Calculus III: Practice Problems, Methods, and Solutions [1 ed.] 9783031474828, 9783031474835, 9783031474859

Citation preview

Mehdi Rahmani-Andebili

Calculus III

Practice Problems, Methods, and Solutions

Calculus III

Mehdi Rahmani-Andebili

Calculus III Practice Problems, Methods, and Solutions

Mehdi Rahmani-Andebili Department of Electrical Engineering Arkansas Tech University Russellville, AR, USA

ISBN 978-3-031-47482-8 ISBN 978-3-031-47483-5 https://doi.org/10.1007/978-3-031-47483-5

(eBook)

# The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland Paper in this product is recyclable.

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 are 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 courses. 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

v

vi

Preface

Calculus 2: Practice Problems, Methods, and Solution • • • •

Applications of Integration Sequences and Series and Their Applications Polar Coordinate System Complex Numbers

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 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 Operation of Plug-in Electric Vehicles: Technical, Geographical, and Social Aspects, Springer Nature, 2019.

Contents

1

Problems: Linear Algebra and Analytical Geometry . . . . . . . . . . . . . . . . . . . .

1

2

Solutions of Problems: Linear Algebra and Analytical Geometry . . . . . . . . . .

9

3

Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

23

Solutions of Problems: Lines, Surfaces, and Vector Functions in ThreeDimensional Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

29

5

Problems: Multivariable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43

6

Solutions of Problems: Multivariable Functions . . . . . . . . . . . . . . . . . . . . . . .

53

7

Problems: Double Integrals and Their Applications . . . . . . . . . . . . . . . . . . . . 7.1 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Applications of Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75 75 80

8

Solutions of Problems: Double Integrals and Their Applications . . . . . . . . . . . 8.1 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Applications of Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83 83 96

9

Problems: Triple Integrals and Their Applications . . . . . . . . . . . . . . . . . . . . . 105 9.1 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 9.2 Applications of Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

10

Solutions of Problems: Triple Integrals and Their Applications . . . . . . . . . . . 109 10.1 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 10.2 Applications of Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

11

Problems: Line Integrals and Their Applications . . . . . . . . . . . . . . . . . . . . . . 117 11.1 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 11.2 Applications of Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

12

Solutions of Problems: Line Integrals and Their Applications . . . . . . . . . . . . . 123 12.1 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 12.2 Applications of Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

4

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

1

Problems: Linear Algebra and Analytical Geometry

Abstract

In this chapter, the basic and advanced problems of linear algebra and analytical geometry are presented. The subjects include matrix, inverse matrix, determinant of matrix, transpose matrix, eigenvectors of matrix, eigenvalues of matrix, rank of matrix, linearly dependent vectors, characteristics equation, rotation matrix, transpose matrix, and coefficient matrix. 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. A and B are two matrices. Which one of the following choices is equivalent to (AB - BA)T [1–5]? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)

(BA - AB)T ATBT - BTAT BTAT - ATBT (BA)T - (AB)T

1.2. Calculate the determinant of the matrix below.

A=

3

7

-1

4 10

1 15

5 3

Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)

0 -1 1 3

1.3. Which one of the following vectors is parallel to the vector of (1, -1, 1)? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus III, https://doi.org/10.1007/978-3-031-47483-5_1

1

2

1 →

→

→

→

→

→

→

→

1) - 2 i þ 2 j þ 2 k 2) - 3 i þ 3 j - 3 k →

3) - 3 i þ 3 j - k →

→

→

4) - 3 i þ j - 3 k

1.4. Calculate the matrix of A100 if the matrix of A is as follows. 2 1 -1 0

A=

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)

2100

1

1 0 200 100 - 100 0 101 100 - 100 - 99 201 101 99

100

1.5. Calculate the value of A12 if: p

2 2 A= p 2 2

p

2 2 p 2 2

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)

-1 0 0 -1 0 1 1 0 1 0 0 1 0 -1 0 -1

1.6. Calculate the determinant of the matrix below?

45

17

3

- 45

- 17

12 -3

21 -9

6 - 11

- 12 3

21 9

6

10

9

-6

10

-7

-8

3

7

8

Problems: Linear Algebra and Analytical Geometry

1

Problems: Linear Algebra and Analytical Geometry

3

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)

-118 0 118 1568

1.7. Calculate the volume of parallelepiped resulted from the vectors below. →

→

→

→

→

→

→

→

a= i þ j þ k b= i - j þ k →

→

→

c =- i þ j

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)

4 2 p

2

1

1.8. Solve the equation below. x2 4 9

x 2

1 1 =0

-3 1

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2) 3) 4)

-1, 2 2, - 3 1, - 3 There is no real answer for the equation.

1.9. For what value of m, the matrix below is not invertible?

A=

-1 2 1 3

mþ1 2m

-1 2

0

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

4

1

1) 2) 3) 4)

Problems: Linear Algebra and Analytical Geometry

-2 -1 1 2

1.10. For what value of m, the system of equations below has infinite answers? x - y þ mz = 0 3x - y þ 2z = 0 6x - 4y þ 5z = 0 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2) 3) 4)

-1 1 0 2

1.11. For what value of m, the vectors below are linearly dependent? V 1 = ð2, m, 1Þ V 2 = ð- 1, 3, 2Þ V 3 = ð1, 4, 3Þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2) 3) 4)

-2 -1 1 2 →

→

→

→

→

→

→

→

1.12. Calculate the image of the vector of A = - i þ 2 j þ 3 k on the vector of B = 2 i - j þ 2 k . Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1→ 1→ 1→ i - j þ k 2 4 2 4→ 2→ 4→ 2) i - j þ k 9 9 9 2→ 4→ 6→ 3) - i þ j þ k 9 9 9 1→ 1→ 3→ 4) - i þ j þ k 4 4 4

1)

1

Problems: Linear Algebra and Analytical Geometry

5

1.13. For what value of “a”, the vectors of (4, 3, 0), (0, a, 2), and (2, 0, -1) are in one plane. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2) 3) 4)

-1 3 -2 4

1.14. Determine the characteristics equation of the linear function below. f ðx, y, zÞ = ðx - y, x - z, 2x - y - zÞ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2) 3) 4)

λ3 - λ = 0 λ2 + λ = 0 λ3 + λ2 - λ = 0 λ3 - λ2 + λ = 0

1.15. Calculate the amount of angle between the two crossing lines below. L1 : L2 :

x-1 y-2 = ,z=3 3 4

x-1 y-2 z-3 = = 3 4 5

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large π 1) 6 π 2) 4 π 3) 3 π 4) 2 1.16. Calculate the distance between the point of P0(1, 2, 3) from the following plane. Plane: 2x - y þ 2z þ 9 = 0 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 5 p 17 2) 3 p 3) 5 15 4) p 2 2

6

1

Problems: Linear Algebra and Analytical Geometry

1.17. Determine the rank of the matrix below. 3

2 5

7

12

A= 1 3

1 2 3 6

3 9

5 15

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)

0 1 2 3

1.18. For what value of “a” and “b”, the vectors below are the eigenvectors of the following matrix. 1 V1 =

1 1

1 , V2 =

A=

1 1

1

a b 1 1

1 1

0 -1

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 2) 3) 4)

a = 1, b = 1 a = 1, b = - 1 a = - 1, b = 1 a = - 1, b = - 1

1.19. Calculate the distance between the point of P(-1, 3, -1) from the following line.

L:

x = 2t þ 1 y=1 z=t

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 2) 3) 4)

1 2 3 4

References

References 1. Rahmani-Andebili, M. (2023). Calculus II – Practice Problems, Methods, and Solutions, Springer Nature. 2. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 3. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature. 4. Rahmani-Andebili, M. (2023). Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 5. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature.

7

2

Solutions of Problems: Linear Algebra and Analytical Geometry

Abstract

In this chapter, the problems of the first chapter are fully solved, in detail, step-by-step, and with different methods. 2.1. The problem can be solved as follows [1–5]: ðAB
BAÞT ¼ ðABÞT
ðBAÞT ¼ BT AT
AT BT Herein, AT is the transpose matrix of A and BT is the transpose matrix of B. Choice (3) is the answer. In this problem, the rule below has been used. ðABÞT ¼ BT AT 2.2. Based on the information given in the problem, we have:

A¼

3 4

7 1


1 5

10

15

3

Therefore:

detðAÞ ¼ jAj ¼

3 4

7 1

10 15


1 5 ¼ 3ð1  3
5  15Þ
7ð3  4
5  10Þ þ ð
1Þð4  15
1  10Þ 3 ) jAj ¼ 3ð
72Þ þ 7  38
50 ) jAj ¼ 0

Choice (1) is the answer.

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus III, https://doi.org/10.1007/978-3-031-47483-5_2

9

10

2 Solutions of Problems: Linear Algebra and Analytical Geometry

In this problem, the rule below has been used.

A¼

) detðAÞ ¼ jAj ¼ ð
1Þ1þ1 a11

a22

a23

a32

a33

a11

a12

a13

a21 a31

a22 a32

a23 a33

þ ð
1Þ1þ2 a12

a21

a23

a31

a33

þ ð
1Þ1þ3 a13

a21

a22

a31

a32

) detðAÞ ¼ ð
1Þ1þ1 a11 ða22 a33
a23 a32 Þ þ ð
1Þ1þ2 a12 ða21 a33
a23 a31 Þ þ ð
1Þ1þ3 a13 ða21 a32
a22 a31 Þ 2.3. Based on the information of the problem, the vector below is given. →

V ¼ ð1,
1, 1Þ ! ! If the vectors of A1 ¼ ða1 , a2 , a3 Þ and A2 ¼ ðb1 , b2 , b3 Þ are in parallel, then we have: a1 b 1 c 1 ¼ ¼ a2 b2 c 2 As can be noticed, for Choice (2), we have: )


3 3
3 ¼ ¼ 1
1 1

Choice (2) is the answer. In this problem, the rule below has been used. ! ! If the vectors of A1 ¼ ða1 , a2 , a3 Þ and A2 ¼ ðb1 , b2 , b3 Þ are in parallel, then: a1 b 1 c 1 ¼ ¼ a2 b2 c 2 ! ! A2 ¼ λ A1 ! ! A1  A2 ¼ 0

ðouter productÞ

2.4. Based on the information given in the problem, we have: A¼

2
1

1 0

First, let us calculate A2 as follows. A2 ¼ AA ¼ Then:

2
1

1 0

2 1 3 ¼
1 0
2

2
1

2

Solutions of Problems: Linear Algebra and Analytical Geometry

3
2

A3 ¼ A2 A ¼

11

2
1

2 1 4 ¼
1 0
3

3
2

As can be noticed, there is a specific rule for An which is as follows. An ¼

nþ1

n


n


nðn
1Þ

Therefore, for n ¼ 100, we have: A100 ¼

101

100


100


99

Choice (3) is the answer. 2.5. Based on the information given in the problem, we have: p 2 2 p 2 2

A¼

p

2
2 p 2 2

First, the matrix of A can be written as follows. π 4 π sin 4

π 4 π cos 4


sin

cos

A¼

As can be noticed, the matrix is a rotation matrix with the angle of θ ¼ π4. Therefore:

A12 ¼

π 4 π sin 4

cos

π 4 π cos 4

12


sin

¼

) A12 ¼


1 0

cos 3π


sin 3π

sin 3π

cos 3π

0
1

Choice (1) is the answer. In this problem, the rules below have been used. We know that for a rotation matrix, we have: cos θ


sin θ

sin θ

cos θ

n

¼

cos nθ


sin nθ

sin nθ

cos nθ

cos 3π ¼ cos π ¼
1 sin 3π ¼ sin π ¼ 0

12

2 Solutions of Problems: Linear Algebra and Analytical Geometry

2.6. Based on the information given in the problem, we have: 45

17

3


45


17

12
3

21
9

6
11


12 3

21 9

6

10

9


6

10


7


8

3

7

8

As can be noticed, the fourth column of the matrix is a factor of the first column. Therefore, the determinant of the matrix is zero. Choice (2) is the answer. In this problem, the rules below have been used. Rule 1: If one row or one column of a matrix is zero, the determinant of the matric is zero. Rule 2: If one row (or one column) of a matrix is a factor of another row (or another column) of the matrix, the determinant of the matric is zero. Rule 3: If one row (or one column) is equal to sum or subtract of two other rows (or two other columns), the determinant of the matric is zero. 2.7. Based on the information given in the problem, we have: →

→

→

→

→

→

→

→

a ¼ i þ j þ k b ¼ i
j þ k →

→

→

c ¼
i þ j

The volume of parallelepiped resulted from three vectors can be calculated by using each of the relations below. ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! Volume ¼ A1 : A2  A3 ¼ A1 : A3  A2 ¼ A2 : A3  A1 ¼ A2 : A1  A3 ¼ A3 : A1  A2 ! ! ! ¼ A3 : A2  A1 Therefore:

→

→

→

Volume ¼ a : b  c

¼

1

1

1

1
1


1 1

1 0

) Volume ¼
2 Choice (2) is the answer.

2

Solutions of Problems: Linear Algebra and Analytical Geometry

13

In this problem, the rule below has been used. a11 a21

a12 a22

a13 a23 ¼ ð
1Þ1þ1 a11 ða22 a33
a23 a32 Þ þ ð
1Þ1þ2 a12 ða21 a33
a23 a31 Þ þ ð
1Þ1þ3 a13 ða21 a32
a22 a31 Þ

a31

a32

a33

2.8. Based on the information given in the problem, we have: x2 4

x 2

9


3

1 1 ¼0 1

The problem can be solved as follows. ) x2 ð2  1
1  ð
3ÞÞ
xð4  1
1  9Þ þ 1ð4  ð
3Þ
2  9Þ ) 5x2 þ 5x
30 ¼ 0 ) x 2 þ x
6 ¼ 0 ) ð x þ 3Þ ð x
2 Þ ¼ 0 ) x ¼ 2,
3 Choice (2) is the answer. In this problem, the rule below has been used.

det

a11 a21

a12 a22

a13 a23

a31

a32

a33

¼ ð
1Þ1þ1 a11 ða22 a33
a23 a32 Þ þ ð
1Þ1þ2 a12 ða21 a33
a23 a31 Þ þ ð
1Þ1þ3 a13 ða21 a32
a22 a31 Þ

2.9. Based on the information given in the problem, we have:

A¼


1

2

mþ1

1
1

3 2

2m 0

As we know, a matrix is not invertible if its determinant is zero. For this problem, we have: jAj ¼ ð
1Þð0
4mÞ
2ð0 þ 2mÞ þ ðm þ 1Þð2 þ 3Þ ¼ 0 ) 4m
4m þ 5m þ 5 ¼ 0 ) m ¼
1 Choice (2) is the answer.

14

2 Solutions of Problems: Linear Algebra and Analytical Geometry

In this problem, the rule below has been used.

det

a11

a12

a13

a21 a31

a22 a32

a23 a33

¼ ð
1Þ1þ1 a11 ða22 a33
a23 a32 Þ þ ð
1Þ1þ2 a12 ða21 a33
a23 a31 Þ þ ð
1Þ1þ3 a13 ða21 a32
a22 a31 Þ

2.10. Based on the information given in the problem, we have: x
y þ az ¼ 0 3x
y þ 2z ¼ 0 6x
4y þ 5z ¼ 0 The system of equations can be written in the matrix form as follows. 1


1

α

x

3


1

2

y

6


4

5

z

0 ¼

0 0

The system of equations or the presented matrix equation has infinite answers if the determinant of the coefficient matrix is zero. Therefore: 1 3


1
1

α 2 ¼0

6


4

5

) 1ð
5 þ 8Þ þ ð15
12Þ þ αð
12 þ 6Þ ¼ 0 ) 3 þ 3
6α ¼ 0 )α¼1 Choice (2) is the answer. In this problem, the rule below has been used.

det

a11

a12

a13

a21 a31

a22 a32

a23 a33

¼ ð
1Þ1þ1 a11 ða22 a33
a23 a32 Þ þ ð
1Þ1þ2 a12 ða21 a33
a23 a31 Þ þ ð
1Þ1þ3 a13 ða21 a32
a22 a31 Þ

2.11. Based on the information given in the problem, we have: V 1 ¼ ð2, m, 1Þ V 2 ¼ ð
1, 3, 2Þ V 3 ¼ ð1, 4, 3Þ If the determinant of the vectors of V1, V2, and V3 is zero, then they are linearly dependent. In other words:

2

Solutions of Problems: Linear Algebra and Analytical Geometry

15

V1 V2 ¼ 0 V3 Therefore: 2

m

1


1 1

3 4

2 ¼0 3

) 2ð9
8Þ
mð
3
2Þ þ 1ð
4
3Þ ¼ 0 ) 2 þ 5m
7 ¼ 0 ) 5m ¼ 5 )m¼1 Choice (3) is the answer. In this problem, the rule below has been used.

det

a11

a12

a13

a21 a31

a22 a32

a23 a33

¼ ð
1Þ1þ1 a11 ða22 a33
a23 a32 Þ þ ð
1Þ1þ2 a12 ða21 a33
a23 a31 Þ þ ð
1Þ1þ3 a13 ða21 a32
a22 a31 Þ

2.12. Based on the information given in the problem, we have: →

→

→

→

A ¼
i þ2 j þ3k →

→

→

→

B ¼2 i
j þ2k

→

→

The image of the vector of A on the vector of B can be calculated as follows. →

Proj →A B

→ →

¼

A:B

→ 2

→

B

B

Therefore, for this problem, we have: →

Proj →A ¼ B

→
2
2 þ 6 → → 2 i
j þ2k 4þ1þ4 →

Proj →A ¼ B

Choice (2) is the answer. In this problem, the rules below have been used.

4→ 2→ 4→ i
j þ k 9 9 9

16

2 Solutions of Problems: Linear Algebra and Analytical Geometry →

→

Image of vector A on vector of B : →

Proj →A B

→ →

¼

A:B

→ 2

→

B

B

Inner product of two vectors: → → → → → → ! ! A1 : A2 ¼ a1 i þ b1 j þ c1 k : a2 i þ b2 j þ c2 k ¼ a1 a2 þ b1 b2 þ c1 c2

Magnitude of a vector: →

A

→

→

→

¼ a i þb j þck ¼

a 2 þ b2 þ c 2

2.13. Based on the information given in the problem, we have: ! A1 ¼ ð4, 3, 0Þ ! A2 ¼ ð0, a, 2Þ ! A3 ¼ ð2, 0,
1Þ ! ! ! The three vectors of A1 , A2 , and A3 are in one plane if: ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! A1 : A2  A3 ¼ A1 : A3  A2 ¼ A2 : A3  A1 ¼ A2 : A1  A3 ¼ A3 : A1  A2 ! ! ! ¼ A3 : A2  A1 ¼ 0 Therefore, for this problem, we can write: 4 ! ! ! A1 : A2  A3 ¼ 0 2

3 a 0

0 2 ¼0
1

) 4ð
a
0Þ
3ð0
4Þ þ 0 )
4a þ 12 ¼ 0 )a¼3 Choice (2) is the answer. In this problem, the rule below has been used. a11

a12

a13

a21 a31

a22 a32

a23 ¼ ð
1Þ1þ1 a11 ða22 a33
a23 a32 Þ þ ð
1Þ1þ2 a12 ða21 a33
a23 a31 Þ þ ð
1Þ1þ3 a13 ða21 a32
a22 a31 Þ a33

2

Solutions of Problems: Linear Algebra and Analytical Geometry

17

2.14. Based on the information given in the problem, we have: f ðx, y, zÞ ¼ ðx
y, x
z, 2x
y
zÞ The characteristics equation of a linear function can be determined by solving the equation below. jA
λI j ¼ 0 For this problem, we have:

A¼

1


1

0

1

0


1

2


1


1

Therefore:

jA
λI j ¼

1
λ 1


1 0
λ

0
1

2


1

ð
1
λÞ

¼0

) ð1
λÞðλð1 þ λÞ
1Þ þ 1ðð
1
λÞ þ 2Þ ¼ 0 ) ð1
λÞ λ2 þ λ
1 þ 1
λ ¼ 0 ) λ2 þ λ
1
λ3
λ2 þ λ þ 1
λ ¼ 0 ) λ3
λ ¼ 0 Choice (1) is the answer. In this problem, the rule below has been used. a11 a21

a12 a22

a13 a23 ¼ ð
1Þ1þ1 a11 ða22 a33
a23 a32 Þ þ ð
1Þ1þ2 a12 ða21 a33
a23 a31 Þ þ ð
1Þ1þ3 a13 ða21 a32
a22 a31 Þ

a31

a32

a33

2.15. Based on the information given in the problem, we have: L1 : L2 :

x
1 y
2 ¼ ,z ¼ 3 3 4

x
1 y
2 z
3 ¼ ¼ 3 4 5

! ! The amount of angle between two crossing lines can be calculated as follows, where V 1 and V 2 are the direction vectors of the lines. cos θ ¼

! ! V1 : V2 ! ! V1 V2

18

2 Solutions of Problems: Linear Algebra and Analytical Geometry

Therefore: ! V 1 ¼ ð3, 4, 0Þ ! V 2 ¼ ð3, 4, 5Þ ) cos θ ¼

p 25 25 2 p ¼ p ¼ ¼p 2 2 2 2 25  50 25 2 3 þ4 þ5

33þ44þ05 3 2 þ 42 þ 0

)θ¼

π 4

Choice (2) is the answer. In this problem, the rules below have been used. Inner product of two vectors: → → → → → → ! ! A1 : A2 ¼ a1 i þ b1 j þ c1 k : a2 i þ b2 j þ c2 k ¼ a1 a2 þ b1 b2 þ c1 c2

Magnitude of a vector: →

A

→

→

→

¼ a i þb j þck ¼

a 2 þ b2 þ c 2

p cos
1

2 π ¼ 2 4

2.16. Based on the information given in the problem, the location of point and the equation of plane are as follows. P0 ð1, 2, 3Þ Plane : 2x
y þ 2z þ 9 ¼ 0 The distance of the point P0(x0, y0, z0) from the plane of ax + by + cz + d ¼ 0 can be calculated as follows. D¼

jax0 þ by0 þ cz0 þ dj a2 þ b 2 þ c 2

Therefore: D¼

j2  1
1  2 þ 2  3 þ 9j 15 p ¼ 3 4þ1þ4 )D¼5

Choice (1) is the answer.

