Integral Calculus for Engineers 9811947929, 9789811947926

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Gavriil Paltineanu Ileana Bucur Mariana Zamfir

Integral Calculus for Engineers

Integral Calculus for Engineers

Gavriil Paltineanu · Ileana Bucur · Mariana Zamfir

Integral Calculus for Engineers

Gavriil Paltineanu Department of Mathematics and Computer Science Technical University of Civil Engineering Bucharest Bucharest, Romania

Ileana Bucur Department of Mathematics and Computer Science Technical University of Civil Engineering Bucharest Bucharest, Romania

Mariana Zamfir Department of Mathematics and Computer Science Technical University of Civil Engineering Bucharest Bucharest, Romania

ISBN 978-981-19-4792-6 ISBN 978-981-19-4793-3 (eBook) https://doi.org/10.1007/978-981-19-4793-3 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 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 Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore

Preface

The basic ideas of integral calculus have been outlined since antiquity, consenting in development of methods for computing different areas and volumes. To understand the basic idea of the notion of integral, let us start from the following example. To compute the area of the circle portion lying the upper half-plane, we can proceed as follows: divide this plane set into rectangles of equal widths, with one of the bases on the axis O x and at least one end of the other base on the curve, and then add their areas. It is clear that if the number of rectangles is big, then the sum of their areas approximates well the area of the semicircle. In other words, the area of the semicircle is the limit toward which the sum of the areas of the rectangles tends, when their number tends to infinity. This limit is called integral. Archimedes (287–212 BC) used various variants of this method to calculate the areas of the circle, sphere, cone, and so on, as well as the volumes of the sphere, cone, revolution ellipsoid, and so on. The methods developed by Archimedes were, two millennia later, the basis of integral calculus. At the end of the seventeenth century, Isaac Newton (1643–1727) and Gottfried Wilhelm von Leibniz (1646–1716) formulated the basic notions and theorems of integration theory, including the fundamental theorem known as the “Leibniz–Newton formula” which allows the computation of the integrals using the antiderivatives. The full clarification of the notion of integral was reached, a century later, by the contributions of the French mathematician Augustin Cauchy (1789–1857) and the German mathematician Georg Bernhard Riemann (1826–1866). The paper is based on the lectures delivered by the authors at the Department of Mathematics and Computer Science of Technical University of Civil Engineering Bucharest, Romania. This book, which is an improved version, in English, of the book in Romanian [7], includes the following chapters: Indefinite integrals, definite integrals, improper integrals, integrals depending on parameter, line integrals, double and triple integrals, and surface integrals.

v

vi

Preface

The basic elements of integral calculus are presented, indispensable for students in higher technical education to successfully approach other theoretical or technical disciplines, such as: Mechanics, physics, strength of materials, statistics and dynamics of constructions, theory of elasticity, finite element method, mathematical geodesy, compensation of measurements and statistics, and so on. The main concern of the authors was to maintain a balance between accessibility and scientific rigor. The book contains many examples, completely solved exercises, drawings and comments designed to facilitate the understanding of notions, theorems and computation algorithms. The book is useful primarily for students in higher technical education, but also for engineers, for Ph.D. students, as well as for other specialists who use different types or methods of integration. Bucharest, Romania March 2022

Gavriil Paltineanu Ileana Bucur Mariana Zamfir

Contents

1 Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 The Notion of Primitive Function (Antiderivative) of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Basic Properties of Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . 1.3 Primitives of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Primitives of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Primitives of Irrational Functions. Binomial Integrals . . . . . . . . . . .

1 1 5 11 18 21

2 Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Area of a Curvilinear Trapezoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Darboux Sums. Definition of Definite Integral . . . . . . . . . . . . . . . . . 2.3 Integrability of Continuous and Monotonic Functions . . . . . . . . . . . 2.4 Riemann Sums. Riemann Criterion for Integrability . . . . . . . . . . . . 2.5 Lebesgue Criterion for Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Properties of Integrable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Area of a Plane Figure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Approximating Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

25 25 28 33 34 41 42 49 60

3 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Convergence and Divergence of Improper Integrals . . . . . . . . . . . . . 3.2 Convergence Criteria for Improper Integrals . . . . . . . . . . . . . . . . . . .

71 71 76

4 Integrals Depending on Parameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 4.1 Proper Integrals Depending on a Parameter . . . . . . . . . . . . . . . . . . . . 87 4.2 Improper Integrals Depending on a Parameter . . . . . . . . . . . . . . . . . 93 4.3 Euler Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 5 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Parameterized Paths. Definition of a Curve . . . . . . . . . . . . . . . . . . . . 5.2 Rectifiable Paths and Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Natural Parameterization of a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Line Integrals of the First Kind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Line Integrals of the Second Kind . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Independence on the Path of Line Integral of the Second Kind . . .

113 113 118 128 132 140 149 vii

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Contents

6 Double and Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Double Integral. Definition and Properties . . . . . . . . . . . . . . . . . . . . 6.2 Basic Properties of the Double Integral . . . . . . . . . . . . . . . . . . . . . . . 6.3 Reducing a Double Integral to an Iterated Single Integral . . . . . . . . 6.4 Change of Variables in Double Integral . . . . . . . . . . . . . . . . . . . . . . . 6.5 Applications of the Double Integral in Geometry and Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5.1 Mass of a Lamina . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5.2 Coordinates of the Center of Mass of a Lamina . . . . . . . . . 6.5.3 Moments of Inertia of a Lamina . . . . . . . . . . . . . . . . . . . . . . 6.6 Riemann–Green Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 Improper Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 Volume of a Space Figure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9 Triple Integrals. Definition and Basic Properties . . . . . . . . . . . . . . . 6.10 Computing Triple Integral. Change of Variables in Triple Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.11 Applications of the Triple Integral in Geometry and Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.11.1 Volume of a Space Domain . . . . . . . . . . . . . . . . . . . . . . . . . . 6.11.2 Mass of a Solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.11.3 Coordinates of the Center of Mass of a Solid . . . . . . . . . . . 6.11.4 Moment of Inertia of a Solid . . . . . . . . . . . . . . . . . . . . . . . . .

159 159 165 166 173

7 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Parameterized Surface Canvases. Definition of a Surface . . . . . . . . 7.2 The Area of a Smooth Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Surface Integral of the First Kind . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Surface Integral of the Second Kind . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Integral Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

215 215 223 230 234 242

180 180 181 183 185 192 195 199 202 210 210 211 211 212

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

Chapter 1

Indefinite Integrals

1.1 The Notion of Primitive Function (Antiderivative) of a Function To understand the need to introduce the notion of antiderivative of a function we will start from a simple problem of mechanics. It is known that the law of the rectilinear motion of a mobile is given by an equation of the form s = f (t) where t represents the time, and s is the distance traveled by the mobile until the moment t. The derivative f ' (t) represents the instantaneous speed of movement of the mobile at the moment t. In practice, we encounter more often the inverse problem: knowing the speed of movement of the mobile v = v(t), find the law of motion of the mobile, i.e. the relationship between the distance traveled s to time t, and the time t. Because v(t) = f ' (t), we are led to the following problem: knowing the derivative f ' (t) = v(t), find the function f (t). It is obvious that this problem is the inverse of the fundamental problem of differential calculus (which consists in finding the derivative of a given function). Here, on the contrary, we are asked to find a function knowing its derivative. Definition 1.1.1 Let I ⊂ R be an interval and f : I → R. A function F : I → R is said to be a primitive function (antiderivative) of the function f on the interval I if F is differentiable function on I and F ' (x) = f (x), ∀ x ∈ I. Theorem 1.1.1 If F : I → R is an antiderivative of the function f : I → R, then for any real constant C, the function G : I → R given by G(x) = F(x) + C, ∀ x ∈ I © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Integral Calculus for Engineers, https://doi.org/10.1007/978-981-19-4793-3_1

1

2

1 Indefinite Integrals

is also an antiderivative of f on the interval I . Moreover, any another antiderivative of f on I has this form. Proof If G = F + C on I , then G ' = F ' + 0 = F ' = f on I , so G is a primitive function of f on I . Conversely, let G be another primitive function of f on I and let H = G − F on I . Obviously, we have: H ' (x) = G ' (x) − F ' (x) = f (x) − f (x) = 0, ∀ x ∈ I. Let a ∈ I be an arbitrary fixed interior point. From Lagrange Theorem it follows that for any x ∈ I , there is a point ξ in the open interval of ends a and x such that: H (x) − H (a) = H ' (ξ )(x − a) = 0 whence it results that H (x) = H (a). Definition 1.1.2 The set of all antiderivatives of the function f : I → R is called the indefinite integral of the function f on the interval I and it is denoted by ∫ f (x) dx. If F : I → R is an antiderivative of f , then from Theorem 1.1.1, it results that ∫ f (x) dx = F(x) + C , ∀x ∈ I where C is the set of all real constant functions on I . Remark 1.1.1 If the function f : I → R admits antiderivatives on I and F : I → R is such a function, then F ' (x) = f (x) or dF(x) = f (x) dx, whence it results ∫ ∫ f (x) dx = dF(x) = F(x) + C , ∀x ∈ I. Remark 1.1.2 (1) Chapter 2 will prove that any continuous function on a real interval admits antiderivatives on that interval (Theorem 2.6.3). (2) If f ∈ C 1 (I ), then ∫

f ' (x) dx = f (x) + C , ∀x ∈ I.

1.1 The Notion of Primitive Function (Antiderivative) of a Function

Next, we recall the table of antiderivatives of the basic elementary functions: ∫ 0 dx = C ∫

x a+1 + C , x ∈ R, a /= −1 a+1  ∫ 1 ln x + C , x ∈ (0, ∞) dx = ln|x| + C = ln(−x) + C , x ∈ (−∞, 0) x ∫ ax + C , x ∈ R, a > 0, a /= 1 a x dx = ln a ∫ ex dx = ex + C , x ∈ R x a dx =

∫ sin x dx = − cos x + C , x ∈ R ∫ cos x dx = sin x + C , x ∈ R ∫ sinhx dx = coshx + C , x ∈ R ∫ coshx dx = sinhx + C , x ∈ R ∫

{π } 1 dx = tanx + C , x ∈ R\ + kπ ; k ∈ Z 2 cos x 2 ∫ 1 dx = −cotx + C , x ∈ R\{kπ ; k ∈ Z} sin2 x ∫ } {π + kπ ; k ∈ Z tanx dx = − ln|cosx| + C , x ∈ R\ 2 ∫ cotx dx = ln|sinx| + C , x ∈ R\{kπ ; k ∈ Z} ∫

| | 1 || x − a || 1 +C dx = ln x 2 − a2 2a | x + a |   x−a 1 ln x+a + C , x ∈ (−∞, −a) ∪ (a, ∞) 2a = 1  a−x , a /= 0 ln x+a + C , x ∈ (−a, a) 2a

3

4

1 Indefinite Integrals

∫

x  1 1 + C , x ∈ R, a /= 0 dx = tan−1 2 +a a a ∫

x  1 + C , x ∈ (−a, a), a > 0 dx = sin−1 √ a a2 − x 2 ∫ 

√ 1 dx = ln x + x 2 + a 2 + C , x ∈ R, a /= 0 √ x 2 + a2 ∫ | | √ 1 | | dx = ln|x + x 2 − a 2 | + C √ 2 2 x −a ⎧

 √ ⎨ ln x + x 2 − a 2 + C , x ∈ (a, ∞)

 = , a > 0. √ ⎩ ln −x − x 2 − a 2 + C , x ∈ (−∞, −a) x2

To demonstrate these formulas it is sufficient to derive the functions from the right side to obtain the functions below the integral sign; e.g.: (

x a+1 +C a+1

)'

= x a , a /= −1, (tanx + C )' =

1 and so on. cos2 x

So for now we know how to calculate the antiderivatives of a small number of elementary functions that actually appear in the table of derivatives of elementary functions. The above table of antiderivatives does not exhaust even the class of the simplest elementary functions, such as, for example, the functions ln x or tan−1 x. Indeed, so far we have not encountered any function whose derivative is ln x or tan−1 x. The problem of integration (antiderivation) is considerably more difficult than the problem of derivation. The derivation of a function is much facilitated by the definition of the derivative, which has a constructive (algorithmic) character. We recall that: f ' (x) = lim

h→0

f (x + h) − f (x) . h

E.g.:  2 ' (x + h)2 − x 2 x = lim h→0 h 2xh + h 2 = lim (2x + h) = 2x. = lim h→0 h→0 h Therefore, starting from the definition, we can find the derivatives of the basic elementary functions. In the example above the derivative of the function f (x) = x 2 is f ' (x) = 2x.

1.2 Basic Properties of Indefinite Integrals

5

Things are completely different in the case of integration.∫ The definition of the antiderivative gives us no clue how to calculate, for example ln x dx. To this difficulty is added the fact that, in the case of integration, we do not have the chain of derivation rules that allow us to start from the derivatives of several functions, to find the derivatives of their different combinations: sums, products, ratios, composed functions, inverse functions, and so on. Finally, we also mention the fact that there are functions that cannot be integrated; more precisely, there are functions that have antiderivatives, but they cannot be expressed by elementary functions. Here are some examples of such primitive functions: ∫ ∫ √ ∫ x sin x e dx, dx, 1 − ε2 · sin2 x dx, |ε| < 1, x x ∫ ∫ ∫  1 2 sin x 2 dx, e−x dx, dx. ln x Emphasizing the difficulties of integral computation, we should not be discouraged, because, even in the case of integration there are rules and methods, truly less than in the case of derivation and with a smaller area of applicability, which allow the computation of antiderivatives of some important classes of functions. In the following we will present the general computation methods for primitive functions.

1.2 Basic Properties of Indefinite Integrals The first property, very useful in applications, is the linearity property. In particular, it allows us to integrate polynomial functions. Proposition 1.2.1 (Linearity Property) Let f, g : I → R be two functions that admit antiderivatives on I and let α, β ∈ R such that α 2 + β 2 > 0. Then α · f + β · g admits antiderivative on I and ∫ ∫ ∫ f (x) dx + β · g(x) dx. (1.1) (α · f (x) + β · g(x)) dx = α · Proof The statement results from the linearity property of the derivation operation: (α · F(x) + β · G(x))' = α · F ' (x) + β · G ' (x) = α · f (x) + β · g(x), x ∈ I, where F and G are primitive functions of f and g, respectively.

6

1 Indefinite Integrals

∫ Example 1.2.1 Compute 2x 3 − 4x 2 + 5x − 7 dx. Taking into account Proposition 1.2.1, Formula (1.1), we have: ∫

∫ ∫ ∫ 2x 3 dx + −4x 2 dx + 5x dx + −7 dx ∫ ∫ ∫ ∫ 3 2 = 2 x dx − 4 x dx + 5 x dx − 7 1 dx

 3 2x − 4x 2 + 5x − 7 dx =

∫

x3 x2 x4 −4· +5· − 7x + C 4 3 2 x4 4x 3 5x 2 = − + − 7x + C . 2 3 2  ∫ 2 √ − 3 x + x14 − √4x dx. Example 1.2.2 Compute x Applying Proposition 1.2.1 again and taking into account the table of antiderivatives of the usual elementary functions we have: =2·

∫ (

2 √ 1 4 − 3x+ 4−√ x x x

)

∫ dx = 2

1 dx − x

∫

1

x 3 +1 = 2 ln x − 1 + +1 3 3√ 3 = 2 ln x − x4 − 4 1

∫

x 3 dx +

x −4 dx − 4

∫

x − 2 dx 1

x − 2 +1 x −4+1 −4· 1 +C −4 + 1 −2 + 1 √ 1 − 8 x + C , x > 0. 3 3x 1

Proposition 1.2.2 (The change of variable formula. First method) Let I, J ⊂ R be two intervals, let F : J → R be an antiderivative of the function f : J → R and let u : I → J be a differentiable function on I . Then the function ( f ◦ u) · u ' : I → R admits antiderivatives on I and ∫

f (u(x)) · u ' (x) dx = F(u(x)) + C , ∀x ∈ I.

(1.2)

Proof The statement results immediately from the rule of derivation of composed functions: (F(u(x)))' = F ' (u(x)) · u ' (x) = f (u(x)) · u ' (x), ∀x ∈ I. Remark 1.2.1 From Proposition 1.2.2, it results that in order to calculate the antiderivative of the function we can proceed as follows: If we make the substitution t = u(x), x ∈ I , then we have dt = du(x) = u ' (x) dx and further ∫ ∫ f (u(x)) · u ' (x) dx = f (t) dt = F(t) + C = F(u(x)) + C , x ∈ I.

1.2 Basic Properties of Indefinite Integrals

7

∫ ∫ We specify that equality f (u(x)) · u ' (x) dx = f (t) dt is a formal equality, because the function from the left side is defined on the interval I and the function from the right side is defined on the interval J . Example 1.2.3 Compute the following indefinite integrals: ∫ ∫ ∫ 2x − 1 u ' (x) 2 u(x) ' dx, dx, e · u (x) dx, 2xex dx, u(x) x2 − x + 4 ∫ ∫ ' cos u(x) · u (x) dx, 3x 2 cos x 3 dx.

∫

If we denote by t = u(x), then dt = u ' (x) dx, and we have: ∫

u ' (x) dx = u(x)

∫

dt = ln|t| + C = ln|u(x)| + C . t

In particular, we obtain: ∫

| 2 |  2x − 1 |x − x + 4| + C = ln x 2 − x + 4 + C . dx = ln x2 − x + 4

In the same way, we get: ∫ u(x)

e ∫

∫

'

· u (x) dx = e

u(x)

+ C and

2

2

2xex dx = ex + C ∫

'

cos u(x) · u (x) dx = sin u(x) + C and

3x 2 cos x 3 dx = sin x 3 + C .

∫ 1 ∫ √ 1 dx. Example 1.2.4 Compute x 2 +a 2 dx and a 2 −x 2 x If we denote by t = a , then dx = a dt, and thus ∫

∫ ∫ 1 1 a 1 dt = dt  x 2 2 2+1 2+1 a t a t + 1 a

x  1 −1 1 = tan t + C = tan−1 + C , x ∈ R, a /= 0. a a a

1 1 dx = 2 x 2 + a2 a

∫

1

dx =

Similarly, we have ∫

∫ ∫ 1 1 a 1 / dx = dt dx = √ √  2 2 2 a a 1 − t2 a −x 1 − ax ∫ 1 = dt = sin−1 t + C √ 1 − t2

x  + C , x ∈ (−a, a), a > 0. = sin−1 a 1

8

1 Indefinite Integrals

∫ ∫ 1 1 dx and √5−2x−x dx. Example 1.2.5 Compute √x 2 −2x+5 2 Using the canonical form of the second degree function, we have: x 2 − 2 x + 5 = (x − 1)2 + 4 and 5 − 2x − x 2 = 6 − (x + 1)2 . For the first integral, using the change of variable t = x − 1, we get dt = dx and ∫ √

1 x2

− 2x + 5

∫

1

∫

1 √ dt dx = √ 2 2 t +4 (x − 1) + 4



 √ √ = ln t + t 2 + 4 + C = ln x − 1 + x 2 − 2x + 5 + C .

dx =

Similarly, with the change of variable t = x + 1, we have dt = dx and ∫ √

1 5 − 2x − x 2

∫ dx =

√

) ( x +1 t + C. dt = sin−1 √ + C = sin−1 √ 6 6 6 − t2 1

Proposition 1.2.3 (The change of variable formula. Second method) Let I, J ⊂ R be two intervals and let u : I → J be a bijective function with the properties: u ∈ C 1 (I ), u ' (x) /= 0, ∀x ∈ I , and its inverse u −1 : J → I is a function of C 1 — class on J . Furthermore, if f : J →  R is' a continuous function and H : J → R is an antiderivative of the function f · u −1 : J → R, then the function f ◦u : I → R admits antiderivative on the interval I and ∫ f (u(x)) dx = H (u(x)) + C , ∀x ∈ I. (1.3) Proof If we compute the derivative of the identity u −1 (u(x)) = x, ∀x ∈ I , it results:  −1 ' u (u(x)) · u ' (x) = 1, ∀x ∈ I. Therefore we have: ∫ ∫  ' f (u(x)) dx = f (u(x)) · u −1 (u(x)) · u ' (x) dx ∫

 '  = f · u −1 (u(x)) · u ' (x) dx.  ' Since H is an antiderivative of the function f · u −1 , from Proposition 1.2.2 it results that ∫

 '  f · u −1 (u(x)) · u ' (x) dx = H (u(x)) + C .

1.2 Basic Properties of Indefinite Integrals

9

Remark∫1.2.2 According to Proposition 1.2.3, for computation of the indefinite integral f (u(x)) dx we proceed as follows: We make the change of variable t = u(x), x ∈ I , and we accept the following formal computation:  ' x = u −1 (t), dx = u −1 (t) dt, hence ∫

∫

 ' f (t) · u −1 (t) dt = H (t) + C = H (u(x)) + C .

f (u(x)) dx =

∫  Example 1.2.6 Compute tan6 x dx, x ∈ − π2 , π2 . 1 If we denote by t = tan x, then x = tan−1 t and dx = 1+t 2 dt. Further we have: ) ∫ ∫ ( ∫ t6 1 4 2 6 t −t +1− dt dt = tan x dx = 1 + t2 1 + t2 t5 t3 tan3 x tan5 x = − + t − tan−1 t + C = − + tanx − x + C . 5 3 5 3 Proposition 1.2.4 (The integral by parts formula) If f, g : I → R are two functions of C 1 —class on I , then the functions f · g, f · g ' and f ' · g admit antiderivatives on I and we have the formula: ∫

∫

'

f (x) · g (x) dx = f (x) · g(x) −

f ' (x) · g(x) dx.

(1.4)

Proof According to the differentiation rule for the product of two functions, we have: ( f · g)' = f ' · g + f · g ' , hence f · g ' = ( f · g)' − f ' · g. From Proposition 1.2.1 and Remark 1.1.2, it results that: ∫

∫

'

∫

f ' (x) · g(x) dx ( f (x) · g(x)) dx − ∫ = f (x) · g(x) − f ' (x) · g(x) dx.

f (x) · g (x) dx =

'

∫ Example 1.2.7 Compute ln x dx, x ∈ (0, ∞). If we denote by f (x) = ln x and by g ' (x) = 1, then we have f ' (x) = g(x) = x. From Proposition 1.2.4, it results that: ∫

∫ ln x dx = x ln x −

x·

1 dx = x ln x − x

∫ 1 dx = x ln x − x + C .

1 x

and

10

1 Indefinite Integrals

∫ Example 1.2.8 Compute tan−1 x dx. If we denote by f (x) = tan−1 x and by g ' (x) = 1, then it follows f ' (x) = and g(x) = x. Further we have: ∫ ∫ x −1 −1 dx. tan dx = x tan x − 1 + x2

1 1+x 2

For the computation of the last integral we make the change of variable t = 1+ x 2 and we obtain dt = 2x dx, and thus: ∫ ∫ 1 1  1 x 1 dt = ln|t| + C = ln 1 + x 2 + C . dx = 2 1+x 2 t 2 2 Therefore, we have: ∫ 1  tan−1 x dx = xtan−1 x − ln 1 + x 2 + C . 2 Example 1.2.9 Compute the integrals First, we notice that:

∫√ ∫√ a 2 − x 2 dx and x 2 + a 2 dx.

∫ ∫ √

x  ∫ x a2 − x 2 − x·√ a 2 − x 2 dx = dx = a 2 sin−1 dx. √ a a2 − x 2 a2 − x 2 To compute the last integral we denote by f (x) = x, g ' (x) = √a 2x−x 2 , and we √ obtain f ' (x) = 1, g(x) = − a 2 − x 2 . Further, we have: ∫ √ ∫ √ x 2 2 dx = −x a − x + a 2 − x 2 dx. x·√ a2 − x 2 Thus, we have: ∫ √ ∫ √

x  √ + x a2 − x 2 − a 2 − x 2 dx = a 2 sin−1 a 2 − x 2 dx, a whence it results that: ∫ √

x  √ + x a2 − x 2 + C 2 a 2 − x 2 dx = a 2 sin−1 a and further: ∫ √

x  x √ a2 sin−1 + a 2 − x 2 dx = a2 − x 2 + C . 2 a 2

1.3 Primitives of Rational Functions

11

Similarly, it is shown that: ∫ √

x 2 + a 2 dx =

 x √ √ a2

x 2 + a2 + C . ln x + x 2 + a 2 + 2 2

∫√ ∫√ x 2 + 4x + 8 dx and −x 2 + 4x − 3 dx. Example 1.2.10 Compute Using the canonical form of the second degree function, we obtain: ∫ √

∫ / x2

+ 4x + 8 dx =

(x + 2)2 + 4 dx.

If we consider the change of variable t = x + 2, and according to Example 1.2.9, it results that: ∫ / ∫ √ ∫ √ x 2 + 4x + 8 dx = t 2 + 4 dt (x + 2)2 + 4 dx =

 t √ √ = 2 ln t + t 2 + 4 + t2 + 4 + C 2

 √ = 2 ln x + 2 + x 2 + 4x + 8 x + 2 √ 2 + x + 4x + 8 + C . 2 In the same manner, we have: ∫ / ∫ √ −x 2 + 4x − 3 dx = 1 − (x − 2)2 dx =

1 −1 x − 2 √ 2 sin (x − 2) + −x + 4x − 3 + C . 2 2

1.3 Primitives of Rational Functions A rational function is a ratio of two polynomials (polynomial functions), meaning P(x) , x ∈ I , where I ⊂ R is an interval, P, Q are a function of the form R(x) = Q(x) two real polynomial functions on I and Q(x) /= 0, ∀x ∈ I . If the degree of the numerator P is greater than or equal to the degree of the denominator Q, then we can apply Euclid’s division algorithm and we obtain: P1 (x) P(x) = C(x) + Q(x) Q(x) where C and P1 are polynomial functions and the degree of P1 is less than the degree of Q. Since C(x) is a polynomial function, it is easy to compute a primitive. Now,

12

1 Indefinite Integrals

one can reduce everything to the computation of a primitive of the proper rational 1 (x) , i.e. a rational function which has the property that the degree of P1 is function PQ(x) less than the degree of Q. If Q(x) has the following decomposition into irreducible polynomials: l1 ln   Q(x) = (x − a1 )k1 · . . . · (x − am )km · x 2 + b1 x + c1 · . . . · x 2 + bn x + cn where ai ∈ R, i = 1, m and b j , c j ∈ R, b2j − 4c j < 0, j = 1, n, then, according 1 (x) to a basic theorem of Algebra, the ratio PQ(x) can be uniquely written as a sum of partial fractions of the form: ) m ( Ai1 P1 (x) Σ Ai2 Aiki = + + ··· + Q(x) x − ai (x − ai )2 (x − ai )ki i=1 ( ) n Σ B jl j · x + C jl j B j1 · x + C j1 B j2 · x + C j2 + + 2 + · · · +  l x2 + bj · x + cj x2 + bj · x + cj x2 + bj · x + cj j j=1 where Ai p , B js , C js are real constants that will be determined, p = 1, ki , s = 1, l j , i = 1, m, j = 1, n. 1 (x) Therefore, to compute the primitive of the rational function PQ(x) it is enough to know how to compute primitives of the following types: ∫ A ∫ Bx+C ∗ dx, b2 − 4c < 0, k ∈ N∗ . 1. (x−a) k dx, x / = a, k ∈ N ; 2. (x 2 +bx+c)k ∫ A ∗ Type 1. (x−a) k dx, x / = a, k ∈ N ∫

A dx = (x − a)k

∫ ∫ 

=  = Example 1.3.1 Compute

∫

1 x−3

A dx, x−a A dx, (x−a)k

if k = 1 if k /= 1

+ C , if k = 1 ∫A ln|x − a| −k A(x − a) dx, if k /= 1 A ln|x − a| + C , if k = 1 1−k + C , if k /= 1. A (x−a) 1−k

dx and

∫

4 (x−2)5

dx.

∫

∫

4 (x − 2)5

1 dx = ln|x − 3| + C . x −3 ∫ dx = 4 (x − 2)−5 dx =4

1 (x − 2)−5+1 + C. +C =− −5 + 1 (x − 2)4

1.3 Primitives of Rational Functions

13

∫ 1 Example 1.3.2 Compute x 2 −a 2 dx, x ∈ R\{±a}, a / = 0. Because we have the following decomposition into partial fractions ( ) 1 1 1 1 − = x 2 − a2 2a x −a x +a it follows that: ∫

Type 2.

∫

Type 2.1.

(∫ ) ∫ 1 1 1 1 dx − dx dx = x 2 − a2 2a x −a x +a 1 = (ln|x − a| − ln|x + a|) + C 2a | | 1 || x − a || + C. ln = 2a | x + a |

Bx+C

(x 2 +bx+c)k ∫ 1 x 2 +bx+c

∫

dx, b2 − 4c < 0, k ∈ N∗ dx, b2 − 4c < 0 1 dx = x 2 + bx + c

∫ 

1

x+

b 2 2

+

4c−b2 4

dx.

If we consider the change of variable t = x + b2 and we denote by a 2 = then we obtain: ( ) ∫ ∫ 1 1 −1 t 1 dx = +C dt = tan x 2 + bx + c t 2 + a2 a a ( ) 2x + b 2 + C. =√ tan−1 √ 2 4c − b 4c − b2 ∫ 1 dx. Example 1.3.3 Compute x 2 +2x+5 Proceeding as above, we have: ∫

1 dx = x 2 + 2x + 5

∫

( ) 1 1 −1 x + 1 + C. dx = tan 2 2 (x + 1)2 + 4

4c−b2 , 4

14

1 Indefinite Integrals

∫ Type 2.2. x 2Bx+C dx, b2 − 4c < 0 +bx+c We have successive: ∫

∫ ∫ x + CB −b 2x + b + 2C B Bx + C B dx = B dx = dx x 2 + bx + c x 2 + bx + c 2 x 2 + bx + c ∫ ∫ B 2C − Bb 2x + b 1 = dx + dx. 2 2 2 x + bx + c 2 x + bx + c

For the first integral, we make the change of variable t = x 2 + bx + c and we obtain dt = (2x + b) dx, and further: ∫ ∫ 2x + b 1 dx = dt = ln|t| + C x 2 + bx + c t  2 = ln x + bx + c + C . The second integral is obviously of Type 2.1. ∫ dx. Example 1.3.4 Compute x 23x+1 +x+1 Proceeding as above, we have: ∫

Type 2.3.

∫ ∫ x + 13 2x + 1 + 23 − 1 3 3x + 1 dx = 3 dx = dx x2 + x + 1 x2 + x + 1 2 x2 + x + 1 ∫ ∫ 3 1 2x + 1 1 = dx − dx  1 2 2 x2 + x + 1 2 x + 2 + 43 ) ( 2x + 1 1 3  + C. = ln x 2 + x + 1 − √ tan−1 √ 2 3 3 ∫

We have:

Bx+C

(x 2 +bx+c)k ∫ 

dx, b2 − 4c < 0, k ∈ N∗

Bx + C x2

+ bx + c

∫ k dx =



Bx + C k dx. 2 2 x + b2 + 4c−b 4 2

, then If we make the change of variable t = x + b2 and we denote by a 2 = 4c−b 4 it results that: ∫ ∫ ∫ B 2C − Bb Bx + C 2t 1 dx = dt + k   k  k dt. 2 2 2 2 2 2 x + bx + c t +a t + a2  ' The first indefinite integral can be compute very easy because t 2 + a 2 = 2t. It is obvious that:

1.3 Primitives of Rational Functions

∫

15

  ln t 2 + a 2 + C , if k = 1  k dt = (t 2 +a 2 )1−k + C , if k /= 1. t 2 + a2 1−k 2t

For the other integral, using the integral by parts formula, we establish the following recurrence relation: ∫

1

1  k dt = 2 2 2 a t +a

Ik =

) ( ∫ 1 t2 t 2 + a2 − t 2  k dt = 2 Ik−1 −  k dt . a t 2 + a2 t 2 + a2

∫

Indeed, if we denote by f (t) = t and g ' (t) = 2 t 2 k , then we get f ' (t) = 1 and (t +a ) ∫ 2t 1 dt = − , whence it results that: g(t) = 21 k−1 2(k−1)(t 2 +a 2 ) (t 2 +a 2 )k ∫

t 1 t2 Ik−1 .  k dt = − k−1 +  2 2 2 2 2(k − 1) t +a 2(k − 1) t + a

Furthermore, we have: ( ) t 1 1 Ik−1 Ik = 2 Ik−1 + k−1 −  a 2(k − 1) 2(k − 1) t 2 + a 2 or ( ) t 1 2k − 3 Ik−1 . +   k dt = 2 a 2(k − 1) t 2 + a 2 k−1 2(k − 1) t 2 + a2

∫

1

Ik =

Example 1.3.5 Compute

∫

x+1

(x 2 +x+1)2 Proceeding as above, we have:

∫

x +1

1  2 dx = 2 2 x +x +1

∫ 

dx.

2x + 2 x2

(1.5)

+x +1

2 dx =

1 2

∫ 

2x + 1 x2

+x +1

+

3 4

2 dx +

1 I2 2

where it was denoted by: ∫ I2 =



∫

1 x2

+x +1

2 dx =



x+

1 1 2 2

2 dx.

