we know from Euler’ formula: 
eθi cos θ i sin θ 
Therefore: 
eπi ¼ cos π þ i sin π ¼ - 1þ 0i 
eπi ¼ - 1 
Choice (3) is the answer. 
In this problem, the rules below were used. 
cos π ¼ - 1 
sin π ¼ 0 
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96 8 Solutions of Problems: Complex Numbers
8.3. The complex conjugate of a complex number can be achieved by changing the sign of imaginary part of the complex 
number while the other parts of the complex number are left intact. In other words: 
aþ ibð Þ ¼ aþ ibð Þ� ¼ a- ib 
Therefore: 
i ¼ i� ¼ - i 
Choice (4) is the answer. 
8.4. All the relations are correct except Choice (3). Its correct relation is as follows. 
a þ ibð Þ a- ibð Þ ¼ a2 - aibþ iba- ibð Þ2 
) aþ ibð Þ a- ibð Þ ¼ a2 þ b2 
Choice (3) is the answer. 
In this problem, the rule below was used. 
i2 ¼ - 1 
8.5. All the relations are correct except Choice (3). Its correct relation is as follows. 
z1z2 ¼ x1 þ iy1ð Þ x2 þ iy2ð Þ ¼ x1x2 þ x1iy2 þ iy1x2 þ i2 y1y2 
) z1z2 ¼ x1x2 - y1y2ð Þ þ i x1y2 þ y1x2ð Þ 
) Im z1z2ð Þ ¼ x1y2 þ y1x2 
Choice (3) is the answer. 
In this problem, the rule below was used. 
i2 ¼ - 1 
8.6. All the relations are correct except Choice (4). Its correct relation is as follows. 
z1 þ z2 ¼ z1j jeiθ1 þ z2j jeiθ2 
Choice (4) is the answer. 
8.7. Based on the information given in the problem, we have: 
z ¼ 1þ 2ið Þ 1þ 3ið Þ 1þ 4ið Þ 
2- 3ið Þ 2- 4ið Þ 
) z ¼ 1- 2ið Þ 1- 3ið Þ 1- 4ið Þ 
2þ 3ið Þ 2þ 4ið Þ
8 Solutions of Problems: Complex Numbers 97
) zz ¼ 1þ 4ð Þ 1þ 9ð Þ 1þ 16ð Þ 
4 þ 9ð Þ 4 þ 16ð Þ ¼ 
5� 10� 17 
13� 20 
) zz ¼ 85 
26 
Choice (3) is the answer. 
In this problem, the rules below were used. 
aþ ibð Þ ¼ a- ib 
z1z2 
z3z4 
¼ z1 z2 
z3 z4 
aþ ibð Þ a- ibð Þ ¼ a2 þ b2 
8.8. The problem can be solved by transferring from Cartesian coordinate to polar coordinate as follows. 
i ¼ 1 
p 
ei tan
- 11 
0 ¼ ei tan - 11 ¼ e π 2i 
Then: 
ii ¼ e π 2i i ¼ e π 2i2 ¼ e- π 2 
Choice (2) is the answer. 
In this problem, the rules below were used. 
aþ ib ¼ a2 þ b2 ei tan - 1 b aj j if a> 0, b> 0 
aþ ib ¼ a2 þ b2 ei π- tan - 1 b aj jð Þ if a< 0, b> 0 
a þ ib ¼ a2 þ b2 ei πþ tan - 1 b aj jð Þ if a< 0, b< 0 
aþ ib ¼ a2 þ b2 e- i tan - 1 b aj j if a> 0, b< 0 
eia 
b ¼ eiab 
i2 ¼ - 1 
8.9. Based on the information given in the problem, we have: 
x2 - y2 ¼ 1 
As we know: 
z ¼ x þ iy ) z2 ¼ x þ iyð Þ2 ¼ x2 - y2 þ i2xy
98 8 Solutions of Problems: Complex Numbers
z ¼ x- iy ) zð Þ2 ¼ x2 - y2 - i2xy 
) z2 þ zð Þ2 ¼ 2 x2 - y2 
Therefore: 
z2 þ zð Þ2 ¼ 2 
Choice (2) is the answer. 
In this problem, the rules below were used. 
i2 ¼ - 1 
aþ ibð Þ ¼ a- ib 
8.10. Based on the information given in the problem, we have: 
1þ ið Þn 
1- ið Þn- 2 ð1Þ 
The problem can be solved as follows. 