2

Solutions of Problems: Linear Algebra and Analytical Geometry

19

2.17. Based on the information given in the problem, we have:

A¼

3

2

5

7 12

1 3

1 3

2 6

3 5 9 15

As we know, the rank of a matrix is equal to the number of linearly independent rows or the number of linearly independent columns. Moreover, for a nonzero matrix in the form of Am  n, we have: 1 ≤ rank ðAmn Þ ≤ minðm, nÞ Therefore, for this problem, we know that: 1 ≤ rank ðAÞ ≤ 3 On the other hand, it can be noticed that the third row is a factor of second row; however, the first row is not a factor of second row. Therefore, the matrix has only two linearly independent rows. Thus: rank ðAÞ ¼ 2 Choice (3) is the answer. 2.18. Based on the information given in the problem, we have: 1 V1 ¼

1 , V2 ¼

1

0
1

1

A¼

1 a

1 b

1 1

1

1

1

Since the vector of V1 is an eigenvector of the matrix of A, we must have: AV 1 ¼ λ1 V 1

)

1 a

1 b

1 1

1 1

1

1

1

1

¼ λ1

1 1 1

Herein, λ1 is the eigenvalue corresponding to the eigenvector of V1. By solving the equations, we have: λ1 ¼ 3 )aþb¼2 a þ b þ 1 ¼ λ1 Likewise, since V2 is an eigenvector of the matrix of A, we must have: AV 2 ¼ λ2 V 2

ð1Þ

20

2 Solutions of Problems: Linear Algebra and Analytical Geometry

)

1 a

1 b

1 1

1 0

1

1

1


1

¼ λ2

1 0
1

where, λ2 is the eigenvalue corresponding to the eigenvector of V2. λ2 ¼ 0 a
1¼0)a¼1

ð2Þ

Solving (1) and (2): b¼1 Choice (1) is the answer. In this problem, the rule below has been used. The vector of V1 is an eigenvector of the matrix of A if the equation below is held. AV 1 ¼ λ1 V 1 Herein, λ1 is the eigenvalue corresponding to the eigenvector of V1. 2.19. Based on the information given in the problem, the point and line are given as follows. Pð
1, 3,
1Þ x ¼ 2t þ 1 L:

y¼1 z¼t →

The distance of the point of P(x, y, z) from the line of V can be calculated as follows.

D¼

! → PP0  V →

V

→ ! where, P0(x0, y0, z0) is an arbitrary point on the line, PP0 is the direction vector between the two points, and V is the direction vector of the line.

The problem can be solved as follows. ! First, the arbitrary point of P0 (1,1,0) on the line is selected which can achieved by considering t ¼ 0. Then: ! PP0 ¼ ð2,
2, 1Þ Next: →

i

PP0  V ¼ 2 2

→

j


2 0

→

k → → 1 ¼
2 i þ 4 k 1

References

21

And finally, we have: →

→

D¼


2 i þ 4 k →

→

2i þ k

p p 4 þ 16 20 ¼ p ¼ p 4þ1 5

)D¼2 Choice (2) is the answer.

References 1. Rahmani-Andebili, M. (2023). Calculus II – Practice Problems, Methods, and Solutions, Springer Nature. 2. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 3. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature. 4. Rahmani-Andebili, M. (2023). Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 5. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature.

3

Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

Abstract

In this chapter, the basic and advanced problems of lines, surfaces, vector functions, and parametric vector functions in three-dimensional coordinate systems are presented. The subjects include polar coordinate system, cartesian coordinate system, equation of surface, surface type, surface area, definite vector integral, indefinite vector integral, unit tangent vector, unit normal vector, general equation of different surfaces such as cylinder, ellipsoid, one-sheet hyperboloid, two-sheet hyperboloid, cone, elliptic paraboloid, and hyperbolic paraboloid, circular curve, curve type, and amount of curvature of a curve. 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. Determine the equation of surface resulted from the rotation of the following hyperbola (on xoy plane) around x-axis [1–5]. x2 y 2 - ¼1 9 4 Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large x 2 y2 z 2 - þ ¼1 1) 9 4 9 2 y2 z 2 x - ¼1 2) 9 4 9 2 2 y z2 x - þ ¼1 3) 9 4 4 x2 y2 z 2 - ¼1 4) 9 4 4 3.2. Determine the equation of surface resulted from the rotation of the curve of y ¼ ex around x-axis. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 2 2 1) z ¼ ex þy 1 2) z ¼ ln x2 þ y2 2 1 3) z ¼ ln x2 þ z2 2 1 4) x ¼ ln y2 þ z2 2

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus III, https://doi.org/10.1007/978-3-031-47483-5_3

23

24

3 Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

3.3. The equation of a surface in polar coordinate system is as follows. Determine its surface equation in cartesian coordinate system. r ¼ cosec θ cot θ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) y2 ¼ x 2) x2 ¼ y2 3) y2 ¼ z 4) x2 ¼ z 3.4. For what value of “a”, the following vector function is continuous for t ¼ 0? →

→

F ðt Þ ¼

→

sin t i þ aðt þ 2Þ j þ →

→

tan t → k t

j þ k

t≠0 t¼0

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 0 1 2) 2 3) 1 3 4) 2 3.5. Calculate the vector integral of the vector function below for 1 ≤ t ≤ 2. →

→

→

F ðt Þ ¼ ðet Þ i þ ðln t Þ j

ð 0 < t < 1Þ

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large →

→

1) ðe2 - eÞ i þ ð2 ln2 - 1Þ j →

→

2) - ðe2 - eÞ i þ ð2 ln2 - 1Þ j →

→

3) ðe2 - eÞ i - ð2 ln2 - 1Þ j →

→

4) - ðe2 - eÞ i - ð2 ln2 - 1Þ j

3.6. Calculate the length of part of the following circular curve which is between (a, 0, 0) and (a, 0, 2πb). →

→

→

→

r ðt Þ ¼ a cos t i þ a sin t j þ bt k

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1)

π 2 3) π

2)

4) 2π

a2 þ b 2 a2 þ b 2 a2 þ b2 a2 þ b 2

3

Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

25

3.7. Calculate the unit tangent vector of the vector function below. →

→

→

r ðt Þ ¼ ðcos t þ t sin t Þ i þ ðsin t - t cos t Þ j

ðt ≥ 0 Þ

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large →

→

→

→

1) cos t i þ sin t j 2) cos t i - sin t j →

→

→

→

3) - cos t i þ sin t j 4) - cos t i - sin t j

3.8. Calculate the amount of curvature of the vector function below. →

→

→

r ðt Þ ¼ t i þ cosh t j

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1 1) cosh t t 2) cosh 2 t 1 3) cosh 2t 1 4) cosh 2 t 3.9. Calculate the amount of curvature of the vector function below for t ¼ 0. →

→

→

→

r ðt Þ ¼ et sin 2t i þ et cos 2t j þ 2et k

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large p 3 1) 2p 2 3 2) 3 p 2 5 3) 9 p 2 5 4) 3 3.10. Calculate the amount of curvature of the function below at (e, e). yðxÞ ¼ x ln x Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

26

3 Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

1 e 1 2) 3 e52 5 3) 3 e2 e 4) 3 5 1)

3.11. Calculate the amount of curvature of the function below at y ¼ 3x ¼ y2 þ 2

p

2.

3 2

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 9 2 2) 9 1 3) 3 2 4) 3 3.12. What is the surface type of the equation below in three-dimensional space? x 2 þ y 2 ¼ a2 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) Sphere 2) Cylinder 3) Ellipsoid 4) Cone 3.13. What is the surface type of the following equation? 16x2 þ 9y2 - 16z2 - 32x - 36y þ 36 ¼ 0 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) Cone 2) One sheet hyperboloid 3) Ellipsoid 4) Two-sheet hyperboloid 3.14. What is the curve type of the equation below? p p 3x2 þ 2 3xy þ y2 - 8x þ 8 3y ¼ 0 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large

References

1) 2) 3) 4)

27

Hyperbola Ellipse Parabola Two parallel lines

3.15. Calculate the surface area enclosed inside the curve below. x2 þ xy þ y2 ¼ 1 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large p π 2 1) 2 p π 3 2) 3 2π 3) p 6 2π 4) p 3 1 3.16. Calculate the unit normal vector for the vector function below for t ¼ . 2 →

→

→

r ðt Þ ¼ t 3 - 3t i þ 3t 2 j

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 4→ 3→ 1) i þ j 5 5 3→ 4→ 2) i þ j 5 5 4→ 3→ 3) i - j 5 5 4→ 3→ 4) - i - j 5 5

References 1. Rahmani-Andebili, M. (2023). Calculus II – Practice Problems, Methods, and Solutions, Springer Nature. 2. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 3. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature. 4. Rahmani-Andebili, M. (2023). Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 5. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature.

4

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

Abstract

In this chapter, the problems of the third chapter are fully solved, in detail, step-by-step, and with different methods. 4.1. Based on the information given in the problem, we have [1–5]: x2 y 2 - ¼1 9 4 )

x 2 y2 -1 ¼ 0 9 4

ð1Þ

Table 4.1 presents the equation of surfaces resulted from the rotation around different axes. As can be seen in (1), the function is in the form of f (x, y) ¼ 0. Therefore, after the rotation of the curve around x-axis, the equation of surface will be as follows. f x, In other words, y will be replaced by

y2 þ z2 ¼ 0

y2 þ z2 . Therefore: x2 9 )

y2 þ z2 4

2

¼1

x2 y2 z 2 - ¼1 9 4 4

Choice (4) is the answer.

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus III, https://doi.org/10.1007/978-3-031-47483-5_4

29

30

4

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

Table 4.1 The equation of surfaces resulted from the rotation around x-axis, y-axis, and z-axis Curve equation

Rotation axis y

f (x,y) ¼ 0

f (x,z) ¼ 0

f (z,y) ¼ 0

Equation of surface resulted from rotation p f x 2 þ z2 , y ¼ 0

x

f x,

z

f

x y

f x, z2 þ y2 ¼ 0 p f z2 þ x 2 , y ¼ 0

z

f

y2 þ z2 ¼ 0 x2 þ y2 , z ¼ 0

z, y2 þ x2 ¼ 0

4.2. Based on the information given in the problem, we have: y ¼ ex ) y - ex ¼ 0

ð1Þ

Table 4.2 presents the equation of surfaces resulted from the rotation around different axes. As can be seen in (1), the function is in the form of f (x, y) ¼ 0. Therefore, after the rotation of the curve around x-axis, the equation of surface will be as follows. y2 þ z2 ¼ 0

f x, In other words, y needs to be replaced by

y2 þ z2 . Therefore: y2 þ z 2 ¼ ex ) x ¼ ln )x¼

y2 þ z 2

1 ln y2 þ z2 2

Choice (4) is the answer. Table 4.2 The equation of surfaces resulted from the rotation around x-axis, y-axis, and z-axis Curve equation f (x,y) ¼ 0

f (x,z) ¼ 0

f (z,y) ¼ 0

Rotation axis y

Equation of surface resulted from rotation p f x 2 þ z2 , y ¼ 0

x

f x,

z

f

x y

f x, z2 þ y2 ¼ 0 p f z2 þ x 2 , y ¼ 0

z

f

In this problem, the rules below have been used. ln eb ¼ b ln ab ¼ b ln a

y2 þ z2 ¼ 0 x2 þ y2 , z ¼ 0

z, y2 þ x2 ¼ 0

4

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

31

4.3. Based on the information given in the problem, we have: r ¼ cosec θ cot θ )r¼

1 cos θ sin θ sin θ

)r¼

cos θ sin 2 θ

To transfer from polar coordinate system to cartesian coordinate system, we need to change the parameters as follows. x ¼ r cos θ y ¼ r sin θ z¼z Therefore: r sin 2 θ ¼ cos θ ) r2 sin 2 θ ¼ r cos θ ) y2 ¼ x Choice (1) is the answer. In this problem, the rules below have been used. cosec θ ¼ cot θ ¼

1 sin θ

cos θ sin θ

4.4. Based on the information given in the problem, we have: →

F ðt Þ ¼

→

sin t i þ aðt þ 2Þ j þ

→

→

→

tan t → k t

j þ k

t≠0 t¼0

→ ! → ! → ! → A vector function with the form of F ðt Þ ¼ F 1 ðt Þ i þ F 2 ðt Þ j þ F 3 ðt Þ k is continuous for the given parameter of t ¼ t0 if:

! ! ! limþ F 1 ðt Þ ¼ lim- F 1 ðt Þ ¼ F 1 ðt 0 Þ

t → t0

t → t0

! ! ! lim F 2 ðt Þ ¼ lim- F 2 ðt Þ ¼ F 2 ðt 0 Þ

t → t0 þ

t → t0

! ! ! lim F 3 ðt Þ ¼ lim- F 3 ðt Þ ¼ F 3 ðt 0 Þ

t → t0 þ

t → t0

32

4

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

For this problem, we have: lim sin t ¼ lim- sin t ¼ 0 ) 0 ¼ 0 ¼ 0 t→0

t → 0þ

lim aðt þ 2Þ ¼ lim- aðt þ 2Þ ¼ 1 ) 2a ¼ 2a ¼ 1 ) a ¼ t→0

t → 0þ

lim

t → 0þ

1 2

tan t tan t ¼ lim¼1)1¼1¼1 t t t→0

Choice (2) is the answer. In this problem, the rules below have been used. sin 0 ¼ 0 lim

t→0

tan x x  lim ¼ 1 x t→0 x

4.5. Based on the information given in the problem, we need to calculate the vector integral for the following vector function for the interval of 1 ≤ t ≤ 2. →

→

→

F ðt Þ ¼ ðet Þ i þ ðln t Þ j

ð 0 < t < 1Þ

Therefore: 2→

I¼

2

F ðt Þdt ¼

et dt

1

2

→

i þ

ln tdt

1

1

2→

→

→

j

) I ¼ ½et 1 i þ ½tln t - t 21 j →

→

) I ¼ e2 - e i þ ð2ln 2 - 2 - ðln 1 - 1ÞÞ j →

→

) I ¼ e2 - e i þ ð2ln 2 - 1Þ j Choice (1) is the answer. In this problem, the rules below have been used. ex dx ¼ ex þ c ln x dx ¼ x ln x - x þ c ln 1 ¼ 0

4

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

33

4.6. Based on the information given in the problem, we need to calculate the length of part of the following circular curve which is between (a, 0, 0) and (a, 0, 2πb). →

→

→

→

r ðt Þ ¼ a cos t i þ a sin t j þ bt k

As can be noticed from the three-dimensional curve, x ¼ a cos t, y ¼ a sin t, and z ¼ bt. Moreover, from z ¼ bt and the points of (a, 0, 0) and (a, 0, 2πb), it is noticed that 0 ≤ t ≤ 2π. On the other hand, we know that the length of a parametric vector function can be calculated as follows. b

l¼

b

→0

r ðt Þ dt ¼

a

a

x0t

2

þ y0t

2

2

þ z0t dt

Therefore: 2π

l¼

2π

ð- a sin t Þ2 þ ðacos t Þ2 þ b2 dt ¼

0

a2 þ b2 dt ¼

0

) l ¼ 2π

a2 þ b2

In this problem, the rules below have been used. d cos x ¼ - sin x dx d sin x ¼ cos x dt ðsin xÞ2 þ ðcos xÞ2 ¼ 1 4.7. Based on the information given in the problem, we have: →

→

r ðt Þ ¼ ðcos t þ t sin t Þ i þ ðsin t - t cos t Þ j

ðt ≥ 0Þ

The unit tangent vector of a vector function can be calculated as follows: →

T ¼

→0

r ðt Þ

→0

r ðt Þ

Therefore, for this problem, we can write: →

→0

→

r ðt Þ ¼ ð- sin t þ sin t þ t cos t Þ i þ ðcos t - cos t þ t sin t Þ j →0

→

→

) r ðt Þ ¼ ðtcos t Þ i þ ðtsin t Þ j →0

) r ðt Þ ¼

ðtcos t Þ2 þ ðtsin t Þ2 ¼ t

dt 0

Choice (4) is the answer.

→

2π

a2 þ b2

ðcos t Þ2 þ ðsin t Þ2 ¼ t

34

4

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System →

→

ðtcos t Þ i þ ðtsin t Þ j ) T ¼ t →

→

→

→

) T ¼ cos t i þ sin t j Choice (1) is the answer. In this problem, the rules below have been used. d ðuvÞ ¼ u0 v þ v0 u dx d cos x ¼ - sin x dx d sin x ¼ cos x dt ðsin xÞ2 þ ðcos xÞ2 ¼ 1 4.8. Based on the information given in the problem, we have: →

→

→

r ðt Þ ¼ t i þ cosh t j

The amount of curvature of a vector function can be calculated as follows. →0

→ 00

r ðt Þ  r ðt Þ

k¼

→0

r ðt Þ

3

Therefore, for this problem, we have: →

→0

→

r ðt Þ ¼ i þ sinh t j

→

→ 00

) r ðt Þ ¼ cosh t j →0

r ðt Þ ¼ →

i r  r ¼ 1

→0

→ 00

0

1 þ sinh 2 t ¼

→

cosh 2 t ¼ cosh t

→

j sinh t

k 0

cosh t

0

→

→

→

→

¼ 0 i - 0 j þ cosh t k ¼ cosh t k

) jr0  r 00 j ¼ cosh t Therefore, the amount of curvature of the vector function is as follows.

4

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System →0

→ 00

r ðt Þ  r ðt Þ

)k¼

35

→0

r ðt Þ

)k¼

3

cosh t cosh 3 t

¼

1 cosh 2 t

Choice (4) is the answer. In this problem, the rules below have been used. d n ðx Þ ¼ nxn - 1 dx d cosh x ¼ sinh x dx d sinh x ¼ cosh x dx cosh 2 t - sinh 2 t ¼ 1 4.9. Based on the information given in the problem, we have: →

→

→

→

r ðt Þ ¼ et sin 2t i þ et cos 2t j þ 2et k

The amount of curvature of a vector function can be calculated as follows. →0

→ 00

r ðt Þ  r ðt Þ

k¼

→0

r ðt Þ

3

Therefore, for this problem, we have: →

→0

→

→

r ðt Þ ¼ ðet sin 2t þ 2et cos 2t Þ i þ ðet cos 2t - 2et sin 2t Þ j þ 2et k →

→ 00

→

→

) r ðt Þ ¼ ð- 3et sin 2t þ 4et cos 2t Þ i þ ð- 3et cos 2t - 4et sin 2t Þ j þ 2et k For t ¼ 0, we have: →0

→

→ 00

→

→

→

r ð 0Þ ¼ 2 i þ j þ 2 k →

→

r ð 0Þ ¼ 4 i - 3 j þ 2 k →

→0

→ 00

i

) r ð 0Þ  r ð 0Þ ¼ 2 4

→

→

1

2

-3

2

j

k

→

→

→

¼ 8 i þ 4 j - 10 k

Therefore, the amount of curvature of the vector function is as follows.

36

4

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System →0

)k¼

→ 00

r ð 0Þ  r ð 0Þ →0

r ð 0Þ

3

82 þ 42 þ 102

¼

22 þ 1 þ 2 2

3

p 180 6 5 ¼ ¼ 27 27 p

p 2 5 )k¼ 9 Choice (3) is the answer. In this problem, the rules below have been used. d ðuvÞ ¼ u0 v þ v0 u dx d x e ¼ ex dx d cos x ¼ - sin x dx d sin x ¼ cos x dt 4.10. Based on the information given in the problem, we have: yðxÞ ¼ x ln x If a curve is presented in the form of y ¼ f(x), the amount of curvature of the curve can be calculated as follows. jy00 j

k¼

1 þ ðy00 Þ2

3 2

Moreover, if a curve is presented in the form of x ¼ f( y), the amount of curvature of the curve can be calculated as follows. x00y

k¼

1 þ x0 y

2

3 2

Therefore, for this problem, we have: y0 ðxÞ ¼ ln x þ x

1 ¼ ln x þ 1 x

y00 ðxÞ ¼

1 x

For (x, y) ¼ (e, e), we have: y0 ðeÞ ¼ ln e þ 1 ¼ 2

4

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

y00 ðeÞ ¼

37

1 e

Therefore, the amount of curvature of the curve is as follows. 1 e

k¼

3 2

1 þ 22 )k¼

¼

1 e 3

52

1 3 e52

Choice (2) is the answer. In this problem, the rules below have been used. d ðuvÞ ¼ u0 v þ v0 u dx d n ðx Þ ¼ nxn - 1 dx d 1 ln x ¼ dx x ln e ¼ 1 4.11. Based on the information given in the problem, we have: 3 2

3x ¼ y2 þ 2 )x¼

1 2 y þ2 3

3 2

If a curve is presented in the form of y ¼ f(x), the amount of curvature of the curve can be calculated as follows. k¼

jy00 j 1 þ ðy00 Þ2

3 2

Moreover, if a curve is presented in the form of x ¼ f( y), the amount of curvature of the curve can be calculated as follows.

k¼

x00y 1 þ x0 y

2

3 2

Therefore, for this problem, we have: 1 x0y ¼ ð2yÞ y2 þ 2 2 1

1 2

¼ y y2 þ 2

) x″y ¼ y2 þ 2 2 þ y2 y2 þ 2

1 2

- 12

38

4

For y ¼

p

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

2, we have: p x ′y ¼ 2 2 x″y ¼ 3

Therefore, the amount of curvature of the curve is as follows.

k¼

x00y 1 þ x0 y

2

3 2

¼

)k¼

3 ð 1 þ 8Þ

3 2

¼

3 27

1 9

Choice (1) is the answer. In this problem, the rules below have been used. d n ðu Þ ¼ nu0 un - 1 dx d ðuvÞ ¼ u0 v þ v0 u dx 4.12. Based on the information given in the problem, we have: x 2 þ y 2 ¼ a2 ) x2 þ y 2 - a2 ¼ 0 Suppose that the general equation of a surface is in the following form. Ax2 þ By2 þ Cz2 þ Dx þ Ey þ Fz þ G ¼ 0 Case 1: If the general equation of a surface can be written in the form of A(x - α)2 + B(y - β)2 + C(z - γ)2 ¼ h, then: Case 1.1: If A  B  C ¼ 0, the surface is a cylinder. Case 1.2: If A  B  C ≠ 0; A, B, and C have the same sign, the surface is an ellipsoid. Case 1.3: If A  B  C ≠ 0; A, B, and C do not have the same sign; A  B  C > 0; and h > 0, the surface is a two-sheet hyperboloid. Case 1.4: If A  B  C ≠ 0; A, B, and C do not have the same sign; A  B  C < 0; and h > 0, the surface is a one-sheet hyperboloid. Case 1.5: If A  B  C ≠ 0; A, B, and C do not have the same sign; and h ¼ 0, the surface is a cone. Case 2: If the general equation of a surface can be written in the form of A(x - α)2 + B(y - β)2 ¼ C(z - γ), then: Case 2.1: If A  B ¼ 0, the surface is a cylinder. Case 2.2: If A  B > 0, the surface is an elliptic paraboloid. Case 2.3: If A  B < 0, the surface is a hyperbolic paraboloid. Now, by comparing the equation of surface with the equation of Case 1, we see that C ¼ 0; therefore, A  B  C ¼ 0. Hence, the surface is a cylinder (Case 1.1). Choice (2) is the answer.