For the first integral, we make the change of variable t = x 2 + x + 1 and we find: ∫ 

2x + 1 x2 + x + 1

∫ 2 dx =

1 1 1 + C. dt = − + C = − 2 2 t t x +x +1

16

1 Indefinite Integrals

For the second integral, we make the change of variable t = x + the recurrence relation (1.5), for k = 2. Thus, we have:

1 2

and we use

) ( ∫ t 4 1 1  + dt I2 = 2 dt =  3 2 t 2 + 43 2 t 2 + 43 t 2 + 43 ( ) 2t 2t 4 =  2 3 + √ tan−1 √ +C 3 t +4 3 3 3 ( ) 2x + 1 4 −1 2x + 1  = + √ tan + C. √ 3 x2 + x + 1 3 3 3 ∫

1

Finally, we have: ∫ 

x +1

x2

1 2x + 1 +  2 dx = −  2 2+x +1 2 x + x + 1 6 x +x +1 ) ( 2 2x + 1 + √ tan−1 √ 3 3 3 ) ( x −1 2x + 1 2 + √ tan−1 + C. =  2 √ 3 x +x +1 3 3 3

The integration of rational functions is well highlighted by the following example: ∫ 7 6 +4x 5 −5x 4 +4x 3 −5x 2 −x Example 1.3.6 Compute xx 6−2x dx. −2x 5 +3x 4 −4x 3 +3x 2 −2x+1 It is easy to see that the polynomial at the denominator has the double root x = 1 and then it admits the following decomposition:  2 x 6 − 2x 5 + 3x 4 − 4x 3 + 3x 2 − 2x + 1 = (x − 1)2 x 2 + 1 . By polynomial remainder theorem, it follows that: x 5 − x 4 + x 3 − 3x 2 − 2x x 7 − 2x 6 + 4x 5 − 5x 4 + 4x 3 − 5x 2 − x = x +  2 x 6 − 2x 5 + 3x 4 − 4x 3 + 3x 2 − 2x + 1 (x − 1)2 x 2 + 1 thus: ∫

The function

x 7 − 2x 6 + 4x 5 − 5x 4 + 4x 3 − 5x 2 − x dx x 6 − 2x 5 + 3x 4 − 4x 3 + 3x 2 − 2x + 1 ∫ x 5 − x 4 + x 3 − 3x 2 − 2x x2 + = dx.  2 2 (x − 1)2 x 2 + 1

x 5 −x 4 +x 3 −3x 2 −2x 2 (x−1)2 (x 2 +1)

rational functions:

admits the following decomposition into partial

1.3 Primitives of Rational Functions

17

B Ex + F x 5 − x 4 + x 3 − 3x 2 − 2x A Cx + D + + = + 2  2 2 . 2 2 2 x − 1 x + 1 − 1) (x x2 + 1 (x − 1) x + 1 To determine the coefficients A, B, C, D, E, F can be done as follows: we multiply both members of the above equality by (x − 1)2 , we put x = 1 and it 1 follows B = −1. Then, we pass the term − (x−1) 2 to the left side and by using elementary computations, it results the following equality: Cx + D Ex + F x 4 + x 3 + 2x 2 + x − 1 A + 2 + = 2 2 .  2 x − 1 x + 1 x2 + 1 (x − 1) x + 1 Similarly, if we multiply both members of the last equality by x − 1 and making 1 x = 1, we find A = 1. Next, we pass the term x−1 to the left side and we deduce the identity: Ex + F x2 + x + 2 Cx + D +  2 = 2 2 x +1 x2 + 1 x2 + 1 whence x 2 + x + 2 = C x 3 + Dx 2 + (C + E)x + D + F. ⎧ C ⎪ ⎪ ⎨ D From the last identity we can write the system ⎪ C +E ⎪ ⎩ D+F solution C = 0, D = 1, E = 1, F = 1. Therefore, we have: ∫

x 5 − x 4 + x 3 − 3x 2 − 2x dx =  2 (x − 1)2 x 2 + 1

∫

= = = =

0 1 , which has the 1 2

∫

1 dx (x − 1)2 ∫ ∫ x +1 1 dx + +  2 dx x2 + 1 x2 + 1 1 + tan−1 x = ln|x − 1| + x − 1 ∫ ∫ 1 1 2x dx + + 2 2 dx   2 2 2 x +1 x +1 1 dx − x −1

= ln|x − 1| + From relation (1.5), it results that:

1 1 + I2 . + tan−1 x −  2 x −1 2 x +1

18

1 Indefinite Integrals

∫

x 1 + 2 dx =  2 2 2 x +1 x2 + 1 x 1 + tan−1 x + C . =  2 2 2 x +1

I2 =

1



∫ x2

1 dx +1

Finally, we have: ∫

x 7 − 2x 6 + 4x 5 − 5x 4 + 4x 3 − 5x 2 − x dx x 6 − 2x 5 + 3x 4 − 4x 3 + 3x 2 − 2x + 1 x2 1 = + ln|x − 1| + 2 x −1 x −1 3 + tan−1 x + C . +  2 2 2 x +1

1.4 Primitives of Trigonometric Functions ∫ In this section we consider the integrals of the form R(cos x, sin x) dx, where R is P(u,v) ,P a rational function of two variables, i.e. a function of the form R(u, v) = Q(u,v) and Q being polynomials of two variables written as follows: P(u, v) =

m Σ n Σ

ai j u i v j , Q(u, v) =

p r Σ Σ

bks u k v s .

k=0 s=0

i=0 j=0

We suppose that I ⊂ (−π, π ) is an ∫interval and Q(cos x, sin x) /= 0, ∀x ∈ I . In general, to compute the integral R(cos x, sin x) dx, we consider the substitution t = tan x2 , x ∈ I . Inversing the function, we obtain x = 2 tan−1 t and 2 dx = 1+t 2 dt. On the other hand, we have: cos x =

1 − tan2 1 + tan2

x 2 x 2

=

2 tan x2 1 − t2 and sin x = 1 + t2 1 + tan2

x 2

=

2t . 1 + t2

On account of the above change of variable, we deduce that: ∫

∫ R(cos x, sin x) dx =

(

2t 1 − t2 , R 1 + t2 1 + t2

)

2 · dt = 1 + t2

where R1 is a rational function with respect to the variable t.

∫ R1 (t) dt

1.4 Primitives of Trigonometric Functions

19

Remark 1.4.1 The interval I can be replaced by any other interval J on which the function x → tan x2 is strictly monotonous and Q(cos x, sin x) /= 0, ∀x ∈ J . ∫ 1 Example 1.4.1 Compute 2−sin dx, x ∈ (−π, π ). x Putting the change of variable t = tan x2 , it results that: ∫

∫

∫

2 dt 2t 2 − 2t + 2 ) ( ∫ 2 dt −1 2t − 1 +C = tan = √ √  2 3 3 t − 21 + 43 ( ) 2 tan x2 − 1 2 = √ tan−1 + C. √ 3 3

1 dx = 2 − sin x

2 1 dt = · 2t 1 + t2 2 − 1+t 2

In the following, we present three particular cases of other changes of variable, which lead to simple computations of integrals of trigonometric functions. ∫ ∫  ∫ Case 1 R(cos x, sin x) dx = R1 cos2 x, sin2 x dx = R2 (tan x) dx where R1 and R2 are rational functions.  Moreover, we assume that I ⊂ − π2 , π2 and Q(cos x, sin x) /= 0, ∀x ∈ I . In this particular case, the substitution is t = tan x, x ∈ I . Inverting the function, 1 we get x = tan−1 t and dx = 1+t 2 dt. Using the trigonometric formulas cos2 x =

1 t2 1 tan2 x 2 = = and sin x = 1 + tan2 x 1 + t2 1 + tan2 x 1 + t2

it results that: ( ∫ ∫  2 2 R1 R1 cos x, sin x dx =

1 t2 , 1 + t2 1 + t2

) ·

1 dt. 1 + t2

∫  1 Example 1.4.2 Compute 2−sin dx, x ∈ − π2 , π2 . 2 x  If we consider the change of variable t = tan x, x ∈ − π2 , π2 , we obtain: ∫

∫ 1 1 dt = · dt 2 2+2 t2 1 + t t 2 − 1+t 2 ) ) ( ( 1 t tanx 1 + C = √ tan−1 √ + C. = √ tan−1 √ 2 2 2 2 ∫  Example 1.4.3 Compute 2−sin1x cos x dx, x ∈ − π2 , π2 . ∫

1 dx = 2 − sin2 x

1

20

1 Indefinite Integrals

First, we notice that: ∫

∫

∫

1 + tan2 x dx. 2 tan2 x − tan x + 2  If we make the change of variable t = tan x, x ∈ − π2 , π2 , it follows that: 1 dx = 2 − sin x cos x

∫

Case 2

1 dx = 2 − tan x cos2 x

∫

∫ 1 + t2 1 1 · dt dt = 2 2 2 2t − t + 2 1 + t 2t − t + 2 ∫ 1 1 = dt  1 2 2 t − 4 + 15 16 ( ) 4t − 1 4 1 +C = · √ tan−1 √ 2 15 15 ( ) 4 tan x − 1 2 + C. = √ tan−1 √ 15 15

1 dx = 2 − sin x cos x

∫

R(cos x, sin x) dx =

∫

R1 (sin x) cos x dx, x ∈ (−π, π )

where R1 is a rational function. Here we use the substitution sin x = t. Then cos x dx = dt and the integral becomes: ∫ ∫ ∫ R(cos x, sin x) dx = R1 (sin x) cos x dx = R1 (t) dt. ∫ Example 1.4.4 Compute cos3 x dx. If we make the change of variable t = sin x, then dt = cos x dx, and further: ∫

∫ cos3 x dx =

∫ =

Case 3

∫

∫ cos2 x cos x dx =

 1 − sin2 x cos x dx

 sin3 x t3 + C. 1 − t 2 dt = t − + C = sin x − 3 3

R(cos x, sin x) dx =

∫

R1 (cos x) sin x dx, x ∈ (−π, π )

where R1 is a rational function. In this case, it is recommended the following change of variable: cos x = t. As sin x dx = −dt, we have: ∫ ∫ ∫ R(cos x, sin x) dx = R1 (cos x) sin x dx = − R1 (t) dt.

1.5 Primitives of Irrational Functions. Binomial Integrals

21

∫ Example 1.4.5 Compute sin5 x dx. If we make the change of variable cos x = t, then sin x dx = −dt and we have: ∫

∫

∫  2 sin4 x sin x dx = 1 − cos2 x sin x dx ∫ ∫ 2   2t 3 t5 1 − 2t 2 + t 4 dt = −t + =− 1 − t 2 dt = − − +C 3 5 5 3 cos x 2 cos x − + C. = − cos x + 3 5

sin5 x dx =

1.5 Primitives of Irrational Functions. Binomial Integrals ∫ The binomial integrals have the form x m (ax n + b) p dx, where a, b ∈ R, a /= 0 and m, n, p ∈ Q. The Russian mathematician Pafnuty Lvovich Chebyshev (1821–1894) showed that these integrals can be reduced to the computation of the integrals of rational functions, only in the following three cases: Case 1 p ∈ Z. If we denote by r the common denominator of the numbers m and n, and make the natural substitution x = t r , we obtain: ∫ ∫   p p x m ax n + b dx = t mr at nr + b r t r −1 dt. Since m r ∈ Z and n r ∈ Z, it follows that the function under the integral sign is a rational function. ∫ Example 1.5.1 Compute √ √41 6 dx, x ∈ (0, ∞). x ( x+2) The integral can be written as follows: ∫

1 6 dx = √ √ 4 x x +2

Thus, we have m = − 21 , n =

1 4

∫

1 −6 1 dx. x− 2 x 4 + 2

and p = −6 ∈ Z.

22

1 Indefinite Integrals

Therefore r = 4 and we make the substitution x = t 4 . It results dx = 4t 3 dt and ∫ ∫ ∫ t 4t 3 1 dt = 4 dt dx = 6 √ √ 2 (t + 2)6 4 t + 2)6 (t x x +2 ∫ ∫ ∫ 1 1 t +2−2 dt = 4 dt − 8 dt =4 6 5 (t + 2) (t + 2)6 (t + 2) 1 1 8 =− + · +C 4 5 (t + 2)5 (t + 2) 8 1 = − √ 4 + √ 5 + C . 4 4 x +2 5 x +2 Case 2 m+1 ∈ Z and p ∈ /Z n In this case, we make the substitution ax n + b = t r , where r is the denominator of the number p. We get: ( x=

tr − b a

) n1

( )1 r t r − b n −1 r −1 , dx = t dt na a

and further ∫

 p r x ax n + b dx = na

∫ (

m

r = na

∫ (

tr − b a tr − b a

(

) mn t

rp

) m+1 n −1

tr − b a

) n1 −1

t r −1 dt

t r ( p+1)−1 dt.

− 1 ∈ Z and r p ∈ Z, it results that the function under the integral sign Since m+1 n is a rational function with respect to t. ∫ / 1 dx, x ∈ (−∞, −1) ∪ (1, ∞). Example 1.5.2 Compute 3 x (x 2 −1) The integral can be written as follows: ∫ x

/

1 x2 − 1

∫ 3 dx =

 − 3 x −1 x 2 − 1 2 dx.

Therefore, we have m = −1, n = 2, p = − 23 .

1.5 Primitives of Irrational Functions. Binomial Integrals

23

Since m+1 = 0 ∈ Z and p = − 23 ∈ / Z, let us make the substitution x 2 − 1 = t 2 . n√ We get x = t 2 + 1, dx = √t 2t +1 dt, thus the last integral becomes ∫ x

/

∫

1 x2 − 1

− 1 2 t t + 1 2 t −3 ·  1 dt t2 + 1 2 ∫ ∫ ∫ 1 1 1  dt = dt = dt − t2 t2 + 1 t2 t2 + 1 1 = − − tan−1 t + C t

√  1 − tan−1 x 2 − 1 + C . = −√ x2 − 1

3 dx =

Case 3 m+1 + p ∈ Z and m+1 ∈ / Z, p ∈ /Z n n n It can be shown, that if we make the substitution axx n+b = t r , x /= 0, where r is the denominator of p, the problem is reduced to the computation of the primitive function of a rational function. ∫ / 1 Example 1.5.3 Compute dx, x ∈ (1, ∞). 3 x 2 (x 2 −1) The integral can be written as follows: ∫ x2

/

∫

1

3 dx = 2 x −1

 − 3 x −2 x 2 − 1 2 dx.

Therefore, we have m = −2, n = 2, p = − 23 . Since m+1 + p = −2 ∈ Z, then n 1 t / √ dt, and further the obvious substitution is x = 1−t 2 . Thus it follows dx = (1−t 2 )3 ∫ x2

/

1 x2 − 1

∫

 3  t 1 − t 2 1 − t 2 2 t −3 ·  3 dt 1 − t2 2 ∫ ∫ ∫ 1 − t2 1 1 = dt = dt − dt = − − t + C 2 2 t t t √ 2 x x −1 = −√ + C. − x x2 − 1

3 dx =

Chapter 2

Definite Integrals

2.1 Area of a Curvilinear Trapezoid The notion of definite integral appeared out of necessity to solve some important problems in geometry and physics, related mainly to the computation of areas and volumes, respectively to the computation of the mass of an inhomogeneous body and of mechanical work done by a variable force, and so on. In the following, we will study the problem of computation the area of a plane figure bounded by a closed curve. Elementary geometry teaches us to calculate only the areas of plane figures bounded by segments of straight lines and by arcs of circles. The problem of finding the area of a plane region bounded by a closed curve remains open. It is easily seen that a plane figure bounded by a closed curve can be divided up into a finite number of “curvilinear trapezoids” (Fig. 2.1), using several straight lines parallel to the coordinate axes. A curvilinear trapezoid is a plane figure bounded from three parts by straight lines, the two of them are parallel to one of axis and the third is perpendicular to them, and the fourth side of the trapezoid is the graph of a function (Fig. 2.2). More precisely, if f : [a, b] → R+ is a continuous positive function, then the set: { } Γf = (x, y) ∈ R2 ; a ≤ x ≤ b, 0 ≤ y ≤ f (x) is a curvilinear trapezoid. Let Δ : a = x0 < x1 < . . . < xi−1 < xi < . . . < xn = b be an arbitrary partition of the interval [a, b]. Let us denote by m i the greatest lower ⎡bound, respectively ⎤ by Mi the least upper bound of the function f on the interval xi−1 , xi , i = 1, n. Furthermore, we consider the sums: sΔ =

n v i=1

m i (xi − xi−1 ) and SΔ =

n v

Mi (xi − xi−1 ).

i=1

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Integral Calculus for Engineers, https://doi.org/10.1007/978-981-19-4793-3_2

25

26

2 Definite Integrals

Fig. 2.1 The division of a plane figure bounded by a closed curve in “curvilinear trapezoids”

Fig. 2.2 A curvilinear trapozid

We notice that sΔ is the sum of the areas of “inscribed rectangles”: ⎤ ⎡ di = xi−1 , xi × [0, m i ] (Fig. 2.3) while SΔ represents the sum of the areas of “circumscribed rectangles”: ⎡ ⎤ Di = xi−1 , xi × [0, Mi ] (Fig. 2.4). If we denote by E Δ =

∪n i=1

di and by FΔ =

∪n i=1

Di , then E Δ ⊂ Γf ⊂ FΔ and

sΔ = area(E Δ ), SΔ = area(FΔ ). From Figs. 2.3 and 2.4, we notice that: sΔ ≤ area(Γf ) ≤ SΔ . If a partition Δ' is a refinement of the partition Δ, i.e. Δ' contains the points of the partition Δ and maybe some addition other points, then: sΔ ≤ sΔ' ≤ area(Γf ) ≤ SΔ' ≤ SΔ . We know intuitively that: area(Γf ) = sup sΔ = inf SΔ . Δ

Δ

Firstly, we specify that for the time being we do not have a rigorous definition of the notion of area of the set Γf , but only an intuitive concept. On the one hand, the

2.1 Area of a Curvilinear Trapezoid

27

y

Fig. 2.3 The sum of the areas of “inscribed rectangles”

mi

x O

a x1

x i –1 x i

b

y

Fig. 2.4 The sum of the areas of “circumscribed rectangles”

Mi

x O

a x1

x i –1 x i

b

above considerations suggest to introduce a new mathematical concept, namely the definite integral: ∫b

def

f (x) dx = sup sΔ = inf SΔ , Δ

a

Δ

and, on the other hand, to define: ∫b f (x) dx.

area(Γf ) = a

Next, we will rigorously present these fundamental notions in mathematical analysis.

28

2 Definite Integrals

2.2 Darboux Sums. Definition of Definite Integral Definition 2.2.1 A partition of the compact interval [a, b] is any finite subset of distinct points having the form: Δ : a = x0 < x1 < . . . < xi−1 < xi < . . . < xn = b. The norm of ⎡the partition ⎤ Δ is denoted by ∥Δ∥, and it is the maximum length of the subintervals xi−1 , xi , i = 1, n: ∥Δ∥ = max (xi − xi−1 ). 1≤i≤n

The partition Δ' is said to be a refinement of the partition Δ, and denoted by Δ ≺ Δ' , if Δ' contains the points of the partition Δ and maybe some additional other points. ∥ ∥ Obviously, if Δ ≺ Δ' , then ∥Δ' ∥ ≤ ∥Δ∥. Example 2.2.1 (1) Δ : a = x0 < x1 = b, ∥Δ∥ = b − a; < x2 = b, ∥Δ∥ = b−a ; (2) Δ : a = x0 < x1 = a+b 2 2 (3) Δ : a = x0 < x1 < . . . < xi < . . . < xn = b, where xi = a + i b−a , i = 0, n, n ∥Δ∥ = b−a . n Furthermore, for any bounded function f : [a, b] → R, we shall denote by: m = inf{ f (x); x ∈ [a, b]}, M = sup{ f (x); x ∈ [a, b]} ⎤} ⎤} { ⎡ { ⎡ m i = inf f (x); x ∈ xi−1 , xi , Mi = sup f (x); x ∈ xi−1 , xi , 1 ≤ i ≤ n. It is easily seen that the following inequalities hold: m ≤ m i ≤ Mi ≤ M, ∀1 ≤ i ≤ n.

(2.1)

Definition 2.2.2 We define the lower Darboux sum of the function f with respect to the partition Δ by sΔ =

n v

m i (xi − xi−1 ),

i=1

and the upper Darboux sum of the function f with respect to the partition Δ by SΔ =

n v i=1

Mi (xi − xi−1 ).

2.2 Darboux Sums. Definition of Definite Integral

29

Geometrically, the Darboux sums represent the shaded areas in Figs. 2.5 and 2.6. From (2.1), it results that for any partition Δ of the interval [a, b], we have: m · (b − a) ≤ sΔ ≤ SΔ ≤ M · (b − a).

(2.2)

Lemma 2.2.1 For any two partitions Δ and Δ' of the interval [a, b] such that Δ ≺ Δ' , we have the inequalities sΔ ≤ sΔ' ≤ SΔ' ≤ SΔ . Proof Let us consider an arbitrary partition of the interval [a, b]: Δ : a = x0 < x1 < . . . < xi−1 < xi < . . . < xn = b. We assume that the partition Δ' of [a, b] contains the points of the partition Δ and only one addition point, namely the point c, situated between xi−1 and xi (i fixed). Let us denote by: { ⎡ ⎤} m i' = inf f (x); x ∈ xi−1 , c and m i'' = inf{ f (x); x ∈ [c, xi ]}.

Fig. 2.5 Lower Darboux sum

Fig. 2.6 Upper Darboux sum

30

2 Definite Integrals

Since m i ≤ m i' and m i ≤ m i'' , it results that: sΔ' − sΔ = m i' (c − xi−1 ) + m i'' (xi − c) − m i (xi − xi−1 ) ≥ m i (c − xi−1 + xi − c) − m i (xi − xi−1 ) = 0. Therefore, we proved that sΔ ≤ sΔ' . Obviously, if we assume that the division Δ' has all the points of the partition Δ and some addition distinct points c1 , . . . , c p , the proof is similar. The proof of inequality SΔ' ≤ SΔ is very similar and the details are left as an exercise. Lemma 2.2.2 For any two arbitrary partitions Δ' and Δ'' of the interval [a, b] we have sΔ' ≤ SΔ'' . Proof Let Δ = Δ' ∪ Δ'' be the partition of [a, b] that includes every point that is either in Δ' , or in Δ'' (or in both). Obviously, we have Δ' ≺ Δ and Δ'' ≺ Δ. From Lemma 2.2.1, it follows that: sΔ' ≤ sΔ ≤ SΔ ≤ SΔ'' . From the inequalities (2.2), we deduce that the set {sΔ }Δ of lower Darboux sums is upper bounded by the real number M · (b − a) and the set {SΔ }Δ of upper Darboux sums is lower bounded by the real number m · (b − a). Let us denote by I∗ = sup sΔ Δ

the lower Darboux integral of the function f on [a, b], and by I ∗ = inf SΔ Δ

the upper Darboux integral of the function f on [a, b]. Lemma 2.2.3 I∗ ≤ I ∗ . Proof From Lemma 2.2.2, it results that sΔ' ≤ SΔ'' , whatever the partitions Δ' and Δ'' . Firstly, if the partition Δ'' is fixed, we get I∗ = supΔ' sΔ' ≤ SΔ'' . Since Δ'' is arbitrary, therefore we have I∗ ≤ inf Δ'' SΔ'' = I ∗ . Definition 2.2.3 A bounded function f : [a, b] → R is said to be (D)—integrable (integrable in the Darboux sense) on [a, b] if I∗ = I ∗ = I . ∫b The common value I is denoted a f (x) dx, and this number is called the definite integral of the function f on the interval [a, b]; the real number I is uniquely determined by the function f and by the interval [a, b].

2.2 Darboux Sums. Definition of Definite Integral

31

Example 2.2.2 Any constant function f : [a, b] → R, f (x) = c, ∀x ∈ [a, b], is (D)—integrable on [a, b]. Indeed, if f (x) = c, ∀x ∈ [a, b], then sΔ = SΔ = c · (b − a), for any partition Δ of the interval [a, b], thus: ∫b

∗

I∗ = I = c · (b − a) =

c dx. a

Lemma 2.2.4 For any ε > 0, there is δε > 0 such that for any partition Δ of the interval [a, b], with ∥Δ∥ < δε , we have: I∗ − ε ≤ sΔ ≤ SΔ ≤ I ∗ + ε.

(2.3)

Proof We will prove the inequality I∗ − ε ≤ sΔ . The proof of the other inequality from (2.3) is left as an exercise to the readers. Since I∗ = supΔ sΔ , it results that for any ε > 0, there is a partition Δ0 of [a, b] such that I∗ −

ε ≤ s Δ0 . 2

Let us suppose that Δ0 : a = c0 < c1 < . . . < ck−1 < ck < . . . < c p = b. Let μ = min1≤k≤ p (ck − ck−1 ) and let Δ : a = x0 < x1 < . . . < xi < . . . < xn = b be a ⎡partition⎤of [a, b], with ∥Δ∥ < μ. If we denote by Δ = Δ ∪ Δ0 , then in the interval xi−1 , xi there is at most one point ck of the partition Δ0 . If we assume that xi−1 < ck < xi , then we denote by: ⎤} { ⎡ and m i'' = inf{ f (x); x ∈ [ck , xi ]}. m i' = inf f (x); x ∈ xi−1 , ck Furthermore, we have: m i' (ck − xi−1 ) + m i'' (xi − ck ) − m i (xi − xi−1 ) ) ) ( ( = m i' − m i (ck − xi−1 ) + m i'' − m i (xi − ck ) ≤ (M − m)(xi − xi−1 ) ≤ (M − m)∥Δ∥. Since the partition Δ has at most p − 1 interior points ck , it follows that: sΔ − sΔ ≤ ( p − 1)(M − m)∥Δ∥.

(2.4)

32

2 Definite Integrals

} { ε Let δε = min μ, 2( p−1)(M−m) , let Δ be a partition of the interval [a, b], with ∥Δ∥ < δε and Δ = Δ ∪ Δ0 . Since δε ≤ μ, it results that ∥Δ∥ < μ, and according to (2.4), we have: sΔ − sΔ ≤ ( p − 1)(M − m)∥Δ∥ ≤ ( p − 1)(M − m)

ε ε = . 2( p − 1)(M − m) 2

Therefore, we have: I∗ −

ε ε ≤ s Δ0 ≤ s Δ ≤ s Δ + , 2 2

hence I ∗ − ε ≤ sΔ . Theorem 2.2.1 (Darboux criterion for integrability) Let f : [a, b] → R be a bounded function. The necessary and sufficient condition that the function f to be (D)—integrable on the interval [a, b] is that for every ε > 0 there is δε > 0 such that for any partition Δ of [a, b], with the property ∥Δ∥ < δε the following inequality is verified SΔ − sΔ < ε. Proof First, suppose that the function f is (D)—integrable on [a, b]. According to Definition 2.2.3, we have I∗ = I ∗ = I . From Lemma 2.2.4, it results that for every ε > 0 there is δε > 0 such that for any partition Δ, with ∥Δ∥ < δε , we have I−

ε ε ≤ sΔ ≤ SΔ ≤ I + . 2 2

Obviously, from the last inequalities we deduce that: ( ε) ε) ( − I− = ε. SΔ − sΔ < I + 2 2 Conversely, suppose that for every ε > 0 there is δε > 0 with the property that for any partition Δ, with ∥Δ∥ < δε , we have SΔ − sΔ < ε. Then, since sΔ ≤ I∗ ≤ I ∗ ≤ SΔ , it results that 0 ≤ I ∗ − I∗ ≤ SΔ − sΔ < ε. As ε > 0 is arbitrary, it follows that I ∗ − I∗ = 0, hence that f is (D)—integrable on [a, b].

2.3 Integrability of Continuous and Monotonic Functions

33

2.3 Integrability of Continuous and Monotonic Functions Theorem 2.3.1 A continuous function f : [a, b] → R on the compact interval [a, b] is (D)—integrable on [a, b]. Proof Let Δ : a = x0 < x1 < . . . < xi−1 < xi < . . . < xn = b be an arbitrary partition of the interval [a, b]. Taking into account to Theorem 5.6.2 (see Differential Calculus, Part I), since the function f is continuous on the compact interval [a, b], values, m i and Mi , on then f is bounded and it ⎡ attains⎤its minimum and ⎤ ⎡ maximum every compact interval xi−1 , xi . Let μi , ηi ∈ xi−1 , xi such that m i = f (μi ) and Mi = f (ηi ). On the other hand, from Theorem 5.6.3 (see Differential Calculus, Part I), it results that f is uniformly continuous on [a, b], thus, | for every | ε > 0 there is δε > 0 such that for all x ' , x '' ∈ [a, b], with the property |x ' − x '' | < δε , it follows that: | ( ') ( )| | f x − f x '' |
0, we denoted by δε = f (b)−ε f (a) . If we suppose that ∥Δ∥ < δε , then the upper and lower Darboux sums of f satisfy:

34

2 Definite Integrals

SΔ − sΔ < δε ·

n v

( f (xi ) − f (xi−1 ))

i=1

=

ε · ( f (b) − f (a)) = ε. f (b) − f (a)

Therefore, we proved that for any ε > 0 there is δε > 0 such that for any partition Δ of [a, b], with ∥Δ∥ < δε , it follows that SΔ − sΔ < ε. From Theorem 2.2.1, we deduce that f is (D)—integrable on the interval [a, b].

2.4 Riemann Sums. Riemann Criterion for Integrability Definition 2.4.1 Let f : [a, b] → R be a given real function on the interval [a, b], let Δ : a = x0 < x1 < . . . < xi−1 < xi < . . . < xn = b be an arbitrary partition ⎡ ⎤of [a, b], and let ξi be an arbitrary intermediate point in the subinterval xi−1 , xi , i = 1, n. The Riemann sum for the function f , associated to the partition Δ and the chosen intermediate points ξi is denoted by σΔ ( f ; ξ ), and it is defined as: σΔ ( f ; ξ ) =

n v

f (ξi )(xi − xi−1 )

i=1

where ξ is the vector ξ = (ξ1 , ξ2 , . . . , ξn ). Furthermore, suppose that f is a bounded and positive function on [a, b]. If we respectively, by Mi the least upper bound of denote by m i the greatest lower bound, ⎡ ⎤ the function f on the subinterval xi−1 , xi , i = 1, n, then: ⎤ ⎡ m ≤ m i ≤ f (ξi ) ≤ Mi ≤ M, ∀ξi ∈ xi−1 , xi and m · (b − a) ≤ sΔ ≤ σΔ ( f ; ξ ) ≤ SΔ ≤ M · (b − a), where m = inf{ f (x); x ∈ [a, b]} and M = sup{ f (x); x ∈ [a, b]}. On the other hand, geometrically, σΔ ( f ; ξ ) is the sum of the areas of the hatched rectangles in Fig. 2.7 (rectangles with height f (ξi ), 1 ≤ i ≤ n and width xi − xi−1 ), which approximates the area of the curvilinear trapezoid { } Γf = (x, y) ∈ R2 ; a ≤ x ≤ b, 0 ≤ y ≤ f (x) . We notice that the approximation is better if we refine Δ. Intuitively, we are led to believe that if ∥Δ∥ → 0, then σΔ ( f ; ξ ) “converges” to a finite number I , whatever the intermediate points ξi are.

2.4 Riemann Sums. Riemann Criterion for Integrability

35

Fig. 2.7 Riemann sum

In other words, “there exists” lim∥Δ∥→0 σΔ ( f ; ξ ) = I . We notice that this limit is not an ordinary limit, because the sum σΔ ( f ; ξ ) is not a function of ∥Δ∥. Indeed, for a value of the “independent variable” ∥Δ∥, there are an infinite number of partitions Δ' which have the same norm as Δ, and implicitly there are an infinity of values for σΔ ( f ; ξ ). To this we must add the fact that we can choose the points ξi in an infinite number of ways. We are led to consider a notion of limit in a large sense. The exact sense of the limit lim∥Δ∥→0 σΔ ( f ; ξ ) = I is given by Definition 2.4.2. Definition 2.4.2 It is said that the function f is (R)—integrable (integrable in the Riemann sense) on [a, b] if there is a finite real number I such that for any ε > 0 there is δε > 0 with⎡the property that for any partition Δ of [a, b], with ∥Δ∥ < δε ⎤ and any points ξi ∈ xi−1 , xi , i = 1, n, we have: |σΔ ( f ; ξ ) − I | < ε. In the following, the real number I is called the definite integral (in the sense of Riemann) of the function f on [a, b] and is denoted by: ∫b I =

f (x) dx = lim σΔ ( f ; ξ ). ∥Δ∥→0

a

Remark 2.4.1 In a less precise writing, in which we do not go into details regarding points ξi , the Riemann sum σΔ ( f ; ξ ) = the ∑npartition Δ and the choice of intermediate∑ b f − x can be noted as well (ξ )(x ) i i i−1 a f (x) Δx. Replacing the Greek letter ∑i=1 by the Latin letter S, the Riemann sum can be∫written as Sab f (x) Δx. is denoted by , i.e. a kind of elongated form (The limit of this expression ) ∫b b S lim∥Δ∥→0 Sa f (x) Δx = a f (x) dx . This is the historical origin of notation ∫b a f (x) dx.

36

2 Definite Integrals

Theorem 2.4.1 An (R)—integrable function f : [a, b] → R on the compact interval [a, b] is bounded on [a, b]. Proof By hypothesis, if f is (R)—integrable on [a, b], then there is I ∈ R such that for ε = 1, there is δ1 > 0 with the property that for any partition Δ, with ∥Δ∥ < δ1 and for any points ξi we have: I − 1 < σΔ ( f ; ξ ) < I + 1.