) 1þ ið Þ
n 
1- ið Þn- 2 ¼ 
1þ ið Þn- 2 
1- ið Þn- 2 1 þ ið Þ
2 ¼ 1 þ i 
1- i 
n- 2 
1 þ 2i- 1ð Þ ¼ 1 þ i 
1- i 
n- 2 
2ið Þ ð2Þ 
On the other hand: 
1þ i 
1- i 
¼ 1þ i 
1- i
� 1 þ i 
1 þ i ¼ 
1 þ 2iþ i2 
1- i2 
¼ 2i 
2 
¼ i ð3Þ 
Solving (1)–(3): 
1þ ið Þn 
1- ið Þn- 2 ¼ i
n- 2 2ið Þ 
) 1þ ið Þ
n 
1- ið Þn- 2 ¼ 2i
n- 1 
Choice (1) is the answer. 
In this problem, the rules below were used. 
i2 ¼ - 1 
aþ ib 
cþ id ¼ 
aþ ib 
cþ id �
c- id 
c- id
8 Solutions of Problems: Complex Numbers 99
8.11. Based on the information given in the problem, we know that: 
Re 
3þ 2i sin θ 
1- 2i sin θ 
¼ 0 ð1Þ 
The problem can be solved as follows. 
3þ 2i sin θ 
1- 2i sin θ 
¼ 3þ 2i sin θ 
1- 2i sin θ
� 1þ 2i sin θ 
1þ 2i sin θ 
¼ 3þ 6i sin θ þ 2i sin θ- 4 sin
2 θ 
1 þ 4 sin 2 θ ¼ 
3- 4 sin 2 θ þ 8i sin θ 
1þ 4 sin 2 θ 
¼ 3- 4 sin
2 θ 
1þ 4 sin 2 θ þ 
8 sin θ 
1þ 4 sin 2 θ i ð2Þ 
Solving (1) and (2): 
3- 4 sin 2 θ 
1 þ 4 sin 2 θ ¼ 0 ) 3- 4 sin
2 θ ¼ 0 ) sin 2 θ ¼ 3 
4 
) sin θ ¼ ± 3
p 
2 
) θ ¼ ± π 
3 
Choice (3) is the answer. 
In this problem, the rules below were used. 
aþ ib 
cþ id ¼ 
a þ ib 
cþ id �
c- id 
c- id 
i2 ¼ - 1 
8.12. The problem can be solved by transferring from Cartesian coordinate to polar coordinate as follows. 
1- i ¼ 1þ - 1ð Þ2 e- iπ 4 ¼ 2 
p 
e- i
π 
4 
1þ i 3 
p 
¼ 1ð Þ2 þ 3 
p 2 
ei
π 
3 ¼ 2eiπ 3 
Therefore: 
z ¼ 1- ið Þ 1þ i 3 
p 
¼ 2 
p 
e- i
π 
4 2ei
π 
3 ¼ 2 2 
p 
ei 
π 
3-
π 
4ð Þ ¼ 2 2 
p 
ei
π 
12 
) z ¼ 2 2 
p 
cos 
π 
12 
þ i sin π 
12 
Choice (1) is the answer. 
In this problem, the rules below were used. 
aþ ib ¼ a2 þ b2 ei tan - 1 b aj j if a> 0, b> 0
100 8 Solutions of Problems: Complex Numbers
aþ ib ¼ a2 þ b2 ei π- tan - 1 b aj jð Þ if a< 0, b> 0 
a þ ib ¼ a2 þ b2 ei πþ tan - 1 b aj jð Þ if a< 0, b< 0 
aþ ib ¼ a2 þ b2 e- i tan - 1 b aj j if a> 0, b< 0 
tan - 1 - 1ð Þ ¼ - π 
4 
tan - 1 3 
p 
¼ π 
3 
z1z2 ¼ z1j jeiθ1 z2j jeiθ2 ¼ z1j j z2j jei θ1þθ2ð Þ 
zj jeiθ ¼ zj j cos θ þ i sin θð Þ 
8.13. The problem can be solved by transferring from Cartesian coordinate to polar coordinate as follows. 
1 þ ið Þ15 
1- ið Þ13 ¼ 
2
p 
ei
π 
4 
15 
2
p 
e- i
π 
4 
13 
¼ 2e
i15π 4 
e- i
13π 
4 
¼ 2ei 15π 4 - - 13π 4ð Þð Þ 
¼ 2ei28π 4 ¼ 2ei7π 
¼ 2 cos 7π þ i sin 7πð Þ ¼ 2 cos π þ i sin πð Þ ¼ 2 - 1þ 0ið Þ 
Thus: 
1þ ið Þ15 
1- ið Þ13 ¼ - 2 
Choice (2) is the answer. 