4

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

39

4.13. Based on the information given in the problem, we have: 16x2 þ 9y2 - 16z2 - 32x - 36y þ 36 ¼ 0 ) 16x2 - 32x þ 9y2 - 36y - 16z2 þ 36 ¼ 0 ) 16 x2 - 2x þ 1 - 16 þ 9 y2 - 4y þ 4 - 36 - 16z2 þ 36 ¼ 0 ) 16ðx - 1Þ2 þ 9ðy - 2Þ2 - 16z2 ¼ 16 As can be noticed from the equation of surface and the equation of Case 1, we have: • • • •

A  B  C ≠ 0, A, B, and C do not have the same sign, A  B  C < 0, and h>0

Therefore, the surface is a one-sheet hyperboloid (Case 1.4). Choice (2) is the answer. In this problem, the rules below have been used. For the general equation of a surface in the following form. Ax2 þ By2 þ Cz2 þ Dx þ Ey þ Fz þ G ¼ 0 Case 1: If the general equation of a surface can be written in the form of A(x - α)2 + B(y - β)2 + C(z - γ)2 ¼ h, then: Case 1.1: If A  B  C ¼ 0, the surface is a cylinder. Case 1.2: If A  B  C ≠ 0; A, B, and C have the same sign, the surface is an ellipsoid. Case 1.3: If A  B  C ≠ 0; A, B, and C do not have the same sign; A  B  C > 0; and h > 0, the surface is a two-sheet hyperboloid. Case 1.4: If A  B  C ≠ 0; A, B, and C do not have the same sign; A  B  C < 0; and h > 0, the surface is a one-sheet hyperboloid. Case 1.5: If A  B  C ≠ 0; A, B, and C do not have the same sign; and h ¼ 0, the surface is a cone. Case 2: If the general equation of a surface can be written in the form of A(x - α)2 + B(y - β)2 ¼ C(z - γ), then: Case 2.1: If A  B ¼ 0, the surface is a cylinder. Case 2.2: If A  B > 0, the surface is an elliptic paraboloid. Case 2.3: If A  B < 0, the surface is a hyperbolic paraboloid. 4.14. If the equation of a surface does not include the variable of “z”, the surface will change to a curve. Suppose the equation is written in the following form. ax2 þ bxy þ cy2 þ dx þ ey þ f ¼ 0 Then, the discriminant is calculated as follows. Δ ¼ b2 - 4ac Case 1: If Δ < 0, the equation is related to an ellipse; however, in some specific cases it can be a circle, a point, or even an empty set.

40

4

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

Case 2: If Δ ¼ 0, the equation is related to a parabola; however, in some specific cases it can be a pair of parallel lines, a single line, or even an empty set. Case 3: If Δ > 0, the equation is related to a hyperbola; however, in some specific cases it can be two crossing lines. For this problem, we have: p p 3x2 þ 2 3xy þ y2 - 8x þ 8 3y ¼ 0 p )Δ¼ 2 3

2

-4  3  1

)Δ¼0 Therefore, based on Case 2, the equation is related to a parabola. Choice (3) is the answer. 4.15. First, we should identify the curve type of the equation. As we know, if the equation of a surface does not include the variable of “z”, the surface will change to a curve. For an equation in the form of ax2 + bxy + cy2 + dx + ey + f ¼ 0, the discriminant is as follows. Δ ¼ b2 - 4ac Case 1: If Δ < 0, the equation is related to an ellipse; however, in some specific cases it can be a circle, a point, or even an empty set. Case 2: If Δ ¼ 0, the equation is related to a parabola; however, in some specific cases it can be a pair of parallel lines, a single line, or even an empty set. Case 3: If Δ > 0, the equation is related to a hyperbola; however, in some specific cases it can be two crossing lines. For this problem, we have: x2 þ xy þ y2 - 1 ¼ 0 ) Δ ¼ 12 - 4  1  1 ) Δ ¼ -3 Therefore, the equation is concerned with an ellipse. On the other hand, we know that the surface area enclosed inside an ellipse can be calculated as follows. 2π S¼p -Δ Thus, for this problem, we have: 2π S¼p 3 Choice (4) is the answer.

4

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

4.16. Based on the information given in the problem, we have: →

→

→

r ðt Þ ¼ t 3 - 3t i þ 3t 2 j

The unit normal vector of a vector function can be calculated as follows: →

N ¼

→0

T ðt Þ

→0

T ðt Þ

where, →

T ¼

→0

r ðt Þ

→0

r ðt Þ

Therefore, for this problem, we have: →

→0

→

r ðt Þ ¼ 3t 2 - 3 i þ 6t j

→0

) r ðt Þ ¼ →0

) r ðt Þ ¼

ðð3t 2 - 3ÞÞ2 þ ð6t Þ2 ¼

9t 4 - 18t 2 þ 9 þ 36t 2

9ð ð t 2 þ 1Þ Þ 2 ¼ 3 t 2 þ 1

9t 4 þ 18t 2 þ 9 ¼

The unit tangent vector of the vector function can be calculated as follows. →

→

) T ¼

→

ð3t 2 - 3Þ i þ 6t j t2 - 1 → 2t → ¼ 2 i þ 2 j 2 t þ1 3 ð t þ 1Þ t þ1

! ) T ′ ðt Þ ¼

→ 4t 2 - 2t 2 → i þ j ð t 2 þ 1Þ 2 ð t 2 þ 1Þ 2

For t ¼ 12, we have: ! 1 32 → 24 → ¼ T′ i þ j 2 25 25 ! 1 ) T′ 2

¼

8 5

Now, the unit normal vector of the vector function can be calculated as follows. →

32

) N ¼ 25 →

)N ¼ Choice (1) is the answer.

→

→

i þ 24 25 j 8 5

4→ 3→ i þ j 5 5

41

42

4

Solutions of Problems: Lines, Surfaces, and Vector Functions in Three-Dimensional Coordinate System

In this problem, the rules below have been used. d n ðx Þ ¼ nxn - 1 dx d u u0 v - v 0 u ¼ dx v v2

References 1. Rahmani-Andebili, M. (2023). Calculus II – Practice Problems, Methods, and Solutions, Springer Nature. 2. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 3. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature. 4. Rahmani-Andebili, M. (2023). Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 5. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature.

5

Problems: Multivariable Functions

Abstract

In this chapter, the basic and advanced problems of multivariable functions are presented. The subjects include domain and range of multivariable functions, Jacobian determinant, gradient, directional derivative, unit vector, tangent plane, curl, divergence, Laplacian, critical point, type of critical point, relative maximum and minimum points, and saddle point. 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. Calculate the limit of the multivariable function below at (0, 0) [1–5]. f ðx, yÞ ¼

x xþy

○ Easy ● Normal ○ Hard Difficulty level Calculation amount ● Small ○ Normal ○ Large 1) The limit is not available 2) 0 3) 1 4) 2 5.2. Calculate the limit of the multivariable function below at (0, 0) through the path of y ¼ 2x. f ðx, yÞ ¼

y2 þ 2x2 x2 þ 2y2

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1 1) 3 2) 1 2 3) 3 1 4) 2

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus III, https://doi.org/10.1007/978-3-031-47483-5_5

43

44

5

Problems: Multivariable Functions

5.3. What should be the value of multivariable function below at (0, 0) so that it is continuous everywhere on the plane. f ðx, yÞ ¼

x2 þ y2 - x2 y2 , x2 þ y2

ðx, yÞ ≠ ð0, 0Þ

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) f (0, 0) ¼ 1 2) f (0, 0) ¼ 0 3) f (0, 0) ¼ -1 4) f (0, 0) ¼ 2 5.4. Calculate the value of the following term if u ¼ ð1 - 2xy þ y2 Þ x

- 12

.

∂u ∂u -y ∂x ∂y

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) (y2 - 2xy)u 2) (2xy - y2)u 3) y2u3 4) xyu3 5.5. Determine the domain of the multivariable function below. z¼

4x2 þ y2 - 8x

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) The outer region of a half ellipse 2) The inner region of a half ellipse 3) The outer region of an ellipse including the ellipse 4) The inner region of an ellipse including the ellipse 2

5.6. Calculate the value of

∂ f at (1, 1) for the following multivariable function. ∂x∂y f ðx, yÞ ¼

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 4 2) 2 3) -2 4) -4

x2 y 2 þ y x

5

Problems: Multivariable Functions

45

r s ∂z 5.7. If z ¼ u2 + v2 + uv, u ¼ , and v ¼ , calculate the value of for r ¼ 2 and s ¼ 1. s r-s ∂s Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) -1 2) -2 3) -3 4) -4 5.8. Calculate the value of the Jacobian determinant of x and y with respect to r and θ if x ¼ r cos θ and y ¼ r sin θ. In other words, calculate the value of the term below. ∂ðx, yÞ ∂ðr, θÞ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) r 1 2) r 3) 0 4) r cos 2θ 1 5.9. Calculate the value of the Jacobian determinant of x and y with respect to r and θ if u ¼ x - 3y and uv ¼ 2y. 2 In other words, calculate the value of the operation below. ∂ðx, yÞ ∂ðu, vÞ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) u 2) v 3) u + v 4) u - v 5.10. At which points, the gradient of the multivariable function below is equal to f ðx, yÞ ¼ ln Difficulty level Calculation amount 3 7 3 1 1) , , , 4 3 4 3 3 7 3 1 2) - , , , 4 3 4 3 3 7 3 , , - ,3) 4 3 4 3 7 3 4) - , , , 4 3 4

○ Easy ● Normal ○ Hard ○ Small ● Normal ○ Large

1 3 1 3

1 þy x

16 ,1 ? 9

46

5

Problems: Multivariable Functions

5.11. Calculate the directional derivative of the multivariable function of f (x, y, z) ¼ z(x + y) - x2 - y2 in the direction of the →

→

→

vector of 2 i - 2 j þ k at (2, 3, 4). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 9 2) 3 3) 4 4) 5 5.12. Calculate the directional derivative of the multivariable function of f (x, y, z) ¼ x3 + 2xy2 + z2x in the direction of the →

→

→

unit vector of p13 i þ p13 j þ p13 k at (1, 1, 1). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large p 10 3 1) 3p 11 3 2) p3 3) 4 3 p 13 3 4) 3 5.13. Calculate the directional derivative of the multivariable function of f (x, y) ¼ e-xy in the direction of the vector of θ ¼ 2π 3 at (1, -1). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large p 1þ 2 1) e 2 p 1- 2 2) e 2 p 1þ 3 e 3) 2 p 1- 3 4) e 2 5.14. Calculate the maximum directional derivative of the multivariable function of f (x, y, z) ¼ x2 - 2y2 + 4z2 at (1, 1, -1). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 21 p 2) 2 21 p 3) 21 p 4) 3 21 →

→

→

5.15. Calculate the curl of the multivariable vector function below (curl f ¼ ∇  f ). →

→

→

f ðx, y, zÞ ¼ x2 i þ ð2x - yÞ j þ yz k Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

5

Problems: Multivariable Functions →

→

→

→

→

→

→

1) z j þ 2 k

47

2) z i - y j þ x k →

3) 2x i - j þ y k →

→

4) z i þ 2 k

5.16. Determine the critical point of the following multivariable function and the type of the critical point. f ðx, y, zÞ ¼ x2 þ 2y2 - x Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 0, , relative maximum point 2 1 2) 0, , relative minimum point 2 1 3) , 0 , relative minimum point 2 1 4) , 0 , saddle point 2 5.17. Which one of the following choices is correct about the multivariable function of f (x, y) ¼ 4xy - x4 - y4 . Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) The critical point of (0, 0) is a relative maximum and the critical points of (1, 1) and (-1, -1) are saddle points. 2) The critical point of (0, 0) is a saddle point and the critical points of (1, 1) and (-1, -1) are relative minimum points. 3) The critical point of (0, 0) is a relative minimum and the critical points of (1, 1) and (-1, -1) are saddle points. 4) The critical point of (0, 0) is a saddle point and the critical points of (1, 1) and (-1, -1) are relative maximum points. 5.18. For the multivariable function below: x2 1 f ðx, yÞ ¼ p e - 4y y

calculate the value of the following operation. ∂f ∂y 2

∂ f ∂x2

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ○ Normal ● Large 1) 1 2) y p 3) y 4) xy

48

5

5.19. Determine the type of the point of

p1 , p1 3 3

Problems: Multivariable Functions

for the multivariable function of f (x, y) ¼ x4 + y4 + (1 - x2 - y2)2.

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ○ Normal ● Large 1) Relative minimum 2) Saddle point 3) Relative maximum 4) Regular point 5.20. Calculate the value of the limit below. lim

ðx, yÞ → ð0, 0Þ

tan - 1 ðxyÞ xy

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) -1 2) 0 3) 1 4) The limit is unavailable. 5.21. Determine the range of the following multivariable function. f ðx, yÞ ¼

23 þ 6x þ 4y - x2 - y2

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) [1, 6] 2) [0, 5] 3) [0, 6] 4) [1, 5] 5.22. Calculate the value of fx(0, 0) and fy(0, 0) if:

f ðx, yÞ ¼

3x4 - 5y4 x3 þ y3 0

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 3, - 5 2) -3, 5 3) 12, - 15 4) Unavailable, unavailable

x≠ -y x ¼ -y

5

Problems: Multivariable Functions

49

∂z for the multivariable function of xyz + 2z2 - 3y ¼ 1 at (2, 1, -2). ∂y Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 5 1) 6 7 2) 6 4 3) 3 5 4) 3

5.23. Calculate the value of

5.24. Determine the equation of the tangent plane on the surface of ez + 3xy2 + 2zx ¼ 4 at (1, -1, 0). Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 3x + 6y + 3z ¼ - 3 2) 2x - y + 2z ¼ 3 3) x - 2y + z ¼ 3 4) 3x - 6y + 2z ¼ 9 5.25. Calculate the equation of the tangent plane on the surface of exy + ey - ez - e ¼ 0 at (1, 1, 1). Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) x + y + z ¼ 0 2) x + y + z ¼ e 3) x - z ¼ 2 4) x + 2y - z ¼ 2 5.26. Which of one of the following choices is parallel to the tangent line on the common part of the two crossing surfaces of z ¼ 4x2 + 4y2 and z ¼ 4 - 4x2 at (0, 1, 4). Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large →

1) j

→

2) i

→

→

→

→

3) i - j

4) i þ j

5.27. By changing the variables, simplify the equation below. uxx - 2uxy þ uyy ¼ 0 Hint: z ¼ x + y and v ¼ x. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) uvv ¼ 0 2) uvz ¼ 0 3) uzz ¼ 0 4) uvv ¼ 1

50

5

Problems: Multivariable Functions

5.28. Calculate the equation of the tangent line on the common part of the two crossing surfaces of x2 + 2y2 + 3z2 ¼ 6 and xyz ¼ 1 at (1, 1, 1). Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large x-1 y-1 z-1 1) ¼ ¼ -1 -2 -1 x-1 y-1 z-1 2) ¼ ¼ 2 1 1 x-1 y-1 z-1 3) ¼ ¼ 2 -1 1 x-1 y-1 z-1 ¼ ¼ 4) 1 -2 1 5.29. Calculate the amount of angle between the cylinder of x2 + y2 ¼ 1 and the sphere of (x - 1)2 + y2 + z2 ¼ 1 at p 1 3 ,0 . , 2 2 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large π 1) 6 π 2) 4 π 3) 3 π 4) 2 →

→ →

5.30. Calculate the divergence of the multivariable function vector function below (div f ¼ ∇: f ) that in which m > 0, r2 ¼ x2 + y2 + z2, and r ≠ 0. f ðx, y, zÞ ¼

→ → m → x i þy j þzk 3 r

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) 0 6 2) 2 r 1 3) - 2 r 4 4) - 2 r →

5.31. Solve the equation of div ∇f

¼ 0 for the following multivariable function. f ðx, y, zÞ ¼ eðx þy 2

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) xfx + yfy + zfz ¼ - 6f 2) xfx + yfy + zfz ¼ 6f 3) xfx + yfy + zfz ¼ - 3f 4) xfx + yfy + zfz ¼ 3f

2

þz2 Þ

References

References 1. Rahmani-Andebili, M. (2023). Calculus II – Practice Problems, Methods, and Solutions, Springer Nature. 2. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 3. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature. 4. Rahmani-Andebili, M. (2023). Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 5. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature.

51

6

Solutions of Problems: Multivariable Functions

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, the limit below is requested [1–5]. lim

ðx, yÞ → ð0, 0Þ

f ðx, yÞ ¼

x x þ y ðx, yÞ → ð0, 0Þ lim

Let us calculate the limit of the multivariable function through the path of y ¼ mx as follows. lim f ðx, yÞ ¼ lim

x→0 x

x→0

x x 1 ¼ ¼ lim þ mx x → 0 xð1 þ mÞ m þ 1

y¼mx

As can be seen, the value of limit depends on the slope of the path; therefore, the limit is unavailable. Choice (1) is the answer. 6.2. Based on the information given in the problem, the limit of the multivariable function below at (0, 0) through the path of y ¼ 2x is requested. f ðx, yÞ ¼

y2 þ 2x2 x2 þ 2y2

Therefore: lim f ðx, yÞ ¼ lim

x→0 y¼2x

x→0 y¼2x

y2 þ 2x2 4x2 þ 2x2 6x2 6 ¼ lim 2 ¼ lim 2 ¼ 2 2 2 9 x → 0 x þ 8x x → 0 9x x þ 2y

lim f ðx, yÞ ¼

x→0 y¼2x

2 3

Choice (3) is the answer.

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus III, https://doi.org/10.1007/978-3-031-47483-5_6

53

54

6

Solutions of Problems: Multivariable Functions

6.3. Based on the information given in the problem, we have: f ðx, yÞ ¼

x2 þ y2  x2 y2 , x2 þ y2

ðx, yÞ ≠ ð0, 0Þ

For continuity of a multivariable function at (x0, y0), the limit and the value of the multivariable function at the point must be equal. In other words: f ðx0 , y0 Þ ¼

f ðx, yÞ

lim

ðx, yÞ → ðx0 , y0 Þ

The limit of the multivariable function through the path of y ¼ mx is as follows: lim f ðx, yÞ ¼ lim

x→0 y¼mx

x→0 y¼mx

x2 þ y2  x2 y2 x2 þ y2

1

¼ lim

x→0 y¼mx

x 2 y2 x2 þ y2

¼ lim

x→0

1

x2 ðmxÞ2 x2 þ ðmxÞ2

Therefore: ) f ð0, 0Þ ¼ 1 Choice (1) is the answer. 6.4. Based on the information given in the problem, we have: u ¼ 1  2xy þ y2

12

The problem can be solved as follows. ∂u 1 ¼  ð2yÞ 1  2xy þ y2 2 ∂x

32

¼ yu3

∂u 1 ¼  ð2x þ 2yÞ 1  2xy þ y2 2 ∂y

32

¼ ðx  yÞu3

) )

)x

∂u ∂u y ¼ xyu3  yðx  yÞu3 ∂x ∂y )x

∂u ∂u  y ¼ y2 u3 ∂x ∂y

Choice (3) is the answer. In this problem, the rule below has been used. d n ðu Þ ¼ nu0 un1 dx 6.5. Based on the information given in the problem, we have: z¼

4x2 þ y2  8x

¼ lim

x→0

1

m2 x2 1 þ m2

¼1

6

Solutions of Problems: Multivariable Functions

55

To determine the domain of the multivariable function, the term under radical must be equal and greater than zero. In other words: 4x2 þ y2  8x ≥ 0 ) 4x2  8x þ y2 ≥ 0 ) 4 x2  2x þ y2 ≥ 0 ) 4 ð x  1Þ 2  1 þ y 2 ≥ 0 ) 4 ð x  1Þ 2  y 2 ≥ 4 ) ð x  1Þ 2 þ

y2 ≥1 4

Therefore, the domain of the multivariable function is the outer region of an ellipse including the ellipse. Choice (3) is the answer. In this problem, the rule below has been used. The general equation of an ellipse with the center located at (x0, y0) is as follows. ð x  x0 Þ 2 ð y  y0 Þ 2 þ ¼1 a2 b2 2

6.6. Based on the informat`ion given in the problem, the value of

∂ f at (1, 1) for the following multivariable function is ∂x∂y

requested. f ðx, yÞ ¼

x2 y2 þ y x

Therefore: 2

∂ x2 y 2 þ x ∂x∂y y ¼

¼

∂ ∂ x2 y2 þ x ∂x ∂y y

∂ x2 2y þ x ∂x y2

¼

2x 2y  2 y2 x

For (x, y) ¼ (1, 1), we have: 2 2  ¼ 2  2 12 12 2

)

∂ f ¼ 4 ∂x∂y

Choice (4) is the answer. In this problem, the rules below have been used. d n ðx Þ ¼ nxn1 dx

56

6

Solutions of Problems: Multivariable Functions

d u u0 v  v 0 u ¼ dx v v2 6.7. Based on the information given in the problem, the value of

∂z for r ¼ 2 and s ¼ 1 is requested assuming that: ∂s

z ¼ u2 þ v2 þ uv u¼ v¼

r s

s rs

The problem can be solving by using chain rule as follows. ∂z ∂z ∂u ∂z ∂v ¼ þ ∂s ∂u ∂s ∂v ∂s )

)

∂z r rsþs ¼ ð2u þ vÞ  2 þ ð2v þ uÞ s ∂s ðr  sÞ2

∂z 2r s r 2s r  2 þ ¼ þ þ s rs rs s s ∂s

r ðr  sÞ2

For r ¼ 2 and s ¼ 1, we have: )

∂z 4 1 ¼ þ 1 2  1 ∂s )



2 12

þ

2 2 þ 21 1

2 ð2  1Þ2

∂z ¼ ð5Þð2Þ þ ð4Þð2Þ ∂s )

∂z ¼ 2 ∂s

Choice (2) is the answer. In this problem, the rules below have been used. d n ðx Þ ¼ nxn1 dx d u u0 v  v 0 u ¼ dx v v2 6.8. Based on the information given in the problem, the Jacobian determinant of x and y with respect to r and θ is requested while we know that: x ¼ r cos θ y ¼ r sin θ

6

Solutions of Problems: Multivariable Functions

57

Therefore: ∂x ∂ðx, yÞ ¼ ∂r ∂ðr, θÞ ∂y ∂r )

cos θ ∂ðx, yÞ ¼ ∂ðr, θÞ sin θ )

∂x ∂θ ∂y ∂θ

r sin θ ¼ r cos 2 θ þ r sin 2 θ r cos θ ∂ðx, yÞ ¼r ∂ðr, θÞ

Choice (1) is the answer. In this problem, the rules below have been used. d sin θ ¼ cos θ dθ d cos θ ¼  sin θ dθ cos 2 θ þ sin 2 θ ¼ 1 6.9. Based on the information given in the problem, the Jacobian determinant of x and y with respect to u and v is requested and we have: 1 u ¼ x  3y 2 uv ¼ 2y The dependent variables can be defined based on the independent variables as follows. x ¼ 2u þ 3uv 1 y ¼ uv 2 Therefore: ∂x ∂ðx, yÞ ∂u ¼ ∂ðu, vÞ ∂y ∂u

∂x 2 þ 3v ∂v ¼ 1 ∂y v 2 ∂v )

Choice (1) is the answer.