(2.5)

Let us consider the partition⎡Δ : a =⎤ x0 < x1 < . . . < xi−1 < xi < . . . < xn = b, with ∥Δ∥ < δ1 and let ξi ∈ xi−1 , xi , i = 1, n, be some arbitrary points. Let us ⎤ ⎡ fix a number j = 1, n and let us consider an arbitrary point ξ 'j ∈ x j−1 , x j and the ) ( vector ξ ' = ξ1 , ξ2 , . . . , ξ j−1 , ξ 'j , ξ j+1 , . . . , ξn . We have: |( ( ) ( ))( )| | ( ) ( )| | f ξ j − f ξ ' x j − x j−1 | = |σΔ f ; ξ j − σΔ f ; ξ ' | j j ≤ (I + 1) − (I − 1) = 2 whence, it results that: | ( ) ( )| | f ξj − f ξ' | ≤ j

⎡ ⎤ 2 , for any ξ 'j ∈ x j−1 , x j , x j − x j−1

⎡ ⎤ thus f is bounded on x j−1 , x j , for all j = 1, n. Hence f is bounded on [a, b]. Remark 2.4.2 From Theorem 2.4.1 we deduce that the study of (R)—integrability can be reduced to the study of the integrability of bounded functions. Remark 2.4.3 The reciprocal statement of Theorem 2.4.1 is not generally true. There are bounded functions which are not (D)—integrable. A such example is Dirichlet function: ⎧ 1 if x ∈ Q f : [0, 1] → R, f (x) = . 0 if x ∈ [0, 1]\Q Indeed, any subinterval of an arbitrary partition Δ of [0, 1] contains both rational points and irrational points, thus: m i = 0, Mi = 1, sΔ = 0, SΔ = 1, I∗ = 0, I ∗ = 1 whence, it results that f is not (D)—integrable on [0, 1]. Remark 2.4.4 If f : [a, b] → R is continuous on [a, b], then f is bounded on on any compact subinterval of [a, b]. [a, b] and it attains its supremum ⎤ ⎡ and infimum Therefore, there are μi , ηi ∈ xi−1 , xi such that m i = f (μi ) and Mi = f (ηi ). It results that Darboux sums sΔ and SΔ are Riemann sums, namely:

2.4 Riemann Sums. Riemann Criterion for Integrability

sΔ =

n v

f (μi )(xi − xi−1 ) and SΔ =

i=1

37 n v

f (ηi )(xi − xi−1 ).

i=1

If f is bounded, the link between Riemann sums and Darboux sums is given by the following lemma: Lemma 2.4.1 Let f : [a, b] → R be a bounded function, let ⎤Δ be an arbitrary ⎡ partition of [a, b] and ε > 0. Then there are αi , βi ∈ xi−1 , x i , 1 ≤ i ≤ n, such that: SΔ − ε


n v

f (βi )(xi − xi−1 ) = σΔ ( f ; β)

i=1

where α = (α1 , . . . , αn ) and β = (β1 , . . . , βn ). ⎤} { ⎡ Proof Let Mi = sup f (x); x ∈ xi−1 , x⎡i . According to the definition of the least ⎤ upper bound, it results that there is αi ∈ xi−1 , x i such that: Mi −

ε < f (αi ). b−a

(2.6)

Multiplying the inequality (2.6) by xi − xi−1 and making the sum, it results that: SΔ − ε
f (βi ). b−a

(2.7)

Multiplying the inequality (2.7) by xi − xi−1 and making the sum, it follows that: sΔ + ε >

n v

f (βi )(xi − xi−1 ) = σΔ ( f ; β).

i=1

Remark 2.4.5 Between Darboux sums and Riemann sums associated to a bounded function f and a partition Δ we have the following relations: ⎤ } { ⎡ sΔ = inf σΔ ( f ; ξ ); ξi ∈ xi−1 , x i , 1 ≤ i ≤ n ⎤ } { ⎡ SΔ = sup σΔ ( f ; ξ ); ξi ∈ xi−1 , x i , 1 ≤ i ≤ n .

38

2 Definite Integrals

The following theorem prove that the both definitions of integrability are equivalent. Theorem 2.4.2 Let f : [a, b] → R be a bounded function. Then the function f is (D)—integrable on [a, b] if and only if f is (R)—integrable on [a, b]. Proof If we assume that f is (D)—integrable on [a, b], then I∗ = I ∗ = I . On the other hand, from Theorem 2.2.1, it results that ∀ε > 0, ∃ δε > 0 such that SΔ − sΔ < ε, for any partition Δ, with ∥Δ∥ < δε . Since sΔ ≤ I ≤ SΔ and sΔ ≤ σΔ ( f ; ξ ) ≤ SΔ , ∀ξ , it follows that: |σΔ ( f ; ξ ) − I | < SΔ − sΔ < ε for any Δ, with ∥Δ∥ < δε and for any vector ξ , whence, it results that f is (R)— integrable on [a, b]. Conversely, let us assume that f is (R)—integrable. Then there is a finite number I ∈ R with the property: ∀ε > 0, ∃ δε > 0 such that ∀Δ, with ∥Δ∥ < δε and ∀ξ we have: I−

ε ε < σΔ ( f ; ξ ) < I + . 4 4

(2.8)

Let Δ : a = x0 < x1 < . . . < xi−1 < xi < . . . < xn = b be an arbitrary partition with ∥Δ∥ < δε . ⎤ ⎡ From Lemma 2.4.1, it results that there is αi ∈ xi−1 , x i , 1 ≤ i ≤ n, such that: SΔ −

ε < σΔ ( f ; α), 4

where α = (α1 , . . . , αn ). According to (2.8), we get: ε SΔ < I + . 2

(2.9)

Similarly, using again Lemma 2.4.1 and inequalities (2.8), it is shown that: ε sΔ > I − . 2

(2.10)

From (2.9) and (2.10), it follows that SΔ − sΔ < ε, for any Δ, with ∥Δ∥ < δε , therefore f is (D)—integrable (according to Theorem 2.2.1). Theorem 2.4.3 (Riemann criterion for integrability) The necessary and sufficient condition that the function f : [a, b] → R to be integrable on the interval [a, b] is there is a real number I such that for any sequence {Δn } of partitions of [a, b], with limn→∞ ∥Δn ∥ = 0 and any choice of intermediate points ξ (n) , we have

2.4 Riemann Sums. Riemann Criterion for Integrability

39

( ) lim σΔn f ; ξ (n) = I.

n→∞

Proof First, from Definition 2.4.2, it results that there is I ∈ R such that ∀ε > 0, ∃ δε > 0 with the property that for any Δ, with ∥Δ∥ < δε and for any ξ , we have: |σΔ ( f ; ξ ) − I | < ε.

(2.11)

Let {Δn } be a sequence of partitions of [a, b] with the property limn→∞ ∥Δn ∥ = 0. Then there is n 0 ∈ N∗ such that ∥Δn ∥ < δε , for any| n ≥( n 0 . ) | Taking into account to (2.11), we deduce that |σΔn f ; ξ (n) − I | < ε, for any n ≥ n 0 and any intermediate point system ξ (n) corresponding to the partition Δn , whence, it results that: ( ) lim σΔn f ; ξ (n) = I.

n→∞

Conversely, ( let us)suppose that there is a real number I ∈ R with the property that limn→∞ σΔn f ; ξ (n) = I , for any sequence of partitions {Δn }, such that ∥Δn ∥ → 0 and for any intermediate point system ξ (n) . Furthermore, if we suppose that the function f is not integrable on [a, b], then ∀I ∈ R, ∃ ε0 > 0 such that ∀δ > 0, there is a partition Δδ , with ∥Δδ ∥ < δ and an intermediate point system ξ δ , with the property | | ( ) |σΔ f ; ξ δ − I | ≥ ε0 . δ Δ| n , with ∥Δn ∥ < n1 and In particular, for δ = n1 , it results that there | is( a partition ) (n) (n) | − I | ≥ ε0 . This means that an intermediate ( point ) system ξ such that σΔn f ; ξ (n) /= I , although ∥Δn ∥ → 0, which contradicts the hypothesis. limn→∞ σΔn f ; ξ Remark 2.4.6 Let f : [a, b] → R be an integrable function and let Δn be an equidistant partition of [a, b], i.e.: (n) Δn : a = x0(n) < x1(n) < . . . < xi−1 < xi(n) < . . . < xn(n) = b

, i = 0, n. where xi(n) = a + i b−a n = 0. Since f is an inteIt is easily seen that limn→∞ ∥Δn ∥ = limn→∞ b−a n we can arbitrarily choose the intermediate points ξi(n) in the interval ⎤ ⎡ grable function, (n) xi−1 , then the associated , xi(n) , i = 1, n. Thus, if we choose ξi(n) = xi(n) = a +i b−a n Riemann sum becomes:

σΔn

(

f ; xi(n)

)

) ( n b−a v b−a . = f a+i n i=1 n

From Theorem 2.4.3, it results that:

40

2 Definite Integrals

) ( n b−a v b−a . f a+i n→∞ n i=1 n

∫b

f (x)dx = lim a

(2.12)

∫1 Example 2.4.1 Compute 0 x dx. Replacing in (2.12) a = 0, b = 1, f (x) = x, it follows that: ∫1

n 1 + 2 + ··· + n n(n + 1) 1vi 1 = lim = . = lim 2 2 n→∞ n→∞ n→∞ n n n 2n 2 i=1

x dx = lim 0

∫π Example 2.4.2 Compute 02 cos x dx. To compute this integral we will use the identity: cos x + cos 2x + · · · + cos nx = Replacing in (2.12) a = 0, b = π

∫2

π , 2

sin

nx 2

cos (n+1)x 2 , x /= 2kπ, k ∈ Z. sin 2x

f (x) = cos x, we obtain:

n π v πi cos n→∞ 2n 2n i=1 ( ) π 2π nπ π cos + cos + · · · + cos = lim n→∞ 2n 2n 2n 2n

cos x dx = lim 0

π sin · = lim n→∞ 2n π sin · = lim n→∞ 2n π 4n

nπ 2n

2

· cos sin

π 4

π 2n

(n+1)π 2n

2

2

· cos (n+1)π 4n π sin 4n

π π · cos π · sin sin 4n 4 4 √ √ 2 2 · = 1. =2· 2 2

= lim 2 · n→∞

Remark 2.4.7 If f : [a, b] → R is an integrable function and {Δn } is a sequence of partitions such that limn→∞ ∥Δn ∥ = 0, then limn→∞ sΔn = limn→∞ SΔn = ∫b a f (x) dx. Indeed, according to Lemma 2.4.1, it results that ∀n ∈ N, ∃ α (n) , β (n) such that: ( ) ( ) 1 1 and 0 ≤ σΔn f ; β (n) − sΔn < . 0 ≤ SΔn − σΔn f ; α (n) < n n Passing to the limit in these inequalities, we obtain:

2.5 Lebesgue Criterion for Integrability

(

lim SΔn = lim σΔn f ; α

n→∞

41

(n)

)

n→∞

∫b =

f (x) dx, respectively a

(

lim sΔn = lim σΔn f ; β

n→∞

(n)

n→∞

)

∫b =

f (x) dx. a

2.5 Lebesgue Criterion for Integrability Definition 2.5.1 A set A ⊂ R is said to be null set if for any ε > 0, there is a sequence {In } of open intervals with the properties: (a) A ⊂ (b)

∞ ∑

∞ ∪

In ;

n=1

l(In ) < ε.

n=1

where l(In ) is the length of the interval In . (We note that some of these intervals In may be the empty set). Proposition 2.5.1 Any singleton subset of R is a null set. ) ( Proof Let A = {x0 }. We can choose I1 = x0 − 3ε , x0 + 3ε and In = ∅, for any ∑∞ ∪∞ n ≥ 2. Obviously, we have A ⊂ n=1 In and n=1 l(In ) = l(I1 ) = 2ε3 < ε. The following statement is obvious. Proposition 2.5.2 If A ⊂ B and B is a null set, then A is also a null set. Proposition 2.5.3 Any countable union of null sets is also a null set. Proof Let An ⊂ R be a null set, for any n ∈ N∗ . According to Definition 2.5.1, it results that for any ε > 0, there is a sequence {In m } of open intervals such that: An ⊂

∞ ⊔ m=1

In m and

∞ v m=1

l(In m )
0. ∫b ∫b f (x)·g(x) dx Dividing (2.14) by a g(x) dx, we obtain m ≤ a ∫ b ≤ M. If we denote by μ = proved.

∫b a

a

f (x)·g(x) dx ∫b , a g(x) dx

g(x) dx

then m ≤ μ ≤ M and the relation (2.13) is

Corollary 2.6.3 Let f, g : [a, b] → R be two functions as in Theorem 2.6.1. If we additionally assume that f is continuous on [a, b], then there is ξ ∈ [a, b] such that:

46

2 Definite Integrals

∫b

∫b f (x) · g(x) dx = f (ξ ) ·

a

g(x) dx.

(2.15)

a

Proof Because f is continuous on the compact interval [a, b], it results that there are α, β ∈ [a, b] such that m = f (α) and M = f (β). According to Theorem 2.6.1, ∫b ∫b there is m = f (α) ≤ μ ≤ f (β) = M such that a f (x) · g(x) dx = μ · a g(x) dx. On the other hand, f has the Darboux property on [a, b], whence it results that there is a point ξ between α and β, so in [a, b], such that μ = f (ξ ). Therefore, the relation (2.15) is verified. Corollary 2.6.4 If f : [a, b] → R is integrable, then there is m ≤ μ ≤ M such that: ∫b f (x) dx = μ · (b − a). a

Proof The statement holds immediately from Theorem 2.6.1, for the particular case g = 1. Corollary 2.6.5 If f : [a, b] → R is continuous, then there is ξ ∈ [a, b] such that: ∫b f (x) dx = f (ξ ) · (b − a). a

Proof The statement follows immediately from Corollary 2.6.2, for the particular case g = 1. Theorem 2.6.2 (The fundamental theorem of integral calculus) Any function f : [a, b] → R continuous on the interval (a, b) has an antiderivative on (a, b). One of the antiderivatives of f is the function ∫x F(x) =

f (t) dt, ∀x ∈ (a, b) c

where c is an arbitrary fixed point from (a, b). Proof Taking into account to Proposition 2.6.6, Remark 2.6.1 and Corollary 2.6.4, for any x ∈ (a, b) and for any h ∈ R such that x + h ∈ (a, b), we have: F(x + h) − F(x) = h =

∫ x+h c

∫x c

∫x f (t) dt − c f (t) dt h ∫ x+h ∫x f (t) dt + x f (t) dt − c f (t) dt h

2.6 Properties of Integrable Functions

47

∫ x+h =

x

f (t) dt f (ξx )h = = f (ξx ) h h

where ξx is a point between x and x + h. Because f is continuous at the point x, further we have: lim

h→0

F(x + h) − F(x) = lim f (ξx ) = f (x). h→0 h

Therefore F ' (x) = f (x), hence F is an antiderivative of f . Theorem 2.6.3 (Newton–Leibniz formula) Let f : [a, b] → R be an integrable function that admits antiderivatives on [a, b]. If F : [a, b] → R is an arbitrary antiderivative of f on the interval [a, b], then: ∫b

| | f (x) dx = F(x)|b = F(b) − F(a). a

(2.16)

a

Proof Let us take Δ : a = x0 < x1 < . . . < xi−1 < xi < . . . < xn = b an arbitrary partition of the interval [a, b]. We notice that: F(b) − F(a) =

n v

(F(xi ) − F(xi−1 )).

i=1

On the other hand, from Lagrange Theorem, it results that there is ξi ∈ (xi−1 , xi ) such that: F(xi ) − F(xi−1 ) = F ' (ξi )(xi − xi−1 ) = f (ξi )(xi − xi−1 ). If we denote by ξ = (ξ1 , ξ2 , . . . , ξn ), then: F(b) − F(a) =

n v

f (ξi )(xi − xi−1 ) = σΔ ( f ; ξ ).

i=1

Let {Δn } be a sequence of partitions of [a, b], such that limn→∞ ∥Δn ∥ = 0, and let ξ (n) be the intermediate point system of Δn , that follows from Lagrange Theorem. Then we have: ( ) F(b) − F(a) = σΔn f ; ξ (n) and therefore:

48

2 Definite Integrals

∫b

( ) f (x) dx = lim σΔn f ; ξ (n) = F(b) − F(a). n→∞

a

∫1( ) Example 2.6.1 Compute 0 2x 3 − 4x 2 + 5x − 7 dx. From Example 1.2.1, an antiderivative function of f (x) = 2x 3 − 4x 2 + 5x − 7 4 3 2 on the interval [0, 1] is F(x) = x2 − 4x3 + 5x2 − 7x. According to Newton–Leibniz formula, we have: ∫1

| ( 4 )| | | ( 3 ) x 4x 3 5x 2 2 2x − 4x + 5x − 7 dx = F(x)||1 = − + − 7x ||1 0 0 2 3 2

0

16 1 4 5 − + −7−0=− . 2 3 2 3

=

Theorem 2.6.4 (The integral by parts formula) If f, g : [a, b] → R are two functions of C 1 —class on [a, b], then: ∫b

∫b | |b f (x) · g (x) dx = f (x) · g(x)| − f ' (x) · g(x) dx. '

a

a

(2.17)

a

Proof Formula (2.17) is obtained immediately if we apply Proposition 2.6.2 and Theorem 2.6.3 to the relation: ( f (x) · g(x))' = f ' (x) · g(x) + f (x) · g ' (x), x ∈ [a, b]. ∫1 Example 2.6.2 Compute 0 2x ex dx. If we denote by f (x) = 2x and by g ' (x) = ex , it results that f ' (x) = 2 and g(x) = ex . From Theorems 2.6.3, 2.6.4 and Proposition 2.6.2, it follows that: ∫1

| | | ∫1 | | | x x |1 x |1 2x e dx = 2x e | − 2e dx = 2x e | − 2e | 0 0 0 x |1

x

0

0

= 2e − 0 − 2e + 2 = 2. Theorem 2.6.5 (The change of variable formula) If J ⊂ R is an interval, f : J → R is a continuous function and u : [a, b] → J is a function of C 1 —class, then: ∫b

'

∫u(b)

f (u(x)) · u (x) dx = a

f (t) dt. u(a)

(2.18)

2.7 Area of a Plane Figure

49

Proof From Theorem 2.6.2, it results that f admits antiderivative functions. If F is an antiderivative function of f , then F ' (t) = f (t), ∀t ∈ J , and according to Theorem 2.6.3, we have: ∫u(b)

∫u(b) f (t) dt =

u(a)

F ' (t) dt = F(u(b)) − F(u(a)).

u(a)

On the other hand, it results that: (F(u(x)))' = F ' (u(x)) · u ' (x) = f (u(x)) · u ' (x), ∀x ∈ [a, b]. Applying Theorem 2.6.3, we obtain: ∫b

∫b

'

f (u(x)) · u (x) dx = a

| | (F(u(x)))' dx = F(u(x))|b a

a

∫u(b) = F(u(b)) − F(u(a)) =

f (t) dt. u(a)

∫2 dx. Example 2.6.3 Compute 0 x 22x−1 −x+4 If we denote by t = x 2 − x + 4, then dt = (2x − 1) dx and t ∈ [4, 6]. From (2.18), we obtain: ∫2 0

2x − 1 dx = x2 − x + 4

∫6

| | 1 3 dt = ln|t|||6 = ln 6 − ln 4 = ln . 4 t 2

4

2.7 Area of a Plane Figure We are familiar with the concept of area of a polygonal plane set (plane figure bounded by a polygon) from the course of elementary mathematics. In what follows, we shall introduce the concept { of squareable plane figure. } If D = [a, b]×[c, d] = (x, y) ∈ R2 ; a ≤ x ≤ b, c ≤ y ≤ d is a rectangle with sides parallel to the coordinate axes, then: def

area(D) =(b − a)(d − c).

50

2 Definite Integrals

The same formula of area remains true even if the rectangle does not contain one or more sides. Definition 2.7.1 An elementary plane set E ⊂ R2 is a finite union of rectangular plane sets with sides parallel to the coordinate axes, which may or may not contain one or more sides and which have in common two by two only at most one side, i.e.: E=

p ⊔

◦

◦

Di , Di ∩ D j = ∅, for i /= j.

i=1

By definition, the area of the elementary set E ⊂ R2 is: def

area(E) =

p v

area(Di ) =

i=1

p v

(bi − ai )(di − ci ).

(2.19)

i=1

Furthermore, we shall denote by E the family of all elementary plane sets. Remark 2.7.1 The family of elementary plane sets has the following properties: ∪p (1) The representation of an elementary set by the form E = i=1 Di is not unique. (2) If E 1 , E 2 ∈ E , then E 1 ∪ E 2 , E 1 ∩ E 2 and E 1 \E 2 ∈ E . 2 (3) The area of an its representation, ∪ pelementary set E∪⊂q R does not depend by ∑ p G , then area(E) = i.e. if E = i=1 Di and E = j j=1 i=1 area(Di ) = ( ) ∑q j=1 area G j . ◦

◦

(4) If E 1 , E 2 ∈ E and E 1 ∩ E 2 = ∅, then area(E 1 ∪ E 2 ) = area(E 1 ) + area(E 2 ). (5) If E 1 , E 2 ∈ E and E 1 ⊂ E 2 , then area(E 1 ) ≤ area(E 2 ) and area(E 2 \E 1 ) = area(E 2 ) − area(E 1 ). We recall that a subset A ⊂ R2 is bounded if there is r > 0 such that A ⊆ B(O, r ) (the ball with center at origin and radius r ). If A ⊂ R2 is a bounded set, then we denote by S∗ ( A) = sup{area(E); E ∈ E , E ⊂ A} the interior area of the set A respectively by S ∗ (A) = inf{area(F); F ∈ E , F ⊃ A} the exterior area of the set A. If the set A does not contain any elementary set, we define S∗ (A) = 0. Clearly, if the set A is bounded, then the both numbers S∗ ( A) and S ∗ (A) exist and S∗ (A) ≤ S ∗ (A).

2.7 Area of a Plane Figure

51

Definition 2.7.2 We say that a bounded set A ⊂ R2 is squareable (has area) if S∗ ( A) = S ∗ (A) = S(A). In this case, the number S( A) is denoted by area(A) and it is called the area of A. Remark 2.7.2 Any elementary plane set is squareable and its area is equal to the area defined in (2.19). Remark 2.7.3 Any polygonal set has area defined by Definition 2.7.2, and this area is the same as the known area of elementary geometry. Indeed, since any polygonal set is a finite union of triangular sets and any triangle is the union or the difference of two right triangles, it is enough to show that any plane set whose boundary is a right triangle has area. Let us take the right triangle ABC with: A = 90◦ , AB = a, AC = b (Fig. 2.8). Let us divide the cathetus AB into n equal parts and let us consider rectangles of type M N Q P, where M N = an and M P is parallel to AC. Let us suppose that B M = i · an . Since the triangles B M P and B AC are similar, it follows that: BM MP b = , hence M P = i · . a b n Therefore, the area of the rectangle M N Q P is i · ab . n2 If we denote by E the union of these rectangles, then E ∈ E , the set E is included in the triangle ABC and we have the equality: area(E) =

ab ab(n − 1) . (1 + 2 + · · · + n − 1) = n2 2n

Similarly, if we denote by F the union of rectangles of type M N S R, then F is an elementary set that includes the triangle ABC and: Fig. 2.8 Any right triangle is squarable

B

M

P

N

Q

A

R

S

C

52

2 Definite Integrals

area(F) =

ab ab(n + 1) . (1 + 2 + · · · + n) = 2 n 2n

Further, we have: ab ab ab(n − 1) ab(n + 1) = sup ≤ S∗ ( ABC) ≤ S ∗ (ABC) ≤ inf = n 2 2n 2n 2 n hence S∗ ( ABC) = S ∗ (ABC) =

ab . 2

Therefore, the triangular set ABC has the area defined by Definition 2.7.2, and this area is the known area of a right triangle in elementary geometry. Remark 2.7.4 There are plane sets⎧that are non-squareable. 1 if x ∈ Q Indeed, if f : R → R, f (x) = is the Dirichlet function and 0 if x ∈ R\Q { } A = (x, y) ∈ R2 ; 0 ≤ x ≤ 1, 0 ≤ y ≤ f (x) , then S∗ (A) = 0 and S ∗ (A) = 1, whence, it results that A is non-squareable set. The following result gives us examples of squareable sets. More precisely, we will show that the curvilinear trapezoids corresponding to the integrable functions are squareable sets. Proposition 2.7.1 If f : [a, b] → R+ is integrable, then the curvilinear trapezoid { } Γf = (x, y) ∈ R2 ; a ≤ x ≤ b, 0 ≤ y ≤ f (x) is squareable and we have: ∫b f (x) d x.

area(Γf ) =

(2.20)

a

Proof Let us take Δ : a = x0 < x1 < . . . < xi−1 < xi < . . . < xn = b an arbitrary partition of [a, b] and let ⎤} ⎤} { ⎡ { ⎡ m i = inf f (x); x ∈ xi−1 , xi , Mi = sup f (x); x ∈ xi−1 , xi , 1 ≤ i ≤ n. If we denote by E Δ =

∪n i=1

⎡

⎤ xi−1 , xi × [0, m i ], then E Δ ∈ E , E Δ ⊂ Γf and

area(E Δ ) =

n v i=1

It results that sΔ ≤ S∗ (Γf ) (Fig. 2.9).

m i (xi − xi−1 ) = sΔ .

2.7 Area of a Plane Figure

53

Fig. 2.9 The subgraph of an integrable function is squarable

Analogously, if we denote by FΔ = FΔ ⊃ Γf and

⎡

∪n i=1

⎤ xi−1 , xi × [0, Mi ], then FΔ ∈ E ,

area(FΔ ) = SΔ ≥ S ∗ (Γf ). Therefore, we have the inequalities: sΔ ≤ S∗ (Γf ) ≤ S ∗ (Γf ) ≤ SΔ .

(2.21)

The fact that f is integrable on [a, b] involves: ∗

∫b

I∗ = sup sΔ = inf SΔ = I = Δ

f (x) dx.

Δ

a

Finally, from (2.21), it results that: ∗

∫b f (x) dx.

area(Γf ) = S∗ (Γf ) = S (Γf ) = a

Corollary 2.7.1 Let f, g : [a, b] → R be two integrable functions on [a, b] such that f (x) ≤ g(x), ∀x ∈ [a, b]. If we denote by Γ f g the intergraph of these functions (Fig. 2.10), i.e. the set { } Γ f g = (x, y) ∈ R2 ; a ≤ x ≤ b, f (x) ≤ y ≤ g(x) , then Γ f g is a squareable set and we have the formula: ( ) area Γ f g =

∫b (g(x) − f (x))dx. a

54

2 Definite Integrals

Fig. 2.10 The area of an intergraph

Example 2.7.1 Compute the area of the ellipse. 2 2 The ellipse equation is ax 2 + by2 − 1 = 0. For symmetry reasons, it is sufficient to compute a quarter of the area of the ellipse, for example the hatched area from Fig. 2.11. ∩

The arc B A is the graph of the function: f (x) =

b √ 2 a − x 2 , x ∈ [0, a]. a

Applying Proposition 2.7.1 and Example 1.2.9, we obtain: ∫a

b √ 2 a − x 2 dx a 0 )| ( b x √ 2 a 2 −1 ( x ) ||a 2 = a − x + sin a 2 2 a |0 2 πab b a π · = . = · a 2 2 4

1 · Area (ellipse) = 4

Therefore, the area of the ellipse with the semiaxes a and b is equal to πab. Fig. 2.11 The area of an ellipse

2.7 Area of a Plane Figure

55

Fig. 2.12 The area of a circulars ector

In particular, for a = b = R, we obtain the area of the circle: π R 2 . Example 2.7.2 Compute the area of a circular sector. We intend to calculate the area of the circular sector O AB in Fig. 2.12. 2 From elementary geometry, it is known that the area of O AB is R 2·α . We will show that the circular sector O AB is squareable, in the sense of Definition 2.7.2, and its area is equal to the above area. Since the equation of the circle of radius R centered at the origin is x 2 + y 2 = R 2 , ∩

we deduce that the arc AB is the graph of the function: f (x) =

√

R 2 − x 2 , x ∈ [0, R sin α].

On the other hand, the straight line segment O B is the graph of the function: g(x) = x · tan

(π 2

) − α = x · ctan α, x ∈ [0, R sin α].

According to Corollary 2.7.1, we have: R∫sin α

(√

Area (O AB) =

) R 2 − x 2 − x · ctan α dx

0

)| ( √ | x R 2 −1 ( x ) x 2 sin · ctan α || R sin α − = R2 − x 2 + 0 2 2 R 2 2 2 2 2 R ·α R sin α cos α R2 · α R sin α cos α + − · = . = 2 2 2 sin α 2 Therefore, any circular sector is squareable and its area is equal to the known area of elementary geometry. Theorem 2.7.1 The necessary and sufficient condition that a bounded subset A ⊂ R2 to be squareable is that for any ε > 0, there are two elementary sets E ε and Fε such that E ε ⊂ A ⊂ Fε and area(Fε ) − area(E ε ) < ε.

56

2 Definite Integrals

Proof First, if S∗ (A) = S ∗ (A) = S( A), on account of the definition of the least upper bound and the greatest lower bound, it follows that there are E ε , Fε ∈ E , E ε ⊂ A ⊂ Fε such that: S( A) −

ε ε < area(E ε ) and area(Fε ) < S( A) + . 2 2

Therefore we have: area(Fε ) − area(E ε ) < ε. Conversely, if for any ε > 0, there are E ε , Fε ∈ E such that E ε ⊂ A ⊂ Fε and area(Fε ) − area(E ε ) < ε, then we have: 0 ≤ S ∗ ( A) − S∗ (A) < area(Fε ) − area(E ε ) < ε. Since ε > 0 is arbitrary, it results that S∗ (A) = S ∗ (A), thus A is squareable. Definition 2.7.3 A set Γ ⊂ R2 is said to be of area zero if for ∀ε > 0, ∃ Fε ∈ E such that Γ ⊂ Fε and area(Fε ) < ε. In particular, we have S ∗ (Γ) = 0, and taking into account that 0 ≤ S∗ (Γ) ≤ ∗ S (Γ), it results that the set Γ is squareable and area(Γ) = 0. Using this definition, Theorem 2.7.1 can be reformulated as follows: Theorem 2.7.2 The necessary and sufficient condition that a bounded subset A ⊂ R2 to be squareable is that its boundary to be of area zero. Proof First, if A is squareable, then ∀ε > 0, ∃ E ε , Fε ∈ E such that: E ε ⊂ A ⊂ Fε and area(Fε \E ε ) = area(Fε ) − area(E ε ) < ε. ◦

◦

Since Γ = ∂ A ⊂ Fε \ E ε and Fε \ E ε is also an elementary set, it results that Γ is of area zero. Conversely, we assume that for any ∀ε > 0, ∃ Fε ∈ E such that Γ = ∂ A ⊂ Fε and area(Fε ) < ε. Let D = [a, b] × [c, d] be a rectangle such that A ⊂ D. Obviously D\Fε ∈ E , hence D\Fε =

p ⊔

◦

◦

Di , Di ∩ D j = ∅, for i /= j.

i=1 ◦

◦

◦

For any i ≤ p, we have either Di ⊂ A, either Di ⊂ D\A, because otherwise, Di ◦

being a convex set, Di ∩ ∂ A /= ∅. But ∂ A ⊂ Fε and Fε ∩ Di = ∅. We have:

2.7 Area of a Plane Figure

57

⎛

⎞

⎜ ⊔ ◦⎟ A ⊂ A∪∂A ⊂ ⎜ Di ⎟ ⎝ ⎠∪ ∂A ◦

◦ Di ⊂A

⎛

⎛

⎞

⎞

◦ ⎜ ⊔ ◦⎟ ⎜ ⊔ ◦⎟ ⎜ Di ⊂ A ⊂ ∂ A ∪ ⎜ Di ⎟ Di ⎟ ⎝ ⎠ ⊂ Fε ∪ ⎝ ⎠ ◦ ◦ ◦ Di ⊂A Di ⊂A Di ⊂A

⊔

⎛

⎞ ⎛

⎞

⎜ ⊔ ◦⎟ ⎜ ⊔ ◦⎟ ⎜ Fε ∪ ⎜ Di ⎟ Di ⎟ ⎝ ⎠\⎝ ⎠ ⊂ Fε ⎛

◦ Di ⊂A

◦ Di ⊂A

⎛

⎞⎞

⎛

⎞

⎜ ⊔ ◦ ⎟⎟ ⎜ ⎜ ⊔ ◦⎟ ⎜ ⎟ ⎜ Area⎜ Di ⎟ Di ⎟ ⎝ Fε ∪ ⎝ ⎠⎠ − Area ⎝ ⎠ < ε. ◦ Di ⊂A

◦ Di ⊂A

The assertion follows now from Theorem 2.7.1. ◦

◦

Proposition 2.7.2 Let A1 and A2 be two squareable sets such that A1 ∩ A2 = ∅. Then their union A = A1 ∪ A2 is also a squareable set and: area(A) = area(A1 ) + area( A2 ). Proof The set A = A1 ∪ A2 is squareable because we use Theorem 2.7.2 and the fact that the boundary of A is embedded in the set Fr A1 ∪ Fr A2 , and this set is of area zero. According to Theorem 2.7.1, for any ε > 0, there are E i , Fi ∈ E , i = 1, 2, with the properties (Fig. 2.13): E 1 ⊂ A1 ⊂ F1 , E 2 ⊂ A2 ⊂ F2 area(F1 ) − area(E 1 ) < ε area(F2 ) − area(E 2 ) < ε. Further, we have: area(E 1 ) + area(E 2 ) ≤ area(A) ≤ area(F1 ∪ F2 ) ≤ area(F1 ) + area(F2 ) Fig. 2.13 The union of two squarable sets is a squarable set

A2

A1 A

58

2 Definite Integrals

Fig. 2.14 A plane curve defined by a polar equation is squarable

area(E 1 ) + area(E 2 ) ≤ area(A1 ) + area(A2 ) ≤ area(F1 ) + area(F2 ). From these inequalities, we deduce: | area(A) − (area(A1 ) + area(A2 ))| ≤ area(F1 ) − area(E 1 ) + area(F2 ) − area(E 2 ) < 2ε hence area(A) = area(A1 ) + area(A2 ). Remark 2.7.5 Every finite union of squareable sets with no common interior points is a squareable set and its area is equal to the sum of the areas of the sets of the family; in the case of an infinite union, the property is no longer preserved. Suppose now that a plane curve is defined by the polar equation: ρ = ρ(θ ), θ ∈ [α, β]. The plane figure bounded by this curve and two rays which start at the pole and make the angles α and β, respectively, with the polar axis is called a curvilinear sector (Fig. 2.14). Proposition 2.7.3 If ρ = ρ(θ ), θ ∈ [α, β] is a continuous positive function, then the corresponding curvilinear sector A = {(θ, ρ); α ≤ θ ≤ β, 0 ≤ ρ ≤ ρ(θ )} is squareable and has the area given by the formula: 1 area(A) = 2

∫β ρ 2 (θ )dθ .