In this problem, the rules below were used. 
a þ ib ¼ a2 þ b2 ei tan- 1 b aj j if a> 0, b> 0 
aþ ib ¼ a2 þ b2 ei π- tan - 1 b aj jð Þ if a< 0, b> 0 
a þ ib ¼ a2 þ b2 ei πþ tan - 1 b aj jð Þ if a< 0, b< 0 
aþ ib ¼ a2 þ b2 e- i tan - 1 b aj j if a> 0, b< 0 
tan - 1 1 ¼ π 
4 
tan - 1 - 1ð Þ ¼ - π 
4
p
p
p
8 Solutions of Problems: Complex Numbers 101
eia 
b ¼ eiab 
z1 
z2 
¼ z1j je
iθ1 
z2j jeiθ2 ¼ 
z1j j 
z2j j e
i θ1 - θ2ð Þ 
zj jeiθ ¼ zj j cos θ þ i sin θð Þ 
8.14. The problem can be solved by transferring from Cartesian coordinate to polar coordinate as follows. 
3
p þ i 
3
p
- i 
10 
¼ 2e
iπ 6 
2e- i
π 
6 
10 
¼ eiπ 3 10 ¼ ei10π 3 
¼ ei 2πþ4π 3ð Þ ¼ ei4π 3 
¼ cos 4π 
3 
þ i sin 4π 
3 
¼ - 1 
2
- i 
3 
2 
Therefore: 
3
p þ i 
3
p
- i 
10 
¼ - 1- i 3
p 
2 
Choice (4) is the answer. 
In this problem, the rules below were used. 
aþ ib ¼ a2 þ b2 ei tan - 1 b aj j if a> 0, b> 0 
aþ ib ¼ a2 þ b2 ei π- tan - 1 b aj jð Þ if a< 0, b> 0 
a þ ib ¼ a2 þ b2 ei πþ tan - 1 b aj jð Þ if a< 0, b< 0 
aþ ib ¼ a2 þ b2 e- i tan - 1 b aj j if a> 0, b< 0 
tan - 1 
3 
3 
¼ π 
6 
tan - 1 -
3 
3 
¼ - π 
6 
z1 
z2 
¼ z1j je
iθ1 
z2j jeiθ2 ¼ 
z1j j 
z2j j e
i θ1 - θ2ð Þ 
eia 
b ¼ eiab 
ei 2πþθð Þ ¼ eiθ 
zj jeiθ ¼ zj j cos θ þ i sin θð Þ
p
þ ¼ þ
102 8 Solutions of Problems: Complex Numbers
cos 
4π 
3 
¼ - 1 
2 
sin 
4π 
3 
¼ - 3 
2 
8.15. The problem can be solved by transferring from Cartesian coordinate to polar coordinate as follows. 
1þ cos 2π 
3 
þ i sin 2π 
3 
120 
¼ 1- 1 
2
þ i 3
p 
2 
120 
¼ 1 
2
þ i 3
p 
2 
120 
¼ eiπ 3 120 ¼ ei40π ¼ ei 20�2πþ0ð Þ 
¼ cos 0ð Þ þ i sin 0ð Þ ¼ 1þ 0i 
Therefore: 
1þ cos 2π 
3 
þ i sin 2π 
3 
120 
¼ 1 
Choice (2) is the answer. 