3u

3 3 1 ¼ u þ 2 uv  2 uv u 2

∂ðx, yÞ ¼u ∂ðu, vÞ

58

6

Solutions of Problems: Multivariable Functions

6.10. Based on the information given in the problem, we have: →

∇f ðx, yÞ ¼ 

16 ,1 9

where, f ðx, yÞ ¼ ln

1 þy x

The gradient of a multivariable function can be calculated as follows. →

∇f ðx, yÞ ¼ grad f ðx, yÞ ¼

∂f ∂f , ∂x ∂y

¼

∂f → ∂f → i þ j ∂x ∂y

Therefore, the problem below needs to be solved. ∂f → ∂f → i þ j ∂x ∂y

ln

16 → → 1 þy ¼ i þ j x 9

1 2 → 1 → 16 → → ) x j ¼ i þ j i þ 1 1 9 þy þy x x

)

1 x2 ¼ 16 1 9 9 3 3 7 1 1 16 þy ) x2 ¼ )x¼ , )y¼ , ) 2 ¼ x 9 16 4 4 3 3 x 1 þy¼1 x 3 7 3 1 ) ðx, yÞ ¼  , , ,  4 3 4 3

Choice (4) is the answer. In this problem, the rule below has been used. d u0 ln uðxÞ ¼ u dx 6.11. Based on the information given in the problem, the directional derivative of the multivariable function below in the →

→

→

direction of the vector of 2 i  2 j þ k at (2, 3, 4) is requested. f ðx, y, zÞ ¼ zðx þ yÞ  x2  y2 →

The directional derivative of the multivariable function of f (x, y, z) in the direction of the unit vector of u at the point of p(x0, y0, z0) can be calculated as follows. →

→

D →u f ðpÞ ¼ ∇f ðpÞ: u

6

Solutions of Problems: Multivariable Functions

59

Therefore, for this problem, we have: →

u ¼

→

∇f ¼

→

→

2 i 2 j þ k

→

22 þ ð2Þ2 þ 12

∂ → ∂ → ∂ → i þ j þ k ∂x ∂y ∂z

¼

2→ 2→ 1→ i  j þ k 3 3 3

zðx þ yÞ  x2  y2 ¼ ðz  2x, z  2y, x þ yÞ

→

→

→

) ∇f ð2, 3, 4Þ ¼ ð0, 2, 5Þ ¼ 2 j þ 5 k →

→

→

→

) ∇f ð2, 3, 4Þ: u ¼ 2 j þ 5 k :

2→ 2→ 1→ 4 5 9 i  j þ k ¼0þ þ ¼ 3 3 3 3 3 3 →

→

) D →u f ð2, 3, 4Þ ¼ ∇f ð2, 3, 4Þ: u ¼ 3 Choice (2) is the answer. 6.12. Based on the information given in the problem, the directional derivative of the multivariable function below in the →

→

→

direction of the unit vector of p13 i þ p13 j þ p13 k at (1, 1, 1) is requested. f ðx, y, zÞ ¼ x3 þ 2xy2 þ z2 x →

The directional derivative of the multivariable function of f(x, y, z) in the direction of the unit vector of u at the point of p(x0, y0, z0) can be calculated as follows. →

→

D →u f ðpÞ ¼ ∇f ðpÞ: u Therefore, for this problem, we have:

1 → 1 → 1 → u ¼p i þp j þp k 3 3 3

→

→

∇f ¼

∂ → ∂ → ∂ → i þ j þ k ∂x ∂y ∂z

x3 þ 2xy2 þ z2 x ¼ 3x2 þ 2y2 þ z2 , 4xy, 2xz

→

→

→

→

) ∇f ð1, 1, 1Þ ¼ ð6, 4, 2Þ ¼ 6 i þ 4 j þ 2 k

→ → → → 1 → 1 → 1 → → ) ∇f ð1, 1, 1Þ: u ¼ 6 i þ 4 j þ 2 k : p i þ p j þ p k 3 3 3

¼

6 þ 4 þ 2 12 p ¼p 3 3

p → → ) D →u f ð1, 1, 1Þ ¼ ∇f ð1, 1, 1Þ: u ¼ 4 3 Choice (3) is the answer. 6.13. Based on the information given in the problem, the directional derivative of the multivariable function below in the 2π direction of the vector of θ ¼ at (1, 1) is requested. 3 f ðx, yÞ ¼ exy

60

6

Solutions of Problems: Multivariable Functions →

The directional derivative of the multivariable function of f(x, y, z) in the direction of the unit vector of u at the point of p(x0, y0, z0) can be calculated as follows. →

→

D →u f ðpÞ ¼ ∇f ðpÞ: u

Thus, for this problem, the unit vector in the direction of the vector of θ ¼ 2π 3 is calculate as follows. →

u ¼ cos

→

∇f ¼

p 3→ 2π → 2π → 1→ j i þ sin j ¼ i þ 2 3 3 2

∂ → ∂ → xy i þ j e ¼ ðyexy , xexy Þ ∂x ∂y →

→

→

∇f ð1, 1Þ ¼ ðe, eÞ ¼ e i  e j →

→

→

→

) ∇f ð1, 1Þ: u ¼ e i  e j : 

p 3→ 1→ j i þ 2 2

p 1þ 3 ) D →u f ð1, 1Þ ¼ ∇f ð1, 1Þ: u ¼  e 2 →

→

Choice (3) is the answer. In this problem, the rules below have been used. cos

2π 1 ¼ 3 2

p 2π 3 sin ¼ 3 2 d u e ¼ u0 e u du 6.14. Based on the information given in the problem, the maximum directional derivative of the following multivariable function at (1, 1, 1) is requested. f ðx, y, zÞ ¼ x2  2y2 þ 4z2 The maximum directional derivative of a multivariable function in the direction of gradient vector of the function is achieved and it is equal to the magnitude of the gradient vector of the function. →

∇f ¼

∂ → ∂ → ∂ → i þ j þ k ∂x ∂y ∂z

→

→

→

→

→

) ∇f ð1, 1, 1Þ ¼ 2 i  4 j  8 k →

→

→

x2  2y2 þ 4z2 ¼ 2x i  4y j þ 8z k

) ∇f ð1, 1, 1Þ ¼

22 þ ð4Þ2 þ ð8Þ2

6

Solutions of Problems: Multivariable Functions

61

p → ) ∇f ð1, 1, 1Þ ¼ 2 21 Choice (2) is the answer. In this problem, the rule below has been used. d n x ¼ nxn1 dx 6.15. Based on the information given in the problem, the curl of multivariable vector function below is requested. →

→

→

f ðx, y, zÞ ¼ x2 i þ ð2x  yÞ j þ yz k The curl of a vector function can be calculated as follows. →

i → → → ∂ curl f ¼ ∇  f ¼ ∂x Pðx, y, zÞ

→

j ∂ ∂y Qðx, y, zÞ

→

k ∂ ∂z Rðx, y, zÞ

where, →

→

→

f ðx, y, zÞ ¼ Pðx, y, zÞ i þ Qðx, y, zÞ j þ Rðx, y, zÞ k Therefore, for this problem, we have:

→

→

→

→

→

→

curl f ¼ curl x2 i þ ð2x  yÞ j þ yz k

→

) curl f ¼

∂ ∂ ðyzÞ  ð2x  yÞ ∂y ∂z

→

→

→

¼ ∇  x2 i þ ð2x  yÞ j þ yz k

→

i 

→

∂ ∂ 2 ðyzÞ  x ∂x ∂z →

→

→

j þ

i ¼ ∂ ∂x x2

→

→

j k ∂ ∂ ∂y ∂z ð2x  yÞ yz

∂ ∂ 2 ð2x  yÞ  x ∂x ∂y

→

k

→

) curl f ¼ ðz  0Þ i  ð0  0Þ j þ ð2  0Þ k →

→

→

) curl f ¼ z i þ 2 k Choice (4) is the answer. In this problem, the rule below has been used. d n x ¼ nxn1 dx

6.16. Based on the information given in the problem, the critical point of the following multivariable function and the type of the critical point are requested. f ðx, y, zÞ ¼ x2 þ 2y2  x

62

6

Solutions of Problems: Multivariable Functions

To determine the critical point, the system of equations below must be solved. fx ¼ 0 fy ¼ 0 The discriminant of the multivariable function of f(x, y, z) is defined as follows. Δ ¼ f xx f yy  f 2xy Case 1: If, at point p(x0, Case 2: If, at point p(x0, Case 3: If, at point p(x0, Case 4: If, at point p(x0,

y 0, y 0, y 0, y 0,

z0), Δ > 0 and fxx > 0, the critical point is a relative minimum point. z0), Δ > 0 and fxx < 0, the critical point is a relative maximum point. z0), Δ < 0, the critical point is a saddle point. z0), Δ ¼ 0, no conclusion can be made.

Therefore, for this problem, we have: f x ¼ 2x  1 ¼ 0 f y ¼ 4y ¼ 0

)

1 2 y¼0 x¼

Thus, the critical point of the multivariable function is follows. p

1 ,0 2

The discriminant of the multivariable function can be calculated as follows. Δ ¼ f xx f yy  f 2xy ¼ 2  4  0 ¼ 8 ) Δ>0 Hence, based on Case 1, since Δ > 0 and fxx > 0, the critical point is a relative minimum point. Choice (3) is the answer. 6.17. Based on the information given in the problem, the type of critical points of the following multivariable function needs to be investigated. f ðx, yÞ ¼ 4xy  x4  y4 The discriminant of the multivariable function of f(x, y, z) is defined as follows. Δ ¼ f xx f yy  f 2xy Case 1: If, at point p(x0, Case 2: If, at point p(x0, Case 3: If, at point p(x0, Case 4: If, at point p(x0,

y 0, y 0, y 0, y 0,

z0), Δ > 0 and fxx > 0, the critical point is a relative minimum point. z0), Δ > 0 and fxx < 0, the critical point is a relative maximum point. z0), Δ < 0, the critical point is a saddle point. z0), Δ ¼ 0, no conclusion can be made.

6

Solutions of Problems: Multivariable Functions

63

The discriminant of the multivariable function is calculated as follows. Δ ¼ 12x2 12y2  42 ¼ 144x2 y2  16 Regarding the critical point of (0, 0): Δ ¼ 0  16 ¼ 16 < 0 Since Δ < 0, the critical point is a saddle point (Case 4). Regarding the critical point of (1, 1): Δ ¼ 144  16 ¼ 128 > 0 f xx ¼ 12x2 ¼ 12 < 0 Since Δ > 0 and fxx < 0, the critical point is a relative maximum point (Case 2). Regarding the critical point of (1, 1): Δ ¼ 144  16 ¼ 128 > 0 f xx ¼ 12x2 ¼ 12 < 0 Since Δ > 0 and fxx < 0, the critical point is a relative maximum point (Case 2). Choice (4) is the answer. 6.18. Based on the information given in the problem, the value of the operation below is requested for the following multivariable function. x2 1 f ðx, yÞ ¼ p e 4y y

∂f ∂y 2 ∂ f ∂x2 Therefore: x2 x2 x2 x2 ∂f 1 x2 1 1 x2 ¼ p e4y þ 2  p e 4y ¼ p e4y þ 2 p e4y y 2y y 4y 4y y ∂y 2y y x2 x2 ∂f 2x 1 x ¼  p e4y ¼ p e4y 4y y 2y y ∂x

2

)

x2 x2 x2 x2 ∂ f 1 2x x 1 x2 ¼ p e4y   p e 4y ¼ p e4y þ 2 p e4y 2 2y y 4y 2y y 2y y 4y y ∂x

64

6

Solutions of Problems: Multivariable Functions

x2 x2 1 x4y2 ∂f p e þ 2 p e 4y 2y y 4y y ∂y ) 2 ¼ 2 x x2 1 x2 ∂ f p e4y þ 2 p e 4y 4y y ∂x2 2y y

∂f ∂y ) 2 ¼1 ∂ f ∂x2 Choice (1) is the answer. In this problem, the rules below have been used. d n ðx Þ ¼ nxn1 dx d ðuvÞ ¼ u0 v þ v0 u dx d u e ¼ u0 e u du 6.19. Based on the information given in the problem, the type of the point of

p1 , p1 3 3

for the following multivariable

function needs to be investigated. f ðx, yÞ ¼ x4 þ y4 þ 1  x2  y2

2

To determine the critical point, the system of equations below must be solved. fx ¼ 0 fy ¼ 0 The discriminant of the multivariable function of f(x, y, z) is defined as follows. Δ ¼ f xx f yy  f 2xy Case 1: If, at point p(x0, Case 2: If, at point p(x0, Case 3: If, at point p(x0, Case 4: If, at point p(x0,

y 0, y 0, y 0, y 0,

z0), Δ > 0 and fxx > 0, the critical point is a relative minimum point. z0), Δ > 0 and fxx < 0, the critical point is a relative maximum point. z0), Δ < 0, the critical point is a saddle point. z0), Δ ¼ 0, no conclusion can be made.

Therefore, for this problem, first, we need to know if the point is a critical point or not as follows. f x ¼ 4x3  4x 1  x2  y2 f y ¼ 4y3  4y 1  x2  y2

)

f x ¼ 8x3  4x þ 4xy2 f y ¼ 8y3  4y þ 4yx2

6

Solutions of Problems: Multivariable Functions

)

1 1 fx p , p 3 3

1 ¼8 p 3

1 1 p ,p 3 3

1 ¼8 p 3

fy

65 3

3

4 4 1 p þp p 3 3 3 1 4 4 p þp p 3 3 3 1 fx p , 3 1 fy p , 3

)

p1 , p1 3 3

Therefore, the point of

2

1 p 3 1 p 3

1 fx p , 3 1 fy p , 3

)

1 p 3 1 p 3

8 4 4 ¼ p p þ p 3 3 3 3 3 8 4 4 ¼ p p þ p 3 3 3 3 3

¼0 ¼0

is a critical point of the multivariable function.

2

f xx ¼

2

∂ x4 þ y4 þ 1  x 2  y2 ∂x2

2

¼

∂ 4x3  4x 1  x2  y2 ∂x

) f xx ¼ 12x2  4 1  x2  y2  4xð2xÞ ¼ 24x2  4 þ 4y2 1 1 ) f xx p , p 3 3

1 ¼ 24 p 3

2

2

1 4þ4 p 3

4 16 ¼84þ ¼ 3 3

1 1 ) f xx p , p > 0 3 3 Δ ¼ f xx f yy  f 2xy ¼ 24x2  4 þ 4y2 24y2  4 þ 4x2  ð8xyÞ2 1 1 )Δ p ,p 3 3

¼

1 24 p 3

1 1 )Δ p ,p 3 3

2

1 4þ4 p 3

¼ 84þ

4 3

2

1 24 p 3

84þ

4 8  3 3

2

1 4þ4 p 3 2

¼

2

1 1  8p p 3 3

16 16 64 192   ¼ >0 3 3 9 9

Therefore, since Δ > 0 and fxx > 0, the critical point is a relative minimum point (Case 1). Choice (1) is the answer. 6.20. Based on the information given in the problem, the limit below is requested. lim

ðx, yÞ → ð0, 0Þ

tan 1 ðxyÞ xy

As we know: lim tan 1 ðf ðxÞÞ  f ðxÞ

f ð xÞ → 0

Therefore: lim

ðx, yÞ → ð0, 0Þ

tan 1 ðxyÞ xy  lim xy ðx, yÞ → ð0, 0Þ xy

2

66

6

)

lim

ðx, yÞ → ð0, 0Þ

Solutions of Problems: Multivariable Functions

tan 1 ðxyÞ ¼1 xy

Choice (3) is the answer. 6.21. Based on the information given in the problem, we have: f ðx, yÞ ¼ ) f ðx, yÞ ¼ ) f ðx, yÞ ¼

23 þ 6x þ 4y  x2  y2 x2 þ 6x þ 23  y2 þ 4y

ðx2  6x þ 9Þ  ðy2  4y þ 4Þ þ 36

) f ðx, yÞ ¼

ðx  3Þ2  ðy  2Þ2 þ 36

) f ðx, yÞ ¼

36  ðx  3Þ2 þ ðy  2Þ2

The multivariable function is a radical function; therefore, f (x, y) ≥ 0. In other words, the minimum value of f (x, y) is zero. On the other hand, the maximum value of 36  [(x  3)2 + (y  2)2] is 36 since the minimum value of [(x  3)2 + (y  2)2] is zero. Thus, the maximum value of f (x, y) is 6. Therefore: 0 ≤ f ðx, yÞ ≤ 6 Choice (3) is the answer. 6.22. Based on the information given in the problem, the value of fx (0, 0) and fy (0, 0) for the following multivariable function is requested.

f ðx, yÞ ¼

3x4  5y4 x3 þ y3 0

x≠  y x ¼ y

In this problem, it is better to solve the problem using the definition of limit as follows. f x ð0, 0Þ ¼ lim

h→0

f ðh, 0Þ  f ð0, 0Þ h

3h4 0 3 ) f x ð0, 0Þ ¼ lim h h h→0 ) f x ð0, 0Þ ¼ 3 f ð0, hÞ  f ð0, 0Þ h h→0

f y ð0, 0Þ ¼ lim

6

Solutions of Problems: Multivariable Functions

67

5h4 3 ) f y ð0, 0Þ ¼ lim h h h→0 ) f y ð0, 0Þ ¼ 5 Choice (1) is the answer. 6.23. Based on the information given in the problem, the value of

∂z for the following multivariable function at (2, 1, 2) is ∂y

requested. xyz þ 2z2  3y ¼ 1 The given multivariable function in the form of f(x, y, z) ¼ 0 is as follows. f ðx, y, zÞ ¼ xyz þ 2z2  3y  1 ¼ 0 The problem can be solved by using the method of implicit differentiation as follows. f 0y ∂z ¼ 0 ∂y fz )

∂z xz  3 ¼ xy þ 4z ∂y

For (x, y, z) ¼ (2, 1, 2), we have: )

ð2Þð2Þ  3 ∂z ¼ ð2Þð1Þ þ 4  ð2Þ ∂y )

∂z 7 ¼ 6 ∂y

Choice (2) is the answer. 6.24. Based on the information given in the problem, the equation of the tangent plane on the surface below at (1, 1, 0) is requested. ez þ 3xy2 þ 2zx ¼ 4 ) f ðx, y, zÞ ¼ ez þ 3xy2 þ 2zx  4 ¼ 0 The equation of a tangent plane on the surface of f(x, y, z) ¼ 0 at p(x0, y0, z0) is as follows: f ′ x ðx0 , y0 , z0 Þðx  x0 Þ þ f ′ y ðx0 , y0 , z0 Þðy  y0 Þ þ f ′ z ðx0 , y0 , z0 Þðz  z0 Þ ¼ 0 Therefore: →

f ′ x , f ′ y , f ′ z ¼ ∇f ¼

∂ → ∂ → ∂ → i þ j þ k ∂x ∂y ∂z

ez þ 3xy2 þ 2zx  4

68

6

Solutions of Problems: Multivariable Functions

) f 0x , f 0y , f 0z ¼ 3y2 þ 2z, 3xy, ez þ 2x ) f 0x ð1, 1, 0Þ, f 0y ð1, 1, 0Þ, f 0z ð1, 1, 0Þ ¼ ð3, 6, 3Þ Hence, the equation of the tangent plane on the surface at the given point is as follows: 3ðx  1Þ  6ðy þ 1Þ þ 3ðz  0Þ ¼ 0 ) 3x  6y þ 3z ¼ 9 ) x  2y þ z ¼ 3 Choice (3) is the answer. In this problem, the rules below have been used. d n x ¼ nxn1 dx d x e ¼ ex dx 6.25. Based on the information given in the problem, the equation of the tangent plane on the surface below at (1, 1, 0) is requested. exy þ ey  ez  e ¼ 0 ) f ðx, y, zÞ ¼ exy þ ey  ez  e ¼ 0 The equation of a tangent plane on the surface of f(x, y, z) ¼ 0 at the point of p(x0, y0, z0) is as follows: f ′ x ðx0 , y0 , z0 Þðx  x0 Þ þ f ′ y ðx0 , y0 , z0 Þðy  y0 Þ þ f ′ z ðx0 , y0 , z0 Þðz  z0 Þ ¼ 0 Therefore: →

f ′ x , f ′ y , f ′ z ¼ ∇f ¼

∂ → ∂ → ∂ → i þ j þ k ðexy þ ey  ez  eÞ ∂x ∂y ∂z

) f 0x , f 0y , f 0z ¼ ðyexy , xexy þ e, ez Þ ) f 0x ð1, 1, 1Þ, f 0y ð1, 1, 1Þ, f 0z ð1, 1, 1Þ ¼ ðe, 2e, eÞ ) eðx  1Þ þ 2eðy  1Þ  eðz  1Þ ¼ 0 ) x þ 2y  z ¼ 2 Choice (4) is the answer.