(2.22)

α

Proof Let us consider Δn : α = θ0 < θ1 < . . . < θi−1 < θi < . . . < θn = β be an arbitrary equidistant partition of the interval [α, β].

2.7 Area of a Plane Figure

59

We denote by m i (respectively, Mi ) the greatest ⎡ lower⎤ bound (respectively, the least upper bound) of the function ρ = ρ(θ ), θ ∈ θi−1 , θi . The area of the circular sector O Ri Pi = {(θ, ρ); θi−1 ≤ θ ≤ θi , 0 ≤ ρ ≤ ρ(θ )} is 21 m i2 (θi − θi−1 ), and the area of the circular sector O Q i Ri−1 is 21 Mi2 (θi − θi−1 ) (Fig. 2.14). If we denote by Pn (respectively, Q n ) the union of these n circular sectors O Ri Pi (respectively, O Q i Ri−1 ), then Pn ⊂ A ⊂ Q n and we have: area(Pn ) =

n v 1 i=1

2

n v 1

m i2 (θi − θi−1 ), area(Q n ) =

i=1

2

Mi2 (θi − θi−1 ).

We note that the both sums are in fact the Darboux sums of the function 21 ρ 2 (θ ), θ ∈ [α, β] with respect to the partition Δn . Taking into account that limn→∞ ∥Δn ∥ = limn→∞ β−α = 0 and the function 21 ρ 2 (θ ) is integrable on [α, β], it results that: n 1 lim area(Pn ) = lim area(Q n ) = n→∞ n→∞ 2

∫β ρ 2 (θ ) dθ .

(2.23)

α

On the other hand, we know that Pn and Q n are squareable sets (Example 2.7.2), whence it results that for any ε > 0, there are two elementary sets E n and Fn with the properties: E n ⊂ Pn ⊂ A ⊂ Q n ⊂ Fn and area(Pn ) − area(E n )
0 such that | f (n+1) (x)| < M, ∀x ∈ [a, b], then: E( f ; x) =

|E( f ; x)| ≤

M · |Un+1 (x)|, ∀x ∈ [a, b]. (n + 1)!

Example 2.8.2 Evaluate the approximation error of the function f (x) = ln x, x ∈ (0, ∞) at the point x = 0.6, using the Lagrange interpolation polynomial with respect to the nodes x0 = 0.4; x1 = 0.5; x2 = 0.7; x3 = 0.8. In this case, we have: f (4) (x) = −

| | 6 6 and | f (4) (x)| ≤ ≈ 234.4, x ∈ [0.4; 0.8]. 4 x (0.4)4

From Corollary 2.8.1, we deduce: |E(ln x; x)| ≤

234.4 · |(x − 0.4)(x − 0.5)(x − 0.7)(x − 0.8)|. 4!

In particular, for x = 0.6, we have: |E(ln x; 0.6)| ≤

234.4 · 0.0004 ≈ 0.0039. 24

According to Formulas (2.25), (2.27) and (2.28), we obtain the expression of the approximation error for equidistant nodes, i.e.: (n+1) (ξt ) ~ f ; t) = f · πn+1 (t)h n+1 . E( (n + 1)!

(2.29)

64

2 Definite Integrals

Let L n be the Lagrange polynomial interpolating the function f : [a, b] → R with respect to the equidistant nodes xi = x0 + i h, i = 0, n, where h = b−a . n By integrating the relation f (x) = L n (x) + E( f ; x), x ∈ [a, b], we obtain: ∫b

∫b f (x) dx =

a

∫b L n (x) dx +

a

E( f ; x) dx.

(2.30)

a

If in the right member of the equality (2.30) we make the change of variable x = a + t · h and take into account the relations (2.26) and (2.29), it results that: ∫b

∫n f (x) dx =

a

~ L n (t) · h dt +

0

=h

n v

E( f ; t) · h dt 0

⎛ ⎝(−1)

i n−i C n

n!

i=0

+

∫n

∫n · 0

h n+2 (n + 1)!

∫n

⎞ πn+1 (t) ⎠ dt · f (xi ) (t − i )

πn+1 (t) · f (n+1) (ξt ) dt.

(2.31)

πn+1 (t) dt, i = 0, n (t − i )

(2.32)

0

Next, we use the notations: di = (−1)

n−i

∫n

Cni n!

0

Ai = h · di , i = 0, n h n+2 R( f ) = (n + 1)!

∫n

πn+1 (t) f (n+1) (ξt ) dt.

(2.33)

(2.34)

0

Taking into account in (2.31) the above notations, we obtain: ∫b f (x) dx = a

n v

Ai · f (xi ) + R( f ).

(2.35)

i=0

The numbers di are called the Newton–Côtes coefficients and Formula (2.35) is called the Newton–Côtes quadrature formula. For n ∈ {1, 2, 3}, the Newton–Côtes quadrature formula have special names: trapezoidal formula (n = 1), Simpson formula (n = 2) and Newton 3/8 formula (n = 3). Next, we present the trapezoidal formula and Simpson formula.

2.8 Approximating Definite Integral

65

Trapezoidal formula (n = 1) In this case we have only two nodes, x0 = a, x1 = b, and h = b − a. From (2.32) and (2.33), we obtain the following Newton–Côtes coefficients: (−1)1 C10 d0 = · 1!

∫1

t · (t − 1) dt = − t

0

(−1) 1!

0

d1 =

C11

0

(t − 1) dt =

h 1 b−a , A0 = = 2 2 2

0

∫1 ·

∫1

t · (t − 1) dt = t −1

∫1 t dt =

b−a 1 h , A1 = = . 2 2 2

0

Trapezoidal formula is: ∫b f (x) dx =

b−a ( f (a) + f (b)) + R( f ). 2

(2.36)

a

∫b Therefore, a f (x) dx is approximated by b−a f (a) + f (b)). 2 ( From a geometrical point of view, the trapezoidal formula approximates the area of the plane figure bounded by the graph of the function f , the axis O x and the straight lines x = a, x = b to the area of the hatched trapezoid in Fig. 2.16. From (2.34), we deduce that the error is: h3 R( f ) = 2

∫1

t(t − 1) f '' (ξt ) dt.

0

| } {| If we denote by M2 = sup | f '' (x)|; x ∈ [a, b] , then it results that:

y

Fig. 2.16 The approximation of the area of a subraph of a function by a trapezoid

O

f

a

b x

66

2 Definite Integrals

(b − a)3 M2 |R( f )| ≤ 2

∫1

(b − a)3 M2 |t(t − 1)| dt = 2

0

∫1 t(1 − t) dt 0

(b − a)3 M2 . = 12 Therefore, if we approximate (b−a) M2 . 12 3

∫b a

f (x) dx by

b−a 2 (

f (a) + f (b)), then the absolute

error is bounded by Obviously, such error is too big, so in practice is used the repeated trapezoidal method. The repeated trapezoidal method consists in dividing the interval [a, b] into n equal subintervals and approximating the integral on each subinterval by the trapezoidal formula. Therefore, we have: ∫b f (x) dx ≈ a

n v xi − xi−1 ( f (xi−1 ) + f (xi )). 2 i=1

Since x0 = a, xn = b and xi − xi−1 = ∫b

b−a f (x) dx ≈ 2n

b−a , n

the repeated trapezoidal formula is:

( f (a) + f (b) + 2

a

n−1 v

) f (xi ) .

(2.37)

i=1

The error that is obtained admits the following upper bound: |R( f )| ≤ n ·

M2 (b − a)3 (b − a)3 M2 · = . 12 n3 12n 2

(2.38)

∫2 x2 Example 2.8.3 Compute the definite integral −2 e− 2 dx using the repeated trapezoidal formula, with a smaller error than ε = 10−3 . We notice that this integral cannot be computed by Leibniz–Newton formula, x2 because the antiderivative of the function f (x) = e− 2 is not expressed by elementary functions. Using the above algorithm we have: x2

a = −2, b = 2, f (x) = e− 2

) x2 ( ) x2 ( x2 f ' (x) = −xe− 2 , f '' (x) = x 2 − 1 e− 2 , f ''' (x) = x 3 − x 2 e− 2 .

2.8 Approximating Definite Integral x

−∞

f ''' (x)

+

f '' (x)

0

67 √ − 3

−2 +

+

↗

3e−2

+

−

0

↗

2e

− 23

√ 3

0 0

↘

+

−1

↗

0 2e

− 23

+∞

2 −

−

−

−

↘

3e−2

↘

0

Since 2 < e < 3, it results that e32 < 43 < 1 and e√2 e < 2√2 2 < 1. From the table of variations of the function f '' , we deduce that: | | } | {| M2 = sup | f '' (x)|; x ∈ [−2, 2] = | f '' (0)| = 1. M2 = If we put (b−a) 12 n 2 −3 10 . On the other hand, 3

43 12 n 2

< ε = 10−3 , then the absolute error becomes |R( f )|
· 103 = 5333, (3) ⇒ n > 73.03. 12 n 2 3 Therefore, in Formula (2.37) we take n = 74 and we find: ∫2

x2

e− 2 dx ≈

−2

2 − (−2) ( f (−2) + f (2) + 2( f (x1 ) + · · · + f (x73 ))) = 2.3924 2 · 74

where xi = −2 + i ·

4 74

= −2 + i ·

2 , 37

i = 0.74.

Simpson Formula (n = 2) Simpson formula corresponds to three nodes, x0 = a, x1 = . h = b−a 2 The Newton–Côtes coefficients are: (−1)2 C20 d0 = 2!

∫2

1 t(t − 1)(t − 2) dt = t 2

0

(−1)1 C21 d1 = 2!

∫2 0

(−1) C22 2! 2

d2 =

∫2 0

∫2

a+b , 2

x2 = b, and

(2 ) 1 t − 3t + 2 dt = , 3

A0 =

b−a 6

0

t(t − 1)(t − 2) dt = − t −1 1 t(t − 1)(t − 2) dt = t −2 2

∫2

(2 ) 4 t − 2t dt = , 3

A1 =

4(b − a) 6

0

∫2 0

) (2 1 t − t dt = , 3

A2 =

b−a . 6

68

2 Definite Integrals

Simpson formula is: ∫b f (x) dx ≈

b−a 6

(

( f (a) + 4 f

a+b 2

)

) + f (b) .

(2.39)

a

If f : [a, b] → R is a function of C 4 —class on [a, b], then it can be shown that: |R( f )| ≤

(b − a)5 M4 2880

(2.40)

where: | } {| M4 = sup | f (4) (x)|; x ∈ [a, b] . From a geometrical point of view, the Simpson formula approximates the area of the subgraph of the function f to the hatched area in Fig. 2.17, i.e. the area of the plane figure bounded by the parabola P2 (the 2nd degree Lagrange polynomial that , b), the axis O x and the interpolates the function f with respect to the nodes a, a+b 2 straight lines x = a, x = b. In the same manner as the trapezoidal method, the repeated Simpson method is used for the best possible approximation. In this case, we divide the interval [a, b] ⎡ ⎤ into 2n equal subintervals, and we apply Formula (2.39) for each interval x2i−2 , x2i . Thus, we have: ∫b a

) ( ) ( n v x2i − x2i−2 x2i−2 + x2i f (x2i−2 ) + 4 f + f (x2i ) . f (x) dx ≈ 6 2 i=1

Since x2i − x2i−2 =

b−a , n

Fig. 2.17 The approximation of a subraph of a function f by a subgraph of a parabola

we deduce that the repeated Simpson formula is:

y

f

P2

O

a

a+b 2

b x

2.8 Approximating Definite Integral

∫b

b−a f (x) dx ≈ 6n

69

( f (a) + f (b) + 4

n v

f (x2i−1 ) + 2

i=1

a

n−1 v

) f (x2i ) .

(2.41)

i=1

From (2.40), it follows that: |R( f )| ≤

(b − a)5 M4 . 2880 n 4

(2.42)

∫2 x2 Example 2.8.4 Compute the definite integral −2 e− 2 dx using the repeated Simpson formula, with a smaller error than ε = 10−3 . x2

We have: a = −2, b = 2, f (x) = e− 2 ; ( ) x2 ( ) x2 f (4) (x) = x 4 − 6x 2 + 3 e− 2 , f (5) (x) = −x x 4 − 10x 2 + 15 e− 2 . √ √ − 5 − 10

√ √ 5 − 10

x

−2

f (5)

−

−

0

+

0

−

0

+

+

f (4)

−5e−2

↘

m

↗

3

↘

m

↗

−5e−2

0

2

We remark that: √ (/ ( / ) ) √ √ 8 − 4 10 not = m < 0 and √ f (4) − 5 − 10 = f (4) 5 − 10 = 5− 10 e 2 | | | 8 − 4√10 | 4√10 − 8 | | √ √ |m| = | < 3. |= 5− 10 | e 5−2 10 | e 2 Indeed, using the known inequality ex > x + 1, x /= 0, it results that: 3e

√ 5− 10 2

( >3 1+

5−

√ ) √ √ 10 21 3 10 − > 4 10 − 8, = 2 2 2

whence, we deduce that: √ 4 10 − 8 e

√ 5− 10 2

< 3.

Therefore | | } | {| M4 = sup | f (4) (x)|; x ∈ [−2, 2] = | f (4) (0)| = 3

70

2 Definite Integrals

According to (2.42), to find the number of nodes we put the condition: 3 · 45 < 10−3 2880 n 4 whence, it results that: / n≥

4

3 · 45 · 1000 = 2880

/ 4

3200 √ 4 ≈ 1067 ≈ 5.715. 3

It can be taken n = 6 and applying Formula (2.41), we obtain: ∫2

2

e −2

− x2

1 dx ≈ 18

( f (−2) + f (2) + 4

6 v i=1

f (x2i−1 ) + 2

5 v

) f (x2i )

i=1

≈ 2.3925 4 = −2 + i · 13 , i = 0.12. where xi = −2 + i · 12 We notice that in this example the computations for the Simpson method are less than for the trapezoidal method. Indeed, the Simpson method computes the values of function in 12 nodes, while the trapezoidal method calculates the values of the function in 74 nodes. At the end of this paragraph we mention the fact that, in practice, the approximate computation of integrals is done using specialized program packages, of which the most used are MATLAB, Maple, Mathcad, and so on.

Chapter 3

Improper Integrals

3.1 Convergence and Divergence of Improper Integrals The definite integral theory was made for bounded functions defined on closed and bounded intervals. In the following, we will generalize the notion of integral for situations in which only one or ∫both of these ∫conditions are∫no longer fulfilled, giving ∞ a ∞ a sense of integrals of the form a f (x)dx, −∞ f (x)dx, −∞ f (x)dx, respectively, ∫b a f (x)dx, where a and b are finite and f is unbounded on [a, b) or (a, b] or (a, b). The first integrals, for which the integration interval is infinite, are called improper integrals of the first kind, while the integrals for which a and b are finite and the function f is not bounded (i.e. lim x→b | f (x)| = ∞ or lim x→a | f (x)| = ∞) are called x>a x 1, then: ∫u lim

u→∞ 1

1 1 u 1−α − 1 =− ∈ R, dx = lim α u→∞ x 1−α 1−α

whence we deduce that the improper integral value: ∫∞ 1

∫∞ 1

1 dx xα

is convergent and has the

1 1 . dx = xα α−1

∫∞ ∫u If α ≤ 1, then limu→∞ 1 x1α dx = ∞, thus 1 x1α dx is divergent. ∫∞ 1 In particular, the improper integral 1 x 2 dx is convergent and its the value is ∫∞ 1 ∫∞ 1 √ dx is divergent. 1 x 2 dx = 1, and the improper integral 1 x ∫b ∫b 1 Example 3.1.4 Prove that the improper integrals a (x−a) α dx and a convergent if α < 1 and divergent if α ≥ 1, where a and b are finite.

1 dx (b−x)α

are

74

3 Improper Integrals

For any a < v < b, we have: ∫b v

1 dx = (x − a)α

∫

b−a v−a (b−a)1−α −(v−a)1−α 1−α

ln

if α = 1 if α /= 1.

We notice that if α < 1, then limv→a (v − a)1−α = 0, which implies that the v>a ∫b 1 integral a (x−a) α dx is convergent and has the value: ∫b a

1 (b − a)1−α , α dx = 1−α (x − a)

∫b 1 and if α ≥ 1, the improper integral a (x−a) α dx is divergent. ∫2 ∫2 1 In particular, 1 √x−1 dx is convergent and 1 (x−1)1√x−1 dx is divergent. ∫b 1 Similarly, it is shown that a (b−x) α dx is convergent if α < 1 and divergent if ∫1 1 ∫1 1 dx is divergent. α ≥ 1. In particular, 0 √1−x dx is convergent and 0 1−x Definition 3.1.4 Let f : [a, c) ∪ (c, b] → R be a locally integrable function on [a, c) ∪ (c, b] (i.e. integrable on any compact interval [u, v] ⊂ [a, c) ∪ (c, b]), where a and b are finite. ∫b We say that the improper integral a f (x)dx is convergent if both improper ∫c ∫b integrals a f (x)dx and c f (x)dx are convergent, that is, if the limit: ⎞ ⎛ c−ε ∫ ∫b lim+ ⎝ f (x)dx + f (x)dx ⎠

ε→0 η→0+

a

c+η

exists and it is finite. In this case, the value of the integral is given by: ∫b

∫u f (x)dx = u→c lim uc

f (x)dx. v

Definition 3.1.5 Let f : [a, c) ∪ (c, b] → R be a locally integrable function on [a, c) ∪ (c, b]. We say that f is Cauchy integrable on [a, b] if there is the finite limit ⎞ ⎛ c−ε ∫ ∫b ∫b lim ⎝ f (x)dx + f (x)dx ⎠ = (v.p.) f (x)dx.

ε→0 ε>0

a

c+ε

a

The limit is called the principal value of the improper integral in Cauchy sense.

3.1 Convergence and Divergence of Improper Integrals

Example 3.1.5 The improper integral

∫2

1 −1 x dx

75

is divergent because:

⎛ −ε ⎞ ∫ ∫2 1 1 2ε lim+ ⎝ dx + dx ⎠ = lim+ (ln ε + ln 2 − ln η) = lim+ ln x x η ε→0 ε→0 ε→0 + + + η→0

−1

η→0

η

η→0

and this limit does not exists. On the other hand, we have: ∫2 (v.p.) −1

⎞ ⎛ −ε ∫ ∫2 1 1 1 dx = lim+ ⎝ dx + dx ⎠ = lim (ln ε + ln 2 − ln ε) = ln 2. ε→0 ε→0 x x x ε

−1

Definition 3.1.5 Let f : R → R be a locally integrable function on R (i.e. f is integrable on any ∫ ∞compact interval [u, v] ⊂ R). ∫u We say that −∞ f (x)dx is convergent if the limu→∞ v f (x)dx exists and it v→−∞ is finite. Therefore, if the integral is convergent, we have: ∫∞

∫u f (x)dx = lim

u→∞ v→−∞

−∞

f (x)dx. v

The following limit is called the principal value of the integral in Cauchy sense. ∫∞ (v.p.)

∫u f (x)dx = lim

u→∞ −u

−∞

f (x)dx.

Convergence in Cauchy sense ∫ ∞does not imply the convergence of improper integral. Sometimes the integral −∞ f (x)dx is divergent, but its principal value is finite. ∫∞ Example 3.1.6 Study the nature of the improper integral −∞ sin x dx. ∫u Since limu→∞ v sin x dx = limu→∞ (− cos u + cos v) does not exists, it v→−∞ ∫ v→−∞ ∞ results that −∞ sin x dx is divergent. On the other hand, we have: ∫∞ (v.p.) −∞

∫u sin x dx = lim

u→∞ −u

sin x dx = lim (− cos u + cos u) = 0. u→∞

76

3 Improper Integrals

3.2 Convergence Criteria for Improper Integrals Theorem 3.2.1 Let f : [a, b) → R be a locally integrable function on [a, b), where b is finite or not. The necessary and sufficient condition for the improper integral ∫b a f (x)dx to be convergent is for any ε > 0, there is a < δε < b such that | | u '' | |∫ | | | f (x)dx | < ε, for all u ' , u '' ∈ [δε , b). | | | |' u

∫u Proof For any a < u < b, we shall denote by F(u) = a f (x)dx. ∫b According to Definition 3.1.2, the integral a f (x)dx is convergent if and only if the limit limu→b F(u) exists and it is finite. u 0, there is a neighborhood Vε of b such that: | ( ') ( )| | F u − F u '' | < ε, for any u ' , u '' ∈ Vε ∩ [a, b). If b is finite, we can assume that Vε has the form Vε = (b − ηε , b + ηε ), where a < b − ηε < b, and we shall choose δε = b − ηε . If b = ∞, we can assume that Vε has the form Vε = (δε , ∞), where δε > a. In both cases, if u ' , u '' ∈ (δε , b), it results that u ' , u '' ∈ Vε ∩ [a, b), hence | ( ') ( )| | F u − F u '' | < ε. On the other hand, it is easily seen that: | u' | | u '' | |∫ | |∫ | ∫u '' | | | | ( ') ( '' )| | | | | F u − F u | = | f (x)dx − f (x)dx | = | f (x)dx ||. | | | |' | a

a

u

∫b Therefore, a f (x)dx is convergent if and only if ∀ε > 0, ∃a < δε < b such that for any u ' , u '' ∈ (δε , b), we have: | | u '' | |∫ | | | f (x)dx | < ε. | | | |' u

∫b Definition 3.2.1 We say that the improper integral a f (x)dx is absolutely ∫b convergent if the improper integral a | f (x)|dx is convergent.

3.2 Convergence Criteria for Improper Integrals

77

Theorem 3.2.2 Any absolutely convergent improper integral is convergent. Proof The statement follows immediately from Theorem 3.2.1 and from the inequality: | | | u '' | u '' | | |∫ |∫ | | | | | f (x)dx | ≤ | | f (x)|dx |. | | | | | | |' |' u

u

Remark 3.2.1 The reciprocal statement of Theorem 3.2.2 is not generally true. There are improper convergent integrals without being absolutely convergent; a such ∫∞ example is Dirichlet integral 0 sinx x dx which is convergent (Example 3.2.9), but it is not absolutely convergent. Theorem 3.2.3 (The first comparison test) Let f, g: [a, b) → R+ be two locally integrable functions on [a, b), where b is finite or not, such that 0 ≤ f (x) ≤ g(x), ∀x ∈ [a, b). ∫b ∫b (1) If a g(x)dx is convergent, then a f (x)dx is convergent and: ∫b 0≤

∫b f (x)dx ≤

a

g(x)dx. a

∫b f (x)dx is divergent, then a g(x)dx is divergent. ∫u ∫u Proof Let F(u) = a f (x)dx and G(u) = a g(x)dx, where a < u < b. Since 0 ≤ f (x) ≤ g(x), ∀x ∈ [a, b), from Proposition 2.6.3 and Proposition 2.6.4, it results that: (2) If

∫b a

0 ≤ F(u) ≤ G(u), ∀a < u < b. On the other hand, the functions F and G are increasing functions, because F ' (u) = f (u) ≥ 0 and G ' (u) = g(u) ≥ 0, ∀a < u < b. ∫b (1) If we suppose now that a g(x)dx is convergent, it follows that the limit L = limu→b G(u) exists and it is finite. u a such that g(x) > ε, for any x ∈ ∫b ∫b [b − δε , b). Because the integrals a g(x)dx and b−δε ε g(x)dx are divergent, then from inequality f (x) > ε g(x), x ∈ [b − δε , b) and from Theorem 3.2.3, ∫b ∫b we deduce that b−δε f (x)dx is divergent, hence a f (x)dx is divergent.

Example 3.2.2 Study the nature of the improper integrals: ∫ ∞ −1 ∫1 x √ dx. (1) 1 tanx x dx and (2) 0 cos x (1) Let us denote by f (x) =

tan−1 x x

> 0 and by g(x) =

1 x

> 0, ∀x ∈ [1, ∞).

−1

f (x) = lim x→∞ tanx x · x = lim x→∞ tan−1 x = π2 ∈ (0, ∞) and Since lim x→∞ g(x) ∫∞ 1 ∫ ∞ tan−1 x dx 1 x dx is divergent (Example 3.1.3), from Theorem 3.2.4, it results that 1 x is divergent.

(2) If we denote by f (x) = lim

x→0 x >0

cos √x x

> 0 and by g(x) =

√1 x

> 0, ∀x ∈ (0, 1], then

f (x) cos x √ = lim √ · x = lim cos x = 1 ∈ (0, ∞). x→0 x→0 g(x) x x>0

x>0

∫1 ∫1 Since the integral 0 √1x dx is convergent (Example 3.1.4), it results that 0 is convergent, according to Theorem 3.2.4.

cos √ x dx x

Theorem 3.2.5 Let f : [a, ∞) → R+ be a locally integrable function on [a, ∞). ∫∞ (1) If there is α > 1 such that lim x→∞ x α · f (x) /= ∞, then the integral a f (x)dx is convergent. ∫∞ (2) If there is α ≤ 1 such that lim x→∞ x α · f (x) > 0, then the integral a f (x)dx is divergent. ∫∞ In particular, if there is lim x→∞ f (x) and a f (x)dx is convergent, then lim x→∞ f (x) = 0. Proof (1) Let α > 1 and l = lim x→∞ x α · f (x) < ∞. On account of the definition of the limit of a function, it results that for ∀ ε > 0, ∃δε > 0 such that: l − ε < x α · f (x) < l + ε, ∀x > δε and further: l −ε l +ε < f (x) < α , ∀x > δε . xα x ∫∞ As δε l+ε dx is convergent (Example 3.1.3), according to Theorem 3.2.3, xα ∫∞ it results that the integral δε f (x)dx is convergent. From Remark 3.1.1, we ∫∞ deduce now that the integral a f (x)dx is convergent.

80

3 Improper Integrals

(2) Let α ≤ 1 and l = lim x→∞ x α · f (x) ∈ (0, ∞). Since l > 0, we can assume that there is ε > 0 with the property 0 < ε < l. For such ε > 0, there is δε > a such that: l − ε < x α · f (x) < l + ε, ∀x > δε and further: l −ε l +ε < f (x) < α , ∀x > δε . α x x ∫∞ Because the integral δε l−ε dx is divergent (Example 3.1.3), from Theorem 3.2.3, ∫ x∞α we deduce that the integral δε f (x)dx is divergent, and from Remark 3.1.1, we have ∫∞ that the integral a f (x)dx is divergent. If there is α ≤ 1 such that l = lim x→∞ x α · f (x) = ∞, then for ∀ε > 0, ∃ δε > a with the property x α · f (x) > ε, ∀x ∈ (δε , ∞). Therefore f (x) >

ε , ∀x > δε . xα

∫∞ ∫∞ Since the integral δε xεα dx is divergent, it follows that δε f (x)dx is divergent, ∫∞ hence a f (x)dx is divergent (Remark 3.1.1). ∫ ∞ P(x) dx, where P and Q are Example 3.2.3 Study the convergence of the integral a Q(x) degree(Q) ≥ degree(P)+2 and Q(x) /= 0, ∀x > a. polynomial functions, |such that | ∫ ∞ P(x) 2 | P(x) | Since lim x→∞ x · | Q(x) | < ∞, from Theorem 3.2.5, it results that a Q(x) dx is absolutely convergent, so it is convergent, according to Theorem 3.2.2. ∫∞ 2 Example 3.2.4 Prove that the Euler–Poisson integral −∞ e−x dx is convergent. ∫ ∞ −x 2 ∫ ∞ −x 2 It is obvious that −∞ e dx = 2 0 e dx. 2 2 Since lim x→∞ x 2 · e−x = lim x→∞ exx 2 = 0, according to Theorem 3.2.5, it results ∫∞ 2 that the integral 0 e−x dx is convergent. Theorem 3.2.6 Let f : [a, b) → R+ , be a locally integrable function on [a, b), with b finite or not. (1) If there is α < 1 such that lim x→b (b − x)α · f (x) < ∞, then the integral x 0, then the integral xa ∫b a f (x)dx is divergent. Proof We proceed in the same manner as in the proof of Theorem 3.2.5, using the ∫b 1 fact that a (x−a) α dx is convergent for α < 1 and divergent for α ≥ 1 (Example 3.1.4). Example 3.2.6 Study the nature of the following improper integrals: ∫2 √ 0

∫2

1 x(x + 1)3

√

dx and 0

1 x 3 (x + 1)

dx.

Using Theorem 3.2.7, it results that the first integral is convergent because: 1

1

lim x 2 · √

x→0 x>0

x(x + 1)

3

= lim √ x→0 x>0

1 (x + 1)

3

= 1 < ∞ and α =

1 < 1, 2

while the second integral is divergent because: 3

lim x 2 · √

x→0 x>0

1 x 3 (x

+ 1)

= lim √ x→0 x>0

1 x +1

= 1 > 0 and α =

3 ≥ 1. 2

82

3 Improper Integrals

The following theorem establishes a strong connection between an improper integral with infinite limits and the numerical series that can be constructed using the integrated function. Theorem 3.2.8 (Cauchy integral criterion) If ∫f : [1, ∞) → R+ is a monotonically ∞ decreasing function, then the improper integral 1 f (x)dx and the numerical series Σ ∞ n=1 f (n) have the same nature. Proof Since f (n) ≤ f (x) ≤ f (n − 1), for any x ∈ [n − 1, n], we have: ∫n f (n) ≤

f (x)dx ≤ f (n − 1), ∀n ≥ 2. n−1

Adding these inequalities, we get: m Σ

∫m f (n) ≤

n=2

f (x)dx ≤

m−1 Σ

f (n), ∀m ≥ 2.

(3.1)

n=1

1

Σ If we assume that the series ∞ n=1 f (n) is convergent, it results that there is M > 0 Σm−1 such that n=1 f (n) < M, ∀m ≥ 2. Using (3.1), we deduce that: ∫m f (x)dx ≤

m−1 Σ

f (n) < M, ∀m ≥ 2.

n=1

1

Let u > 1 be arbitrary and let m ∈ N∗ , m > u. Since f ≥ 0, it results that: ∫u

∫m f (x)dx ≤

1

f (x)dx < M. 1

∫u ∫∞ Therefore, exists limu→∞ 1 f (x)dx ≤ M,Σ hence 1 f (x)dx is convergent. ∞ If we assume is divergent, then it results Σmnow that the numerical series n=1 f∫(n) m that limm→∞ n=1 f (n)∫ = ∞ and so that limm→∞ 1 f (x)dx = ∞, whence we ∞ deduce that the integral 1 f (x)dx is divergent. ∫∞ Σ 1 Example 3.2.7 The improper integral 1 x1α dx and the numerical series ∞ n=1 n α have the same nature, i.e. are convergent for α > 1 and divergent for α ≤ 1. Theorem 3.2.9 (Abel–Dirichlet criterion) Let f, g: [a, b) → R, with b finite or not, be two functions having the properties:

3.2 Convergence Criteria for Improper Integrals

83

| |∫ u (1) f is continuous and there is M > 0 such that | a f (x)dx | ≤ M, ∀a < u < b. (2) g is positive, monotonically decreasing, of C 1 —class on [a, b) and lim x→b g(x) = 0. x 0 such that | f (x, t)| < M, ∀(x, t) ∈ D. Furthermore, we have: | | | β(t | | ∫ 0) | | | | f (x, t) − f (x, t0 )|dx | |F(t) − F(t0 )| ≤ | | | |α(t0 ) | + M|β(t) − β(t0 )| + M|α(t) − α(t0 )|. On the other hand, the function f is uniformly continuous on D (see ) ( 5.6.3, ) ( Theorem > 0,| ∃δε' > 0 such that ∀ x ' , t ' , x '' , t '' ∈ Part 1, Differential Calculus), hence ∀ε | | ' | D, with the property |x − x '' | < δε' , |t ' − t '' | < δε' , we have: | ( ' ') ( )| | f x , t − f x '' , t '' |
0 such that ∀t ∈ [c, d], with |t − t0 | < δε'' , we have: |α(t) − α(t0 )|
0, ∃ a < δt,ε < b such that | u ' f (x, t) dx | < ε, ∀ u ' , u '' ∈ δt, ε , b . There is another type of convergence, with properties better than pointwise convergence, in which δ depends only on ε and does not depend on t. This type of convergence is called uniform convergence. Specifically, we have: Definition 4.2.2 Let the function f : [a, b) × [c, d] → R, with b finite or not. ∫b The improper integral depending on a parameter a f (x, t) dx is |called uniformly | |∫ b | convergent on the interval [c, d] if ∀ ε > 0, ∃ a < δε < b such that | u f (x, t) dx | < ε, ∀ u ≥ δε and ∀ t ∈ [c, d]. Remark 4.2.2 Let the function f : [a, b) × [c, d] → R, with b finite or not. ∫b Then a f (x, t) dx |is uniformly convergent on [c, d] if and only if ∀ ε > 0, | |∫ u '' | ' ∃ a < δε < b such that | u ' f (x, t) dx | < ε, ∀ u , u '' ∈ (δε , b), ∀ t ∈ [c, d]. Remark 4.2.3 Uniform convergence implies pointwise convergence, according to Remarks 4.2.2 and 4.2.1; the reciprocal statement is not generally true. Theorem 4.2.1 (Weierstrass criterion) Let the function f : [a, b) × [c, d] → R, with b finite or not. If there is a function ϕ: [a, b) → R+ with the properties: (1) | f (x, t)| ≤ ϕ(x), ∀(x, t) ∈ [a, b) × [c, d] ∫b (2) a ϕ(x)dx is convergent then the improper integral depending on a parameter ∫b a f (x, t) dx is uniformly convergent on the interval [c, d]. ∫b

ϕ(x) dx by Theorem 3.2.1, it | |∫ is'' convergent, | | u follows that ∀ ε > 0, ∃ a < δε < b such that | u ' ϕ(x) dx | < ε, ∀ u ' , u '' ∈ (δε , b). Therefore, ∀ u ' , u '' ∈ (δε , b), ∀ t ∈ [c, d], we have:

Proof Since the improper integral

a

| | u '' | | u '' | | u '' | |∫ | |∫ | |∫ | | | | | | | f (x, t) dx | < | | f (x, t)| dx | < | ϕ(x) dx | < ε | | | | | | | |' | |' | |' u

whence it results that the integral

u

u

∫b a

f (x, t) dx is uniformly convergent on [c, d].