In this problem, the rules below were used. 
cos 
2π 
3 
¼ - 1 
2 
sin 
2π 
3 
¼ 3
p 
2 
aþ ib ¼ a2 þ b2 ei tan - 1 b aj j if a> 0, b> 0 
aþ ib ¼ a2 þ b2 ei π- tan - 1 b aj jð Þ if a< 0, b> 0 
a þ ib ¼ a2 þ b2 ei πþ tan - 1 b aj jð Þ if a< 0, b< 0 
a ib a2 b2 e- i tan
- 1 b 
aj j if a> 0, b< 0 
tan - 1 3 
p 
¼ π 
3 
eia 
b ¼ eiab 
ei 2πþθð Þ ¼ eiθ 
zj jeiθ ¼ zj j cos θ þ i sin θð Þ 
cos 0ð Þ ¼ 1 
sin 0ð Þ ¼ 0
8 Solutions of Problems: Complex Numbers 103
8.16. Based on the information given in the problem, we have: 
eaþib ¼ 1- i 3 
p 
ð1Þ 
The problem can be solved by transferring from Cartesian coordinate to polar coordinate as follows. 
1- i 3 
p 
¼ 2e- π 3i ð2Þ 
As we know: 
eaþib ¼ ea eib ð3Þ 
Solving (1)–(3): 
ea eib ¼ 2e- π 3i ) e
a ¼ 2 
eib ¼ e- π 3i ) a ¼ ln 2, b ¼ -
π 
3 
Choice (1) is the answer. 
In this problem, the rules below were used. 
aþ ib ¼ a2 þ b2 ei tan - 1 b aj j if a> 0, b> 0 
aþ ib ¼ a2 þ b2 ei π- tan - 1 b aj jð Þ if a< 0, b> 0 
a þ ib ¼ a2 þ b2 ei πþ tan - 1 b aj jð Þ if a< 0, b< 0 
aþ ib ¼ a2 þ b2 e- i tan - 1 b aj j if a> 0, b< 0 
tan - 1 - 3 
p 
¼ - π 
3 
eaþb ¼ ea eb 
8.17. Based on the information given in the problem, we have: 
zm ¼ cos π 2m þ i sin 
π 
2m 
Therefore: 
∏ 
1 
m¼1 
zm ¼ ∏ 
1 
m¼1 
e 
π 
2mi ¼ e π 2i e 1 2 π 2ð Þi e 1 4 π 2ð Þi⋯ ¼ e π 2iþ1 2 π 2ið Þþ1 4 π 2ið Þþ⋯ 
) ∏ 
1 
m¼1 
zm ¼ exp 
1 
m¼1 
π 
2m 
i ¼ e 
1 
m¼1 
π 
2mi ¼ e 
π 
2i 
1- 1 2 
) ∏ 
1 
m¼1 
zm ¼ eπi ¼ cos π þ i sin π ¼ - 1þ 0i
104 8 Solutions of Problems: Complex Numbers
) ∏ 
1 
m¼1 
zm ¼ - 1 
Choice (4) is the answer. 
In this problem, the rules below were used. 
∏ 
1 
m¼1 
zm ¼ z1z2 . . . z1 
eix ¼ cos xþ i sin x 
z1j jeiθ1 z2j jeiθ2 ¼ z1j j z2j jei θ1þθ2ð Þ 
1 
m¼1 
am ¼ a1 1- q if qj j< 1 
eπi ¼ - 1 
8.18. Based on the information given in the problem, we have: 
z1 - z2 
z1 þ z2 ¼ 1 ð1Þ 
Let us assume: 
z1 ¼ x1 þ iy1 ð2Þ 
z2 ¼ x2 þ iy2 ð3Þ 
Solving (1)–(3): 
x1 þ iy1ð Þ- x2 - iy2ð Þ 
x1 þ iy1ð Þ þ x2 - iy2ð Þ 
¼ 1 ) x1 - x2ð Þ þ i y1 þ y2ð Þ 
x1 þ x2ð Þ þ i y1 - y2ð Þ 
¼ 1 
) x1 - x2ð Þ þ i y1 þ y2ð Þj j 
x1 þ x2ð Þ þ i y1 - y2ð Þj j
¼ 1 
) x1 - x2ð Þ þ i y1 þ y2ð Þj j ¼ x1 þ x2ð Þ þ i y1 - y2ð Þj j 
) x1 - x2ð Þ2 þ y1 þ y2ð Þ2 ¼ x1 þ x2ð Þ2 þ y1 - y2ð Þ2 
) x1ð Þ2 þ x2ð Þ2 - 2x1x2 þ y1ð Þ2 þ y2ð Þ2 þ 2y1y2 ¼ x1ð Þ2 þ x2ð Þ2 þ 2x1x2 þ y1ð Þ2 þ y2ð Þ2 - 2y1y2 
) - 4x1x2 þ 4y1y2 ¼ 0 ) x1x2 - y1y2 ¼ 0 
) Re z1z2ð Þ ¼ 0 
Choice (3) is the answer. 