6

Solutions of Problems: Multivariable Functions

69

In this problem, the rule below has been used. d u e ¼ u0 e u du 6.26. Based on the information given in the problem, a parallel line to the tangent line on the common part of the two crossing surfaces below at (0, 1, 4) is requested. z ¼ 4x2 þ 4y2 ) f ðx, y, zÞ ¼ 4x2 þ 4y2  z ¼ 0 z ¼ 4  4x2 ) gðx, y, zÞ ¼ 4x2 þ z  4 ¼ 0 The common part of the two crossing surfaces of f(x, y, z) ¼ 0 and g(x, y, z) ¼ 0 is a curve that the direction vector of the tangent line on this curve can be calculated as follows. →

→

→

V ¼ ∇f  ∇g A parallel line to the tangent line will have a similar direction vector. Therefore: →

∇f ðx, y, zÞ ¼

∂ → ∂ → ∂ → i þ j þ k ∂x ∂y ∂z

4x2 þ 4y2  z ¼ ð8x, 8y, 1Þ

→

) ∇f ð0, 1, 4Þ ¼ ð0, 8, 1Þ →

∇gðx, y, zÞ ¼

∂ → ∂ → ∂ → i þ j þ k ∂x ∂y ∂z

4x2 þ z  4 ¼ ð8x, 0, 1Þ

→

) ∇gð0, 1, 4Þ ¼ ð0, 0, 1Þ →

i

→

) V ð0, 1, 4Þ ¼ ð0, 8, 1Þ  ð0, 0, 1Þ ¼ 0 0 →

→

→

→

8 0

1 1

j

k

→

) V ð0, 1, 4Þ ¼ 8 i  i Choice (2) is the answer. 6.27. Based on the information given in the problem, we have: uxx  2uxy þ uyy ¼ 0

ð1Þ

z¼xþy

ð2Þ

v¼x

ð3Þ

70

6

Solutions of Problems: Multivariable Functions

The problem can be solved by using the method of changing the variables as follows. ∂u ∂u ∂v ∂u ∂z ¼ þ ¼ uv þ uz ∂x ∂v ∂x ∂z ∂x

ð4Þ

∂ ∂ ∂ ð uv þ uz Þ ¼ u v þ uz ∂x ∂x ∂x

ð5Þ

∂uv ∂v ∂uv ∂z ∂uz ∂v ∂uz ∂z þ þ þ ∂v ∂x ∂z ∂x ∂v ∂x ∂z ∂x

ð6Þ

) uxx ¼ uvv þ uvz þ uzv þ uzz

ð7Þ

) uxx ¼ uvv þ 2uvz þ uzz

ð8Þ

ux ¼

) uxx ¼ ) uxx ¼

By assuming uvz ¼ uzv, we have:

Now, uxy can be calculated as follows. uxy ¼

∂ ∂ ∂ ðuv þ uz Þ ¼ uv þ u z ∂y ∂y ∂y

) uxy ¼

∂uv ∂z ∂uz ∂z þ ¼ uvz þ uzz ∂z ∂y ∂z ∂y

ð9Þ ð10Þ

Moreover, uyy can be calculated as follows. uy ¼

∂u ∂u ∂z ¼ ¼ uz ∂y ∂z ∂y ∂ ð uz Þ ∂y

ð12Þ

∂uz ∂z ¼ uzz ∂z ∂y

ð13Þ

) uyy ¼ ) uyy ¼

ð11Þ

Solving (1), (8), (10), and (13): ðuvv þ 2uvz þ uzz Þ  2ðuvz þ uzz Þ þ uzz ¼ ) uvv ¼ 0 Choice (1) is the answer. 6.28. Based on the information given in the problem, the equation of the tangent line on the common part of the two crossing surfaces below at (1, 1, 1) is requested. x2 þ 2y2 þ 3z2 ¼ 6 ) f ðx, y, zÞ ¼ x2 þ 2y2 þ 3z2  6 ¼ 0 xyz ¼ 1 ) gðx, y, zÞ ¼ xyz  1 ¼ 0

6

Solutions of Problems: Multivariable Functions

71

The common part of the two crossing surfaces of f(x, y, z) ¼ 0 and g(x, y, z) ¼ 0 is a curve that the direction vector of the tangent line on this curve can be calculated as follows. →

→

→

V ¼ ðV 1 , V 2 , V 3 Þ ¼ ∇ f  ∇ g The equation of the tangent line on the common part of the two crossing surfaces at the point of p(x0, y0, z0) is as follows. y  y0 x  x0 z  z0 ¼ ¼ V 1 ðx0 , y0 , z0 Þ V 2 ðx0 , y0 , z0 Þ V 3 ðx0 , y0 , z0 Þ For this problem, we have: ∂ → ∂ → ∂ → i þ j þ k ∂x ∂y ∂z

→

∇f ¼

→

→

→

x2 þ 2y2 þ 3z2  6 ¼ 2x i þ 4y j þ 6z k

→

→

→

→

) ∇f ð1, 1, 1Þ ¼ 2 i þ 4 j þ 6 k →

∇g ¼

→ → → ∂ → ∂ → ∂ → i þ j þ k ðxyz  1Þ ¼ yz i þ xz j þ xy k ∂x ∂y ∂z →

→

→

→

) ∇gð1, 1, 1Þ ¼ i þ j þ k →

→

→

1

1

1

i V ð1, 1, 1Þ ¼ ð2, 4, 6Þ  ð1, 1, 1Þ ¼ 2 →

→

j 4

→

k → → → 6 ¼ i ð4  6Þ  j ð2  6Þ þ k ð2  4Þ

→

→

→

→

→

) V ð1, 1, 1Þ ¼ 2 i þ 4 j  2 k  i  2 j þ k

Therefore, the equation of the tangent line on the common part of the two crossing surfaces at the point of (1, 1, 1) is as follows. x1 y1 z1 ¼ ¼ 1 2 1 Choice (4) is the answer. 6.29. Based on the information given in the problem, the amount of angle between the cylinder and sphere below at p 3 1 , 2 2 ,0

is requested. x2 þ y2 ¼ 1 ) f ðx, y, zÞ ¼ x2 þ y2  1 ¼ 0 ðx  1Þ2 þ y2 þ z2 ¼ 1 ) gðx, y, zÞ ¼ ðx  1Þ2 þ y2 þ z2  1 ¼ 0

The amount of angle between two surfaces at a specific point can be achieved by determining the angle between their normal vectors at the same point. Moreover, the angle between two vectors can be calculated as follows. cos θ ¼

→

→

→

→

∇f :∇g

∇f ∇g

72

6

Solutions of Problems: Multivariable Functions

For this problem, we have: →

N 1 ¼ ∇f ðx, y, zÞ ¼

→

p 1 3 , ,0 2 2

) cos θ ¼

p ¼ 1, 3, 0

→

∂ → ∂ → ∂ → i þ j þ k ∂x ∂y ∂z

) N2

x2 þ y2  1 ¼ ð2x, 2y, 0Þ

p 1 3 ¼ ∇f , ,0 2 2

p 1 3 , ,0 2 2

) N1

N 2 ¼ ∇gðx, y, zÞ ¼

∂ → ∂ → ∂ → i þ j þ k ∂x ∂y ∂z

→

¼ ∇g

ðx  1Þ2 þ y2 þ z2  1 ¼ ð2x  2, 2y, 2zÞ p 1 3 , ,0 2 2

p p 1, 3, 0 : 1, 3, 0 p p 1, 3, 0 1, 3, 0 )θ¼

p ¼ 1, 3, 0

¼

1 þ 3 1 ¼ 22 2

π 3

Choice (3) is the answer. In this problem, the rules below have been used. d n x ¼ nxn1 dx 1 π cos 1 ¼ 2 3 6.30. Based on the information given in the problem, the divergence of the multivariable function below is requested. f ðx, y, zÞ ¼

→ → m → x i þy j þzk r3

where, m>0 r 2 ¼ x2 þ y2 þ z 2 , r ≠ 0 The divergence of a vector function can be calculated as follows. →

→ →

div f ¼ ∇: f ¼

∂ → ∂ → ∂ → → i þ j þ k :f ∂x ∂y ∂z

Therefore, for this problem, we have: →

div f ¼

→ → → ∂ → ∂ → ∂ → my mz mx i þ j þ k : 3 i þ 3 j þ 3 k 2 2 2 2 2 2 2 2 2 ∂x ∂y ∂z 2 2 2 ðx þ y þ z Þ ðx þ y þ z Þ ðx þ y þ z Þ

6

Solutions of Problems: Multivariable Functions

73

3

1

3

1

mðx2 þ y2 þ z2 Þ2  3mx2 ðx2 þ y2 þ z2 Þ2 mðx2 þ y2 þ z2 Þ2  3my2 ðx2 þ y2 þ z2 Þ2 ) div f ¼ þ ð x2 þ y2 þ z 2 Þ 3 ð x2 þ y2 þ z 2 Þ 3 →

3

þ →

) div f ¼

1

mðx2 þ y2 þ z2 Þ2  3mz2 ðx2 þ y2 þ z2 Þ2 ð x2 þ y2 þ z 2 Þ 3 mðx2 þ y2 þ z2 Þ  3mx2 ð x2 þ y2 þ z 2 Þ →

) div f ¼

5 2

þ

mðx2 þ y2 þ z2 Þ  3my2 ð x2 þ y2 þ z 2 Þ

þ

5 2

mðx2 þ y2 þ z2 Þ  3mz2 5

ðx2 þ y2 þ z2 Þ2

mr 2  3mx2 mr 2  3my2 mr 2  3mz2 þ þ r5 r5 r5 →

) div f ¼

3mr 2  3mðx2 þ y2 þ z2 Þ r5 →

) div f ¼

3mr 2  3mr 2 r5 →

) div f ¼ 0 Choice (1) is the answer. In this problem, the rule below has been used. d u u0 v  v 0 u ¼ dx v v2 6.31. Based on the information given in the problem, the equation below is requested to be solved for the following multivariable function. →

div ∇f

¼0

f ðx, y, zÞ ¼ eðx þy 2

ð1Þ

2

þz2 Þ

ð2Þ

The divergence of gradient of a multivariable function is called Laplacian of the function. In other words: →

div ∇f

→

→

¼ ∇ : ∇f

→ ∂ → ∂ → ∂ → i þ j þ k : ∇f ∂x ∂y ∂z

¼

→2

¼∇ f ¼

2

2

2

∂ f ∂ f ∂ f þ þ ¼ f xx þ f yy þ f zz ∂x2 ∂y2 ∂z2

ð3Þ

Therefore, for this problem, we have: →2 2 2 2 ∂ ∂ ðx2 þy2 þz2 Þ ∂ ∂ ðx2 þy2 þz2 Þ ∂ ∂ ðx2 þy2 þz2 Þ ∇ eðx þy þz Þ ¼ þ þ e e e ∂x ∂x ∂y ∂y ∂z ∂z

ð4Þ

2 2 2 2 2 2 2 2 2 ∂ ∂ ∂ 2xex þy þz þ 2yex þy þz þ 2zex þy þz ∂x ∂y ∂z

ð5Þ

→2

) ∇ eðx þy 2

2

þz2 Þ

¼

→ 2 2 2 2 2 2 2 2 2 2 2 2 ) ∇ eðx þy þz Þ ¼ 2 þ 4x2 ex þy þz þ 2 þ 4y2 ex þy þz þ 2 þ 4z2 ex þy þz 2

ð6Þ

74

6

Solutions of Problems: Multivariable Functions

→ 2 2 2 2 2 2 2 2 2 ) ∇ eðx þy þz Þ ¼ 6ex þy þz þ 4 x2 þ y2 þ z2 ex þy þz

ð7Þ

→ 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ) ∇ eðx þy þz Þ ¼ 6ex þy þz þ 2x 2xex þy þz þ 2y 2yex þy þz þ 2z 2zex þy þz

ð8Þ

2

2

From (4)–(5), it is noticed that (8) can be written as follows. →2 2 2 2 2 2 2 2 2 2 2 2 2 ∂ ∂ ∂ 2 2 2 ) ∇ eðx þy þz Þ ¼ 6ex þy þz þ 2x eðx þy þz Þ þ 2y eðx þy þz Þ þ 2z eðx þy þz Þ ∂x ∂y ∂z

ð9Þ

Solving (2) and (9): → 2 2 2 ) ∇ eðx þy þz Þ ¼ 6f þ 2xf x þ 2yf y þ 2zf z 2

Solving (1) and (10): 6f þ 2xf x þ 2yf y þ 2zf z ¼ 0 ) xf x þ yf y þ zf z ¼ 3f Choice (3) is the answer. In this problem, the rules below have been used. d u e ¼ u0 e u du d ðuvÞ ¼ u0 v þ v0 u dx

References 1. Rahmani-Andebili, M. (2023). Calculus II – Practice Problems, Methods, and Solutions, Springer Nature. 2. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 3. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature. 4. Rahmani-Andebili, M. (2023). Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 5. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature.

ð10Þ

7

Problems: Double Integrals and Their Applications

Abstract

In this chapter, the basic and advanced problems of double integrals and their applications are presented. The subjects include definite and indefinite double integrals, density function, mass, center of mass, center of gravity, moment of inertia, variables separation technique, integration by parts, changing order of integration, updating boundaries of variables, and surface area. 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

Double Integrals

7.1. Calculate the value of definite double integral below [1–5]. x þ y - x2 þy2 e dxdy 2 2 R2 x þ y Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 0 π 2) 2 3) e-1 4) e-2 7.2. Calculate the value of definite multiple integral below. e 1

x 0

dydx ðx þ yÞ2

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 2 2) 1 3 3) 2 4) 2

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus III, https://doi.org/10.1007/978-3-031-47483-5_7

75

76

7

Problems: Double Integrals and Their Applications

7.3. Calculate the value of following definite multiple integral where the region of D is defined as 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. dxdy 2 D ðx þ y þ 1Þ Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 5 1) ln 3 7 2) ln 3 3) ln 2 4 4) ln 3 7.4. Calculate the value of following definite multiple integral where the region of D is restricted by the lines of y ¼ x, y ¼ π, and x ¼ 0. cosðx þ yÞdxdy D

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 2) 2 3) -1 4) -2 7.5. Calculate the value of following definite multiple integral where the region of D is confined by the lines of y ¼ x, y ¼ 0, and x ¼ 1. y

ex dA D

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2(e - 1) 1 2) ðe - 1Þ 2 1 3) 2e 2 1 4) 2e þ 2 7.6. Calculate the value of following definite double integral. 1 0

1 0

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

xye - 3ðxþyÞ dxdy

7.1 Double Integrals

77

1 81 1 2) 9 1 3) 8 1 4) 4 1)

7.7. Which one of the following choices is equivalent to the definite multiple integral below. e 1

ln x

f ðx, yÞdydx

0

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)

1 0 1 0 1 0 1 0

ln y 0 f ðx, yÞdxdy 0 ln y f ðx, yÞdxdy e ey f ðx, yÞdxdy ey e f ðx, yÞdxdy

7.8. Which one of the following choices is equivalent to the definite multiple integral below. 4

2 p

1

f ðx, yÞdydx

x

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) 2) 3) 4)

1 0 2 1 1 0 2 1

y2 0 f ðx, yÞdxdy 4 y2 f ðx, yÞdxdy 4 2 4 1 f ðx, yÞdxdy þ 1 y2 y2 1 f ðx, yÞdxdy

f ðx, yÞdxdy

7.9. Calculate the value of following definite multiple integral. 1 0

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) ln 2 2) 3ln 2 3) 2ln 2 4) 1

yþ1 y

1 dxdy x

78

7

7.10. Calculate the value of following definite multiple integral. 1

1

2

ey dydx 0

x

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1 1) ðe - 1Þ 2 1 2) ðe þ 1Þ 2 1 3) e 2 4) e 7.11. Calculate the value of definite integral of

1 0 f ðxÞdx

if we have: x

f ðxÞ ¼

2

ey dy 1

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 + e 2) 1 - e 1 3) ð1 - eÞ 2 1 4) ð1 þ eÞ 2 7.12. Calculate the value of following definite multiple integral. 1

1

3

x3 ey dydx 0

x2

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1 1) ðe - 1Þ 2 1 2) ðe - 1Þ 3 1 3) ðe - 1Þ 6 1 ðe - 1Þ 4) 12 7.13. Calculate the value of following definite multiple integral. 1 0

1 x

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large

sin y dydx y

Problems: Double Integrals and Their Applications

7.1 Double Integrals

1) 2) 3) 4)

79

1 - cos 1 1 + sin 1 cos1 2

7.14. Calculate the value of definite multiple integral below. 1

1 p3 y

0

x4 þ 1 dxdy

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 0 p 2-1 2) 6 p 3 9-1 3) p4 2 2-1 4) 6 7.15. Calculate the value of definite multiple integral below where D ¼ {(x, y) E ℝ2|x2 + y2 ≤ 1, x ≥ 0, y ≥ 0}. 3

xy x2 þ y2 2 dxdy D

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 14 1 2) 12 1 3) 7 1 4) 6 7.16. Calculate the value of definite multiple integral below where R ¼ {(x, y)|x2 + y2 ≤ 1}. sin π x2 þ y2 dxdy R

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 0 2) 1 3) 2 4) 4 7.17. Calculate the value of definite multiple integral below. I¼

sin π 2 ≤ x2 þy2 ≤ 4π 2

x 2 þ y2 x2

þ y2

dydx

80

7

Problems: Double Integrals and Their Applications

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 2π 2) -2π 3) -4π 4) 4π

7.2

Applications of Double Integrals

7.18. Calculate the surface area between y-axis, the line of x ¼ π4, above the curve of y ¼ sin x, and under the curve of y ¼ cos x. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large p 2 1) -1 2 p 2 2) þ1 2 p 3) 2 þ 1 p 4) 2 - 1 7.19. Calculate the volume restricted by the surface of z ¼ x2 + 4y2 and the parabolas of y2 ¼ x and x2 ¼ y on xoy plane. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 3 1) 7 6 2) 7 3 3) 14 9 4) 7 7.20. Calculate the volume inside a cylinder restricted by the curves of y ¼ 2x and y ¼ x2 and under the plane of z ¼ x + 2y. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 24 1) 5 28 2) 5 21 3) 4 15 4) 4 7.21. Calculate the mass of a circular object with the radius of one and the density function of ρ ¼ Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large π 1) 3 2π 2) 3 3 3) π 2 4) 2π

1 - x 2 - y2 .

References

81

7.22. Calculate the center of mass of the region of D ¼ {(x, y)|x2 + y2 ≤ a2, x ≥ 0} with a uniform density. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 4 1) a, 0 3π 4 4 2) a, a 3π 3π 2 a, 0 3) 3π 1 4) a, 0 3π 7.23. Calculate the distance of center of gravity of a uniform thin region from x-axis where the region is restricted by the curve of y ¼ x2, x-axis, and the line of x ¼ 2. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 1.6 2) 1.4 3) 1.5 4) 1.2 7.24. Calculate the moments of inertia of a circular plane with a center at the origin, radius of two, and density of unity with respect to x-axis and the origin. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) 4π, 8π 2) 8π, 4π 3) 4π, 4π 4) 16π, 8π

References 1. Rahmani-Andebili, M. (2023). Calculus II – Practice Problems, Methods, and Solutions, Springer Nature. 2. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 3. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature. 4. Rahmani-Andebili, M. (2023). Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 5. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature.

8

Solutions of Problems: Double Integrals and Their Applications

Abstract

In this chapter, the problems of the seventh chapter are fully solved, in detail, step-by-step, and with different methods.

8.1

Double Integrals

8.1. Based on the information given in the problem, the value of definite double integral below is requested [1–5]. I¼

x þ y - x2 þy2 e dxdy 2 2 R2 x þ y

The problem can be solved as follows. I¼

2

R

I¼ R

2

2 2 2 2 x y e - x þy þ 2 e - x þy dxdy x2 þ y 2 x þ y2

2 2 x e - x þy dxdy þ x2 þ y2

R

2

2 2 y e - x þy dxdy x2 þ y 2

As can be noticed, the function of the first integral is odd with respect to the variable of x and the function of the second integral is odd with respect to the variable of y. Therefore, since the integration regions are symmetric, the value of integrals are zero. In other words: )I¼0 Choice (1) is the answer. In this problem, the rule below has been used. If f (x, y) is an odd function and R2 is a symmetric region, then:

R2

f ðx, yÞdA ¼ 0

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus III, https://doi.org/10.1007/978-3-031-47483-5_8

83

84

8 Solutions of Problems: Double Integrals and Their Applications

8.2. Based on the information given in the problem, we have: e

I¼

x

1

0

dydx ð x þ yÞ 2

The problem can be solved as follows. e

I¼ 1

x 0

e

)I¼

-1 ð x þ yÞ

1

e

-1 1 þ dx ¼ 2x x

1

)I¼

e

dy dx ¼ ð x þ yÞ 2

1 ln x 2

e 1

¼

1

x dx 0

dx 2x

1 1 ln e - ln 1 2 2

)I¼

1 2

Choice (1) is the answer. In this problem, the rules below have been used. du ¼ ln u þ c u ln e ¼ 1 ln 1 ¼ 0 8.3. Based on the information given in the problem, we have: dxdy 2 D ð x þ y þ 1Þ

I¼

D : f0 ≤ x ≤ 1, 0 ≤ y ≤ 1g The problem can be solved as follows. 1

I¼ 0

1 0

dx dy ¼ ð x þ y þ 1Þ 2 1

)I¼ 0

-

1 0

1

-1 dy xþyþ1 0

1 1 dy þ yþ2 yþ1

) I ¼ ½ - lnðy þ 2Þ þ lnðy þ 1Þ10 ) I ¼ - ln 3 þ ln 2 - ð- ln 2 þ ln 1Þ ¼ 2 ln 2 - ln 3 ¼ ln 4 - ln 3 ) I ¼ ln Choice (4) is the answer.