4.2 Improper Integrals Depending on a Parameter

95

Example 4.2.1 Prove that the improper integral convergent on R. Indeed, for any t ∈ R we have

∫∞ 0

e−x cos(t x) dx is uniformly

| | −x |e cos(t x)| ≤ e−x , ∀ x ∈ [0, ∞). On the other hand, the improper integral ∫∞

∫u

−x

e dx = lim

u→∞

0

∫∞ 0

e−x dx is convergent, because:

| | e dx = − lim e | u→∞ 0 −x |u

−x

0

( ) = lim −e−u + 1 = 1 ∈ R. u→∞

According to Theorem 4.2.1, the improper integral uniformly convergent on R..

∫∞ 0

e−x cos(t x) dx is

Lemma 4.2.1 Let the function f : [a, b) × [c, d] → R, with b finite or not, let {bn } be a real number sequence such that a < bn < b and limn→∞ bn = b and ∫b ∫b let us consider Fn (t) = a n f (x, t) dx, ∀ t ∈ [c, d]. If a f (x, t) dx is uniformly convergent on [c, d], then the sequence of functions {Fn } is uniformly convergent on [c, d] to the function F, where: ∫b F(t) =

∫u f (x, t) dx = lim

u→b u 0, ∃ a < δε < b such that ∀ u ' , u '' ∈ (δε , b) and ∀ t ∈ [c, d], we have: | | u '' | |∫ | | | f (x, t) dx | < ε. | | | |'

(4.7)

u

Since limn→∞ bn = b, it results that there is n ε ∈ N∗ such that bn ∈ (δε , b), for any n ≥ n ε . If we assume now that n ≥ n ε and m ≥ n ε , from (4.7), we get: |b | |∫ m | | | |Fn (t) − Fm (t)| = || f (x, t) dx || < ε, ∀ t ∈ [c, d] | | bn

(4.8)

96

4 Integrals Depending on Parameter

Therefore, the sequence of functions {Fn } is uniformly Cauchy (fundamental) on [c, d], hence it is uniformly convergent on [c, d]. On the other hand, it is obvious that for any t ∈ [c, d] we have: ∫bm f (x, t) dx = F(t).

lim Fm (t) = lim

m→∞

m→∞ a

Passing to the limit in (4.8) on m → ∞, we get: |Fn (t) − F(t)| ≤ ε, ∀ t ∈ [c, d], hence Fn →u[c,d] F. Theorem 4.2.2 Let f : [a, b) × [c, d] → R be a continuous function, where b is ∫b finite or not. If we assume that the integral a f (x, t) dx is uniformly convergent on ∫b [c, d], then the function F: [c, d] → R, defined by F(t) = a f (x, t) dx, ∀ t ∈ [c, d] is also continuous on [c, d]. ∫b Proof Let a < bn < b, limn→∞ bn = b, and let Fn (t) = a n f (x, t) dx, t ∈ [c, d]. According to Theorem 4.1.1, we deduce that the functions Fn are continuous on [c, d], n ∈ N∗ . On the other hand, from Lemma 4.2.1, it follows that Fn →u[c,d] F. Finally, from Theorem 3.2.1 from [6], we deduce that F is continuous on [c, d]. Remark 4.2.4 Under the conditions of Theorem 4.2.2, the following equality holds: ∫b lim F(t) = lim

t→t0

∫b f (x, t) dx =

t→t0 a

f (x, t0 ) dx = F(t0 ). a

Theorem 4.2.3 Let us consider the function f : [a, b) × [c, d] → R, with b finite or not, which verified the conditions: (i) f and ∂∂tf are continuous on [a, b) × [c, d] ∫b (ii) the integral a f (x, t) dx is uniformly convergent on [c, d] ∫b (iii) the integral a ∂∂tf (x, t) dx is uniformly convergent on [c, d]. ∫b Then the function F: [c, d] → R, F(t) = a f (x, t) dx, ∀ t ∈ [c, d] is differentiable on [c, d] and the following formula holds: ⎞ ⎛ b ∫ ∫b ∂f ∂ ' ⎠ ⎝ f (x, t)dx = F (t) = (x, t) dx, ∀ t ∈ [c, d]. ∂t ∂t a

a

Proof Let the real number sequence a < bn < b, with limn→∞ bn = b and let us ∫b take Fn (t) = a n f (x, t) dx, t ∈ [c, d]. Obviously, the sequence of functions {Fn } is uniformly convergent on [c, d] to the function F.

4.2 Improper Integrals Depending on a Parameter

97

On the other hand, from Theorem 4.1.2, we deduce that the functions Fn are differentiable on [c, d] and: Fn' (t)

∫bn = a

∂f (x, t) dx, ∀t ∈ [c, d], ∀ n ∈ N∗ . ∂t

∫b We also notice that, if we denote by G(t) = a ∂∂tf (x, t)dx, ∀t ∈ [c, d], according to Lemma 4.2.1, it results that Fn' →u[c,d] G. Finally, from Theorem 3.2.2 from [6], it follows that F is differentiable on [c, d] and F ' (t) = G(t), ∀t ∈ [c, d]. ∫π cos x) Example 4.2.2 Compute F(t) = 0 ln(1+t dx, t ∈ (−1, 1). cos x ln(1+t cos x) cos x) π = lim x→ π2 t · ln(1+t = t, i.e. the point First, we remark that lim x→ 2 cos x t cos x cos x) , t ∈ (−1, 1). x = π2 is not a real singularity for the function f (x, t) = ln(1+t cos x ∂f 1 cos x l We have ∂t (x, t) = cos x · 1+t cos x = 1+t cos x . ∫π ∫π We notice that the integral 0 ∂∂tf (x, t)dx = 0 1+t lcos x dx is uniformly convergent on any interval [a,[ b] ⊂] (−1, 1). Indeed, if x ∈ 0, π2 , then cos x ≥ 0 and: 0< ∫

π 2

l l < , ∀ t ∈ [a, b] ⊂ (−1, 1). 1 + t cos x 1 + a cos x

l 1+a cos x

dx is a proper integral, so convergent, from Weierstrass criterion ∫π (Theorem 3.2.2), it results that 02 1+t lcos x dx is uniformly convergent on [a, b]. Similarly, we have: Since

0

[π ] l l < ,x ∈ , π , t ∈ [a, b] 1 + t cos x 1 + b cos x 2 ∫π whence we deduce that the integral π 1+t lcos x dx is uniformly convergent on [a, b]. 2 Furthermore, from Theorem 4.2.3, we have: 0
1, according to ∫ π2 1 1 1 < a 2 −sin Theorem 4.2.1, because t 2 −sin 2 2 , if t > a > 1 and 0 a 2 −sin2 x dx is x x a proper integral. From Theorem 4.2.3, it follows that: π

π

F ' (t) =

∫2 0

∂f (x, t)dx = ∂t

∫2

t2 0

2t dx, t > 1. − sin2 x

4.2 Improper Integrals Depending on a Parameter

99

If we make the change of variable u = tan x2 , then dx = F ' (t) =

∫∞ 0

2t t2

u2 1+u 2

−

∫∞

·

1 du = 2 t 1 + u2

∫∞ 0

2 du, 1+u 2

1 ( ) du t 2 − 1 u2 + t 2

√ 2 · du = t 2 − 1 tan−1 2 2−1 t 2 t u + t 2 −1 0 )|∞ (√ t 2 − 1 || π 2 −1 u | =√ tan . =√ 2 2 | t t −1 t −1 2t = 2 t −1

and we have:

1

(√

t2 − 1 u t

)|∞ | | | |

0

Therefore, we obtain: ∫ ∫ ( ) √ π dt = π ln t + t 2 − 1 + C , t > 1. F(t) = F ' (t)dt = √ t2 − 1 On the other hand, we have: ⎛ π ⎞ ∫2 ) ( √ ( ) ⎜ ⎟ C = lim ⎝ ln t 2 − sin2 x dx − π ln t + t 2 − 1 ⎠ t→∞

⎛ ⎜ = lim ⎝

0 π

∫2

t→∞

⎛

⎞ ( ) ( ) √ sin2 x ⎟ dx − π ln t + t 2 − 1 ⎠ ln t 2 1 − 2 t

0

⎞ π ( ∫2 ( ( ) 2 )) √ sin x ⎜ ⎟ 2 ln t + ln 1 − 2 dx − π ln t + t 2 − 1 ⎠ = lim ⎝ t→∞ t 0

∫

π 2

= lim

t→∞

) ( ( ( )) √ sin2 x dx + lim π ln t − π ln t + t 2 − 1 ln 1 − 2 t→∞ t

0

∫ = 0

π 2

) ) ( 1 sin2 x t dx + lim π ln = π ln . lim ln 1 − 2 √ 2 t→∞ t→∞ t 2 t + t −1 (

Finally, we get: π

∫2 F(t) = 0

( ) √ (2 ) t + t2 − 1 2 ln t − sin x dx = π ln , t > 1. 2

0

100

4 Integrals Depending on Parameter

∫∞ Example 4.2.4 Prove that the Dirichlet integral has the value 0 sinx x dx = π2 . ∫∞ The fact that the Dirichlet integral 0 sinx x dx is convergent was established in the Example 3.2.9. We consider now the following improper integral depending on parameter t ∈ [0, ∞): ∫∞ F(t) =

e−t x

sin x dx, t ∈ [0, ∞) x

0

and we notice that ∫∞ F(0) =

sin x dx. x

0

First, integrating by parts, we obtain: ∫

e−t x sin x dx = −e−t x cos x − t

∫

e−t x cos x dx ( ) ∫ = −e−t x cos x − t e−t x sin x + t e−t x sin x dx

whence we deduce that: ∫ e−t x (t sin x + cos x) + C = ϕ(x, t) + C e−t x sin x dx = − 1 + t2

(4.9)

We notice that the function ϕ, which is an antiderivative of the function e−t x sin x, is bounded. Indeed: |ϕ(x, t)| ≤

1+t ≤ 2, if t ≥ 0. 1 + t2

(4.10)

∫∞ Next, we evaluate the integral u e−t x sinx x dx, for u > 0. Because the integral is convergent, integrating by parts, we obtain: | | | |∞ | | | |∫ ∫ ∞ ∫∞ | | | | | |ϕ(x, t)| | e−t x sin x dx | = | ϕ(x, t) ||∞ + ϕ(x, t) dx | ≤ |ϕ(x, t)| + dx | | x | | 2 u x x u x2 u | | | | u

u

Taking into account to (4.10), it follows that:

4.2 Improper Integrals Depending on a Parameter

101

|∞ | |∫ | ∫∞ | | 2 1 sin x 4 | e−t x | dx = . dx | ≤ + 2 | 2 x x u | | u u

(4.11)

u

∫∞ From (4.11), we deduce that limu→∞ u e−t x sinx x dx = 0, hence the improper ∫ ∞ −t x sin x dx is uniformly convergent on [0, ∞), according to Definition integral 0 e x 4.2.2. We remark that the function under the sign of integral is not defined in x = 0, but has a finite limit in x = 0, namely lim x→0 e−t x sinx x = 1. √ −t x sin x , if x > 0 e x Let us consider the continuous function f (x, t) = . 1, if x = 0 ∫∞ Obviously F(t) = 0 f (x, t)dx and because the improper integral ∫ ∞ −t x sin x e dx is uniformly convergent on [0, ∞) and f is continuous, it results 0 x that F is continuous on [0, ∞) (Theorem 4.2.2). In particular, we have F(0) = limt→0 F(t). t>0 On the other hand, we get: ∫∞ 0

( ) ∫∞ ∂ −t x sin x e dx = − e−t x sin x dx. ∂t x 0

| | Let a > 0 be arbitrary. Since |e−t x sin x | ≤ e−a x , ∀ x ∈ [0, ∞), ∀ t ∈ [a, ∞) ∫ ∞ −a x ∫∞ and 0 e dx = a1 is convergent, it follows that the integral 0 e−t x sin x dx is uniformly convergent on [a, ∞), ∀ a > 0 (Theorem 4.2.1), hence uniformly convergent on (0, ∞). From (4.9) and Theorem 4.2.3, it results that for any t > 0, we have: '

∫∞

F (t) = −

e

−t x

0

(we notice that lim x→∞ Therefore:

| e−t x (t sin x + cos x) ||∞ 1 sin x dx = | 0 = − 1 + t2 1 + t2

t sin x+cos x et x

| | x| ≤ = 0, since | t sin x+cos et x

F(t) = −tan−1 t + C , ∀t > 0. On the other hand, we have: ∫∞ |F(t)| ≤ 0

whence it results that:

e−t x dx =

1 , ∀t > 0, t

t+1 et x

→ 0 if x → ∞).

(4.12)

102

4 Integrals Depending on Parameter

lim F(t) = 0.

(4.13)

t→∞

From (4.12) and (4.13), we deduce that C = F(t) = −tan−1 t +

π , 2

π , ∀ t > 0. 2

Using (4.14), we obtain F(0) = limt→0 F(t) = t>0 We can immediately prove that: ∫∞ 0

and so:

π , 2

hence

(4.14) ∫∞ 0

sin x dx x

=

π . 2

⎧π ⎨ 2 , if α > 0 sin α x dx = 0, if α = 0 . ⎩ π x − 2 , if α < 0

Theorem 4.2.4 Let f : [a, b) × [c, d] → R be a continuous function. If the improper ∫b integral depending on parameter a f (x, t)dx is uniformly convergent on [c, d], ∫b then the function F: [c, d] → R, defined by F(t) = a f (x, t)dx, ∀ t ∈ [c, d] is continuous (thus integrable) on [c, d] and: ∫d

∫b F(t)dt =

c

a

⎞ ⎛ d ∫ ⎝ f (x, t) dt ⎠dx c

which is equivalent to: ∫d c

⎞ ⎞ ⎛ b ⎛ ∫ ∫b ∫d ⎝ f (x, t)dx ⎠dt = ⎝ f (x, t)dt ⎠dx. a

a

c

∫b Proof Let a < bn < b, limn→∞ bn = b and let Fn (t) = a n f (x, t)dx, t ∈ [c, d]. According to Theorem 4.1.3, the functions Fn are continuous on [c, d], ∀ n ∈ N∗ and: ⎛ ⎞ ∫d ∫bn ∫d Fn (t)dt = ⎝ f (x, t)dt ⎠dx. c

a

c

On the other hand, from Lemma 4.2.1, it follows that Fn →u[c,d] F, whence we deduce that: ∫d

∫d Fn (t)dt =

lim

n→∞ c

F(t)dt. c

4.2 Improper Integrals Depending on a Parameter

103

Therefore, we have: ∫d

∫d F(t)dt = lim

n→∞

c

⎞ ⎛ ∫bn ∫d ⎝ f (x, t)dt ⎠dx, Fn (t)dt = lim n→∞

c

a

c

)

( ∫u ∫d

∫d whence, it results that there is limu→b f (x, t)dt dx = F(t)dt and it is finite. u 0, 2 and multiplying both members by e−t . Since dx = t du, we get: 2

I ·e

−t 2

∫∞ =

2 2 t e−t (1+u ) du

0

and further: ∫∞ 0

which is equivalent to:

I · e−t

2

⎞ ⎛ ∫∞ ∫∞ 2 2 dt = ⎝ t e−t (1+u ) du ⎠dt. 0

0

(4.15)

104

4 Integrals Depending on Parameter

∫∞ I·

e−t

2

⎞ ⎛ ∫∞ ∫∞ 2 2 dt = I 2 = ⎝ t e−t (1+u ) du ⎠dt.

0

0

0

According to Theorem 4.2.4, changing the order of integration, we can write: ⎞ ⎛ ∫∞ ∫∞ ∫∞ ( 2 2 I 2 = ⎝ t e−t (1+u ) dt ⎠du = 0

∫∞ = 0

0

0

1 1 ( ) du = 2 2 2 1+u

whence it results that I =

∫∞ 0

| ) 1 2 2 | ) · e−t (1+u ) ||∞ du ( 0 −2 1 + u 2

| 1 1 −1 ||∞ π du = tan u | = 0 1 + u2 2 4

√

π . 2

Therefore, we find: ∫∞

e−x dx = 2

√ π.

−∞

Immediately, it follows that: ∫∞

e−α x dx = 2

0

1 2

√

π , α > 0. α

Remark 4.2.6 Analogously, using similar reasons as above (Examples 4.2.4 and 4.2.5), it is shown that the common value of the Fresnel integrals from Example 3.2.10 is: ∫∞

( ) sin x 2 dx =

0

∫∞

( ) 1 cos x 2 dx = 2

√

π . 2

0

4.3 Euler Integrals Definition 4.3.1 The Euler integral of the first kind or beta-function is the following improper integral: ∫1 B( p, q) =

x p−1 (1 − x)q−1 dx, p, q > 0.

(4.16)

0

The Euler integral of the second kind or gamma-function is the improper integral:

4.3 Euler Integrals

105

∫∞ ┌( p) =

x p−1 e−x dx, p > 0.

(4.17)

0

Both integrals define important non-elementary functions, which prove to be extremely useful in mathematical and physical applications. Their properties have been thoroughly studied, and tables with their values have been drawn up. Theorem 4.3.1 The beta-function defined in (4.16) is convergent and continuous for any p > 0 and q > 0. Proof First, we will show that the integral (4.16) is pointwise convergent for any p > 0 and q > 0. To prove that we will decompose the integral into a sum of two integrals: 1

∫2

∫1 x

p−1

(1 − x)

q−1

0

dx =

∫1 x

0

p−1

(1 − x)

q−1

dx +

x p−1 (1 − x)q−1 dx. 1 2

∫1 If p ≥ 1, then 02 x p−1 (1 − x)q−1 dx is a proper integral, hence it is convergent. If 0 < p < 1, then 1 − p < 1 and lim x→0 x 1− p x p−1 (1 − x)q−1 = 1. x>0 ∫1 By Theorem 3.2.7, it follows that the improper integral 02 x p−1 (1 − x)q−1 dx is convergent. ∫1 If q ≥ 1, then 1 x p−1 (1 − x)q−1 dx is a proper integral, hence it is convergent. 2

If 0 < q < 1, then 1 − q < 1 and lim x→1 (1 − x)1−q x p−1 (1 − x)q−1 = 1. From x 0 and q > 0. Moreover, we note that the beta-function B( p, q) is continuous for any p > 0 and q > 0. Indeed, let p0 > 0 and q0 > 0 be two fixed positive numbers. From the ∫1 above considerations, it results that the improper integral 0 x p0 −1 (1 − x)q0 −1 dx is convergent. On the other hand, since x p−1 (1 − x)q−1 ≤ x p0 −1 (1 − x)q0 −1 , ∀ p ≥ p0 , q ≥ q0 , x ∈ (0, 1), ∫1 from Theorem 4.2.1, it follows that 0 x p−1 (1 − x)q−1 dx is uniformly convergent for any p ≥ p0 , q ≥ q0 . According to Theorem 4.2.2, it results that the beta-function B( p, q) is continuous on the domain [ p0 , ∞) × [q0 , ∞). As p0 > 0 and q0 > 0 are arbitrary, we deduce that the beta-function B( p, q) is continuous for any p > 0 and q > 0. Remark 4.3.1 The beta-function B( p, q) is uniformly convergent on any domain [ p0 , ∞) × [q0 , ∞), with p0 > 0, q0 > 0.

106

4 Integrals Depending on Parameter

Theorem 4.3.2 The beta-function B( p, q) has the following properties: B( p, q) = B(q, p), p > 0, q > 0.

(4.18)

If p > 1, then the following recurrence relation holds: p−1 B( p − 1, q). p+q −1

B( p, q) =

(4.19)

In particular, for any m, n ∈ N∗ , m ≥ 2, we have: B(m, n) =

∫∞ B( p, q) = 0

(m − 1)! · (n − 1)! (m + n − 1)!

t p−1 dt = (1 + t) p+q

∫∞ 0

t q−1 dt. (1 + t) p+q

(4.20)

(4.21)

π

∫2

sin2 p−1 u · cos2 q−1 u du.

B( p, q) = 2

(4.22)

0

Proof Assertion (i) follows immediately if we make the change of variable x = 1−t. (ii) Integrating by parts the relation (4.16), for p > 1 and q > 0, it results that: | ∫1 | 1 p−1 p−1 q |1 B( p, q) = − x (1 − x) | + x p−2 (1 − x)q dx 0 q q 0

=

p−1 q

∫1

) ( x p−2 (1 − x)q−1 − (1 − x)q−1 x dx

0

p−1 p−1 B( p − 1, q) − B( p, q). = q q Further, we have: ) ( p−1 p−1 B( p, q) = B( p − 1, q) 1+ q q or

4.3 Euler Integrals

107

B( p, q) =

p−1 B( p − 1, q). p+q −1

Similarly, it is shown that if q > 1, then: B( p, q) =

q −1 B( p, q − 1). p+q −1

Since B( p, 1) = 1p , for any n ∈ N∗ , we get: n−2 n − (n − 1) n−1 · · ... · B( p, 1) p+n−1 p+n−2 p + n − (n − 1) (n − 1)! = . p · ( p + 1) · . . . · ( p + n − 1)

B( p, n) =

In particular, for m, n ∈ N∗ , m ≥ 2 we have: B(m, n) =

(m − 1)! · (n − 1)! . (m + n − 1)!

(iii) In (4.16) we consider the change of variable x = ∫∞ B( p, q) = 0

t 1+t

and we obtain:

t p−1 1 · dt = p−1 (1 + t) (1 + t)q−1 (1 + t)2

∫∞ 0

t p−1 dt. (1 + t) p+q

Moreover ∫∞ B( p, q) = B(q, p) = 0

t q−1 dt. (1 + t) p+q

(iv) In (4.16) we consider the change of variable x 2 sin u cos u du and we have π

=

∫2 sin2 p−2 u cos2 q−2 u sin u cos u du = 2

0

sin2 u , dx

π

∫2 B( p, q) = 2

=

sin2 p−1 u cos2 q−1 u du. 0

Theorem 4.3.3 The gamma-function (4.17) is convergent and continuous for any p > 0. Proof For the beginning we will show that the gamma-function ┌( p) is pointwise convergent for any p > 0. To prove that we will decompose the integral into a sum of two integrals:

108

4 Integrals Depending on Parameter

∫∞ x

∫1

p−1 −x

dx =

e

0

x

p−1 −x

e

∫∞ dx +

0

x p−1 e−x dx.

1

∫1 If p ≥ 1, then 0 x p−1 e−x dx is a proper( integral, thus it) is convergent. If 0 < p < 1, then 1− p < 1 and lim x→0 x 1− p x p−1 e−x = 1. By Theorem 3.2.7, x>0 ∫1 the improper integral 0 x p−1 e−x dx is convergent. ( ) = 0. From On the other hand, we notice that lim x→∞ x 2 x p−1 e−x ∫ ∞ p−1 −x Theorem 3.2.5, it follows that the improper integral 1 x e dx is convergent. Therefore we proved that the gamma-function ┌( p) is pointwise convergent for any p > 0. Let r > 0, s > 0 be two positive arbitrary numbers, with r < s. If p ∈ [r, s], then. ) ( x p−1 e−x ≤ x r −1 + x s−1 e−x , for any x > 0. ∫∞( ) As the improper integral 0 x r −1 + x s−1 e−x dx is convergent, from ∫ ∞ p−1 −x Theorem 4.2.1, we deduce that ┌( p) = 0 x e dx is uniformly convergent on [r, s]. From Theorem 4.2.2 and from to the fact that the function f (x, p) = x p−1 e−x is continuous on (0, ∞) × (0, ∞), it results that the gamma-function ┌( p) is continuous on the interval [r, s], so on (0, ∞), because r > 0, s > 0 are arbitrary. Remark 4.3.2 The gamma-function ┌( p) is uniformly continuous on any compact interval [r, s] ⊂ (0, ∞), where 0 < r < s < ∞. Theorem 4.3.4 The gamma-function ┌( p) has the following properties: ┌(1) = 1. ┌( p + 1) = p ┌( p), p > 0.

(4.23)

In particular, ┌(n + 1) = n!, n ∈ N∗ . B( p, q) =

┌( p) · ┌(q) , p > 0, q > 0. ┌( p + q)

B( p, 1 − p) = ┌( p) · ┌(1 − p) = In particular, ┌ Proof (i) ┌(1) =

∫∞ 0

(1) 2

=

√

π.

| | e−x dx = −e−x ||∞ = 1. 0

π , p ∈ (0, 1). sin(π p)

(4.24) (4.25)

4.3 Euler Integrals

109

| | x p e−x dx = −x p e−x ||∞ 0 0 ∫ ∞ +p x p−1 e−x dx = p ┌( p). ∫

┌( p + 1) = (ii)

∞

0

In particular, for n ∈ N∗ we have: ┌(n + 1) = n · ┌(n) = n · (n − 1) · ┌(n − 1) = · · · = n · (n − 1) · . . . · 2 · ┌(1). Since ┌(1) = 1, it results that ┌(n + 1) = n!. This formula is interesting because it provides an analytical expression for n!, namely: ∫∞ n! =

x n e−x dx, n ∈ N∗ .

0

(iii) First, we notice that if we make the change of variable x = t y, t > 0, then dx = t dy and we get: ∫∞ ┌( p) = t

p

y p−1 e−t y dy.

(4.26)

0

Substituting in (4.26), p with p + q and t with 1 + t, we get: ┌( p + q)t p−1 = t p−1 (1 + t) p+q

∫∞

y p+q−1 e−(1+t)y dy

0

Taking into account now the formula (4.21), we deduce: ∫∞

┌( p + q)t p−1 dt (1 + t) p+q 0 ⎞ ⎛ ∫∞ ∫∞ = ⎝t p−1 y p+q−1 e−(1+t)y dy ⎠dt

┌( p + q)B( p, q) =

0

∫∞ =

0

⎛ ⎝t p−1

0

0

∫∞ =

e 0

∫∞

−y

y

⎞ y p+q−1 e−(1+t)y dt ⎠dy ⎛∞ ∫

p+q−1 ⎝

⎞ t

0

p−1 −t y

e

dt ⎠dy.

110

4 Integrals Depending on Parameter

From (4.26), it follows that: ∫∞

y p+q−1 e−y · ┌( p)dy yp

┌( p + q) · B( p, q) = 0

∫∞ = ┌( p)

y q−1 e−y dy = ┌( p) · ┌(q).

0 ┌( p)·┌(q) . ┌( p+q)

Therefore, we proved that B( p, q) =

(iv) The proof of the formula (4.25) will not be done because it is too long and requires superior knowledge of mathematics. ( ) ( ) √ From (4.25), for p = 21 , we get ┌ 2 21 = sinπ π = π , whence ┌ 21 = π . 2 ( ) The value for ┌ 21 can also be found by direct computation, using the change the variable x = t 2 and Example 4.2.5. Indeed: √ ( ) ∫∞ ∫∞ √ π 1 2 − 21 −x = x e dx = 2 e−t dt = 2 · ┌ = π. 2 2 0

0

Example 4.3.1 Compute the following integrals using the Euler integrals B and ┌: ∫∞ ∫∞ 2 n −x 2 (1) x 3 e−3 x dx; (2) x e dx, n ∈ N∗ ; −∞

0

(3)

∫1 √

π

x2

−

x3

∫2

7

5

dx; (4) sin 2 x cos 2 x dx; 0

0

(1) If we make the change of variable 3 x = t, then dx = 13 dt, t ∈ [0, ∞), and: ∫∞

3 −3 x

x e

1 dx = 3

∫∞

0

t 3 −t e dt 33

0

1 = 4 3

∫∞

t 3 e−t dt =

2 ┌(4) (4.23) 3! = = . 81 81 27

0

(2)

∫∞

−∞

x 2 n e−x dx = 2 2

∫∞ 0

x 2 n e−x dx. 2

If we make the change of variable x 2 = t ≥ 0, then dx =

1 √ dt, 2 t

t ∈ [0, ∞) and:

4.3 Euler Integrals

111

∫∞ 1 1 x e dx = 2 x e dx = 2 t e √ dt = t n− 2 e−t dt 2 t −∞ 0 0 0 ( ) ( ) ( ) 1 1 1 (4.23) =┌ n− +1 = n− =┌ n+ 2 2 2 ) ( 1 (4.23) (4.23) = ... = ┌ n− 2 ) ( ) ( ) ( ) ( 3 5 3 1 1 1 · n− · n− ··· · ┌ = n− 2 2 2 2 2 2 (2 n − 1) · (2 n − 3) · (2 n − 5) · · · 3 · 1 = 2n √ not (2 n − 1)!! √ · π= · π. 2n ∫∞ √ 2 For n = 0, we obtain the Euler–Poisson integral −∞ e−x dx = π. A new property of the gamma-function can be retained: ∫∞

∫∞

2 n −x 2

2 n −x 2

∫∞

n −t

( ) 1 1 · 3 · · · (2 n − 3) · (2 n − 1) √ (2 n − 1)!! √ π= π , n ∈ N∗ (4.27) ┌ n+ = n 2 2 2n ) ( 3 (4.24) 1 x(1 − x) 2 dx = B 2, 2 0 0 0 (3) (3) (3) ┌ 1! · ┌ 2 ┌(2) · ┌ 2 (3) (7) ) = 5 (25 ) = ( = 5 ┌ 2 ┌ 1+ 2 ┌ 2 2 (3) (3) ┌ ┌ 4 . = 5 ( 2 3) = 5 3 2 (3) = 15 · · ┌ ┌ 1 + 2 2 2 2 2 (4) Using Formula (4.22), we get: ∫1 √

x 2 − x 3 dx =

π

∫2 0

7 sin 2

x

5 cos 2

∫1 √

∫1

x 2 (1 − x) dx =

( ) ( ) 7 9 ) ( 1 9 7 (4.24) 1 ┌ 4 · ┌ 4 = x dx = · B , · 2 4 4 2 ┌(4) ( ) ( ) 3 5 ( ) ( ) 5 3 1 ┌ 1 + 4 · ┌ 1 + 4 (4.23) 1 5 3 = ·┌ · · ·┌ = · 2 3! 12 4 4 4 4 ( ) ( ) ( ) ( ) 1 3 (4.23) 5 1 1 3 5 = ·┌ 1+ ·┌ · ·┌ ·┌ = 64 4 4 64 4 4 4 √ ) ( ) ( 1 1 (4.25) 5 π 5 2π 5 = ·┌ ·┌ 1− · . = = 256 4 4 256 sin π4 256

Chapter 5

Line Integrals

5.1 Parameterized Paths. Definition of a Curve Definition 5.1.1 A parameterized path in R3 of C k -class, k ∈ N∗ is any vector function r : I ⊂ R → R3 of C k -class, defined on an interval I ⊂ R. If we denote by x, y, z the scalar components of the vector function r , then: r (t) = (x(t), y(t), z(t)), ∀t ∈ I. The fact that the vector function r : I → R3 is of C k -class is equivalent to the fact that the scalar functions x, y, z : I → R are of C k -class on the interval I ; i.e. and their derivatives are continuous on I . they are k-times differentiable ⎧ ⎨ x = x(t) The equations y = y(t) , t ∈ I , are called the parametric equations of the ⎩ z = z(t) path or a parametric representation of the path and t is called a parameter. We draw attention to the fact that according to Definition 5.1.1, a parameterized path is a vector function (application) and not a set of points. To better mark this we will note a path parameterized with (I, r ) or (I, r = r (t)). The direct image r (I ) of the interval I through the vector function r , i.e. the subset r (I ) = {r (t); t ∈ I } = {(x(t), y(t), z(t))t ∈ I } ⊂ R3 is called the support of parameterized path r . If I = [a, b] is a compact interval, then its support r (I ) is a compact and connected subset of R3 . In this case, the points r (a) and r (b) are called the ends (extremities) of the path. If r (a) = r (b), then the path is called closed. − → − → − → Let O x yz be an orthogonal coordinate system and let i , j , k be the unit vectors of the axis O x, O y, respectively Oz. If we identify any point M(x, y, z) ∈ R3 −−→ with its position vector O M, then we obtain the vector equation of the path: © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Integral Calculus for Engineers, https://doi.org/10.1007/978-981-19-4793-3_5

113

114

5 Line Integrals

Fig. 5.1 The circle of radius R centered at the origin O

− → − → − → → r (t) = − r (t) = x(t) i + y(t) j + z(t) k , ∀ t ∈ I. Example 5.1.1 Let r : [0, 2π ] → R3 be the path defined by: − → − → − → r (t) = R cos t i + R sin t j , t ∈ [0, 2π ], R > 0. ⎧ ⎨ x = R cos t The parametric equations are: y = R sin t , t ∈ [0, 2π ]. ⎩ z =0 We noticethat for any t ∈ [0, 2π ], the point (x(t), y(t), 0) verifies the equation x 2 + y2 = R2 . of the circle z = 0 It results that the support of the path is the circle of radius R centered at the origin O that lies in the plane O x y. The parameter t is the angle between the position vector −−→ O M and the positive direction of the axis O x (Fig. 5.1). We notice also that this path is closed, because r (0) = r (2π ) = (R, 0, 0). Example 5.1.2 Let r : [0, 2π ] → R3 be the path defined by: − → − → − → − → r (t) = R cos t i + R sin t j + ht k , t ∈ [0, 2π ], R, h > 0. ⎧ ⎨ x = R cos t The parametric equations are: y = R sin t , t ∈ [0, 2π ]. ⎩ z = ht We notice that for any t ∈ [0, 2π ], we have (x(t))2 + (y(t))2 = R 2 , whence, it results that the support of this path is situated on the right circular cylinder of radius R and axis of symmetry Oz given by x 2 + y 2 = R 2 .