In this problem, the rules below were used.
þj j ¼ þ
8 Solutions of Problems: Complex Numbers 105
a þ ib 
cþ id ¼ 
aþ ibj j 
cþ idj j 
a ib a2 b2 
Re z1z2ð Þ ¼ Re x1 þ iy1ð Þ x2 þ iy2ð Þð Þ ¼ x1x2 - y1y2 
i2 ¼ - 1 
8.19. Based on the information given in the problem, we have: 
6z- i 
2þ 3iz ≤ 1 
6z- ij j 
2þ 3izj j ≤ 1 ) 6 xþ iyð Þ- ij j≤ 2þ 3i xþ iyð Þj j 
) 6xþ i 6y- 1ð Þj j≤ 2- 3yð Þ þ 3ixj j 
) 6xð Þ2 þ 6y- 1ð Þ2 ≤ 2- 3yð Þ2 þ 3xð Þ2 
) 36x2 þ 36y2 - 12y þ 1≤ 4- 12yþ 9y2 þ 9x2 
27x2 þ 27y2 ≤ 3 ) x2 þ y2 ≤ 1 
9 
) x2 þ y2 ≤ 1 
3 
) zj j≤ 1 
3 
) max zj jð Þ ¼ 1 
3 
Choice (3) is the answer. 
In this problem, the rules below were used. 
aþ ib 
cþ id ¼ 
aþ ibj j 
cþ idj j 
a þ ibj j ¼ a2 þ b2 
2 i ¼ - 1 
8.20. Based on the information given in the problem, we have: 
1 þ sin θ þ i cos θ 
1þ sin θ- i cos θ 
The problem can be solved as follows.
1 sin cos 1 sin cos 1 sin cos
106 8 Solutions of Problems: Complex Numbers
þ θ þ i θ 
1þ sin θ- i cos θ ¼ 
þ θ þ i θ 
1þ sin θ- i cos θ �
þ θ þ i θ 
1þ sin θð Þ þ i cos θ 
¼ 1þ sin θ þ i cos θð Þ
2 
1 þ sin θð Þ2 - i2 cos 2θ ¼
1 þ sin 2 θ þ icos θð Þ2 þ 2 sin θ þ 2 sin θi cos θ þ 2i cos θ 
1þ sin 2 θ þ 2 sin θ þ cos 2θ 
¼ 1þ sin
2 θ- cos 2θ þ 2 sin θ þ 2i cos θ 1 þ sin θð Þ 
2þ 2 sin θ 
¼ sin
2 θ þ cos 2θ þ sin 2 θ- cos 2θ þ 2 sin θ þ 2i cos θ 1þ sin θð Þ 
2þ 2 sin θ 
¼ 2 sin
2 θ þ 2 sin θ þ 2i cos θ 1þ sin θð Þ 
2þ 2 sin θ 
¼ 2 sin θ 1 þ sin θð Þ þ 2i cos θ 1 þ sin θð Þ 
2 1 þ sin θð Þ 
¼ 2 1þ sin θð Þ sin θ þ i cos θð Þ 
2 1þ sin θð Þ ¼ sin θ þ i cos θ 
Therefore: 
1þ sin θ þ i cos θ 
1þ sin θ- i cos θ ¼ sin θ þ i cos θ 
Choice (1) is the answer. 
In this problem, the rules below were used. 
aþ ib 
c þ id ¼
aþ ib 
c þ id �
c- id 
c- id 
a þ ibð Þ a- ibð Þ ¼ a2 þ b2 
a þ b þ cð Þ2 ¼ a2 þ b2 þ c2 þ 2abþ 2bcþ 2ac 
sin 2 θ þ cos 2 θ ¼ 1 
References 
1. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
2. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021. 
3. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature, 2021.