4 3

8.1 Double Integrals

85

In this problem, the rules below have been used. un du ¼

unþ1 þc nþ1

du ¼ ln u þ c u ln 1 ¼ 0 ln a - ln b ¼ ln

a b

8.4. Based on the information given in the problem, we have: cosðx þ yÞdxdy

I¼ D

Herein, the region of D is restricted by the lines of y ¼ x, y ¼ π, and x ¼ 0. The problem can be solved as follows. π

I¼

y

0

cosðx þ yÞdxdy

0 π

)I¼

sinðx þ yÞ

0 π

)I¼

y 0

dy

ðsinð2yÞ - sinðyÞÞdy

0

)I¼

-1 cos 2y þ cos y 2

π

-1 -1 cos 2π þ cos π cos 0 þ cos 0 2 2

¼

0

)I¼ -

1 -1 -1 þ1 2 2

) I ¼ -2 Choice (4) is the answer. In this problem, the rules below have been used. cos u du ¼ sin u þ c sin u du ¼ - cos u þ c cos 2π ¼ 1 cos π ¼ - 1 cos 0 ¼ 1

86

8 Solutions of Problems: Double Integrals and Their Applications

8.5. Based on the information given in the problem, we have: y

I¼

ex dA D

Herein, the region of D is confined by the lines of y ¼ x, y ¼ 0, and x ¼ 1. The problem can be solved as follows. 1

I¼

x

0

1

y

ex dydx ¼

y

xex

0

0 1

)I¼

x 0

dx

ðxe - xÞdx

0

)I¼

e

x 2 x2 2 2

1

¼ e

0

1 1 - ð 0 - 0Þ 2 2

1 ) I ¼ ð e - 1Þ 2 Choice (2) is the answer. In this problem, the rules below have been used. eu du ¼ eu þ c un du ¼

unþ1 þc nþ1

8.6. Based on the information given in the problem, we have: 1

I¼ 0

1

xye - 3ðxþyÞ dxdy

0

The problem can be solved by variables separation as follows. 1

I¼

1

xe - 3x dx

0

)I¼

-

x - 3x 1 - 3x - e e 3 9 ) I ¼ 0- -

1 0

In this problem, the rules below have been used.

-

1 9

)I¼ Choice (1) is the answer.

ye - 3y dy

0

y - 3y 1 - 3y - e e 3 9

0- 1 81

1 9

1 0

8.1 Double Integrals

87

1 1 xeax ¼ xeax - 2 eax þ c a a lim xe - x ¼ 0

ðIntegration by partsÞ

ðGrowth rate orderÞ

x→1

8.7. Based on the information given in the problem, we have: e

I¼

ln x

1

f ðx, yÞdydx

0

The problem can be solved by changing the order of integration, and consequently updating the boundaries of variables. As can be seen in Fig. 8.1, after changing the order of integration, the boundaries of x change from e y to e. Moreover, the boundaries of y change from 0 to 1. Therefore: 1

I¼

e

f ðx, yÞdxdy

ey

0

Choice (3) is the answer.

Figure 8.1 The graph of problem 8.7

8.8. Based on the information given in the problem, we have: 4

I¼

2 p x

1

f ðx, yÞdydx

The problem can be solved by changing the order of integration, and consequently updating the boundaries of variables. As can be seen in Fig. 8.2, after changing the order of integration, the boundaries of x change from 1 to y2 . Moreover, the boundaries of y change from 1 to 2. Therefore: I¼ 1

Choice (4) is the answer.

y2

2 1

f ðx, yÞdxdy

88

8 Solutions of Problems: Double Integrals and Their Applications

Figure 8.2 The graph of problem 8.8

8.9. Based on the information given in the problem, we have: 1

I¼ 0

yþ1 y

1 dxdy x

The problem can be solved as follows. 1

I¼ 0 1

)I¼

yþ1 y

1

1 dx dy ¼ x

ln x 0 1

½lnðy þ 1Þ - ln ydy ¼

0

yþ1 y

dy 1

lnðy þ 1Þdy -

0

ln ydy 0

) I ¼ ½ðy þ 1Þ lnðy þ 1Þ - ðy þ 1Þ

1 0

- ½yln y - y

1 0

) I ¼ ð2ln 2 - 2Þ - ðln 1 - 1Þ - ðln 1 - 1Þ - lim y → 0þ y ln y - 0 ) I ¼ 2 ln 2 -

H

) I ¼ 2 ln 2 -

lim y → 0þ

lim y → 0

þ

ln y 1 y 1 y

-

1 y2

) I ¼ 2 ln 2 - lim y → 0þ ð- yÞ ¼ 2 ln 2 - 0 ) I ¼ 2 ln 2 Choice (1) is the answer. In this problem, the rules below have been used. du ¼ ln u þ c u ln udu ¼ u ln u - u þ c ln 1 ¼ 0

8.1 Double Integrals

89

8.10. Based on the information given in the problem, we have: 1

I¼

1

2

ey dydx 0

x

Solving the double integral in this form is impossible. Therefore, the order of integration needs to be changed. As can be seen in Fig. 8.3, after changing the order of integration, the boundaries of x change from 0 to y. Moreover, the boundaries of y change from 0 to 1. Therefore: 1

I¼

2

ey dxdy

1

)I¼

y

0

0

2

y

ey x 0

0

1

dy ¼

1 2 ) I ¼ ey 2

2

yey dy 0

1 0

1 ) I ¼ ð e - 1Þ 2 Choice (1) is the answer. In this problem, the rules below have been used. xn dx ¼

xnþ1 þc nþ1

eu du ¼ eu þ c

Figure 8.3 The graph of problem 8.10

90

8 Solutions of Problems: Double Integrals and Their Applications

8.11. Based on the information given in the problem, the value of definite integral of x

f ðxÞ ¼

1 0 f ðxÞdx

is requested while we have:

2

ey dy 1

Therefore: 1

I¼ 0

x

1

2

ey dydx ¼ -

1

2

ey dydx

1

0

x

Solving the multiple integral in this form is impossible. Therefore, the order of integration needs to be changed. As can be seen in Fig. 8.4, after changing the order of integration, the boundaries of x change from 0 to y. Moreover, the boundaries of y change from 0 to 1. Therefore: 1

I¼ -

y

2

ey dxdy 0

0 1

)I¼ -

yey

2

0

)I¼ -

-1 1 y2 1 ¼ ð e - 1Þ e 2 2 0

1 ) I ¼ ð1 - eÞ 2 Choice (3) is the answer. In this problem, the rules below have been used. xn dx ¼

xnþ1 þc nþ1

eu du ¼ eu þ c

Figure 8.4 The graph of problem 8.11

8.1 Double Integrals

91

8.12. Based on the information given in the problem, the value of following definite multiple integral is requested. 1

I¼

1

3

x3 ey dydx x2

0

Solving the integral in this form is impossible. Therefore, the order of integration needs to be changed. As can be seen in Fig. 8.5, after changing the order of integration, the boundaries of x change from 0 to the boundaries of y change from 0 to 1. Therefore: I¼ 1

ey

3

x3 ey dxdy 0

)I¼

p y

1 0

p

1 4 x 4

3

0

y 0

)I¼ )I¼

dy ¼

1 y3 e 12

1 4

1 0

1 0

1 ðe - 1Þ 12

Choice (4) is the answer. In this problem, the rules below have been used. xn dx ¼

3

y2 ey dy

xnþ1 þc nþ1

eu du ¼ eu þ c

Figure 8.5 The graph of problem 8.12

p

y . Moreover,

92

8 Solutions of Problems: Double Integrals and Their Applications

8.13. Based on the information given in the problem, the value of definite multiple integral below is requested. 1

I¼ 0

1 x

sin y dydx y

Solving the integral in this form is impossible. Therefore, the order of integration needs to be changed. As can be seen in Fig. 8.6, after changing the order of integration, the boundaries of x change from 0 to y. Moreover, the boundaries of y change from 0 to 1. Therefore: 1

I¼ 0

y 0

sin y dxdy ¼ y 1

)I¼

1 0

y sin y dy x y 0

sin y dy ¼ - cos y

0

1 0

) I ¼ 1 - cos 1 Choice (1) is the answer. In this problem, the rules below have been used. xn dx ¼

xnþ1 þc nþ1

sin x dx ¼ - cos x þ c cos 0 ¼ 1

Figure 8.6 The graph of problem 8.13

8.1 Double Integrals

93

8.14. Based on the information given in the problem, the value of definite multiple integral below is requested. 1

I¼

1 p3

0

x4 þ 1dxdy

y

Solving the integral in this form is impossible. Therefore, the order of integration needs to be changed. As can be seen in Fig. 8.7, after changing the order of integration, the boundaries of y change from 0 to x3. Moreover, the boundaries of x change from 0 to 1. Therefore: x3

1

)I¼ 0

x4 þ 1dydx

0 1

)I¼

x3

x4 þ 1dx

0

)I¼

1 2 4  x þ1 4 3

)I¼

3 2

1 0

1 p 2 2-1 6

Choice (4) is the answer. In this problem, the rules below have been used. xn dx ¼

xnþ1 þc nþ1

un du ¼

unþ1 þc nþ1

Figure 8.7 The graph of problem 8.14

94

8 Solutions of Problems: Double Integrals and Their Applications

8.15. Based on the information given in the problem, the value of definite multiple integral below is requested where D ¼ {(x, y) E ℝ2 j x2 + y2 ≤ 1, x ≥ 0, y ≥ 0}. 3

I¼

xy x2 þ y2 2 dxdy D

Since the region of integration is part of a circle, the integral should be solved in polar coordinate system as follows. D ¼ 0 ≤ r ≤ 1, 0 ≤ θ ≤ π 2

)I¼ 0

1

3

ðr cos θÞðr sin θÞ r 2 2 rdrdθ

0 π 2

)I¼ 0 π 2

)I¼

1

cos θ sin θ r6 drdθ

0

ðcos θ sin θ dθÞ

0

)I¼

1 7

π 2

0

π 2

r7 1 7 0

1 1 -1 sin 2θ dθ ¼ cos 2θ 2 14 2

)I¼

-1 ðcos π - cos 0Þ 28 )I¼

1 14

Choice (1) is the answer. In this problem, the rules below have been used. x ¼ r cos θ y ¼ r sin θ r¼

x 2 þ y2

dxdy ¼ rdrdθ xn dx ¼

xnþ1 þc nþ1

cos θ sin θ ¼ sin aθ dθ ¼ -

1 sin 2θ 2 1 cos aθ þ c a

cos π ¼ - 1 cos 0 ¼ 1

π 2 0

8.1 Double Integrals

95

8.16. Based on the information given in the problem, the value of definite multiple integral below is requested where R ¼ {(x, y)jx2 + y2 ≤ 1}. I¼

sin π x2 þ y2 dxdy R

Since the region of integration is a circle, the integral should be solved in polar coordinate system as follows. D ¼ f0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π g 2π

)I¼ 0

1

2π

)I¼

1

dθ 0

) I ¼ ð2π Þ -

sin πr 2 rdrdθ

0

r sin πr 2 dr

0

1 1 cos πr2 2π 0

¼ - cos πr 2

1 ¼ - ½cos π - cos 0 0

)I¼2 Choice (3) is the answer. In this problem, the rules below have been used. x ¼ r cos θ y ¼ r sin θ r¼

x 2 þ y2

dxdy ¼ rdrdθ xn dx ¼

xnþ1 þc nþ1

sin udu ¼ - cos u þ c cos π ¼ - 1 cos 0 ¼ 1 8.17. Based on the information given in the problem, the value of definite multiple integral below is requested. I¼

sin π 2 ≤ x2 þy2 ≤ 4π 2

x 2 þ y2 x2 þ y2

dydx

Since the region of integration is part of a circle, the integral should be solved in polar coordinate system as follows. D ¼ fπ ≤ r ≤ 2π, 0 ≤ θ ≤ 2π g

96

8 Solutions of Problems: Double Integrals and Their Applications 2π

)I¼

sinðr Þ rdrdθ r

2π π

0 2π

)I¼

2π

dθ π

0

sinðr Þdr

) I ¼ 2π ½ - cosðr Þ2π π ¼ ð2π Þðcos π - cos 2π Þ ¼ 2π ð- 1 - 1Þ ) I ¼ - 4π Choice (3) is the answer. In this problem, the rules below have been used. x ¼ r cos θ y ¼ r sin θ x2 þ y2

r¼

dxdy ¼ rdrdθ xn dx ¼

xnþ1 þc nþ1

sin x dx ¼ - cos x þ c cos π ¼ - 1 cos 2π ¼ 0

8.2

Applications of Double Integrals

8.18. Based on the information given in the problem, the surface area between y-axis, the line of x ¼ π4, above the curve of y ¼ sin x, and under the curve of y ¼ cos x is requested. Therefore: S¼

dydx D π 4

)S¼

dydx 0 π 4

)S¼ 0

cos x sin x

ðcos x - sin xÞdx

8.2 Applications of Double Integrals

97

p

p 2 2 - ð 0 þ 1Þ þ 2 2

π ) S ¼ ðsin x þ cos xÞ 4 ¼ 0 )S¼

p

2-1

Choice (4) is the answer. In this problem, the rules below have been used. xn dx ¼

xnþ1 þc nþ1

cos xdx ¼ sin x þ c sin xdx ¼ - cos x þ c p π 2 cos ¼ 4 2 p π 2 sin ¼ 4 2 cos 0 ¼ 1 sin 0 ¼ 0 8.19. Based on the information given in the problem, the volume restricted by the surface of z ¼ x2 + 4y2 and the parabolas of y2 ¼ x and x2 ¼ y on xoy plane is requested. Therefore: f ðx, yÞdA

V¼ D

where, f ðx, yÞ ¼ z ¼ x2 þ 4y2 As can be seen in Fig. 8.8, the boundaries of y change from x2 to 0 to 1. Therefore: 1

)V ¼ 0 1

)V ¼

4y3 x yþ 3

x2 þ 4y2 dydx 1

dx ¼ x2

x and the boundaries of x change from

x2

p x

2

0

p x

p

0

3

5

x2 þ

4x2 4 - x4 - x6 dx 3 3

98

8 Solutions of Problems: Double Integrals and Their Applications

)V ¼

4 7 2 72 2 4 52 x5 x x þ  x 5 21 7 5 3 )V ¼

1

¼ 0

2 8 1 4 þ - 7 15 5 21

3 7

Choice (1) is the answer. In this problem, the rule below has been used. xn dx ¼

xnþ1 þc nþ1

Figure 8.8 The graph of problem 8.19

8.20. Based on the information given in the problem, the volume inside a cylinder restricted by the curves of y ¼ 2x and y ¼ x2 and under the plane of z ¼ x + 2y is requested. Therefore: f ðx, yÞdA

V¼ D

where, f ðx, yÞ ¼ z ¼ x þ 2y As can be seen in Fig. 8.9, the boundaries of y change from x2 to 2x and the boundaries of x change from 0 to 2. Therefore: 2

)V ¼ 0 2

)V ¼ 0

2x x2

2

ðx þ 2yÞdydx ¼

2x2 þ 4x2 - x3 - x4 dx ¼

0

xy þ y2

2x dx x2

2 3 4 3 1 4 4 5 x þ x - x - x 3 4 5 3

2 0

8.2 Applications of Double Integrals

99

)V ¼

2 4 1 4  8 þ  8 -  16 -  32 - ð0Þ 3 3 4 5 )V ¼

28 5

Choice (2) is the answer. In this problem, the rule below has been used. xn dx ¼

xnþ1 þc nþ1

Figure 8.9 The graph of problem 8.20

8.21. Based on the information given in the problem, the mass of a circular object with the radius of one and the density function below is requested. ρ¼

1 - x2 - y2

The mass of a flat object with the density function of ρ(x, y) can be calculated as follows. M¼

ρdxdy D

Thus, for this problem, we have: 1 - ðx2 þ y2 Þdxdy

M¼ D

Since the region of integration is circular, the integral should be solved in polar coordinate system. Therefore: 1p

2π

M¼ 0

1 - r 2 rdrdθ

0

100

8 Solutions of Problems: Double Integrals and Their Applications 1p

2π

)M¼

dθ 0

1p

1p

1 - r 2 rdr ¼ - π

) M ¼ 2π

1 - r 2 rdr

0

0

1 - r 2 ð- 2rdr Þ ¼ - π

0

)M¼ -

2 π 1 - 12 3 )M¼

3 2

2 þ π 1 - 02 3

2 1 - r2 3

3 2

1 0

3 2

2π 3

Choice (2) is the answer. In this problem, the rules below have been used. x ¼ r cos θ y ¼ r sin θ r¼

x2 þ y 2

dxdy ¼ rdrdθ xn dx ¼

xnþ1 þc nþ1

un du ¼

unþ1 þc nþ1

sin xdx ¼ - cos x þ c 8.22. Based on the information given in the problem, the center of mass of the region below with a uniform density is requested. D ¼ ðx, yÞ x2 þ y2 ≤ a2 , x ≥ 0 The center of mass of a flat region can be calculated as follows.

x¼

xρðx, yÞdA D

ρðx, yÞdA D

y¼

yρðx, yÞdA D

ρðx, yÞdA D

The region of D is a half-circle with the characteristics below.

8.2 Applications of Double Integrals

101

π π ≤θ≤ 2 2

D: 0 ≤ r ≤ a, -

The region of D is symmetric with respect to x-axis. Therefore: y¼0 On the other hand, density is uniform. Therefore for x we have: xdA x¼

D

dA D

ðrcos θÞrdrdθ

)x¼

D

¼

r 2 dr cos θdθ D

rdrdθ

rdrdθ

D

1 3 3r

a 0

)x¼ 1 2 2r

a 0

D

π sin θ 2 π 2 π θ 2π 2

1 3 2 3 a ð1 - ð- 1ÞÞ a ¼ 13 2 π ¼ 13 2 π a π a 2 2 2 2

)x¼ Therefore, the center of mass of the region is

4 3π a, 0

4 a 3π

. Choice (1) is the answer.

In this problem, the rules below have been used. x ¼ r cos θ y ¼ r sin θ r¼

x 2 þ y2

dxdy ¼ rdrdθ xn dx ¼

xnþ1 þc nþ1

cos θdθ ¼ sin θ þ c

102

8 Solutions of Problems: Double Integrals and Their Applications

8.23. Based on the information given in the problem, distance of center of gravity of a uniform thin region from x-axis is requested which is y. The region is restricted by the curve of y ¼ x2, x-axis, and the line of x ¼ 2. The center of mass of a flat region can be calculated as follows. xρðx, yÞdA

x¼

D

ρðx, yÞdA D

yρðx, yÞdA

y¼

D

ρðx, yÞdA D

Therefore: 2

x2

2

ydydx y¼

0

¼

0 x2

2

0

2

dydx 0

y

0

0 2

)y¼

ðyÞ2 2

0

x4 2

dx ¼

2

x5 10 x3 3

x2 dx 0

x2 0

dx

x2 dx 0

2 0 32 6 ¼ ¼ 10 8 5 2 3 0

) y ¼ 1:2 Choice (4) is the answer. In this problem, the rule below has been used. xn dx ¼

xnþ1 þc nþ1

8.24. Based on the information given in the problem, the moments inertia of a circular plane with a center at the origin, radius of two (r ¼ 2), and density of unity (ρ(x, y) ¼ 1) with respect to x-axis (Ix) and the origin (Io) are requested. The moments inertia of a flat object with respect to x-axis, y-axis, and the origin can be calculated as follows. Ix ¼

y2 ρðx, yÞdA D

Iy ¼

x2 ρðx, yÞdA D

x2 þ y2 ρðx, yÞdA

Io ¼ D

8.2 Applications of Double Integrals

103

Therefore: 2π

Ix ¼

2

0

ðrsin θÞ2 rdrdθ

0 2π

) Ix ¼

2

sin 2 θdθ

r 3 dr

0 2π

) Ix ¼ 0

) Ix ¼

) Ix ¼

0 2

1 ð1 - cos 2θÞdθ 2

1 sin 2θ θ2 2

r 3 dr 0

2π

 0

r4 4

2 0

24 -0 4

1 sin 4π sin 0 2π -0 þ 2 2 2

1 ¼ ð2π Þð4Þ 2

) I x ¼ 4π 2π

Io ¼

2

r 2 rdrdθ 0

0 2π

) Io ¼

2

r 3 dr

dθ 0

) I o ¼ ½θ2π 0 

0

2

r4 4

¼ ð2π - 0Þ 0

24 -0 4

) I o ¼ 8π Choice (1) is the answer. In this problem, the rules below have been used. y ¼ r sin θ r¼

x 2 þ y2

dxdy ¼ rdrdθ xn dx ¼

xnþ1 þc nþ1

1 sin 2 θ ¼ ð1 - cos 2θÞ 2 cos aθdθ ¼

1 sin θ þ c a

sin 4π ¼ 0 sin 0 ¼ 0

¼ 2π  4

104

8 Solutions of Problems: Double Integrals and Their Applications

References 1. Rahmani-Andebili, M. (2023). Calculus II – Practice Problems, Methods, and Solutions, Springer Nature. 2. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 3. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature. 4. Rahmani-Andebili, M. (2023). Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 5. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature.

9

Problems: Triple Integrals and Their Applications

Abstract

In this chapter, the basic and advanced problems of Triple Integrals and their applications are presented. The subjects include definite and indefinite triple integrals, volume, mass, density function, and three-dimensional coordinate system such as cartesian, cylindrical, and spherical coordinate systems. 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.

9.1

Triple Integrals

9.1. Calculate the value of definite triple integral below if the region of S is restricted by coordinate planes and the plane of x + y + z ¼ 1 [1–5]. z2 dxdydz V

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 30 1 2) 25 1 3) 60 1 4) 20 9.2. Calculate the value of following definite multiple integral. π 2

π 0

0

x

sin x cosðsin yÞdydxdz

0

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) -π sin 1 2) -π cos 1 3) π sin 1 4) π cos 1 # The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus III, https://doi.org/10.1007/978-3-031-47483-5_9

105

106

9

Problems: Triple Integrals and Their Applications

9.3. Calculate the value of following definite triple integral in which the region of V is in the first one-eighth of threedimensional coordinate system (first octant) restricted by the planes of y ¼ 5 and x + z ¼ 2. xdV V

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 3 1) 10 3 2) 20 10 3) 3 20 4) 3 9.4. Calculate the value of following definite triple integral.

x2 þy2 þz2 ≤ a2

ð2 þ x þ sin zÞdxdydz

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) πa3 2) 2πa3 4 3) πa3 3 8 4) πa3 3 9.5. Calculate the value of following definite triple integral in which V ¼ {(x, y, z)|x2 + y2 + z2 ≤ 4, z ≥ 0}. ð3 þ 2xyÞdxdydz V

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large 1) 16π 2) 16π 2 2π 3) 3 4) 32π

9.2

Applications of Triple Integrals

9.6. Calculate the volume restricted by the planes of z ¼ 0, x + z ¼ 1, and x ¼ y2. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large

References

107

4 15 16 2) 15 8 3) 3 8 4) 15

1)

9.7. Calculate the value of following definite triple integral. p

p 1 - z2

1

1 - y2 - z 2

dxdydz 0

0

0

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ● Small ○ Normal ○ Large π 1) 3 π 2) 6 3π 3) 8 3π 4) 4 9.8. Calculate the mass of a spherical star with the radius of “a” and density function below. ρðx, y, zÞ ¼ 30e -

3 x2 þy2 þz2 2 a2

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 40πa3 1 - 1e 2) 20πa3 1 - 1e 3) 40πa3 1 þ 1e 4) 20πa3 1 þ 1e 9.9. Calculate the mass of an object in the region of 1 ≤ x2 + y2 + z2 ≤ 4 and with the density function of ρðx, y, zÞ ¼ x2 þ y2 þ z2 . Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 3π 2 2) 15π 14π 2 3) 3 4) 16π

References 1. Rahmani-Andebili, M. (2023). Calculus II – Practice Problems, Methods, and Solutions, Springer Nature. 2. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 3. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature. 4. Rahmani-Andebili, M. (2023). Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 5. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature.