5.1 Parameterized Paths. Definition of a Curve

115

z

Fig. 5.2 The cylindrical helix

B A

y

x 

The support of this path is the arc AB of the cylindrical helix of radius R and slope h/R, where A(R, 0, 0) and B(R, 0, 2π h) (Fig. 5.2). Example 5.1.3 Let r : R → R3 be the path defined by: − → − → − → − → r (t) = (x0 + lt) i + (y0 + mt) j + (z 0 + nt) k , t ∈ R where x0 , y0 , z 0 , l, m, n are real constants. The parametric equations are: ⎧ ⎨ x = x 0 + lt y = y0 + mt , t ∈ R. ⎩ z = z 0 + nt The support of this path is the straight line passing through the point M0 (x0 , y0 , z 0 ) and having the direction parameters l, m, n (Fig. 5.3). In other words, this straight − → − → − → → → v , where − v =l i +m j +n k . line has the same direction as − Definition 5.1.2 A path (I, r ) is called simple if the vector function r is injective, i.e. ∀t1 , t2 ∈ I, t1 /= t2 ∀t1 , t2 ∈ I, t1 /= t2 , it results that r (t1 ) /= r (t2 ). A closed path is simple if the equality r (t1 ) = r (t2 ) implies t1 = t2 or t1 = a and t2 = b, respectively t1 = b and t2 = a, where a and b are the ends of the interval I . The paths presented in Example 5.1.1, Example 5.1.2 and Example 5.1.3 are simple. Definition 5.1.3 Let (I, r ) be a parameterized path of C 1 -class. A point M0 (x(t0 ), y(t0 ), z(t0 )) is called singular point of (I, r ) if x ' (t0 ) = y ' (t0 ) = z ' (t0 ) = 0.

116

5 Line Integrals

Fig. 5.3 The support of a straight line

A smooth path is a path without singular points, i.e.: x '2 (t) + y '2 (t) + z '2 (t) > 0, ∀t ∈ I. A smooth path has a tangent at each point and the position of the tangent continuously depends of the point of tangency. Every path is oriented with respect to the increasing parameter. For example, in Example 5.1.2 the path is oriented from A to B. Definition 5.1.4 Two parameterized paths (I1 , r1 ) and (I2 , r2 ) of C k -class are called equivalent if there is a function λ : I1 → I2 bijective, of C k -class, with the inverse λ−1 : I2 → I1 of C k -class and with the property λ' (t1 ) /= 0, ∀t1 ∈ I1 such that: r2 (λ(t1 )) = r1 (t1 ), ∀t1 ∈ I1 . We use the notation (I1 , r1 ) ∼ (I2 , r2 ). Such a function λ is called change of parameter. Since λ' is continuous on I1 and λ' /= 0 on I1 , it follows that λ' > 0 or λ' < 0 on I1 . If λ' > 0 on I1 , then λ is strictly increasing and we say that the paths are equivalent to the same orientation. Otherwise, the paths are equivalent to opposite orientation. It is obvious that two equivalent paths have the same support. Example 5.1.4 Let r1 : I1 → R2 and r2 : I2 → R2 be two paths defined by: ( π) − → − → − → ,R >0 r 1 (t1 ) = R sin t1 i + R cos t1 j , t1 ∈ I1 = 0, 2 / − → − → − → r 2 (t2 ) = t2 i + R 2 − t22 j , t2 ∈ I2 = (0, R). 

These paths have the same support, namely the arc AB of the circle of radius R centered at the origin (Fig. 5.4).

5.1 Parameterized Paths. Definition of a Curve

117

Fig. 5.4 Three parametrical representations of the arc AB

We notice that the function λ : I1 → I2 defined by ( π) λ(t1 ) = R sin t1 , ∀t1 ∈ I1 = 0, 2 ) ( is bijective, of C ∞ -class on I1 and λ' (t1 ) = R cos t1 > 0, ∀t1 ∈ 0, π2 . Moreover, for any t1 ∈ I1 we have: − → − → → − → √ − → − → r 1 (t1 ). r 2 (λ(t1 )) = λ(t1 ) i + R 2 − λ2 (t1 ) j = R sin t1 i + R cos t1 j = − It follows that λ is a change of parameter, and therefore the two paths are equivalent to the same orientation. Let us consider now the path r3 : I3 → R2 defined by: ( π) − → − → − → . r 3 (t3 ) = R cos t3 i + R sin t3 j , t3 ∈ I3 = 0, 2 We notice that the function μ : I3 → I2 , defined by μ(t3 ) = R cos t3 , ∀t3 ∈ I3 is a change of parameter. Since ( π) μ' (t3 ) = −R sin t3 < 0, ∀t3 ∈ 0, 2 we deduce that μ is strictly decreasing, hence the paths (I3 , r3 ) and (I2 , r2 ), respectively, the paths (I3 , r3 ) and (I1 , r1 ) are equivalent to opposite orientation. Indeed, the orientation of the paths (I1 , r1 ) and (I2 , r2 ) is from A to B, while the orientation of the path (I3 , r3 ) is from B to A. Definition 5.1.5 A parameterized curve is any class of equivalent parameterized paths.

118

5 Line Integrals

3 ( Therefore, ) γ is a parameterized curve if there is a parameterized path r : I → R 2 r : I → R such that:

( ) } { γ = ρ : I → R3 R2 parameterized path; (I, ρ) ∼ (I, r ) . Because (I, r ) ∼ (I, r ), it follows that r ∈ γ . A parameterized curve is simple (closed, smooth) if the path that defines it is simple (closed, smooth). A simple curve is considered to be positively oriented if the path that defines it is oriented in the direction of increasing the parameter, and negatively orientedotherwise. ( ) Let γ be a simple and smooth parameterized curve and let r : I → R3 R2 be the parameterized path that defines it, oriented in the direction of increasing the parameter. We will denote by γ+ the family of all parameterized paths equivalent to r and having the same orientation as r . Obviously, r ∈ γ+ . We will note with γ− the family of all equivalent parameterized paths with r having the opposite orientation to r . The support of a parameterized curve γ is the support of the path that defines it and, obviously, it coincides with the support of any representative path of γ . Let γ be the parameterized curve defined by the path r1 in Example 5.1.4. Its 



support is the arc AB from Fig. 5.4. The support of the curve γ+ is the arc AB 

(oriented from A to B), while the support of the curve γ− is the arc B A. Obviously, r2 ∈ γ+ and r3 ∈ γ− . Whenever there is no confusion, we will identify a curve γ by any of its representative. Definition 5.1.6 Let r1 : [a, b] → R3 and r2 : [b, c] → R3 be two parameterized paths with the property r1 (b) = r2 (b). It is called the union of the paths and is denoted by r1 ∪ r2 the following path that can be written as:  r1 ∪ r2 : [a, c] → R3 , (r1 ∪ r2 )(t) =

r1 (t) if t ∈ [a, b] . r2 (t) if t ∈ [b, c]

If γi is the curve defined by ri , i = 1, 2, then γ1 ∪ γ2 is the curve defined by the path r1 ∪ r2 . A curve is called piecewise smooth if it is a union of finite number of smooth curves.

5.2 Rectifiable Paths and Curves The notion of curve (path) introduced in paragraph 5.1 is quite general, and therefore, in some cases, especially in the case of curves that admit multiple points, the support of a curve may differ significantly from the intuitive image that we have of a curve.

5.2 Rectifiable Paths and Curves

119

M1

Fig. 5.5 The length of a polygonal line inscribed in a curve

M0

M i –1

Mi

The Italian mathematician Giuseppe Peano proved that two continuous functions x = x(t), y = y(t) can be defined on the interval [0, 1] (thus a path), such that, when the parameter t crosses the interval [0, 1], the corresponding point M(x(t), y(t)) starts to the point (0, 0) (corresponding to the value t = 0), passes through all points of the square [0, 1] × [0, 1] and ends to the point (1, 1) (corresponding to the value t = 1). In other words, the support of this path fills a square. It is clear that the notion of length for such a path does not make sense. In the following, we will introduce the notion of rectifiable path (which has length) and we will show how to compute the length of a rectifiable path using the definite integral. Let r : [a, b] → R3 be a path and let x = x(t), y = y(t), z = z(t), t ∈ [a,] be its parametric equations. Consider an arbitrary partition Δ of the interval [a, b], i.e. Δ : a = t0 < t1 < ... < ti−1 < ti < ... < tn = b and we will denote by Mi the coordinate points (x(ti ), y(ti ), z(ti )), i = 0, n, which lie on the support of the path. n Σ Mi−1 Mi > 0 be the length of the polygonal line inscribed in Let L Δ (r ) = i=1

the support of the path r with vertices at the point Mi corresponding to the values t = ti (Fig. 5.5). The set {L Δ (r )}Δ , when Δ is any possible partition of the interval [a, b], is a set of positive numbers upper bounded or not. Definition 5.2.1 The path r is called rectifiable (has length) if the set {L Δ (r )}Δ is upper bounded, i.e. ∃M > 0such that L Δ (r ) < M, for any partition Δ of [a, b]. If the path r is rectifiable, then its length is defined by: L(r ) = sup{L Δ (r )} < ∞. Δ

Lemma 5.2.1 For any four real numbers a1 , a2 , b1 , b2 the following inequality is established: | |/ / | | 2 | a + a 2 − b2 + b2 | ≤ |a1 − b1 | + |a2 − b2 |. (5.1) 2 1 2| | 1

120

5 Line Integrals

Proof Amplifying with the conjugate and taking into account the inequality of the triangle, we obtain: | 2 | |/ | / |a + a 2 − b2 − b2 | | 2 | 2 1 2 | a + a 2 − b2 + b2 | = / 1 / 2 1 2| | 1 a12 + a22 + b12 + b22 ≤

|a1 − b1 ||a1 + b1 | + |a2 − b2 ||a2 + b2 | / / . a12 + a22 + b12 + b22

(5.2)

On the other hand, we have: / / 2 2 |a1 + b1 | ≤ |a1 | + |b1 | ≤ a1 + a2 + b12 + b22 and similarly: / / 2 2 |a2 + b2 | ≤ |a2 | + |b2 | ≤ a1 + a2 + b12 + b22 . Taking into account these inequalities in (5.2), it results that: |/ | / | 2 | | a + a 2 − b2 + b2 | ≤ |a1 − b1 | + |a2 − b2 |. 2 1 2| | 1 Remark 5.2.1 The inequality (5.1) remains established for any 2n real numbers ai , bi ∈ R, i = 1, n. For example, for n = 3 we have: |/ | / | 2 | | a + a 2 + a 2 − b2 + b2 + b2 | ≤ |a1 − b1 | + |a2 − b2 | + |a3 − b3 |. 2 3 1 2 3| | 1

(5.3)

The proof is the same as the proof of Lemma 5.2.1. Theorem 5.2.1 Let r : [a, b] → R3 , r (t) = (x(t), y(t), z(t)), t ∈ [a, b] be a smooth parameterized path. Then r is rectifiable and its length is given by:

L = L(r ) =

∫b √

x '2 (t) + y '2 (t) + z '2 (t) dt.

a

Proof Let Δ : a = t0 < t1 < ... < ti−1 < ti < ... < tn = b be an arbitrary partition of the interval [a, b] and let L Δ (r ) the length of the polygonal line inscribed in the support of the path r (Fig. 5.5), i.e.:

5.2 Rectifiable Paths and Curves

L Δ (r ) =

121

n / Σ (x(ti ) − x(ti−1 ))2 + (y(ti ) − y(ti−1 ))2 + (z(ti ) − z(ti−1 ))2 . i=1

From Lagrange theorem, it follows that there are ξi , ηi , μi ∈ (ti−1 , ti ), i = 1, n, such that: x(ti ) − x(ti−1 ) = x ' (ξi )(ti − ti−1 ) y(ti ) − y(ti−1 ) = y ' (ηi )(ti − ti−1 ) z(ti ) − z(ti−1 ) = z ' (μi )(ti − ti−1 ). Taking into account to this equalities in L Δ (r ), we get: L Δ (r ) =

n √ Σ

x '2 (ξi ) + y '2 (ηi ) + z '2 (μi ) · (ti − ti−1 ).

(5.4)

i=1

The function g : [a, b] → R given by g(t) =

√

x '2 (t) + y '2 (t) + z '2 (t), t ∈ [a, b]

is continuous on [a, b], because the derivatives functions x ' , y ' , z ' are continuous, hence g is integrable on [a, b]. Consider the Riemann sum: σΔ (g; ξ ) =

n √ Σ

x '2 (ξi ) + y '2 (ξi ) + z '2 (ξi ) · (ti − ti−1 ).

(5.5)

i=1

Since g is integrable on [a, b], it follows that for any ε > 0 there is δε' > 0 such that for any partition Δ of [a, b], with the property Δ < δε' and for any intermediate points ξ = (ξ1 , ξ2 , ..., ξn ) we have: | | | | ∫b | | |σΔ (g; ξ ) − g(t)dt | < ε. | | | |

(5.6)

a

On the other hand, from the inequality (5.3) and the generalized inequality of the triangle, it results that: |L Δ (r ) − σΔ (g; ξ )| 0 such that ∀ t', t'' ∈ [a, b], with |t ' − t ' | < δε'' , we have: | '( ') ( )| | y t − y ' t '' |
0, h > 0. ⎩ z =ht According to Theorem 5.2.1, we have: L=

∫2 π√

R 2 sin2 t + R 2 cos2 t + h 2 dt =

0

∫2 π√

R 2 + h 2 dt = 2π

√

R2 + h2.

0

Remark 5.2.1 Let r : [a, b] → R2 be a smooth parameterized plane path defined by: r (t) = (x(t), y(t)), t ∈ [a, b]. Then r is rectifiable and its length is:

L = L(r ) =

∫b √ a

x '2 (t) + y '2 (t)dt.

124

5 Line Integrals

Example 5.2.2 Compute the length of the circle x 2 + y 2 = R 2 . A parametric representation of the circle is: 

x = R cos t , t ∈ [0, 2 π ], R > 0. y = R sin t

Using Remark 5.2.1, we have: L=

∫2π √

x '2 (t)

+

y '2 (t)dt

=

∫2π √

∫2π R2

sin t + 2

R2

cos2

tdt =

R dt = 2π R. 0

0

0

Example 5.2.3 Compute the length of the ellipse

x2 a2

+

y2 b2

− 1 = 0, a, b > 0.

A  parametric representation of the ellipse is: x = a cos t , t ∈ [0, 2 π ], a, b > 0. y = b sin t It is enough to calculate a quarter of the length of the ellipse. By Remark 5.2.1, we have: π

π

L = 4

∫ 2 √ ∫2 / ) ( 2 2 2 2 a sin t + b cos t dt = a 2 − a 2 − b2 cos2 tdt. 0

0

If we denote by c the focal length and by ε the eccentricity of the ellipse, then a 2 − b2 = c2 and ε = ac ∈ (0, 1).a 2 − b2 = c2 and ε = ac ∈ (0, 1). Next, we have: π π ∫ 2 √ ∫2 / ) ( 2 L c =a 1 − ε2 cos2 t dt cos2 tdt = a 1− a 4

0

0

π

π 2

∫ 2 √ ∫ / ) (π − t dt = a 1 − ε2 sin2 t dt. 1 − ε2 sin2 =a 2 0

0

Remains to calculate the integral: π

∫ 2 √ 1 − ε2 sin2 t dt, 0 < ε < 1. 0

The antiderivative of this function is not an elementary function, and therefore the computation of this integral cannot be done with the Leibniz–Newton formula.

5.2 Rectifiable Paths and Curves

125

Attempting to calculate the length of the ellipse led to an integral that cannot be computed exactly. Such an integral is called an elliptic integral. The following types of elliptic integrals are known: (1) Elliptic integral of the first kind: π

∫2

√

K (k) = 0

1 1 − k 2 sin2 ϕ

dϕ, k ∈ (0, 1).

(2) Elliptic integral of the second kind: π

E(k) =

∫2 /

1 − k 2 sin2 ϕ dϕ, k ∈ (0, 1).

0

(3) Elliptic integral of the third kind: π

∫2 F(k, h) = 0

1 dϕ, k ∈ (0, 1). ( )√ 2 1 + h sin ϕ 1 − k 2 sin2 ϕ

The computation of these integrals is done with approximate methods, and tables have been drawn up with their (approximate) values for different values of the parameters k, respectively, k and h. Remark 5.2.2 Let f : [a, b] → R be a function of C 1 -class. The length of the graph of this function is:

L=

∫b √

1 + f '2 (x) dx.

a

Proof Indeed, the function f defines a smooth parameterized path, namely: 

x=t , t ∈ [a, b]. y = f (t)

The graph of f coincides with the support of this path. The statement results now from Remark 5.2.1. Example 5.2.4 Compute the length of the graph of the function: f : [0, b] → R, f (x) = a ch

(x ) , x ∈ [0, b], a > 0. a

126

5 Line Integrals

y

Fig. 5.6 Catenary

a O

b

x

The graph of this function is called catenary, and it is shown in Fig. 5.6. From Remark 5.2.2, we deduce that: ∫b / ∫b ( ) (x ) 2 x dx = cosh dx L= 1 + sinh a a 0 0 ( ) ( x )|b b | = a sinh . | = a sinh a 0 a Remark 5.2.3 Let ρ : [α, β] → R, ρ = ρ(θ ), be a function of C 1 -class and let r : [α, β] → R2 be the path given by r (θ ) = (ρ(θ ) cos θ, ρ(θ ) sin θ ), θ ∈ [α, β]. Then r is rectifiable and its length is:

L = L(r ) =

∫β √

ρ 2 (θ ) + ρ '2 (θ ) dθ.

α

Proof Indeed, a parametric representation of the path is: 

x = ρ(θ ) cos θ , θ ∈ [α, β]. y = ρ(θ ) sin θ 

The support of this path is the arc AB, represented in Fig. 5.7. According to Theorem 5.2.1, we have: ∫β / L= (ρ ' (θ ) cos θ − ρ(θ ) sin θ )2 + (ρ ' (θ ) sin θ + ρ(θ ) cos θ )2 dθ α

=

∫β √ α

ρ 2 (θ ) + ρ '2 (θ ) dθ.

5.2 Rectifiable Paths and Curves

127

Fig. 5.7 The length of a curve in polar coordinates

Fig. 5.8 The length of the cardioid

Example 5.2.5 Compute the length of the cardioid: ρ(θ ) = a(1 + cos θ ), θ ∈ [0, 2 π ], a > 0. The cardioid is represented in Fig. 5.8. For reasons of symmetry, it is enough to compute only a half of the length of this curve. From Remark 5.2.3 we have: L = 2

∫π / ∫π √ 2 2 2 2 a (1 + cos θ ) + a sin θ dθ = 2a 2 (1 + cos θ )dθ 0

∫π /

= 0

∫π = 2a 0

0

( ) θ 2 2 dθ 2a · 2 cos 2

| ( )|π ( ) θ || θ dθ = 4a sin = 4a. cos 2 2 || 0

Therefore, the length of the cardioid is L = 8a.

128

5 Line Integrals

Remark 5.2.2 According to Theorem 5.2.1, it follows that if r1 : [a, b] → R3 and r2 : [b, c] → R3 are two smooth parameterized paths and if r = r1 ∪ r2 is the path obtained by their union, then r is rectifiable and: L(r ) = L(r1 ) + L(r2 ). Moreover, any piecewise smooth curve is rectifiable and its length is the sum of the lengths of its smooth parts.

5.3 Natural Parameterization of a Curve A curve admits an infinite parametric representations. Next, we will present an important parameterization of the rectifiable curves, namely the natural parameterization. Let r : [a, b] → R3 be a smooth parameterized path defined by: r (t) = (x(t), y(t), z(t)), t ∈ [a, b]. According to Theorem 5.2.1, the path r is rectifiable and its length is:

L = L(r ) =

∫b √

x '2 (t) + y '2 (t) + z '2 (t) dt.

a

For any t ∈ [a, b], we consider the function: s : [a, b] → [0, L], s = λ(t) =

∫ t √

x '2 (u) + y '2 (u) + z '2 (u) du.

a

If A(x(a), y(a), z(a)) and B(x(b), y(b), z(b)) are the ends of the path support 

and M(x(t), y(t), z(t)) belongs to the support AB, then s = λ(t) represents the 

length of the arc √ AM (Fig. 5.9). Since λ' (t) = x '2 (t) + y '2 (t) + z '2 (t) > 0, ∀ t ∈ [a, b], λ(a) = 0, λ(b) = L, then, it results that the function λ : [a, b] → [0, L] is of C 1 -class, strictly increasing and bijective. Its inverse λ−1 : [0, L] → [a, b] is also a function of C 1 -class. ( ) Let the vector function r˜ : [0, L] → R3 , given by r˜ (s) = r λ−1 (s) , s ∈ [0, L]. Since r˜ (λ(t)) = r (t), ∀ t ∈ [a, b]

5.3 Natural Parameterization of a Curve

129

Fig. 5.9 The length of the arc AM

Fig. 5.10 The diagramme of parametric representation of the curve in function of the arc element

it results that the paths ([a, b], r ) and ([0, L], r˜ ) are equivalent to the same orientation, and hence λ is a change of parameter. ) ( ⎧ ˜ = x (λ−1 (s)) ⎨ x(s) Definition 5.3.1 The parametric equations y˜ (s) = y λ−1 (s) , s ∈ [0, L], ) ( ⎩ z˜ (s) = z λ−1 (s) define a new parametric representation of the path ([a, b], r ) called the natural parameterization. Example 5.3.1 Find the natural parametric representation of the curve: x 2 + y 2 = R 2 , x, y ≥ 0. The support of this plane curve is the quarter of circle in the first quadrant, which has the parametric representation: 

[ π] x = R sin t , R > 0. , t ∈ 0, y = R cos t 2

∫2 √ π

Its length is: L =

0

R 2 cos2 t + R 2 sin2 t dt =

πR . 2

 −−→ → r (t), then the length of the arc AM, which we denote by s (Fig. 5.11), If O M = −

is:

130

5 Line Integrals

Fig. 5.11 The parametrical representation of the arc AM in function of S (the arc element)

s = λ(t) =

∫ t √

R 2 sin2 u + R 2 cos2 u du = Rt.

0

] [ ] [ Obviously, λ is a function strictly increasing. It follows that λ : 0, π2 → 0, π2R ] [ ] [ is bijective and its inverse λ−1 : 0, π2R → 0, π2 is defined by: ] [ πR s . λ (s) = t = , s ∈ 0, R 2 −1

The natural representation of this path is: 

( ) ] [ πR x(s) ˜ = R sin (Rs ) . , s ∈ 0, y˜ (s) = R cos Rs 2

Example 5.3.2 Find the natural parameterization of the cylindrical helix. A parametric representation of the circular helix is: ⎧ ⎨ x = R cos t y = R sin t , ⎩ z = ht and its length is L = 2 π s = λ(t) =

∫ t √

√

t ∈ [0, 2π ], R > 0, h > 0

R 2 + h 2 (Example 5.2.1). Then we have:

R 2 sin2 u + R 2 cos2 u + h 2 dt = t

√

R 2 + h 2 , t ∈ [0, 2π ].

0

] [ √ The inverse function is λ−1 : 0, 2 π R 2 + h 2 → [0, 2π ],

5.3 Natural Parameterization of a Curve

λ−1 (s) = √

s R2 + h2

131

] [ √ , s ∈ 0, 2π R 2 + h 2 .

Therefore, the natural parametric representation of the cylindrical helix is: ( ) ⎧ √ s ⎪ x(s) ˜ = R cos ⎪ 2 2 ⎨ ] [ √ ( R +h ) 2 + h2 . , s ∈ 0, 2 π R √ s y ˜ = R sin (s) R 2 +h 2 ⎪ ⎪ ⎩ z˜ (s) = h √ R 2s+h 2 Theorem 5.3.1 Let us consider r˜ (s) = (x(s), ˜ y˜ (s), z˜ (s)), s ∈ [0, L] the natural parametric representation of the path (I, r ). Then: ∥ ∥ ∥ d˜r ∥ ∥ ∥ = 1 and x˜ '2 (s) + y˜ '2 (s) + z˜ '2 (s) = 1. ∥ ds ∥ ( ) Proof Since r˜ (s) = r λ−1 (s) , s ∈ [0, L], it results that: ) ( ) ( dr λ−1 (s) d λ−1 (s) d˜r ( ) = . · ds ds d λ−1 (s) On the other hand, on account of inverse function theorem, we have: ( ) d λ−1 (s) 1 = ' ds λ (t) where t = λ−1 (s). √ Taking into account to the fact that λ' (t) = x '2 (t) + y '2 (t) + z '2 (t), t ∈ I , we deduce: ( ' ) dr 1 d˜r 1 = · ' x (t), y ' (t), z ' (t) = √ ds dt λ (t) x '2 (t) + y '2 (t) + z '2 (t)

(5.12)

hence ∥ ∥ ∥ d˜r ∥ ∥ ∥ = √ ∥ ds ∥ Since

d˜r ds

1 x '2 (t) + y '2 (t) + z '2 (t)

·

√

x '2 (t) + y '2 (t) + z '2 (t) = 1.

( ) = x˜ ' (s), y˜ ' (s), z˜ ' (s) , it follows that: x˜ '2 (s) + y˜ '2 (s) + z˜ '2 (s) = 1.

(5.13)

132

5 Line Integrals

The vector

− → d r˜ ds

→ d− r dt

is collinear to

is known to be tangent to the curve. From (5.12), we deduce that → d− r , dt

and thus it is also tangent to the curve. On the other hand,

from (5.13), it results that

− → d r˜ ds

is a unit vector.

− → − → − → − → Remark 5.3.1 The vector ddsr˜ = x˜ ' (s) i + y˜ ' (s) j + z˜ ' (s) k is the unit tangent ˜ y˜ (s), z˜ (s)). vector to the curve at the point M(x(s),

Remark 5.3.2 Although any smooth curve has a natural parametric representation, it cannot always be found, because we cannot always calculate the definite integral in the right-hand side of the formula: s = λ(t) =

∫ t √

x '2 (u) + y '2 (u) + z '2 (u) du.

a

However, the theoretical importance of the natural representation of a curve is very high, as we will see in the next paragraph.

5.4 Line Integrals of the First Kind The line integral is an extension of the definite integral, in the sense that the integrated interval [a, b] is replaced by a space (or plane) curve. Let γ be a smooth curve and let x = x(t), y = y(t), z = z(t), t ∈ [a, b] be a parametric representation of it. Such a curve is rectifiable and its length is:

L=

∫b √

x '2 (t) + y '2 (t) + z '2 (t) dt.

a

Let x = x(s), ˜ y = y˜ (s), z = z˜ (s), s ∈ [0, L] be its natural parameterization, and let f : Ω ⊂ R3 → R be a continuous function. We assume also that the curve support is included in Ω. Definition 5.4.1 The line integral of the first kind (line integral ∫ with respect to arc length) of the function f along the curve γ is denoted by γ f (x, y, z) ds and is defined by ∫L

∫ f (x, y, z)ds = γ

f (x(s), ˜ y˜ (s), z˜ (s))ds. 0

(5.14)

5.4 Line Integrals of the First Kind

Example 5.4.1 Compute

∫ ( γ

133

) x − y + z 2 ds, where γ is the cylindrical helix:

⎧ ⎨ x = R cos t y = R sin t , t ∈ [0, 2 π ], R > 0, h > 0. ⎩ z = ht According to Example 5.3.2, the natural parametric representation of γ is: ( x˜ = R cos √

)

s R2 + h2

(

) hs , y˜ = R sin √ , z˜ = √ , 2 2 2 R +h R + h2 s

] [ √ where s ∈ 0, 2 π R 2 + h 2 . From Definition 5.4.1, we have: ∫

(

γ

) x − y + z 2 ds

√ 2 π ∫R 2 +h 2(

=

( R cos √

0

=R

√

R2

+

h2

|2 π | h s | )| + ( 2 3 R + h2 | 2 3

( − R sin √

R2 + h2

( ( · sin √ √

)

s

)

s R2 + h2

R 2 +h 2

=

0

)

s R2 + h2

) h2s2 + 2 ds R + h2

))|2π √ R 2 +h 2 | | + cos √ | 2 2 R +h (

s

0

8h π · 3 2

3

√

R2 + h2.

Next, we present the physical interpretation of the line integral of the first kind. Let us suppose that a non-homogeneous wire of negligible thickness is described by the smooth curve γ . 



We denote by AB the support of curve γ and by f : AB → R+ the continuous positive function that expresses the linear density of the wire. Let Δ be an arbitrary partition of the interval [0, L]: Δ : 0 = s0 < s1 < · · · < si− 1 < si < · · · < sn = L 

and let Mi (x(s ˜ i ) , y˜ (si ), z˜ (si )) ∈ AB, i = 0, n, where A = M0 and B = Mn 

(Fig. 5.12). We specify that si represents the length of the arc AMi . If the partition Δ is refining enough, we can assume that the density of the wire is constant on the 

arc Mi−1 Mi ; i.e. the density is equal to the value of the function f at one of the ends of this arc, e.g. 

f (M) = f (Mi ), ∀M ∈ Mi−1 Mi .

134

5 Line Integrals

Fig. 5.12 The mass of a material wire



It follows that the mass of the piece of the wire Mi−1 Mi is approximately equal to f (Mi )(si − si−1 ) and the total mass of the wire is approximated by the sum: n Σ

f (Mi )(si − si−1 ) =

n Σ

i=1

f (x(s ˜ i ), y˜ (si ), z˜ (si ))(si − si−1 ).

i=1

The exact value of the total mass of the wire γ will be: Mass(γ ) = lim

Δ → 0

n Σ

f (x(s ˜ i ), y˜ (si ), z˜ (si ))(si − si−1 ).

i=1

Taking into account to Definition 5.4.1, it results that the mass of the wire is given by: ∫

∫L Mass(γ ) =

f (x(s), ˜ y˜ (s), z˜ (s))ds =

f (x, y, z)ds. γ

0

The exact meaning of the above limit is as follows: ∀ε > 0, ∃δε > 0 such that for any partition Δ of the interval [0, L], with Δ < δε , we have: | | n | | Σ | | f (x(s ˜ i ), y˜ (si ), z˜ (si ))(si − si−1 )| < ε. |Mass(γ ) − | | i=1

∫ In conclusion, the line integral of the first kind γ f (x, y, z)ds is the mass of a non-homogeneous wire of negligible thickness, which is described by the smooth curve γ and has the continuous density function f . If we denote by x G , yG , z G the coordinates of the center of mass G of the non-homogeneous wire γ , then it is shown that:

5.4 Line Integrals of the First Kind

∫ γ

xG = ∫

x f (x, y, z)ds γ

f (x, y, z)ds

135

∫

y f (x, y, z)ds

γ

, yG = ∫

γ

f (x, y, z)ds

∫ γ , zG = ∫

z f (x, y, z)ds γ

f (x, y, z)ds

.