Index 
A 
Angle, 74, 76, 79, 80, 85 
Applications of integration, v, 1–52 
Arc length of a curve, 10–12, 37–45 
Average value of a function, 15 
C 
Cartesian coordinate system, 73–75 
Center of gravity, 13–14, 49–52 
Coefficient, 54, 60 
Complex conjugate, 89, 96 
Complex form, 90, 93 
Complex numbers, 89–94, 96 
Concept of equivalent functions, 60, 66–68 
Concept of growth rate, 59, 61, 64 
Consecutive terms, 63, 69 
Convergence, 55, 63, 65, 71 
Convergence radius, 55, 56, 63–65 
Convergence range, 55, 63–65 
Convergent, 54, 55, 58, 61, 62, 69–71 
Curve type, 75, 76 
D 
Defining a new variable, 19, 22, 46, 47 
Divergent, 54, 55, 57, 61, 62, 69 
E 
Ellipse, 75, 76, 86 
Even function, 19–21, 50 
G 
General term of a sequence, 54, 57 
H 
Harmonic series, 62 
Hyperbola, 75, 76 
I 
Imaginary part, 96 
Integration by parts, 22, 25, 33–35 
Intersection point, 16–18, 76, 85 
Intersection points of the curves, 16–18 
# The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 
M. Rahmani-Andebili, Calculus II, https://doi.org/10.1007/978-3-031-45353-3 
107
L 
Limit, 30, 53, 54, 56, 57, 69, 70 
List of integral of functions, 15, 22, 25, 26, 28–30 
M 
Maclaurin expansion, 54, 60, 65 
Maclaurin series, 60, 65 
Mean value of a function, 1, 15–16 
N 
Necessary criterion of convergence, 71 
O 
Odd function, 50 
Order of growth rate, 59, 61, 64 
P 
Parabola, 75, 76, 83 
Parametric curve, 13, 38, 39 
Parametric relation, 10 
Polar coordinate system, 73–87 
Polar curve, 79, 80, 85 
Polar equation, 74,81 
Polar function, 75, 76 
P-series, 61, 62, 71 
Q 
Quadrant, 19, 75 
R 
Radius of curve, 74, 79, 80, 85 
Range of integration, 47 
Real part, 91 
Rotation of a surface area around x-axis, 28–30, 32–36, 47 
Rotation of a surface area around y-axis, 8, 31 
S 
Sequences, 56, 57, 59–71 
Series, 54–71 
Spiral, 74, 81, 82 
Stirling’s approximation, 64, 65, 69
https://doi.org/10.1007/978-3-031-45353-3#DOI
108 Index
Straight line, 75, 76, 83 
Surface area, 2–9, 12, 13, 16–27, 30, 31, 45–50, 75, 81, 82 
Surface area enclosed by curves, 31, 33 
Surface area of a solid of revolution, 12–13, 45–49 
T 
Tangent line, 74, 79, 80, 85 
Taylor expansion, 65 
Taylor series, 56, 65, 66 
Telescoping series, 62, 63, 68, 69 
Theorem, 62, 68–70 
Transfer from cartesian coordinate to polar coordinate, 77, 78, 97, 
99–103 
Trigonometry, 26, 29 
V 
Volume resulted from rotation of an enclosed region, 7–10, 28–37
	Preface
	Precalculus: Practice Problems, Methods, and Solution
	Calculus 1: Practice Problems, Methods, and Solution
	Calculus 2: Practice Problems, Methods, and Solution
	Calculus 3: Practice Problems, Methods, and Solution
	The Other Works Published by the Author
	Contents
	1: Problems: Applications of integration
	1.1 Mean Value of a Function
	1.2 Surface Area Bounded by Curves
	1.3 Volume Resulted from Rotation of an Enclosed Region
	1.4 Arc Length of a Curve
	1.5 Surface Area of a Solid of Revolution
	1.6 Center of Gravity
	References
	2: Solutions of Problems: Applications of Integration
	2.1 Mean Value of a Function
	2.2 Surface Area Bounded by Curves
	2.3 Volume Resulted from Rotation of an Enclosed Region
	2.4 Arc Length of a Curve
	2.5 Surface Area of a Solid of Revolution
	2.6 Center of Gravity
	References
	3: Problems: Sequences and Series and Their Applications
	References
	4: Solutions of Problems: Sequences and Series and Their Applications
	References
	5: Problems: Polar Coordinate System
	References
	6: Solutions of Problems: Polar Coordinate System
	References
	7: Problems: Complex numbers
	References
	8: Solutions of Problems: Complex Numbers
	References
	Index