Solutions of Problems: Triple Integrals and Their Applications

10

Abstract

In this chapter, the problems of the ninth chapter are fully solved, in detail, step-by-step, and with different methods.

10.1

Triple Integrals

10.1. Based on the information given in the problem, the value of definite triple integral below is requested [1–5]. I=

z2 dxdydz V

Herein, the region of S is restricted by coordinate planes and the plane of x + y + z = 1. Therefore: 1-x

1

I= 0

0

0

0

0 1

)I= 0 1

)I=

-

0

z3 1 - x - y dydx 3 0

1-x

1

)I=

z2 dzdydx

0 1-x

1

)I=

1-x-y

0

-

ð 1 - x - yÞ 3 dydx 3

ð 1 - x - yÞ 4 1 - x dx 12 0

ð 1 - x - ð 1 - x Þ Þ 4 ð 1 - x - 0Þ 4 þ dx 12 12 1

)I= 0

- ð 1 - xÞ 5 1 ð1 - xÞ4 dx = 12 60 0 )I=

- 05 1 5 þ 60 60

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus III, https://doi.org/10.1007/978-3-031-47483-5_10

109

110

10

)I=

Solutions of Problems: Triple Integrals and Their Applications

1 60

Choice (3) is the answer. In this problem, the rule below has been used. un du =

unþ1 þc nþ1

10.2. Based on the information given in the problem, the value of following definite triple integral is requested. π 2

π

I= 0

x

0

π 2

π

)I=

0

0

0

π - cos x 2 y

sin x cosðsin yÞdydx

0

π 2

) I =π π 2

x

dz 0

) I =π

sin x cosðsin yÞdydxdz

0

π 2

cosðsin yÞ sin x dxdy

y π 2

cosðsin yÞdy = π

- cos

0 π 2

) I =π

π þ cos y cosðsin yÞdy 2

cos y cosðsin yÞdy

0

π π ) I = π sinðsin yÞ 2 = π sin sin - sinðsin 0Þ = π ðsin 1 - sin 0Þ 2 0 I = π sin 1 Choice (3) is the answer. In this problem, the rules below have been used. xn dx =

xnþ1 þc nþ1

sin u du = - cos u þ c cos u du = sin u þ c sin

π =1 2

sin 0 = 0

10.1

Triple Integrals

111

10.3. Based on the information given in the problem, the value of following definite triple integral is requested. I=

xdV V

Herein, the region of V is in the first one-eighth of three-dimensional coordinate system (first octant) restricted by the planes of y = 5 and x + z = 2. Therefore: 2

I=

xdzdydx 0

2

)I=

5

x z 0

0

2-x 0

2-x

5 0

0 2

dydx =

5

0 2

)I=

2x - x2

y

0

I = 5x2 -

5 3 x 3

2

xð2 - xÞdydx =

0

0

5 dx = 0

2

5

2x - x2 dydx

0

5 2x - x2 dx

0

2 5 = 5 × 4 - × 8 - ð 0 - 0Þ 3 0 )I=

20 3

Choice (4) is the answer. In this problem, the rule below has been used. xn dx =

xnþ1 þc nþ1

10.4. Based on the information given in the problem, the value of following definite triple integral is requested. I=

x2 þy2 þz2 ≤ a2

ð2 þ x þ sin zÞdxdydz

The problem can be solved as follows: I=

x2 þy2 þz2 ≤ a2

2dxdydz þ

x2 þy2 þz2 ≤ a2

xdxdydz þ

sin zdxdydz x2 þy2 þz2 ≤ a2

As can be noticed, the region is a sphere with the radius of “a” which is a symmetric region. Moreover, the functions of the second and third integrals are odd with respect to their variables. Therefore, those integrals are zero. )I=

x2 þy2 þz2 ≤ a2

2dxdydz = 2

dxdydz x2 þy2 þz2 ≤ a2

112

10

) I =2× )I=

Solutions of Problems: Triple Integrals and Their Applications

4 3 πa 3

8 3 πa 3

Choice (4) is the answer. In this problem, the rule below has been used. If f (x, y, z) is an odd function and V is a symmetric region, then: f ðx, y, zÞdV = 0 V

10.5. Based on the information given in the problem, the value of following definite triple integral is requested. ð3 þ 2xyÞdxdydz

I= V

Herein, V = {(x, y, z)jx2 + y2 + z2 ≤ 4, z ≥ 0}. The problem can be solved as follows: 3dxdydz þ

I=

2xydxdydz

V

V

As can be noticed, the region is the top half-sphere with the radius of two which is a symmetric region with respect to x and y. Moreover, the function of the second integral is odd with respect to its variables. Therefore, that integral is zero. 3dxdydz þ 0 = 3

)I= V

dxdydz = 3 × V

I = 16π Choice (1) is the answer. In this problem, the rule below has been used. If f (x, y, z) is an odd function and V is a symmetric region, then: f ðx, y, zÞdV = 0 V

1 4 × π ð 2Þ 3 2 3

10.2

10.2

Applications of Triple Integrals

113

Applications of Triple Integrals

10.6. Based on the information given in the problem, the volume restricted by the planes of z = 0, x + z = 1, and x = y2 is requested. The problem can be solved as follows: p x

1

V=

)V= 0 1

)V= 0

dzdydx

p - x 0

0 1

1-x

p x

p

1

x

ð1 - xÞ½y - px dx = ) V =2

p

1

½z10 - x dydx = p - x

x

p - x

0

ð1 - xÞdydx

p 2 xð1 - xÞdx = 2

0

1

1

3

x2 - x2 dx

0

2 32 2 52 x - x 3 5

1 0

)V=

=2

2 2 -0 3 5

8 15

Choice (4) is the answer. In this problem, the rule below has been used. xn dx =

xnþ1 þc nþ1

10.7. Based on the information given in the problem, the value of definite triple integral below is requested. I=

p

p 1 - z2

1

1 - y2 - z 2

dxdydz 0

0

0

From the boundaries of the variables, it is noticed that the value of integral is equal to the volume of part of a sphere (with radius of one) located in the first one-eighth of three-dimensional coordinate system (first octant). Therefore: I=

1 4 π × 13 8 3 )I=

π 6

Choice (2) is the answer. 10.8. Based on the information given in the problem, the mass of a spherical star with the radius of “a” and density function below is requested. ρðx, y, zÞ = 30e -

3 x2 þy2 þz2 2 a2

The mass of an object with the density function of ρ(x, y, z) can be calculated as follows.

114

10

M=

Solutions of Problems: Triple Integrals and Their Applications

ρðx, y, zÞdV V

Therefore: 3 x2 þy2 þz2 2 a2

30e -

M=

dxdydz

V

The problem can be easily solved in spherical coordinate system as follows. π

2π

M= 0

a

0

r3

30e - a3 × r 2 dr sin φdφdθ

0 π

2π

) M = 30

dθ 0

a

sin φdφ

0

0

½ - cos φπ0

) M = 30½θ2π 0

r3

r 2 e - a3 dr

-

) M = ð30 × 2π Þð- cos π þ cos 0Þ × -

) M = ð60π Þð2Þ × -

a3 - ar33 e 3

a 0

a3 - aa33 - e0 e 3

a3 - 1 e -1 3

) M = 40πa3 1 -

1 e

Choice (1) is the answer. In this problem, the rules below have been used. r=

x2 þ y 2 þ z 2

xn dx =

xnþ1 þc nþ1

sin x dx = - cos x þ c eu du = eu þ c cos π = - 1 cos 0 = 1 The integral concerned with the volume calculations in the cartesian, cylindrical, and spherical coordinate systems are as follows, respectively. V=

f ðx, y, zÞdxdydz = V

f ðR, φ, zÞRdRdφdz = V

f ðr, φ, θÞr 2 dr sin φdφdθ V

10.2

Applications of Triple Integrals

115

10.9. Based on the information given in the problem, the mass of an object in the region of 1 ≤ x2 + y2 + z2 ≤ 4 and with the density function below is requested. x2 þ y2 þ z 2

ρðx, y, zÞ =

As can be noticed, the region is a spherical shell with the inner and outer radii of one and two. Therefore, applying spherical coordinate system is the best choice. The mass of an object with the density function of ρ(x, y, z) can be calculated as follows. ρðx, y, zÞdV

M= V

Therefore: x2 þ y2 þ z2 dzdydx

M= V π

2π

)M= 0

0

2

π

2π

)M=

r × r 2 dr sin φdφdθ

1

dθ 0

2

sin φdφ

r 3 dr

0

) M = ½θ2π 0

1

½ - cos φπ0

1 4 r 4

) M = ð2π - 0Þð- cos π þ cos 0Þ × ) M = 2π × 2 ×

1 × 15 4

) M = 15π Choice (2) is the answer. In this problem, the rules below have been used. r=

x 2 þ y2 þ z 2

xn dx =

xnþ1 þc nþ1

sin xdx = - cos x þ c cos π = - 1 cos 0 = 1

2 1

1 4 2 - 14 4

116

10

Solutions of Problems: Triple Integrals and Their Applications

The integral concerned with the volume calculations in the cartesian, cylindrical, and spherical coordinate systems are as follows, respectively. f ðx, y, zÞdxdydz =

V= V

f ðR, φ, zÞRdRdφdz = V

f ðr, φ, θÞr 2 dr sin φdφdθ V

References 1. Rahmani-Andebili, M. (2023). Calculus II – Practice Problems, Methods, and Solutions, Springer Nature. 2. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 3. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature. 4. Rahmani-Andebili, M. (2023). Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 5. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature.

Problems: Line Integrals and Their Applications

11

Abstract

In this chapter, the basic and advanced problems of line integrals and their applications are presented. The subjects include parametric curve, work done by a force on a moving object, vector force, conservative force, Green’s theorem, piecewise smooth and simple closed curve, cartesian coordinate system, and polar coordinate system. 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.

11.1

Line Integrals →

11.1. Calculate the value of line integral below from t ¼ 0 to t ¼ π2 where C is a parametric curve with the equation of →

→

ðcos t Þ i þ ðsin t Þ j [1–5]. xyds C

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 12 2) 1 3) 2 4) 3 p p → 11.2. Calculate the value of line integral below from 2, 2, 0 to (0, 0, 2) where C is the common part of the surface of y ¼ x and the sphere of x2 + y2 + z2 ¼ 4 in the first one-eighth of three-dimensional coordinate system (first octant). ðx þ y þ zÞds C

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large p 1) 4 þ 4 2 p 2) 4 2 p 3) 4 þ 2 2 p 4) 2 2 þ 2 # The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus III, https://doi.org/10.1007/978-3-031-47483-5_11

117

118

11

11.3. Calculate the value of line integral of

→ C

Problems: Line Integrals and Their Applications

→

→

F :d r in which C ðt Þ is a straight line from (0, 0, 0) to (1, 1, 1) and we have:

→

→

→

→

F ¼ x2 - y i þ y2 - z j þ z 2 - x k

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 1 2) 2 1 3) 2 4) -1 11.4. Calculate the value of line integral of and we have:

→ C

→

→

F :d r in which C ðt Þ is part of the parabola of y ¼ 2x2 from (0, 0) to (1, 2) →

→

→

F ¼ 3xy i - y2 j

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 25 1) 6 7 2) 6 25 3) 6 7 4) 6 →

11.5. Calculate the value of line integral below in which C is two straight lines from (0, 0) to (π, 0) and from (π, 0) to π, e - x ðcos ydx - sin ydyÞ C

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) -1 + e-π 2) 1 - 2e-π 3) 2 - e-π 4) 1 - e-π 11.6. Calculate the value of line integral of →

→ C

→

→

→

F :d r in which C : 2

→

2

→

→

→

→

F ðx, y, zÞ ¼ yex i þ xey j þ cosh xy2 k

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) ee - 1 2) ee - e 2 3) ee - 1 2 4) ee - e

→

r ð t Þ ¼ e t i þ et j þ 3 k , 0 ≤ t ≤ 1

π 2

.

11.2

Applications of Line Integrals

119 →

11.7. Calculate the value of line integral below in which C is two straight lines from (0, 0) to (0, -1) and from (0, -1) to (1, -1). yx2 dx þ ðx þ yÞdy C

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1 1) 3 1 2) 2 2 3) 3 1 4) 6

11.2

Applications of Line Integrals →

→

→

11.8. Calculate the amount of work done by the force of F ¼ 3x2 i þ xy j on a moving object through the curve of y ¼ 4x2 from (0, 0) to (1, 4). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 12 1) 5 27 2) 5 32 3) 5 37 4) 5 →

→

→

→

11.9. Calculate the amount of work done by the force of F ðx, y, zÞ ¼ ðy2 - z2 Þ i þ 2yz j - y k on a moving object through → the path of r ðt Þ ¼ ðt, t 2 , t 3 Þ from t ¼ 0 to t ¼ 1. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 7 1) 20 1 2) 35 5 3) 4 6 4) 7 11.10. Determine the relations between the parameters of “a”, “b”, and “c” so that the line integral of of the path. →

→

→

→

F ðx, y, zÞ ¼ ay2 þ 2czx i þ yðbx þ czÞ j þ ay2 þ cx2 k

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

B→ → A F :d r

is independent

120

11

1) 2) 3) 4)

Problems: Line Integrals and Their Applications

c¼b¼a c ¼ b ¼ 2a c ¼ 2b ¼ a c ¼ 2b ¼ 2a

11.11. Determine the value of λ so that the line integral below is independent of the path. B

z2 dx þ 2ydy þ λxzdz

A

Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) λ ¼ 0 2) λ ¼ 1 3) λ ¼ 2 4) λ ¼ - 1 →

→

→

11.12. Calculate the amount of work done by the force of F ðx, yÞ ¼ ðx2 þ yÞ i - ðy2 þ 2xÞ j on a moving object through the circular path of x2 + y2 ¼ 9 in counterclockwise direction. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 0 2) 27π 3) -27π 4) 9π →

→

→

11.13. Calculate the amount of work done by the force of F ðx, yÞ ¼ ðsin yÞ i þ xð1 þ cos yÞ j on a moving object through the circular path of x2 + y2 ¼ 9 in counterclockwise direction. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 18π 2) 12π 3) 9π 4) 6π →

11.14. Calculate the line integral below in which C is the circle of x2 + y2 ¼ 9 in counterclockwise direction. 3y - e sin x dx þ 7x þ C

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 12π 2) 24π 3) 36π 4) 48π

1 þ y4 dy

11.2

Applications of Line Integrals

121 →

11.15. Calculate the line integral below. The circular path of C is shown in Fig. 11.1. I¼

y3 dx - x3 dy C

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 35 1) - π 3 2) -12π 45 3) - π 4 4) -11π

Figure 11.1 The graph of problem 11.15 →

11.16. Calculate the line integral below where the circular path of C is shown in Fig. 11.2. 2x2 - y2 dx þ x2 þ y2 dy C

Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 8 1) 3 4 2) 3 2 3) 3 4) 0

Figure 11.2 The graph of problem 11.6

122

11

Problems: Line Integrals and Their Applications

References 1. Rahmani-Andebili, M. (2023). Calculus II – Practice Problems, Methods, and Solutions, Springer Nature. 2. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 3. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature. 4. Rahmani-Andebili, M. (2023). Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 5. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature.

12

Solutions of Problems: Line Integrals and Their Applications

Abstract

In this chapter, the problems of the eleventh chapter are fully solved, in detail, step-by-step, and with different methods.

12.1

Line Integrals

12.1. Based on the information given in the problem, the value of line integral below from t ¼ 0 to t ¼ I¼

π is requested [1–5]. 2

xyds C

→

→

→

Herein, C is a parametric curve with the equation of ðcos t Þ i þ ðsin t Þ j . →

From C , it is seen that: xðt Þ ¼ cos t yðt Þ ¼ sin t →

→

→

The line integral of the scalar function of f (x, y) through the curve of C ðt Þ ¼ xðt Þ i þ yðt Þ j from the point of t ¼ a to t ¼ b can be calculated as follows. f ðx, yÞds ¼

f ðxðt Þ, yðt ÞÞ

C

C

x0t

2

2

þ y0t dt

Therefore: xyds ¼

I¼

cos t sin t

C π 2

)I¼ 0

)I¼

1 1 - cos 2t 2 2

ð- sin t Þ2 þ ðcos t Þ2 dt

C

π 2

0

1 1 sin 2t  1  dt ¼ 2 2 ¼

π 2

sin 2tdt 0

1 1 π 1 1 1 1 þ cosð0Þ ¼ - cos 2  þ 2 2 2 2 2 2 2

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus III, https://doi.org/10.1007/978-3-031-47483-5_12

123

124

12

)I¼

Solutions of Problems: Line Integrals and Their Applications

1 2

Choice (1) is the answer. In this problem, the rules below have been used. d cos t ¼ - sin t dt d sin t ¼ cos t dt cos t sin t ¼

1 sin 2t 2

ðsin t Þ2 þ ðcos t Þ2 ¼ 1 sin atdt ¼ -

1 cos at þ c a

cos π ¼ - 1 cos 0 ¼ 1 12.2. Based on the information given in the problem, the value of line integral below from requested.

p p 2, 2, 0 to (0, 0, 2) is

ðx þ y þ zÞds

I¼ C →

Herein, C ðt Þ is the common part of the surface of y ¼ x and the sphere of x2 + y2 + z2 ¼ 4 in the first one-eighth of threedimensional coordinate system (first octant). →

From C ðt Þ, it is seen that: xð t Þ ¼ yð t Þ ¼

p p

2 cos t 2 cos t

zðt Þ ¼ 2 sin t 0≤t≤

π 2 →

→

→

→

The line integral of the scalar function of f (x, y, z) through the curve of C ðt Þ ¼ xðt Þ i þ yðt Þ j þ zðt Þ k from the point of t ¼ a to t ¼ b can be calculated as follows. f ðx, y, zÞds ¼ C

f ðxðt Þ, yðt Þ, zðt ÞÞ C

x0t

2

þ y0t

2

2

þ z0t dt

12.1

Line Integrals

125

Therefore: I¼

p

ðx þ y þ zÞds ¼ C

2cos t þ

p

p - 2 sin t

2 cos t þ 2 sin t

2

p þ - 2 sin t

2

þ ð2cos t Þ2 dt

C

p 2 2cos t þ 2 sin t

)I¼

4ðsin t Þ2 þ 4ðcos t Þ2 dt

C π 2

)I¼2

p 2 2cos t þ 2 sin t dt

0

p ) I ¼ 2 2 2sin t - 2 cos t

π p p π π 2 ¼ 2 2 2sin - 2 cos - 2 2 2sin 0 - 2 cos 0 2 2 0

p ) I ¼ 2 2 2 - 0 - 2 ð 0 - 2Þ p )I ¼4 2þ4 Choice (1) is the answer. In this problem, the rules below have been used. d cos t ¼ - sin t dt d sin t ¼ cos t dt ðsin t Þ2 þ ðcos t Þ2 ¼ 1 cos tdt ¼ sin t þ c sin tdt ¼ - cos t þ c sin

π ¼1 2

cos

π ¼0 2

sin 0 ¼ 0 cos 0 ¼ 1 12.3. Based on the information given in the problem, the value of line integral of →

→

→

→

F ¼ x2 - y i þ y2 - z j þ z 2 - x k

→

Herein, C ðt Þ is a straight line from (0, 0, 0) to (1, 1, 1).

→ C

→

F :d r is requested, where:

126

12

Solutions of Problems: Line Integrals and Their Applications

Therefore: →

→

→

→

C ðt Þ ¼ ðt, t, t Þ ¼ t i þ t j þ t k , 0 ≤ t ≤ 1 →

→

→

→

→

The line integral of the vector function of F through the curve of C ðt Þ ¼ xðt Þ i þ yðt Þ j þ zðt Þ k from the point of t ¼ a to t ¼ b can be calculated as follows. →

b→

→

F:d r ¼

→0

→

F C ðt Þ : C ðt Þdt

a

C

Therefore: →

b→

→

F:d r ¼

I¼

→

1

→0

F C ðt Þ : C ðt Þdt ¼

a

C

→

→

→

→

→

→

t 2 - t i þ t 2 - t j þ t 2 - t k : i þ j þ k dt

0 1

)I¼

t 2 - t þ t 2 - t þ t 2 - t dt

0 1

)I¼3

t 2 - t dt ¼ 3

0

)I¼3

t3 t2 3 2

1 0

1 1 1 ¼3 3 2 6

)I¼ -

1 2

Choice (3) is the answer. In this problem, the rules below have been used. d n x ¼ nxn - 1 dx xn dx ¼

xnþ1 þc nþ1

12.4. Based on the information given in the problem, the value of line integral of →

→

→ C

→

F : d r is requested, where:

→

F ¼ 3xy i - y2 j

→

In this regard, C ðt Þ is part of the parabola of y ¼ 2x2 from (0, 0) to (1, 2). Therefore: →

→

→

C ðt Þ ¼ t i þ 2t 2 j

→

→

→

→

→

The line integral of the vector function of F through the curve of C ðt Þ ¼ xðt Þ i þ yðt Þ j þ zðt Þ k from the point of t ¼ a to t ¼ b can be calculated as follows.