( )  In the case of a homogeneous wire i.e., f (M) = k = ct. , ∀ M ∈ AB , it results that: ∫ Mass(γ ) = k

1ds = k · L , x G = γ

1 · L

∫ x ds, yG = γ

1 · L

∫ y ds, z G = γ

1 · L

∫ z ds. γ

Using the same argument as above, we deduce that the moments of inertia of the wire γ with respect to the origin O(0, 0, 0) of the axes and with respect to the coordinate axes and the coordinate planes are given by the formulas: ∫

(

IO = γ

∫

IO x =

) x 2 + y 2 + z 2 f (x, y, z) ds

(

) y 2 + z 2 f (x, y, z) ds

γ

∫ IO y =

(

) x 2 + z 2 f (x, y, z) ds

γ

∫ I Oz = γ

(

) x 2 + y 2 f (x, y, z) ds

∫

IO x y =

z 2 f (x, y, z) ds γ

∫ IO x z =

y 2 f (x, y, z) ds γ

∫ I O yz =

x 2 f (x, y, z) ds. γ

If we knew the natural parametric representation of a curve, then Formula (5.14) would be sufficient to calculate the line integral of the first kind along this curve. Usually, a curve is given by a parametric representation in which the parameter t is arbitrary, and its natural representation cannot be found. The following theorem allows the computation of the line integral of the first kind if the parametric representation is arbitrary. Theorem 5.4.1 Let γ be a smooth curve and let x = x(t), y = y(t), z = z(t), t ∈ [a, b] be a parametric representation of it. If Ω ⊂ R3 is a domain that contains the support of the curve γ and f : Ω → R is a real continuous function on Ω, then

136

5 Line Integrals

we have: ∫

∫b

√ f (x(t), y(t), z(t)) x '2 (t) + y '2 (t) + z '2 (t) dt.

f (x, y, z)ds = γ

(5.15)

a

Proof Since f is continuous and the functions x, y, z are of C 1 -class on [a, b], it follows that the integral from the right member of (5.15) exists. According to Definition 5.4.1, we have: ∫

∫L f (x, y, z)ds =

γ

f (x(s), ˜ y˜ (s), z˜ (s)) ds. 0

If we make the change of variable s = λ(t), t ∈ [a, b], it results that: x˜ ◦ λ = x, y˜ ◦ λ = y, z˜ ◦ λ = z. √ ds = λ' (t)dt = x '2 (t) + y '2 (t) + z '2 (t) dt. and further: ∫

∫L f (x, y, z)ds =

f (x(s), ˜ y˜ (s), z˜ (s))ds

γ

0 −1 λ∫ (L)

=

f (x(t), y(t), z(t))λ' (t)dt

λ−1 (0)

∫b f (x(t), y(t), z(t)) ·

=

√

x '2 (t) + y '2 (t) + z '2 (t) dt.

a

Example 5.4.2 Compute

∫( γ

) x − y + z 2 ds, where γ is the cylindrical helix:

x = R cos t, y = R sin t, z = ht, t ∈ [0, 2π ], R, h > 0. According to Theorem 5.4.1, we get: ∫ (

)

x − y + z 2 ds =

γ

∫2 π( 0

=

√

R cos t − R sin t + h 2 t 2

)√

( R 2 + h 2 R sin t + R cos t +

R 2 sin2 t + R 2 cos2 t + h 2 dt

h2t 3 3

)|2π | 8h 2 π 3 √ 2 | · R + h2. | = | 3 0

5.4 Line Integrals of the First Kind

137

Example 5.4.3 Find the mass and the moment of inertia with respect to O z axis for a wire γ lying along the cylindrical helix with the constant density f (x, y, z) = 1. The parametric equations of the wire are: x = R cos t, y = R sin t, z = ht, t ∈ [0, 2 π ], R, h > 0. According to Example 5.2.1, the length of the wire is L = 2π Then the mass of the wire is: ∫ √ Mass(γ ) = 1ds = L = 2 π R 2 + h 2 .

√

R2 + h2.

γ

The moment of inertia with respect to O z axis is calculated as follows: ∫

(

I Oz =

) x 2 + y 2 ds

γ

∫2π =

(

R 2 cos2 t + R 2 sin2 t

)√

R 2 sin2 t + R 2 cos2 t + h 2 dt

0

∫2π =

√ √ R 2 R 2 + h 2 dt = 2π R 2 R 2 + h 2 .

0

Remark 5.4.1 The definition and formula for computing the line integral of the first kind along a space smooth curve are directly transposed to the case when the function is defined on the points of a smooth plane curve having the parametric representation: x = x(t), y = y(t), t ∈ [a, b]. In this case, the computation Formula (5.15) becomes: ∫

∫b f (x, y)ds =

γ

Example 5.4.4 Compute x2 a2

+

y2 b2

(5.16)

a

∫ γ

ellipse

√ f (x(t), y(t)) x '2 (t) + y '2 (t) dt.

x y ds, where γ is the arc of the first quadrant of the

− 1 = 0, a, b > 0. 

A parametric representation of curve γ is Using Remark 5.4.1, we have:

] [ x = a cos t , t ∈ 0, π2 , a, b > 0. y = b sin t

138

5 Line Integrals π

∫2

∫ x y ds = γ

√ ab cos t sin t a 2 sin2 t + b2 cos2 t dt.

0

] [ If we consider the change of variable u = a 2 sin2 t + b2 cos2 t, t ∈ 0, π2 , then ) ( 2 du = 2 a − b2 sin t cos t dt, and further we obtain: ∫ γ

|a 2 ∫a 2 √ √ || ab ab ) ) · u u| u du = ( 2 x y ds = ( 2 |2 2 a − b2 3 a − b2 b b2 ) ) ( 3 ( ab a − b3 ab a 2 + ab + b2 ) = . = ( 2 3(a + b) 3 a − b2

Remark 5.4.2 If the smooth plane curve γ is given by an explicit representation, y = y(x), x ∈ [a, b], then x can be taken as a parameter on the curve and Formula (5.16) becomes: ∫

∫b f (x, y)ds =

γ

√ f (x, y(x)) 1 + y '2 (x) dx.

(5.17)

a

∫ 1 Example 5.4.5 Compute γ y−x ds, where γ is the segment on the straight line y = 2x + 1 from point A(0, 1) to point B(2, 5). According to Remark 5.4.2, we have: ∫ γ

1 ds = y−x =

∫2 0

√

√ √ ∫ 1 1 · 1 + 22 dx = 5 dx 2x + 1 − x x +1 2

|2 √ | 5 ln|x + 1|| = 5 ln 3.

0

0

Remark 5.4.1 If γ is a piecewise smooth curve and the function f is continuous and bounded on each smooth piece of the curve γ , then: ∫ f (x, y, z) ds = γ

p ∫ Σ

f (x, y, z)ds,

i=1 γ i

where γ = γ1 ∪ γ2 ∪ · · · ∪ γ p . Remark 5.4.2 The line integral of the first kind does not depend on the orientation of the curve γ .

5.4 Line Integrals of the First Kind

139

Indeed, a parametric representation of the curve γ− is the following: x = x(L ˜ − s), y = y˜ (L − s), z = z˜ (L − s), s ∈ [0, L]. If we make the change of variable u = L − s, we obtain: ∫

∫L f (x, y, z) ds =

f (x(L ˜ − s), y˜ (L − s), z˜ (L − s)) ds

γ−

0

∫0 f (x(u), ˜ y˜ (u), z˜ (u))du =

=− ∫

∫L f (x(u), ˜ y˜ (u), z˜ (u))du 0

L

=

f (x, y, z)ds. γ+

Remark 5.4.3 Since the line integral of the first kind was defined as a definite integral, it preserves the basic properties of the definite integral. In what follows, we will present the basic properties of the line integral of the first kind, frequently used in computations. Proposition 5.4.1 (Linearity Property) ∫ (α f (x, y, z) + βg(x, y, z)) ds γ

∫

∫ f (x, y, z) ds + β

=α γ

g(x, y, z) ds. γ

Proposition 5.4.2 (Monotony Property) ∫ If f is positive and continuous function, then f (x, y, z) ds ≥ 0. γ

Proposition 5.4.3 (Additivity Property) 





If an arc ABis composed by two arcs AC and C B, then: ∫

∫ f (x, y, z) ds = 

∫ f (x, y, z) ds +



AB

f (x, y, z) ds. 

AC

CB

or ∫

∫ f (x, y, z) ds + 

AB

∫ f (x, y, , z) ds +



BC

f (x, y, z) ds = 0. 

CA

140

5 Line Integrals

Proposition 5.4.4 (Estimation of the modulus of the integral) | | | ∫ |∫ | | | f (x, y, z) ds | ≤ | f (x, y, z)| ds. | | | | γ

γ

Theorem 5.4.2 (The mean value theorem) 



If f is continuous on AB = γ , then there is a point (ξ, η, ζ ) ∈ AB such that: ∫ f (x, y, z) ds = f (ξ, η, ζ ) · L , γ

where L is the length of the curve γ .

5.5 Line Integrals of the Second Kind Let γ be a smooth curve, let us consider x = x(t), y = y(t), z = z(t), t ∈ [a, b] a parametric representation of γ and let x = x(s), ˜ y = y˜ (s), z = z˜ (s), s ∈ [0, L] be its natural parametric representation. 

We denote also by AB the support of γ . → → τ = − τ (M) the unit tangent vector at the point We will note by − 

M(x(s) ˜ , y˜ (s) , z˜ (s)) ∈ AB, oriented in the sense of increasing the parameter s (Fig. 5.13). According to Remark 5.3.1, we have: − → − → − → − → τ = x˜ ' (s) i + y˜ ' (s) j + z˜ ' (s) k . Fig. 5.13 The unit vector of the arc AB

5.5 Line Integrals of the Second Kind

141

− → Let F = (P, Q, R) : Ω ⊂ R3 → R3 be a continuous vector field. We further 

assume that AB ⊂ Ω. We have: − → − → − → − → F (x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k , ∀(x, y, z) ∈ Ω. Definition 5.5.1 The line integral of the second kind (line integral with respect − → to coordinates) of the vector field F = (P, Q, R) along the oriented curve γ+ is denoted by ∫ P(x, y, z) dx + Q(x, y, z)dy + R(x, y, z) dz γ+

and it is given by the following definite integral: ∫ P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz γ+

∫ =

− → − F ·→ τ =

∫L

γ+

∫L =

− → → F (x(s), ˜ y˜ (s), z˜ (s)) · − τ ds

0

(

P(x(s), ˜ y˜ (s), z˜ (s)) · x˜ ' (s) + Q(x(s), ˜ y˜ (s), z˜ (s)) · y˜ ' (s)

0

) + R(x(s), ˜ y˜ (s), z˜ (s)) · z˜ ' (s) ds.

(5.18)

Remark 5.5.1 The line integral of the second kind depends on the orientation of the curve γ .  → Indeed, the unit tangent vector to the curve γ− at the point M ∈ AB is −− τ, whence, it results that:

∫

∫L P(x, y, z) dx + Q(x, y, z)dy + R(x, y, z) dz =

γ−

( →) − → F (x, y, z) · −− τ ds

0

∫L =−

− → → F (x, y, z) · − τ ds = −

P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz. γ+

0

Example 5.5.1 Compute

∫

∫ γ+

x dx + y dy + z dz, where γ is cylindrical helix:

142

5 Line Integrals

⎧ ⎨ x = 3 cos t y = 3 sin t , t ∈ [0, 2 π ]. ⎩ z = 4t According to Example 5.3.2, the natural representation of the curve γ is: ( ) ⎧ ˜ = 3 cos( 5s) ⎨ x(s) y˜ (s) = 3 sin 5s , s ∈ [0, 10π ]. ⎩ z˜ (s) = 45s From Formula (5.18), it results that: ∫ x dx + y dy + z dz γ+

=

∫10 π( ( s ) ( 3 ( s )) (s ) 3 ( s ) 4s 4 ) 3 cos + 3 sin ds · − sin · cos + · 5 5 5 5 5 5 5 5 0

∫10π = 0

|10π | 16 s | 16s ds = · | = 32π 2 . 25 25 2 | 2|

0

For the physical interpretation of the line integral of the second kind, we consider a 

smooth curve γ such that the arc AB is its support. Let x = x(s), ˜ y = y˜ (s), z = z˜ (s), s ∈ [0, L] be the natural parameterization  − → of γ and let F = (P, Q, R) : AB ⊂ R3 → R3 be a continuous vector field. We consider also an arbitrary partition Δ of the interval [0, L]: Δ : 0 = s0 < s1 < · · · < si−1 < si < · · · < sn = L . 

We denote by Mi (x(s ˜ i ), y˜ (si ), z˜ (si )) ∈ AB, i = 0, n (Fig. 5.14). The length of 

the arc Mi−1 Mi is si − si−1 . Fig. 5.14 The mechanical work of a variable force which acts along the curve

5.5 Line Integrals of the Second Kind

143

] [ Let ξi ∈ si−1 , si be an arbitrary point, let Ni (x(ξ ˜ i ), y˜ (ξi ), z˜ (ξi )) be the corre − → sponding point on the arc M M , and let τ be the unit tangent vector to the curve i−1

i

i

γ+ at the point Ni . If the partition Δ is refining enough, we can assume that the vector  − → field F = (P, Q, R), which we interpret as a force, is constant on the arc Mi−1 Mi ; i.e. it is equal to its value at the point Ni . − → Under these conditions, the work done by the force F acting on an object, moving 

it along the arc Mi−1 Mi can be approximated by: − → → F (Ni ) · − τ i (si − si−1 ) − → → where by F (Ni ) · − τ i means the scalar product of the two vectors.  − → The total work done by the force F to move an object along the entire arc AB is approximated by the sum: n n Σ Σ − → − → → → F (Ni ) · − F (x(ξ ˜ i ), y˜ (ξi ), z˜ (ξi )) · − τ i (si − si−1 ) = τ i (si − si−1 ) i=1

i=1

=

n Σ (

P(x(ξ ˜ i ), y˜ (ξi ), z˜ (ξi )) · x˜ ' (ξi )

i =1

+ Q(x(ξ ˜ i ), y˜ (ξi ), z˜ (ξi )) · y˜ ' (ξi ) ) + R(x(ξ ˜ i ), y˜ (ξi ), z˜ (ξi )) · z˜ ' (ξi ) (si − si−1 ). The exact value of the total work will be equal to: W = lim

Δ →0

∫

n Σ − → → F (Ni ) · − τ i (si − si−1 ) i=1

P(x, y, z) dx + Q(x, y, z)dy + R(x, y, z)dz.

= γ+

Therefore, the line integral of the second kind ∫ P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz γ+

− → − → − → − → is the mechanical work done by the variable force F = P i + Q j + R k for moving an object along the oriented curve γ+ . Dependence of line integral of the second kind on the orientation of the integrated path is coherent with physical interpretation of the integral as being the work of a

144

5 Line Integrals

field of force along a path. Indeed, if the direction of tracing the trajectory is reversed the work performed by the force field changes its sign in the opposite. The following theorem allows the computation of the line integral of the second kind when the parametric representation of the curve is arbitrary. Theorem 5.5.1 Let γ be a smooth curve and let us take x = x(t), y = y(t), z = z(t), t ∈ [a, b], one of its parametric representation. We denote by γ+ the curve oriented in the direction of increasing the parameter. If Ω ⊂ R3 is a domain including the − → support of γ and F = (P, Q, R) : Ω → R3 is a continuous vector field, then: ∫ P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz γ+

∫b =

(

P(x(t), y(t), z(t)) · x ' (t) + Q(x(t), y(t), z(t)) · y ' (t)

a

) + R(x(t), y(t), z(t)) · z ' (t) dt.

(5.19)

Proof Clearly, the integral of the right member of (5.19) exists, because x, y, z are of C 1 -class on [a, b] and P, Q, R are continuous. According to Definition 5.5.1, we have: ∫ P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz γ+

∫L =

(

P(x(s), ˜ y˜ (s), z˜ (s)) · x˜ ' (s) + Q(x(s), ˜ y˜ (s), z˜ (s)) · y˜ ' (s)

0

) + R(x(s), ˜ y˜ (s), z˜ (s)) · z˜ ' (s) ds. If we make the change of variable change: s = λ(t) =

∫ t √

x '2 (u) + y '2 (u) + z '2 (u) du, t ∈ [a, b]

a

then we obtain: ( ) x(λ(t)) ˜ = x λ−1 (λ(t)) = x(t) y˜ (λ(t)) = y(t), z˜ (λ(t)) = z(t). Also, taking into account the differential rules for composed and inverse functions, we have:

5.5 Line Integrals of the Second Kind

145

( ) ( ) 1 d ( ( −1 )) dx λ−1 (s) d λ−1 (s) x˜ (s) = = x ' (t) · ' . x λ (s) = ( −1 ) · ds ds λ (t) d λ (s) '

Similarly, we have: y˜ ' (s) = y ' (t) · λ'1(t) ; z˜ ' (s) = z ' (t) · ds = λ' (t) dt =

√

1 λ' (t)

and

x '2 (t) + y '2 (t) + z '2 (t) dt.

Therefore, we obtain: ∫L 0

(

P(x(s), ˜ y˜ (s), z˜ (s)) · x˜ ' (s) + Q(x(s), ˜ y˜ (s), z˜ (s)) · y˜ ' (s)

) + R(x(s), ˜ y˜ (s), z˜ (s)) · z˜ ' (s) ds ∫b

=

(

P(x(t), y(t), z(t)) · x ' (t) + Q(x(t), y(t), z(t)) · y ' (t)

a

) + R(x(t), y(t), z(t)) · z ' (t) dt. ∫ Example 5.5.2 Compute y dx + z dy + x dz, where γ has the parametric repreγ+ ⎧ R ⎪ x = + cos t) (1 ⎨ 2 R sentation: y = 2 (1 − cos t) , t ∈ [0, 2π ], R > 0. ⎪ ⎩ z = √R sin t 2 According to Theorem 5.5.1 (Formula (5.19)), we have: ∫ y dx + z dy + x dz γ+

z

Fig. 5.15 The circle lied in the plane x+y = R, parallel to Oz axis and passes through the points A(R, 0, 0) and B(0, R, 0)

P

A

B

O Q

x

y

146

5 Line Integrals ∫2π( =

R (1 − cos t) · 2

( −

0

=−

R2 4

∫2π sin t dt + 0

R2 4

) ) R R R R R sin t + √ sin t · sin t + (1 + cos t) · √ cos t dt 2 2 2 2 2

∫2π 0

R2 sin t cos t dt + √ 2 2

∫2π 0

R2 dt + √ 2 2

∫2π 0

π R2 cos t dt = √ . 2

We notice that from a geometric point of view, the support of curve γ is the circle: 

x 2 + y2 + z2 = R2 x+y = R.

This circle lies in the plane x + y = R that is parallel to Oz axis and passes through the points A(R, 0, 0) and ( B(0, R, ) 0). The straight line segment [AB] is a diameter of the circle. The point R2 , R2 , 0 is its center and √R2 is its radius. Remark 5.5.2 On an arbitrary curve γ we have two orientations. If we take two parametric representations (I1 , r1 ), (I2 , r2 ) of the curve γ , then the change of parameter λ : I1 → I2 is increasing or decreasing. In the first case, (I1 , r1 ) and (I2 , r2 ) have the same orientation of γ . In the second case, they have different orientations of γ . Which of them is considered as positive is a choice who must be specified. Remark 5.5.3 If the curve γ is given by a parametric representation, then γ+ represents the curve γ oriented in the sense of increasing the parameter. If the curve γ is a closed space curve and it is given as an intersection of two surfaces, then the orientation of the curve cannot be deduced from the context and must be specified by the statement. If the curve γ is a closed plane curve, usually the positive orientation of such a curve is counterclockwise. For the line integral of the second kind on a closed curve, a special symbol is used, namely the integral symbol provided with a circle in the middle of it, a circle on which appears an arrow representing the direction of orientation of the curve. When the arrow does not appear on the circle placed on the integral sign, so when ∮ it is denoted by the symbol , we will understand that the integration is performed on a closed curve oriented in a positive direction. For example, in the case of the closed space curve in Example 5.5.2, it can be specified that the orientation of curve is counterclockwise if we look from the point O− the origin of the axis system. Example 5.5.2 can be reformulated as follows: ∮ Compute y dx + z dy + x dz, where γ+ is the space circle γ+



x 2 + y2 + z2 = R2 x+y =R

traced out in the counterclockwise direction, if we look straight down from the center of the sphere.

5.5 Line Integrals of the Second Kind

147

Remark 5.5.1 For a plane curve γ having the parametric representation: x = x(t), y = y(t), t ∈ [a, b] − → and a vector field F = (P, Q), Formula (5.19) becomes: ∫ P(x, y) dx + Q(x, y) dy γ+

∫b =

(

) P(x(t), y(t)) · x ' (t) + Q(x(t), y(t)) · y ' (t) dt.

(5.20)

a

Example 5.5.3 Compute

∮ γ

x dy − y dx, where γ is the circle x 2 + y 2 = R 2 , R > 0,

traced out in a counterclockwise direction. We use the parametric representation of the circle: 

x = R cos t , t ∈ [0, 2π ]. y = R sin t

and we obtain: ∮

∫2π x dy − y dx =

γ

∫2π R 2 dt = 2π R 2 .

(R cos t · R cos t − R sin t · (−R sin t))dt = 0

0

Remark 5.5.2 If the plane curve γ is given by the explicit representation y = y(x), x ∈ [a, b], then the formula for computing the line integral of the second kind becomes: ∫

∫b P(x, y)dx + Q(x, y)dy =

γ+

(

) P(x, y(x)) + Q(x, y(x)) · y ' (x) dx.

(5.21)

a

Remark 5.5.3 Similarly, if the plane curve γ is given by the explicit representation x = x(y), y ∈ [c, d], then the formula for computing the line integral of the second kind becomes: ∫d

∫ P(x, y)dx + Q(x, y)dy = γ+

c

(

) P(x(y), y) · x ' (y) + Q(x(y), y) dy.

(5.22)

148

5 Line Integrals

Example 5.5.4 Compute

∫ γ+

 ( ) 2 x y dx − x 2 + y dy, where γ+ is the arc AB of the

parabola y = x 2 , with A(1, 1) and B(2, 4). 

The explicit representation of the arc AB is: y(x) = x 2 , x ∈ [1, 2]. From Remark 5.5.2, it results that: ∫

(

∫2

)

2x y dx − x + y dy = 2



( 3 ( 2 ) ) 2x − x + x 2 · 2x dx

1

AB

∫2 =−

|2 15 x 4 || 2x dx = − | = − . 2 1 2 3

1

Remark 5.5.4 If γ is a union of smooth curves, γ = γ1 ∪ γ2 ∪ · · · ∪ γ p , then: ∫ P dx + Q dy + R dz =

p ∫ Σ

P dx + Q dy + R dz.

i=1 (γ ) i +

γ+

Example 5.5.5 Compute

∮ γ

(z − y) dx + (x − z)dy + (y − x) dz, where γ is the

triangle ABC with vertices A(a, 0, 0), B(0, b, 0), C(0, 0, c), a, b, c > 0 and the orientation A → B → C → A. Obviously, the curve is the union of the sides of the triangle: γ = AB ∪ BC ∪C A. According to Remark 5.5.4, we have: ∮

∫

∫ +

(z − y)dx + (x − z)dy + (y − x)dz = γ

Fig. 5.16 The triangle ABC

AB

BC

∫ + CA

5.6 Independence on the Path of Line Integral of the Second Kind

149

The canonical equations of the straight line (AB) are: (AB) :

x −a y−0 z−0 = = = t, 0−a b−0 0−0

hence the parametric representation of the segment AB is: ⎧ ⎨ x = a − at AB : y = bt , t ∈ [0, 1]. ⎩ z =0 Further, we have: ∫ (z − y) dx + (x − z) dy + (y − x) d z AB

∫1

∫1 a b dt = ab.

((−bt) · (−a) + (a − at) · b + (bt − a + at) · 0)dt =

= 0

0

Similarly, it is shown that: ∫ (z − y) dx + (x − z) dy + (y − x) dz = bc BC

∫

(z − y) dx + (x − z) dy + (y − x) dz = ca. CA

Therefore, we have: ∮ (z − y)dx + (x − z)dy + (y − x) dz = ab + bc + ca. γ

5.6 Independence on the Path of Line Integral of the Second Kind In this paragraph we will analyze the case when the value of the line integral of the second kind depends only on the extremities of the curve and does not depend on the shape of the curve itself. This case is interesting both from a mathematical point of view, because the computation of such an integral is simpler, and from a practical point of view, because it has applications in thermodynamics.

150

5 Line Integrals

Definition 5.6.1 Let D ⊂ R3 be an open subset and let P, Q, R : D → R be three arbitrary functions. The following expression: ω = P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz, (x, y, z) ∈ D is called differential 1-form on D with coefficients P, Q, R. Moreover, if the functions P, Q, R are of C k -class on D, then ω is called a differential 1-form of C k -class. Example 5.6.1 If f : D ⊂ R3 → R is differentiable function on D, then the first differential d f = ∂∂ xf dx + ∂∂ yf dy + ∂∂ zf dz is a differential 1-form on D with the coefficients

∂f ∂f ∂f , , . ∂x ∂y ∂z

Definition 5.6.2 A differential 1-form ω = P dx + Q dy + R dz on D is called exact if there is a function f : D → R of C 1 -class on D such that ω = d f , which means that the following equalities on D are verified: P=

∂f , ∂x

Q=

∂f , ∂y

R=

∂f . ∂z

− → Remark 5.6.1 If we consider the vector field V : D ⊂ R3 → R3 , − → − → − → − → V (x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k , ∀(x, y, z) ∈ D, − → then the differential 1-form ω = P dx + Q dy + R dz is exact on D if V is a conservative vector field (potential vector field); i.e. there is a scalar function − → f : D → R of C 1 -class on D such that V = grad f . Definition 5.6.3 Let D ⊂ R∫3 be an open and connected subset. We say that the line integral of the second kind P dx + Q dy + R dz is independent on the path in γ

D if for any two points A , B ∈ D and any piecewise smooth curves γ1 and γ2 , which have the supports included in D and with the same ends A and B, we have: ∫

∫ P dx + Q dy + R dz = γ1

P dx + Q dy + R dz. γ2

Theorem 5.6.1 Let D ⊂ R3 be an open and connected set and let P, Q, R : D → R three continuous functions on D. Then the differential 1-form ω = P dx + ∫Q dy + R dz is exact on D if and only if the line integral of the second kind P d x + Q d y + R d z is independent on the path in D.

γ

Proof First, by hypothesis, there is a function f of C 1 -class on D such that: P=

∂f , ∂x

Q=

∂f , ∂y

R=

∂f . ∂z

(5.23)

5.6 Independence on the Path of Line Integral of the Second Kind

151

Let A and B be two arbitrary points in D and let γ be a piecewise smooth curve 

whose support AB is included in D. If x = x(t), y = y(t), z = z(t), t ∈ [a, b] is a parametric representation of γ , then A has the coordinates (x(a), y(a), z(a)) and B has the coordinates (x(b), y(b), z(b)). Let F : [a, b] → R be the composed function: F(t) = f (x(t), y(t)z(t)), t ∈ [a, b]. Taking into account to the differential rules of the composed functions and the equalities (5.23), it results that: ∂f ∂f (x(t), y(t), z(t)) · x ' (t) + (x(t), y(t), z(t)) · y ' (t) ∂x ∂y ∂f + (x(t), y(t), z(t)) · z ' (t) ∂z = P(x(t), y(t), z(t)) · x ' (t) + Q(x(t), y(t), z(t)) · y ' (t) + R(x(t), y(t), z(t)) · z ' (t).

F ' (t) =

The equality is valid for any point t ∈ [a, b] except a finite number of points, namely those points that correspond to the union points set of the smooth curves that compose the curve γ . Since the equality is true excepting a null set, it follows that: ∫ P(x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz γ+

∫b =

(

P(x(t), y(t), z(t)) · x ' (t) + Q(x(t), y(t), z(t)) · y ' (t)

a

) + R(x(t), y(t), z(t)) · z ' (t) dt ∫b =

F ' (t) dt = F(b) − F(a) = f (B) − f ( A).

a

Therefore, the value of the integral does not depend on the shape of the curve but depends only on its ends. Conversely, let M0 (x0 , y0 , z 0 ) ∈ D be a fixed point, M(x, y, z) ∈ D an arbitrary 

point and γ a piecewise curve whose support M0 M is included in D (Fig. 5.17). Since, by hypothesis, the line integral is independent on the path in D, it results that we can define a function f : D → R such that for any M(x, y, z) ∈ D, we have: ∫ f (x, y, z) =

P dx + Q dy + R dz.  M0 M

152

5 Line Integrals

Fig. 5.17 The independence of the line integral on the path in D

M

N

z

M0 y

x

D

Let N (x + h, y, z) ∈ D be a point such that the straight line segment M N ⊂ D (we use here the fact that D is open set). A parametric representation of M N is: x = t, y = y, z = z, t ∈ [x, x + h]. Further, we have: ∫ P dx + Q dy + R dz

f (x + h, y, z) = M0 M

∫

=



∪ MN



∫

P dx + Q dy + R dz +  M0 M

P dx + Q dy + R dz. 

MN

According to Corollary 2.6.4, it results that: f (x + h, y, z) − f (x, y, z) = h

∫



MN

P dx + Qdy + Rdz

h P(t, y, z) dt = x h P(ξ, y, z)h = P(ξ, y, z), = h ∫ x+h

where ξ is a point between x and x + h. Using again the fact that the function P is continuous, it results that there is lim

h→0

f (x + h, y, z) − f (x, y, z) = P(x, y, z), h

that is: ∂f = P. ∂x Similarly, by substituting the segment M N with a segment parallel to O y axis, respectively, Oz axis, it is shown that ∂∂ yf = Q, respectively, ∂∂ zf = R; hence ω is exact.

5.6 Independence on the Path of Line Integral of the Second Kind

153

Fig. 5.18 The representation of the closed curve γ as the union of two curves γ 1 and γ 2

Theorem 5.6.2 Let D ⊂ R3 be an open and connected set ∫ and let P, Q, R : D → R three continuous functions on D. Then the line integral P dx + Q dy + R dz is ∮ γ independent on path in D if and only if the line integral P dx + Q dy + R dz = 0, γ

for any closed piecewise smooth curve γ whose support is included in D. Proof First, let γ1 , γ2 be two piecewise smooth curves that have the same ends and that have their supports included in D (Fig. 5.18) and let γ = γ1 ∪ (γ2 )− . Obviously γ is a closed piecewise smooth curve, whose support is included in D. From the hypothesis, it follows that: ∮ P dx + Q dy + R dz = 0. γ

On the other hand, we have: ∫

∮ P dx + Q dy + R dz = γ

∫ P dx + Q dy + R dz +

γ1

∫

=

P dx + Q dy + R dz

(γ2 )−

∫

P dx + Q dy + R dz − γ1

It results that: ∫

P dx + Q dy + R dz = 0. γ2

∫ P dx + Q dy + R dz =

γ1

that is the line integral

P dx + Q dy + R dz, γ2

∫ γ

P dx + Q dy + R dz is independent on path in D.

Conversely, let γ be a closed, piecewise smooth curve, whose support is included in D and let r (t) = (x(t), y(t), z(t)), t ∈ [a, b] be an arbitrary parametric representation of γ . For any a < c < b, we denoted by γ1 the curve r = r (t), t ∈ [a, c], and by γ2 the curve r = r (t), t ∈ [c, b]. Obviously, γ = γ1 ∪ γ2 . By hypothesis, we have:

154

5 Line Integrals

∫

∫ P dx + Q dy + R dz = (γ1 )+

P dx + Q dy + R dz, (γ2 )+

whence, we deduce: ∫

∫ P dx + Q dy + R dz = γ

∫ P dx + Q dy + R dz +

(γ1 )+

P dx + Q dy + R dz = 0.

(γ2 )−

Definition 5.6.4 A differential 1-form ω = P dx + Q dy + R dz is called closed in D ⊂ R3 if its coefficients P, Q, R : D → R are functions of C 1 -class on D and for any point in D we have: ∂Q ∂P ∂R ∂Q ∂R ∂P = , = , = . ∂y ∂x ∂z ∂ x ∂z ∂y − → Remark 5.6.2 Let V = (P, Q, R) : D ⊂ R3 → R3 be a vector field of C 1 -class on D. Then the differential 1-form ω = P dx + Q dy + R dz is closed in D if and ⇀

only if the field V is irrotational, i.e.: − → Curl V =

(

) ( ) ( ) ∂Q − ∂P ∂R − ∂Q ∂P − ∂R → → → − → − i + − j + − k = 0 , ∀(x, y, z) ∈ D. ∂y ∂z ∂z ∂x ∂x ∂y

Theorem 5.6.3 If the differential 1-form ω = P dx + Q dy + R dz with coefficients P , Q , R : D → R of C 1 -class is exact in D, then ω is closed in D. Proof By hypothesis, there is a function f : D → R of C 2 -class on D such that P = ∂∂ xf , Q = ∂∂ yf , R = ∂∂ zf . Since the second-order derivatives of the function f are continuous, it follows that the mixed derivatives are equal. Therefore, we have: ∂2 f ∂2 f ∂Q ∂P = = = ∂y ∂ y∂ x ∂ x∂ y ∂x ∂2 f ∂2 f ∂R ∂P = = = ∂z ∂z∂ x ∂ x∂ z ∂x ∂2 f ∂2 f ∂R ∂Q = = = . ∂z ∂ z∂ y ∂ y∂z ∂y Definition 5.6.5 A subset S ⊂ R3 is said to be a star domain (star convex set) if there is a point A ∈ S with the property that the straight line segment [A , M] is included in S, for any M ∈ S. We recall that [A, M] = {(1 − t)A + t B;

t ∈ [0, 1]}.