12.1

Line Integrals

127 →

b→

→

F:d r ¼

→0

→

F C ðt Þ : C ðt Þdt

a

C

Therefore: →

b→

→

F:d r ¼

I¼

1

→0

→

F C ðt Þ : C ðt Þdt ¼

a

C 1

)I¼

→

→

→

→

6t 3 i - 4t 4 j : i þ 4t j dt

0

6t 3 - 16t 5 dt ¼ 6 

0

t4 t6 - 16  4 6

)I¼ -

1

¼ 0

6 16 18 - 32 ¼ 4 6 12

7 6

Choice (2) is the answer. In this problem, the rules below have been used. d n x ¼ nxn - 1 dx xn dx ¼

xnþ1 þc nþ1

12.5. Based on the information given in the problem, the value of line integral below is requested. e - x ðcos ydx - sin ydyÞ

I¼ C →

Herein, C is two straight lines from (0, 0) to (π, 0) and from (π, 0) to π,

π . 2

Therefore: ! C 1 ðt Þ ¼ ð0, 0Þ þ ðπ, 0Þt ¼ ðt, 0Þ, 0 ≤ t ≤ π ! π π C 2 ðt Þ ¼ ðπ, 0Þ þ π, t ¼ ðπ, t Þ, 0 ≤ t ≤ 2 2 e - x ðcos ydx - sin ydyÞ þ

)I¼ C1 π

)I¼

π 2

e - t ðcos 0dt - sin 0ð0ÞÞ þ

0

e - π ðcos t ð0Þ - sin tdt Þ

0 π

)I¼

e 0

) I ¼ - e-t

e - x ðcos ydx - sin ydyÞ C2

-t

π 2

dt þ

e - π ð- sin t Þdt

0

π π π þ e - π cos t 2 ¼ - e - π þ e0 þ e - π cos - e - π cos 0 2 0 0

128

12

Solutions of Problems: Line Integrals and Their Applications

) I ¼ 1 - 2e - π Choice (2) is the answer. In this problem, the rules below have been used. cos 0 ¼ 1 sin 0 ¼ 0 eu du ¼ eu þ c sin xdx ¼ - cos x þ c e0 ¼ 1 cos

π ¼0 2

cos 0 ¼ 1 12.6. Based on the information given in the problem, the value of line integral of →

2

→

2

→ C

→

F : d r is requested, where:

→

→

F ðx, y, zÞ ¼ yex i þ xey j þ cosh xy2 k

→

→

→

C:

→

→

r ð t Þ ¼ e t i þ et j þ 3 k , 0 ≤ t ≤ 1 →

→

→

→

→

The line integral of the vector function of F through the curve of C ðt Þ ¼ xðt Þ i þ yðt Þ j þ zðt Þ k from the point of t ¼ a to t ¼ b can be calculated as follows. →

b→

→

F:d r ¼

→0

→

F C ðt Þ : C ðt Þdt

a

C

Therefore: →

b→

→

F:d r ¼

I¼ C

→

1

→0

F C ðt Þ : C ðt Þdt ¼

a

2t

→

2t

0 1

)I¼

2t

1

2t

et ee et þ et ee et þ 0 dt ¼ 2

0

2t

2

1 0

) I ¼ ee - e

2t

e2t ee dt 0

) I ¼ ee

Choice (4) is the answer.

→

→

→

→

→

et ee i þ et ee j þ cosh et e2t k : et i þ et j þ 0 k dt

12.1

Line Integrals

129

In this problem, the rules below have been used. d x e ¼ ex dx eu du ¼ eu þ c 12.7. Based on the information given in the problem, the value of line integral below is requested. I¼

yx2 dx þ ðx þ yÞdy C

→

Herein, C is two straight lines from (0, 0) to (0, -1) and from (0, -1) to (1, -1). Therefore: ! C1 ðt Þ ¼ ð0, 0Þ þ ð0, - 1Þt ¼ ð0, - t Þ, 0 ≤ t ≤ 1 ! C2 ðt Þ ¼ ð0, - 1Þ þ ð1, 0Þt ¼ ðt, - 1Þ, 0 ≤ t ≤ 1 )I¼

yx2 dx þ ðx þ yÞdy þ

yx2 dx þ ðx þ yÞdy

C1 1

)I¼

C2 1

ð- t Þ02 ð0Þ þ ð0 þ ð- t ÞÞð- dt Þ þ

0

ð- 1Þt 2 dt þ ðt þ ð- 1ÞÞð0Þ

0 1

)I¼

1

tdt þ

0

)I¼

t2 2

1

þ 0

- t 2 dt

0

t3 3

)I¼

1

¼ 0

1 1 -0- þ 0 2 3

1 6

Choice (4) is the answer. In this problem, the rules below have been used. xn dx ¼

xnþ1 þc nþ1

130

12.2

12

Solutions of Problems: Line Integrals and Their Applications

Applications of Line Integrals

12.8. Based on the information given in the problem, the amount of work done by the force below on a moving object through the curve of y ¼ 4x2 from (0, 0) to (1, 4) is requested. →

→

→

F ¼ 3x2 i þ xy j →

→

The work done by the force of F through the curve of C is equal to the line integral of the force fucntion through the curve. In other words: →

b→

→

F:d r ¼

W¼

→0

→

F C ðt Þ : C ðt Þdt

a

C →

Based on the problem, C ðt Þ can be determined as follows. →

→

→

C ðt Þ ¼ t i þ 4t 2 j Therefore: 1

W¼

→

→

→

→

3t 2 i þ 4t 3 j : i þ 8t j dt

0 1

)W¼

3t 2 þ 32t 4 dt ¼ t 3 þ

0

)W¼

1

32 4 t 5

0

¼1þ

32 -0 5

37 5

Choice (4) is the answer. In this problem, the rules below have been used. d n x ¼ nxn - 1 dx xn dx ¼

xnþ1 þc nþ1

12.9. Based on the information given in the problem, the amount of work done by the following force on a moving object → through the path of r ðt Þ ¼ ðt, t 2 , t 3 Þ from t ¼ 0 to t ¼ 1 is requested. →

→

→

→

F ðx, y, zÞ ¼ y2 - z2 i þ 2yz j - y k →

→

The work done by the force of F through the curve of C is equal to the line integral of the force function through the curve. In other words: →

W¼ C →

Based on the problem, C ðt Þ is as follows.

b→

→

F :d r ¼

→

→0

F C ðt Þ : C ðt Þdt

a

12.2

Applications of Line Integrals

131 →

→

→

→

C ðt Þ ¼ t i þ t 2 j þ t 3 k Therefore: 1

W¼

2

t2

- t3

→

2

→

→

→

→

→

i þ 2t 2 t 3 j - t 2 k : i þ 2t j þ 3t 2 k dt

0 1

)W ¼

→

→

→

→

→

→

t 4 - t 6 i þ 2t 5 j - t 2 k : i þ 2t j þ 3t 2 k dt

0 1

)W¼

1

t 4 - t 6 þ 4t 6 - 3t 4 dt ¼

0

- 2t 4 þ 3t 6 dt

0 1

2t 5 3t 7 þ 5 7

)W¼ -

2 3 þ þ0 5 7

¼ 0

)W¼

1 35

Choice (2) is the answer. In this problem, the rules below have been used. d n x ¼ nxn - 1 dx xn dx ¼

xnþ1 þc nþ1

12.10. Based on the information given in the problem, the relations between the parameters of “a”, “b”, and “c” is requested so that the line integral of

→ B→ A F:d r

is independent of the path.

→

→

→

→

F ðx, y, zÞ ¼ ay2 þ 2czx i þ yðbx þ czÞ j þ ay2 þ cx2 k → B→ A F:d r

In order for the line integral of In this condition, we have:

be independent of any path, the vector force must be a conservative force.

→

→

curl F ¼ ∇  F ¼ 0 Therefore, for this problem, we can write: →

i → ∂ curl F ¼ ∂x 2 ay þ 2czx →

→

→

j ∂ ∂y yðbx þ czÞ

→

k ∂ ¼0 ∂z ay2 þ cx2 →

→

) curl F ¼ ð2ay - ycÞ i - ð2cx - 2cxÞ j þ ðyb - 2ayÞ k ¼ 0

132

12

Solutions of Problems: Line Integrals and Their Applications

2ay ¼ cy )

2a ¼ c

2cx ¼ 2cx )

b ¼ 2a

by ¼ 2ay ) c ¼ b ¼ 2a Choice (2) is the answer. In this problem, the rule below has been used.

→

→

i j → → ∂ ∂ curl F ¼ ∇  F ¼ ∇  ðF 1 , F 2 , F 3 Þ ¼ ∂x ∂y F1 F2

→

k ∂ ∂z F3

12.11. Based on the information given in the problem, the value of λ is needs to be determined so that the line integral below is independent of the path. B

I¼

z2 dx þ 2ydy þ λxzdz

A

In order for the line integral of In this condition, we have:

→ B→ A F:d r

be independent of any path, the vector force must be a conservative force.

→

→

curl F ¼ ∇  F ¼ 0 Therefore, for this problem, we can write: →

i → ∂ curl F ¼ ∂x z2

→

→

j k ∂ ∂ ¼0 ∂y ∂z 2y λxz

→

→

) curl F ¼ ð2z - λzÞ j ¼ 0 ) 2z - λz ¼ 0 )λ¼2 Choice (3) is the answer. In this problem, the rule below has been used. →

→

i j → → curl F ¼ ∇  F ¼ ∇  ðF 1 , F 2 , F 3 Þ ¼ ∂ ∂ ∂x ∂y F1 F2

→

k ∂ ∂z F3

12.2

Applications of Line Integrals

133

12.12. Based on the information given in the problem, the amount of work done by the following force on a moving object through the circular path of x2 + y2 ¼ 9 in counterclockwise direction is requested. →

→

→

F ðx, yÞ ¼ x2 þ y i - y2 þ 2x j →

→

The work done by the force of F through the curve of C is equal to the line integral of the force function through the curve. In other words: →

W¼

b→

→

F:d r ¼

→

→0

F C ðt Þ : C ðt Þdt

a

C

For this problem, we have: →

→

→

→

x2 þ y i - y2 þ 2x j : dx i þ dy j

W¼ C

)W ¼

x2 þ y dx - y2 þ 2x dy C

→

Green’s Theorem Assume that C is a piecewise smooth and simple closed curve with counterclockwise direction in →

xoy plane, and D is a flat region bounded by C . If P(x, y) and Q(x, y) are the functions in the region of D and have continuous partial derivatives there, then: Pðx, yÞdx þ Qðx, yÞdy ¼ C

D

∂ ∂ Qðx, yÞ Pðx, yÞ dA ∂x ∂y

→

Since C has the conditions of Green’s theorem, we can write: ∂ - y2 þ 2x ∂x

W¼ D

-

∂ ∂y

x2 þ y

ð- 2 - 1ÞdA ¼ - 3

)W ¼

dA

dA

D

D

) W ¼ - 3 π  32 ) W ¼ - 27π Choice (3) is the answer. 12.13. Based on the information given in the problem, the amount of work done by the force below on a moving object through the circular path of x2 + y2 ¼ 9 in counterclockwise direction is requested. →

→

→

F ðx, yÞ ¼ ðsin yÞ i þ xð1 þ cos yÞ j →

→

The work done by the force of F through the curve of C is equal to the line integral of the force function through the curve. In other words: →

b→

→

F:d r ¼

W¼ C

→

→0

F C ðt Þ : C ðt Þdt

a

134

12

Solutions of Problems: Line Integrals and Their Applications

For this problem, we have: →

W¼

→

→

→

ðsin yÞ i þ xð1 þ cos yÞ j : dx i þ dy j C

)W¼

ðsin yÞdx þ xð1 þ cos yÞdy C

→

Green’s Theorem Assume that C is a piecewise smooth and simple closed curve with counterclockwise direction in →

xoy plane, and D is a flat region bounded by C . If P(x, y) and Q(x, y) are the functions in the region of D and have continuous partial derivatives there, then: Pðx, yÞdx þ Qðx, yÞdy ¼ C

D

∂ ∂ Qðx, yÞ Pðx, yÞ dA ∂x ∂y

→

Since C has the conditions of Green’s theorem, we can write: W¼ D

∂ ∂ ðxð1 þ cos yÞÞ ðsin yÞ dA ∂x ∂y ð1 þ cos y - cos yÞdA ¼

)W¼ D

dA D

) W ¼ π  32 ) W ¼ 9π Choice (3) is the answer. In this problem, the rule below has been used. ∂ ðsin yÞ ¼ cos y ∂y 12.14. Based on the information given in the problem, the line integral below is requested. 3y - e sin x dx þ 7x þ

I¼

1 þ y4 dy

C →

Herein, C is the circle of x2 + y2 ¼ 9 in counterclockwise direction. →

Green’s Theorem Assume that C is a piecewise smooth and simple closed curve with counterclockwise direction in →

xoy plane, and D is a flat region bounded by C . If P(x, y) and Q(x, y) are the functions in the region of D and have continuous partial derivatives there, then: Pðx, yÞdx þ Qðx, yÞdy ¼ C

D

∂ ∂ Qðx, yÞ Pðx, yÞ dA ∂x ∂y

12.2

Applications of Line Integrals

135

→

Since C has the conditions of Green’s theorem, we can write: I¼ D

∂ 7x þ ∂x

1 þ y4 -

∂ 3y - e sin x ∂y

ð7 - 3ÞdA ¼ 4

)I¼ D

dA

dA D

) I ¼ 4 π  32 ) I ¼ 36π Choice (3) is the answer. 12.15. Based on the information given in the problem, the value of line integral below is requested where the related circular →

path of C is shown in Fig. 12.1. I¼

y3 dx - x3 dy C

→

Green’s Theorem Assume that C is a piecewise smooth and simple closed curve with counterclockwise direction in →

xoy plane, and D is a flat region bounded by C . If P(x, y) and Q(x, y) are the functions in the region of D and have continuous partial derivatives there, then: ∂ ∂ Qðx, yÞ Pðx, yÞ dA ∂x ∂y

Pðx, yÞdx þ Qðx, yÞdy ¼ C

D

→

Since C has the conditions of Green’s theorem, we can write: I¼ D

∂ 3 ∂ y - x3 ∂y ∂x

dA

x2 þ y2 dA

I ¼ -3 D

The integral can be easily solved in polar coordinate system as follows. π

I ¼ -3 0

2

π

) I ¼ -3 t4 4

)I¼ Choice (3) is the answer.

2

r 3 dr

dθ 0

) I ¼ - 3½θπ0

r 2  rdrdθ

1

1

2

¼ - 3π 4 1

- 45π 4

1 4

136

12

Solutions of Problems: Line Integrals and Their Applications

In this problem, the rules below have been used. r¼

x2 þ y 2

xn dx ¼

xnþ1 þc nþ1

The integral concerned with the area calculations in the cartesian and polar coordinate systems are as follows, respectively. f ðx, yÞdxdy ¼

A¼

f ðr, θÞrdrdθ

S

S

Figure 12.1 The graph of problem 12.15

12.16. Based on the information given in the problem, the value of line integral below is requested where the related circular →

path of C is shown in Fig. 12.2. I¼

2x2 - y2 dx þ x2 þ y2 dy C

→

Green’s Theorem Assume that C is a piecewise smooth and simple closed curve with counterclockwise direction in →

xoy plane, and D is a flat region bounded by C . If P(x, y) and Q(x, y) are the functions in the region of D and have continuous partial derivatives there, then: Pðx, yÞdx þ Qðx, yÞdy ¼ C

D

∂ ∂ Qðx, yÞ Pðx, yÞ dA ∂x ∂y

→

Since C has the conditions of Green’s theorem, we can write: I¼ D

∂ 2 ∂ x þ y2 2x2 - y2 ∂x ∂y I¼2

ðx þ yÞdA D

The integral can be easily solved in polar coordinate system as follows.

dA

12.2

Applications of Line Integrals

137 π

I¼2

1

0

ðrcos θ þ r sin θÞrdrdθ

0 π

)I¼2 0

) I ¼ 2½sin θ - cos θπ0

t3 3

1

ðcos θ þ sin θÞdθ

r 2 dr 0

1

¼ 2ðsin π - cos π - sin 0 þ cos 0Þ 0

) I ¼ 2ð0 - ð- 1Þ - 0 þ 1Þ I¼

1 -0 3

1 3

4 3

Choice (2) is the answer. In this problem, the rules below have been used. x ¼ r cos θ y ¼ r sin θ xn dx ¼

xnþ1 þc nþ1

sin π ¼ 0 cos π ¼ - 1 sin 0 ¼ 0 cos 0 ¼ 1 The integral concerned with the area calculations in the cartesian and polar coordinate systems are as follows, respectively. A¼

f ðx, yÞdxdy ¼ S

f ðr, θÞrdrdθ S

Figure 12.2 The graph of problem 12.6

138

12

Solutions of Problems: Line Integrals and Their Applications

References 1. Rahmani-Andebili, M. (2023). Calculus II – Practice Problems, Methods, and Solutions, Springer Nature. 2. Rahmani-Andebili, M. (2023). Calculus I (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 3. Rahmani-Andebili, M. (2021). Calculus – Practice Problems, Methods, and Solutions, Springer Nature. 4. Rahmani-Andebili, M. (2023). Precalculus (2nd Ed.) – Practice Problems, Methods, and Solutions, Springer Nature. 5. Rahmani-Andebili, M. (2021). Precalculus – Practice Problems, Methods, and Solutions, Springer Nature.

Index

A Analytical geometry, 1–6, 9–21 Angle, 5, 11, 17, 50, 71 B Boundaries, 87, 89–93, 97, 98, 113 Boundaries of variables, 87 C Cartesian coordinate system, 24, 31 Center of gravity, 81, 102 Center of mass, 81, 100–102 Chain rule, 56 Changing order of integration, 87, 89–93 Changing variables, 49 Characteristics equation, 5, 17 Circular curve, 24, 33 Circular object, 80, 99 Circular path, 120, 121, 133, 135, 136 Circular plane, 81, 102 Coefficient matrix, 14 Columns, 12, 19 Cone, 26, 38, 39 Conservative force, 131, 132 Continuous, 24, 31, 44 Continuous partial derivatives, 133–136 Coordinate planes, 105, 109 Counterclockwise direction, 120, 133–136 Critical points, 47, 61–65 Crossing lines, 5, 17, 40 Crossing surfaces, 49, 50, 69–71 Curl, 46, 61 Curvature, 25, 26, 34–38 Curve, 23, 27, 29, 30, 36–40, 69, 71, 80, 81, 96, 98, 102, 119, 123, 124, 126, 128, 130, 133 Curve type, 26, 40 Cylinder, 26, 38, 39, 50, 71, 80, 98 Cylindrical coordinate system, 114, 116 D Definite multiple integral, 75–79, 91–95, 105 Definite triple integral, 105–107, 109–113 Definite vector integral, 24, 32 Density, 81, 101, 102 Density function, 80, 99, 107, 113, 115 Determinant, 1, 2, 12–14 Directional derivative, 46, 58–60

Direction vectors, 17, 20, 69, 71 Discriminant, 39, 40, 62–64 Distance, 5, 6, 18, 20, 81, 102 Divergence, 50, 72, 73 Domain, 44, 55 Double integral, 75–81, 83–103 E Eigenvalue, 19, 20 Eigenvectors, 6, 19, 20 Ellipse, 27, 39, 40, 44, 55 Ellipsoid, 26, 38, 39 Elliptic paraboloid, 38, 39 Empty set, 39, 40 Equation of surface, 23, 29, 30, 38, 39 Even function, 39, 40 F First octant, 106, 111, 113, 117, 124 Force, 119, 120, 130, 133 G General equation, 38, 39, 55 Gradient, 45, 58, 60, 73 Green’s theorem, 133–136 Growth rate order, 87 H Half ellipse, 44 Hyperbola, 23, 27, 40 Hyperbolic paraboloid, 38, 39 I Image, 4, 15, 16 Implicit differentiation, 67 Independent of path, 119, 120, 131, 132 Infinite answers, 4, 14 Inner product, 16, 18 Integration by parts, 87 Integration regions, 83 Invertible, 3, 13 J Jacobian determinant, 45, 56, 57

# The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Rahmani-Andebili, Calculus III, https://doi.org/10.1007/978-3-031-47483-5

139

140 L Laplacian, 73 Limit, 43, 48, 53, 54, 65, 66 Line, 6, 17, 20, 23–27, 29–41, 76, 80, 81, 85, 86, 96, 102, 118, 119, 125, 127, 129 Linear algebra, 1–6, 9–21 Linearly dependent, 4, 14 Line integral, 117–121, 123–137 M Magnitude, 16, 18, 60 Mass, 80, 99, 107, 113, 115 Matrices, 1–3, 6, 11–14, 19, 20 Maximum directional derivative, 46, 60 Moment of inertia, 81, 102 Moving object, 119, 120, 130, 133 Multiple integral, 90 Multivariable function, 43–50, 53–74 N Nonzero matrix, 19 O Odd function, 83, 112 One-eighth of three-dimensional coordinate system, 106, 111, 113, 117, 124 One-sheet hyperboloid, 26, 38, 39 Order of integration, 87, 89–93 Outer product, 10 P Parabola, 27, 40, 118, 126 Parallel, 1, 10, 49 Parallelepiped, 3, 12 Parallel lines, 27, 40, 69 Parametric curve, 117, 123 Parametric vector function, 33 Path, 43, 53, 54, 119, 120, 130–132 Piecewise smooth and simple closed curve, 133–136 Plane, 5, 16, 18, 23, 44, 80, 97, 98, 105, 106, 109, 111, 113, 133–136 Polar coordinate systems, 24, 31, 94, 95, 99, 135–137 R Range, 48 Rank, 6, 19 Region, 44, 55, 76, 81, 85, 86, 94, 95, 99–102, 105–107, 109, 111, 112, 115, 133–136 Relative minimum points, 47, 62, 64, 65 Rotation, 23, 29, 30

Index Rotation matrix, 11 Row, 12, 19 S Saddle points, 47, 48, 62–64 Scalar function, 123, 124 Spherical coordinate systems, 114–116 Spherical shell, 115 Spherical star, 107, 113 Surface, 23–27, 29–42, 49, 67, 68, 71, 80, 97, 117, 124 Surface area, 80, 96 Surface area enclosed, 27, 40 Surface equation, 24 Surface type, 26 Symmetric, 83, 101 Symmetric region, 83, 111, 112 T Tangent line, 49, 50, 69–71 Tangent plane, 49, 67, 68 Three-dimensional coordinate system, 23–27 Three-dimensional curve, 33 Three-dimensional space, 26 Transpose matrix, 9 Two-sheet hyperboloid, 26, 38, 39 Type of critical points, 62 U Uniform density, 81, 100 Uniform thin region, 81, 102 Unit normal vector, 27, 41 Unit tangent vector, 25, 33, 41 Unit vector, 46, 58–60 V Variables separation, 86 Vector, 1, 3–6, 10, 12, 14–16, 18–20, 27, 41, 46, 58–60, 71 Vector force, 131, 132 Vector function, 23–27, 29–41, 46, 50, 61, 72, 126, 128 Vector integral, 24, 32 Volume, 3, 12, 80, 97, 98, 106, 113, 114, 116 W Work, 119, 120, 130, 133 X xoy plane, 23, 80, 97, 133–136