5.6 Independence on the Path of Line Integral of the Second Kind

155

Remark 5.6.3 Any convex set is a star domain, but the reciprocal statement is not generally true. For example, the set R2 \{(x, 0); x > 0} is a star domain (with respect to the point O(0 , 0)), but it is not convex. Theorem 5.6.4 If D ⊂ R3 is an open star domain, then any closed differential 1-form in D is exact in D. Proof By hypothesis, there is A ∈ D such that [A, M] ⊂ D, ∀M ∈ D. For simplicity, we suppose that A coincides with the origin O(0 , 0 , 0) and M has the coordinates (x , y , z). Let t ∈ [0, 1] be arbitrary and let us take T = (1 − t)O + t M = (t x, t y, t z) the corresponding point on the segment [O, M]. We define the function f : D → R as follows: ∫1 f (x, y, z) =

(P(T ) · x + Q(T ) · y + R(T ) · z)dt 0

∫1 =

(P(t x, t y, t z) · x + Q(t x, t y, t z) · y + R(t x, t y, t z) · z)dt. 0

Taking into account the differentiation theorem of the integral depending on a parameter (Theorem 4.1.2), it results that: ∂f = ∂x

∫1 ( 0

+

∂P (t x, t y, t z) · t · x + P(t x, t y, t z) ∂x

) ∂Q ∂R (t x, t y, t z) · t · y + (t x, t y, t z) · t · z dt. ∂x ∂x

On the other hand, by hypothesis, we have: ∂P ∂R ∂P ∂Q = and = , ∂x ∂y ∂x ∂z Hence: ∂f = ∂x

∫1 ( 0

∂P ∂P (t x, t y, t z) · t · x + P(t x, t y, t z) + (t x, t y, t z) · t · y ∂x ∂y

) ∫1 d ∂P + (t x, t y, t z) · t · z dt = (P(t x, t y, t z) · t)dt ∂z dt 0

= P(t x, t y, t z) · t|10 = P(x, y, z) · 1 − P(0, 0, 0) · 0 = P(x, y, z).

156

5 Line Integrals

Therefore ∂∂ xf = P, and analogously ∂∂ yf = Q, ∂∂zf = R; that is ω is exact. From Theorems 5.6.1, 5.6.2, 5.6.3, 5.6.4, we can summarize the following result: Remark 5.6.1 Let ω = P dx + Q dy + R dz be a differential 1-form of C 1 -class on the open star domain D ⊂ R3 . Then the following assertions are equivalent: 1 (1) ω is exact in D (i.e. ( ∃ f ∈ C (D) such that d f = P dx ) + Q dy + R dz). ∂Q ∂P ∂P ∂R ∂Q ∂R (2) ω is closed in D i.e., ∂ y = ∂ x , ∂ z = ∂ x , ∂z = ∂ y . ∫ (3) P dx + Q dy + R dz is independent on path in D. γ ∮ (4) P dx + Q dy + R dz = 0, for any closed piecewise smooth curve γ , whose γ

support is included in D. Remark 5.6.4 Under the hypotheses from Remark 5.6.1, we have: (1) If the line integral

∫ γ

P dx + Q dy + R dz is independent on the path in D, then

we get the Leibniz–Newton formula for line integral: ∫ P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz γ

∫

=

P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz 

AB B(x∫ 2 ,y2 ,z 2 )

P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz

= A(x1 ,y1 ,z 1 )

= f (B) − f (A) = f (x2 , y2 , z 2 ) − f (x1 , y1 , z 1 ) where f is a differentiable function on D such that d f = P dx + Q dy + R dz, and the points A(x1 , y1 , z 1 ), B(x2 , y2 , z 2 ) ∈ D are the ends of the support of the curve γ. (2) The differentiable function f from remark (1) is easily determined using the formula:

∫x f (x, y, z) =

∫y P(x, y0 , z 0 )dx +

x0

∫z Q(x, y, z 0 )dy +

y0

R(x, y, z)dz z0

for any (x, y, z) ∈ D, where (x0 , y0 , z 0 ) ∈ D is an arbitrary fixed point and under the hypothesis that the corresponding path is included in D.

5.6 Independence on the Path of Line Integral of the Second Kind

Example 5.6.2 Compute

(2,3,4) ∫ (1,1,1)

157

yz dx + zx dy + x y dz.

The differential 1-form ω = yz dx + zx dy + x y dz is closed in R3 , because: P = y z, Q = zx, R = x y and

∂Q ∂P ∂R ∂Q ∂P ∂R = = z, = = x, = = y. ∂x ∂y ∂y ∂z ∂z ∂x

From Remark 5.6.1, we deduce that this line integral is independent on the path in R3 . Then, for the computation of the line integral we can choose the curve γ as the polygonal line determined by the points A(1, 1, 1),B(2, 1, 1), C(2, 3, 1), and D(2, 3, 4) (the polygonal line that joints the points) (1, 1, 1) and (2, 3, 4) by straight line segments parallel to the coordinate axes. ∫

(2,3,4) ∫

yz dx + zx dy + x y dz = (1,1,1)

yz dx + zx dy + x y dz AB

∫

+

yz dx + zx dy + x y dz BC

∫

yz dx + zx dy + x y dz

+ CD

(2,1,1) ∫

yz dx + zx dy + x y dz

= (1,1,1)

(2,3,1) ∫

+

yz dx + zx dy + x y dz

(2,1,1) (2,3,4) ∫

+

yz dx + zx dy + x y dz

(2,3,1)

∫2

∫3 dx +

= 1

∫4 2dy +

1

6dz = 1 + 4 + 18 = 23. 1

Another solution can be given if we notice that ω = d f , where f (x, y, z) = x yz. Then: (2,3,4) ∫

(1,1,1)

|(2,3,4) | | | yz dx + zx dy + x y dz = x yz | = 24 − 1 = 23. | | (1,1,1)

158

5 Line Integrals

Remark 5.6.2 A differential 1-form ω = P dx + Q dy is closed in D ⊂ R2 if: P, Q : D → R are functions of C 1 -class on D and ∂∂Py = ∂∂Qx .

∮ ( ) Example 5.6.3 Compute 4y(x − 1)dx + 2x 2 − 4x + y dy, where γ is the ellipse γ

x2 4

y2 25

+ = 1. We denote by P(x, y) = 4y(x − 1) and Q(x, y) = 2x 2 − 4x + y. Obviously. P, Q : R2 → R are functions of C 1 -class on R2 and ∂∂Py = ∂∂Qx = 4(x − 1). It follows that the differential 1-form ( ) ω = 4y(x − 1)dx + 2x 2 − 4x + y dy.

is closed in R2 , and hence it is exact in R2 . Since γ is a closed curve, from Theorem 5.6.1 and Theorem 5.6.2, we deduce that ∮ ( ) 4y(x − 1)dx + 2x 2 − 4x + y dy = 0 γ

thus no computation is necessary.

Chapter 6

Double and Triple Integrals

In this chapter we will study the integration of the functions of two and three real variables. The obtained results extend without difficulty to functions of n-variables. In the case of functions of one variable, the integrated domain is a closed and bounded interval, and the definition of the integral is made with the help of Riemann or Darboux sums, in which the lengths of the subintervals corresponding to the partition of the integrated interval appear. For functions of two variables, the integrated domain is naturally to be a bounded plane set. In order to be able to define the Riemann or Darboux sums in this case as well, we must assume that the integrated domain is squareable (has area). In Chap. 2, Sect. 2.7, we showed that a bounded plane set is squareable if and only if its boundary is a set of area zero (Theorem 2.7.2).

6.1 Double Integral. Definition and Properties We begin this paragraph by presenting a particular case of a squareable bounded plane set, a very common case in applications. Theorem 6.1.1 The support of a piecewise smooth curve is a set of area zero. Proof Since a finite union of area zero sets is also an area zero set, it is sufficient to show that the support of a smooth curve γ is a set of area zero. Let r : [a, b] → R2 be the representative parameterized path of smooth curve γ , r (t) = (x(t), y(t)), t ∈ [a, b]. We denote by L the length of this path (Theorem 5.2.1) ˜ y = y˜ (s) be its natural parameterization. and let x = x(s), Let us take Δ : 0 = s0 < s1 < · · · < si−1 < si < · · · < sn = L an equidistant partition of the interval [0, L] and let Mi (x(s ˜ i ), y˜ (si )), i = 0, n be points on the ͡

curve support (Fig. 6.1). The length of the arc Mi−1 Mi is Ln . Consider a square Di centered at Mi and with the sides parallel to the coordinate axes, having the length 2 · Ln . © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Integral Calculus for Engineers, https://doi.org/10.1007/978-981-19-4793-3_6

159

160

6 Double and Triple Integrals

Fig. 6.1 The support of a smooth curve

L n Mi Di

M i −1

It is obvious that the curve support is included in area

( n ∪ i=0

) Di

≤

n Σ i=0

area(Di ) =

∪n i=0

Di and that:

4 L2 (n + 1). n2

2

Since lim 4nL2 (n + 1) = 0, for n sufficiently large, the area of the set n→∞ is however small, hence the support of the curve γ is a set of area zero.

∪n i=0

Di

Corollary 6.1.1 Any bounded plane set whose boundary is piecewise smooth is squareable. Proof The statement follows from Theorem 6.1.1 and Theorem 2.7.2. Let A ⊂ R2 be a bounded subset. Then there is a circle (disk) that contains the set A. It follows that the distance between any two points of the set A is less than the diameter of this circle. Therefore, the set of positive real numbers {dist(M, N ); M, N ∈ A} is upper bounded, hence it has a least upper bound (Fig. 6.2). Definition 6.1.1 Let A ⊂ R2 be a bounded subset. The following number is called the diameter of the set A: not

d( A) = diam(A) = sup{dist(M, N ); M, N ∈ A}. In the following we will denote by D a squareable bounded set of R2 . Definition 6.1.2 It is called a partition of D any finite family of subsets {Di }, ◦ ◦ ∪p Di ⊂ D, i = 1, p, with the properties: Di ∩ D j = ∅, for i /= j and D = i=1 Di . If we denote by ρ the partition D1 , D2 , . . . , D p of D, then the norm of this partition is defined as follows: Fig. 6.2 The diamenter of a set

d ( A)

A

6.1 Double Integral. Definition and Properties

161

Fig. 6.3 The partition of the domain D

D1 D2

Dp

Di

∥ρ∥ = max{diam(Di ); 1 ≤ i ≤ p}. From Proposition 2.7.3, we deduce (Fig. 6.3): area(D) =

p Σ

area(Di ).

i=1

Definition 6.1.3 The partition ρ ' of the domain D is said to be a refinement of the partition ρ of D, and denoted by ρ ' ≻ ρ, whether each subdomain of the partition ρ is a finite union of subdomains of the partition ρ ' which have no common interior points. ( ) Therefore, if ρ is the partition (Di )1≤i≤ p , then ρ ' has the form Di' j 1≤i≤ p , where ∥ ∥ 1≤ j≤ni ∪i Di = nj=1 Di' j , ∀ 1 ≤ i ≤ p. Obviously, if ρ ' ≻ ρ, then ∥ρ∥ ≥ ∥ρ ' ∥. Let ρ : D1 , D2 , . . . , D p be a partition of the domain D ⊂ R2 and let f : D → R be a bounded function on D. We denote by: m = inf{ f (x, y); (x, y) ∈ D}, M = sup{ f (x, y); (x, y) ∈ D} m i = inf{ f (x, y); (x, y) ∈ Di }, Mi = sup{ f (x, y); (x, y) ∈ Di }, 1 ≤ i ≤ p. Definition 6.1.4 The lower Darboux sum, respectively the upper Darboux sum associated to the bounded function f and the partition ρ are defined as follows: sρ =

p Σ i=1

m i · area(Di ) and Sρ =

p Σ

Mi · area(Di ).

i=1

Since m ≤ m i ≤ Mi ≤ M, ∀1 ≤ i ≤ p and area(D) = follows: m · area(D) ≤ sρ ≤ Sρ ≤ M · area(D).

Σp i=1

area(Di ), it

(6.1)

162

6 Double and Triple Integrals

Lemma 6.1.1 If ρ ≺ ρ ' , then sρ ≤ sρ ' ≤ Sρ ' ≤ Sρ . Proof We assume that the partition ρ consists of domains (Di )1≤i≤ p and the partition ( ) ρ ' consists of domains Di' j 1≤i≤ p . 1≤ j≤n i ∪i Since ρ ≺ ρ ' , it results that for any 1 ≤ i ≤ p, we have Di = nj=1 Di' j . If } { we denote by m i' j = inf f (x, y); (x, y) ∈ Di' j , then m i ≤ m i' j , ∀ i = 1, p, ∀ j = 1, n i . Further, we have: sρ =

p Σ

m i · area(Di ) =

i=1

≤

p ni Σ Σ

p Σ

⎞ ⎛ ni Σ ( ') m i ·⎝ area Di j ⎠

i=1

j=1

( ) m i' j · area Di' j = sρ ' .

i=1 j=1

Therefore, we proved that sρ ≤ sρ ' . Similarly, it is shown that Sρ ' ≤ Sρ . Lemma 6.1.2 For any two partitions ρ ' and ρ '' of domain D, we have sρ ' ≤ Sρ '' . ( ) Proof Let us suppose that the partition ρ ' consists of the domains Di' 1≤i≤ p and ( ) the partition ρ '' consists of the domains D ''j . If we denote by ρ the partition 1≤ j≤q ( ) consisting of the domains Di' ∩ D ''j 1≤i≤ p , then ρ is finer as ρ ' and ρ '' . According 1≤ j≤q

to Lemma 6.1.1, we have: sρ ' ≤ sρ ≤ Sρ ≤ Sρ '' . Next, let us denote by P the family of all partitions of the domain D and we define the lower (upper) Darboux double integral as follows: { } { } I∗ = sup sρ ; ρ ∈ P and I ∗ = inf Sρ ; ρ ∈ P . The existence of these real numbers, I∗ and I ∗ , results from the inequalities (6.1). On the other hand, from Lemma 6.1.2, we deduce that: I∗ ≤ I ∗ . Definition 6.1.5 Let f : D → R be a bounded function on the squareable bounded domain D ⊂ R2 . We say that the function f is integrable in the Darboux sense on D if: I∗ = I ∗ = I.

6.1 Double Integral. Definition and Properties

The common value I is denoted by

˜

163

f (x, y)dx dy and it is called the double

D

integral of the function f on the domain D; the real number I is uniquely determined by f and D. Example 6.1.1 Let D ⊂ R2 be an arbitrary squareable bounded subset. Prove that any constant function f : D → R, f (x, y) = k, ∀(x, y) ∈ D is integrable on D and: ¨ 1 dx dy = area(D). D

Indeed, for any partition ρ : D1Σ , . . . , D p of D we have m i = Mi = k,∀ i = 1, p, p whence, it results that sρ = Sρ = i=1 k · area(Di ) = k · area(D), and further that: ∗

¨

I∗ = I = k · area(D) =

¨ 1 dx dy = area(D).

k dx dy, hence D

D

Theorem 6.1.2 (Darboux criterion for integrability) Let D ⊂ R2 be a squareable bounded subset and let f : D → R be a bounded function. The necessary and sufficient condition that f to be integrable on D is that for every ε > 0, there is δε > 0 with the property that for any partition ρ of D, with ∥ρ∥ < δε the following inequality is verified Sρ − sρ < ε. The proof is similar to the proof presented in the one-dimensional case (Theorem 2.2.1). Theorem 6.1.3 If D ⊂ R2 is a compact squareable subset, then any continuous function f : D → R is integrable on D. The proof is similar to the proof presented in the one-dimensional case (Theorem 2.3.1). The following theorem can also be proved: Theorem 6.1.4 (Lebesgue criterion for integrability) Let D ⊂ R2 be a compact squareable subset. The necessary and sufficient condition that the function f : D → R to be integrable on D is that f to be bounded and the set of all its points of discontinuity to be a null set. Next, we define the Riemann sums. Definition 6.1.6 Let f : D → R be a bounded function, let ρ : D1 , D2 , . . . , D p be an arbitrary partition of D and let (ξ, η) = (ξi , ηi )1≤i≤ p be such that (ξi , ηi ) ∈ Di , ∀ i = 1, p, an arbitrary system of intermediate points associated to the partition ρ. The Riemann sum of the function f , associated to the partition ρ and the intermediate points (ξi , ηi ) is defined as follows:

164

6 Double and Triple Integrals

σρ ( f ; ξ, η) =

p Σ

f (ξi , ηi ) · area(Di ).

i=1

Since m i ≤ f (ξi , ηi ) ≤ Mi , ∀ i = 1, p, it results that: sρ ≤ σρ ( f ; ξ, η) ≤ Sρ for any intermediate point system (ξ, η). Definition 6.1.7 Let f : D → R be a bounded function. We say that the function f is integrable in the Riemann sense on D if there is a finite real number I with the property that for any ε > 0, there is δε > 0 such that for any partition ρ of D, with ∥ρ∥ < δε and any intermediate points (ξi , ηi ) ∈ Di , the following inequality is verified: | | |σρ ( f ; ξ, η) − I | < ε. The real number I is called the double integral of f on D and it is denoted by: ¨ I =

f (x, y)dx dy. D

Remark 6.1.1 For any ε > 0, there are (αi , βi ) ∈ Di and (λi , μi ) ∈ Di such that: Sρ − σρ ( f ; α, β) < ε and σρ ( f ; λ, μ) − sρ < ε where (α , β) = (αi , βi )1 ≤ i ≤ p , (λ , μ) = (λi , μi )1 ≤ i ≤ p . Indeed, from the definition of the least upper bound, it follows that for any ε > 0, ε . Further, we have: there is (αi , βi ) ∈ Di such that Mi − f (αi , βi ) < area(D) Sρ − σρ ( f ; α, β) =

p Σ

(Mi − f (αi , βi ))area(Di )

i=1


0; 1 + t2

π 1 π , b = , c = √ , d = 1. 2 3 3 =

2t , 1+t 2

according to formula (6.7), we have:

2 1 + t2 · dt = 2t 1 + t2

∫1 √1 3

| | √ 1 1 | dt = ln|t| | 1 = ln 3 = ln 3. | √13 t 2

Next, we present a change of variable formula for the double integral analogous to formula (6.7). Let Ω ⊂ R2 be an open, bounded, squareable subset. To keep the analogy with the one-dimensional case (it is known that intervals are the only connected sets in R), we assume in addition that set Ω is connected. Let F: Ω → R2 , F(u, v) = (x(u, v), y(u, v)), ∀(u, v) ∈ Ω be a vector function with the properties: (i) F is of C 1 -class on Ω (ii) F: Ω → D is bijective (iii) F is a regular transformation on Ω, this means that: | ∂x | D(x, y) det JF (u, v) = (u, v) = || ∂u ∂y D(u, v) ∂u

∂x ∂v ∂y ∂v

| | |(u, v) /= 0, ∀(u, v) ∈ Ω |

and let D = F(Ω) ⊂ R2 . Such a vector function is also called a change of coordinates or a change of variables. Since a regular transformation transforms open and connected sets into open and connected sets, it results that D ⊂ R2 is also an open and connected set. Proposition 6.4.1 By a change of variables, any point on the boundary of D corresponds to a point on the boundary of Ω, and coverse. In other words: F(∂Ω) = ∂ D. Proof Indeed, since Ω and D are open sets and F: Ω → R2 is continuous we have: ( ) Ω = Ω ∪ ∂Ω, D = D ∪ ∂ D, D = F(Ω) ⊂ F Ω ⊂ F(Ω) = D.

6.4 Change of Variables in Double Integral

175

( ) Using the continuity of F and the fact that Ω is compact, we deduce that F Ω ( ) ( ) is compact, and therefore D ⊂ F Ω , i.e. D = F Ω or D ∪ ∂ D = F(Ω ∪ ∂Ω) = F(Ω) ∪ F(∂Ω) = D ∪ F(∂ D). Since D ∩ ∂ D = ∅, we deduce from above that ∂ D ⊂ F(∂Ω). We prove now that F(∂Ω) ⊂ ∂ D. In( the ) contrary case, we choose z 0 ∈ ∂Ω such that F(z 0 ) ∈ / ∂ D. Since F(∂Ω) ⊂ F Ω = D = D ∪ ∂ D, we get F(z 0 ) ∈ D. Since F is an homeomorphism between Ω and D, there is z 1 ∈ Ω such that F(z 1 ) = = F(z 0 ) ∈ D. We consider now r > 0 such that B(z 1 , r ) ⊂ Ω and B(z 1 , r ) ∩ B(z 0 , r ) = ∅. From the above considerations F(B(z 1 , r ))( is(an open )) subset of D containing F(z 0 ). If 0 < r ' < r is sufficiently small, then F B z 0 , r ' ⊂ F(B(z 1 , r )), i.e. we arrive to the contradictory relations: ( ( ) ) B z 0 , r ' ∩ Ω /= ∅, B(z 1 , r ) ∩ B z 0 , r ' ∩ Ω = ∅, ( ( ) ) F(B(z 1 , r )) ∩ B z 0 , r ' ∩ Ω /= ∅. Remark 6.4.1 The statement of Proposition 6.4.1 remains valid even when F is a continuous map from Ω to R2 or F is an homeomorphism between Ω and D. These conditions are verified in all further situations. Taking into account that the Jacobian (u, v) → D(x,y) v): Ω → R is continD(u,v) (u, uous on the connected subset Ω, it results that it keeps constant sign on Ω. Therefore, either D(x, y) (u, v) > 0, ∀(u, v) ∈ Ω D(u, v) either D(x, y) (u, v) < 0, ∀(u, v) ∈ Ω. D(u, v) We notice that we find the situation from the one-dimensional case, with the instead of the derivative u ' . Jacobian D(x,y) D(u,v) The following formula can be proved, analogous to formula (6.7): Theorem 6.4.1 (The change of variables formula in double integral) Let F: Ω → D, F(u, v) = (x(u, v), y(u, v)), ∀(u, v) ∈ Ω be a change of variables and let f : D → R be a continuous function. Then the domain D = F(Ω) is squareable and the following formula holds: ¨

¨ f (x, y)dx dy = D

Ω

| | | | D(x, y) | f (x(u, v), y(u, v))| (u, v)||du dv. D(u, v)

(6.8)

176

6 Double and Triple Integrals

The formula (6.8), as well as formula (6.7), is extremely used in practice and transforming the initial integral into an easier to compute integral. In the case of formula (6.8) a new element intervenes, which is missing in the case of the simple integral: formula (6.8) aims at a simplification not only of the function f , but also the integrated domain. The choice of changing variables makes the domain Ω to have a simpler shape than the domain D (to be, e.g., a rectangle). Then the computation of the initial integral is considerably simplified. Case 1. The most used change of variables is the transition to the polar coordinates: {

x = ρ cos θ , 0 < ρ < ∞, 0 < θ < 2π. y = ρ sin θ

(6.9)

If we denote by: A = {(ρ, θ ); 0 < ρ < ∞, 0 < θ < 2π }, and B = R2 \{(x, 0); x ≥ 0} F(ρ, θ ) = (ρ cos θ, ρ sin θ ), then F: A → B is a regular transformation, because, its Jacobian is: det JF (ρ, θ ) =

| | D(x, y) || cos θ −ρ sin θ || = ρ > 0, ∀(ρ, θ ) ∈ A. =| sin θ ρ cos θ | D(ρ, θ )

Let 0 < α < β < 2π and let ϕ : [α, β] → R be a continuous function. We denote also by: Ω = {(ρ, θ ); α < θ < β, 0 < ρ < ϕ(θ )} and D = F(Ω). Then F: Ω → D is a change of variables. If f : D → R is continuous, then according to formula (6.8) we have: ¨

¨ f (x, y)dx dy =

f (ρ cos θ, ρ sin θ ) · ρdρ dθ Ω

D

∫β = α

⎞ ⎛ ϕ(θ ) ∫ ⎝ f (ρ cos θ, ρ sin θ ) · ρ dρ ⎠dθ . 0

6.4 Change of Variables in Double Integral

Example 6.4.2 Compute

˜ √

177

x 2 + y 2 dx dy if:

D

{ } √ x 2 2 2 D = (x, y); x + y < a , √ < y < x 3, x > 0 . 3 ) ( In this case, Ω = F −1 (D) is the rectangle π6 , π3 × (0, a). Indeed, taking into account the formulas (6.9) in the inequalities that define the domain D, we obtain: { } √ cos θ 2 2 Ω = (ρ, θ ); ρ < a , √ < sin θ < 3 cos θ 3 { } ( √ π π) 1 × (0, a). = (ρ, θ ); 0 < ρ < a, √ < tanθ < 3 = , 6 3 3 Therefore we have (Figs. 6.9 and 6.10): ¨ √

x 2 + y 2 dx dy =

¨ √ Ω

D

π

∫3 = π 6

Example 6.4.3 Compute

˜

ρ 2 cos2 θ + ρ 2 sin2 θ · ρ dρ dθ

π ⎞ ⎛ a ∫3 3 ∫ π · a3 a ⎝ ρ 2 dρ ⎠dθ = dθ = . 3 18

0

π 6

x y dx dy, where:

D

{ } D = (x, y); x 2 + y 2 < 2ax, y > 0, a > 0 . We notice that the equation x 2 + y 2 − 2 a x = 0 is the equation of the circle of radius a centered at the point (a, 0) (Fig. 6.11). Substituting (6.9) in the inequalities of D, we obtain (Fig. 6.12): Fig. 6.9 A circular sector of a circle centered in the origin and of radius a

y

π 3 D O

π 6 a

x

178

6 Double and Triple Integrals

ρ

Fig. 6.10 The variation domain of the polar coordinates θ and ρ

a

Ω O

π

π

6

3

θ

{ } Ω = (ρ, θ ); ρ 2 < 2 a ρ cos θ, ρ sin θ > 0 { π} . = (ρ, θ ); 0 < ρ < 2 a cos θ, 0 < θ < 2 Further, we have: ¨ ¨ x y dx dy = (ρ cos θ · ρ sin θ ) · ρ dρ dθ Ω

D

π

∫2 =

π ⎞ ⎛ 2 a cos θ ∫ ∫2 ⎝ ρ 3 cos θ sin θ dρ ⎠dθ = 4 a 4 cos5 θ sin θ dθ

0

π

= −4 a

0

0

∫2

| 2 a4 cos6 θ || π 2 = cos θ · (cos θ ) dθ = −4 a · . 6 |0 3 5

4

'

4

0

Fig. 6.11 The superior semicircle centered in (a, 0) and of radius a

y D O

Fig. 6.12 The variation domain of the polar coordinates θ and ρ

a

2a x

ρ

ρ ( θ ) = 2 a cos θ

2a

Ω O

π 2

θ

6.4 Change of Variables in Double Integral

179

y

Fig. 6.13 The ellipse centered in the origin and of semi-axis a and b

b

D O

a x

Case 2. Another change of variables is the transition to generalized polar coordinates and it is used if the integrated domain is bounded by an ellipse. If the equation of 2 2 the ellipse is ax 2 + by2 = 1, then the generalized polar coordinate is defined by the relations: { x = a ρ cos θ , 0 < ρ < 1, 0 < θ < 2π. (6.10) y = b ρ sin θ The Jacobian is: | | D(x, y) || a cos θ −a ρ sin θ || = a b ρ > 0. =| b sin θ b ρ cos θ | D(ρ, θ ) Example 6.4.4 Compute

˜

(x + y − 2)dx dy, where (Fig. 6.13):

D

} { x2 y2 D = (x, y) ∈ R2 ; 2 + 2 ≤ 1 a b In this case, the generalized polar coordinates (6.10) are used and it is obtained (Fig. 6.14): ¨ D

⎛ ⎞ ∫2 π ∫1 (x + y − 2)dx dy = ⎝ (a ρ cos θ + b ρ sin θ − 2) · a b ρ dρ ⎠dθ 0

0

∫2π =a b 2

| | ∫2π ρ 3 ||1 ρ 3 ||1 2 sin θ dθ cos θ | dθ + a b 3 0 3 |0 0

0

∫2π − 2a b 0

| ρ |1 dθ = −2π a b. 2 |0 2|

180

6 Double and Triple Integrals

ρ

Fig. 6.14 The variation domain of the polar coordinates θ and ρ

1

Ω O

2π

θ

6.5 Applications of the Double Integral in Geometry and Mechanics 6.5.1 Mass of a Lamina By thin plate or lamina we mean a material flat object of negligible thickness in the shape of a bounded squareable plane domain D ⊂ R2 . The lamina is generally considered non-homogeneous, its density being given by a continuous positive function f : D → R+ . Let ρ : D1 , D2 , . . . , D p be an arbitrary partition of D and let (ξi , ηi ) ∈ Di , i = 1, p be an arbitrary point. The mass of each plate Di may be approximated with the product f (ξi , ηi ) · area(Di ). Obviously, the mass of the entire plate is approximated by: Mass(D) ≈

p Σ

f (ξi , ηi ) · area(Di ).

i=1

The exact value of the mass of lamina is given by the formula: Mass(D) = lim

∥ρ∥→0

p Σ i=1

¨ f (ξi , ηi ) · area(Di ) =

f (x, y)dx dy.

(6.11)

D

Remark 6.5.1 If the lamina D is homogeneous, i.e. it has a constant density at each point ( f (x, y) = k = ct., ∀(x, y) ∈ D), then its mass is equal to: Mass(D) = k · area(D).

6.5 Applications of the Double Integral in Geometry and Mechanics

181

6.5.2 Coordinates of the Center of Mass of a Lamina Let D ⊂ R2 be a non-homogeneous lamina of density f : D → R+ and let (x G , yG ) be the coordinates of the center of mass G. We consider a partition ρ : D1 , D2 , . . . , D p of the domain D and some arbitrary points (ξi , ηi ) ∈ Di . The mass of the plate Di is approximated by the product f (ξi , ηi ) · area(Di ). If we consider the mass of the plate Di concentrated at a single point, namely at the point (ξi , ηi ), then the coordinates of the center of mass will be (Fig. 6.15): p Σ

xG ≈

p Σ

ξi f (ξi , ηi ) · area(Di )

i=1 p Σ

ηi f (ξi , ηi ) · area(Di )

i=1 p Σ

, yG ≈ f (ξi , ηi ) · area(Di )

. f (ξi , ηi ) · area(Di )

i=1

i=1

Assuming that f is continuous on D, at the limit we get the exact coordinates: p Σ

xG =

˜

ξi f (ξi , ηi ) · area(Di )

i=1 lim p ∥ρ∥→0 Σ

= ˜ f (ξi , ηi ) · area(Di )

˜

ηi f (ξi , ηi ) · area(Di )

i=1 lim p ∥ρ∥→0 Σ

f (x, y) dx dy

D

i=1 p Σ

yG =

x f (x, y)dx dy

D

= ˜

y f (x, y)dx dy

D

f (ξi , ηi ) · area(Di )

(6.12)

f (x, y) dx dy

D

i=1

Remark 6.5.2 In the particular case of a homogeneous plate, it results that: ˜ D xG = ˜

˜

x dx dy 1 dx dy

=

D

x dx dy

D

area(D)

˜ D , yG = ˜

˜

y dx dy 1 dx dy

=

y dx dy

D

area(D)

.

(6.13)

D

Example 6.5.1 Find the coordinates of the center of mass of a homogeneous lamina D, with the constant density f (x, y) = 2, where: Fig. 6.15 The center of mass of a non-homogeneous lamina D

y G

D O

x

182

6 Double and Triple Integrals

{ } π D = (x, y) ∈ R2 ; 0 ≤ x ≤ , 0 ≤ y ≤ cos x . 2 Using the formulas (6.13), we have successively (Fig. 6.16): π

∫2

¨ 1 dx dy =

π ⎛ cos x ⎞ ∫ ∫2 ⎝ dy ⎠ dx = cos x dx = 1.

0

D

0

0

π ⎞ ⎛ cos x ¨ ∫ ∫ ∫2 x dx dy = ⎝ x dy ⎠dx = x cos x dx π 2

0

D

0

0 π 2

| ∫ |π π | 2 = x sin x | − sin x dx = − 1. 0 2 0 π ⎞ ⎛ cos x ∫ ∫ ∫2 ¨ cos2 x dx y dx dy = ⎝ y dy ⎠dx = 2 π 2

0

D

∫ =

0 π 2

0

( )| |π 1 sin 2x cos 2x + 1 π dx = + x || 2 = . 0 4 4 2 8

0

Therefore the center of mass has the coordinates (x G , yG ) =

(π 2

) − 1, π8 .

Remark 6.5.3 If the lamina is symmetrical with respect to O x axis (respectively O y axis), then its center of mass G ∈ O x, hence yG = 0 (respectively G ∈ O y, hence x G = 0).

Fig. 6.16 The center of mass of a non-homogeneous lamina D

y 1

cos x

G

O

π 2

x

6.5 Applications of the Double Integral in Geometry and Mechanics

183

6.5.3 Moments of Inertia of a Lamina It is known that the moment of inertia of a material point with respect to a certain axis is equal to the product of the mass of the point and the square of the distance from the point to the axis. In the case of a system of material points, the moment of inertia with respect to an axis is the sum of the moments of inertia of the material points that constituting the system with respect to the same axis. Let D be a lamina with a continuous density f : D → R+ , let ρ : D1 , D2 , . . . , D p be any partition of D and (ξi , ηi ) ∈ Di some arbitrary points. We approximate as before the mass of the plate Di with the product f (ξi , ηi ) · area(Di ) and consider this mass concentrated at the point (ξi , ηi ). The moment of inertia of this of material points with respect to O y axis will be equal to the sum Σ p system 2 ξ f (ξ i , ηi ) · area(Di ). If the norm of the partition ρ is small, this sum can be i=1 i considered as an approximate value of the moment of inertia I y of the lamina D with respect to O y axis. The exact value of the moment of inertia I y with respect to O y is: I y = lim

∥ρ∥→0

p Σ

¨ ξi2 f (ξi , ηi ) · area(Di ) =

x 2 f (x, y)dx dy.

i=1

(6.14)

D

Similarly, the moment of inertia Ix of the lamina D with respect to O x axis is: Ix = lim

∥ρ∥→0

p Σ

¨ ηi2 f (ξi , ηi ) · area(Di ) =

i=1

y 2 f (x, y) dx dy.

(6.15)

D

Taking into account that the moment of inertia of a material ( point )of mass m (placed at the point (x, y)) with respect to the origin O(0, 0) is m· x 2 + y 2 , applying the same arguments we find that: ¨

(

I O = Ix + I y =

) x 2 + y 2 f (x, y)dx dy.

(6.16)

D

Remark 6.5.4 If the plate is homogeneous, namely it has a constant density at each point ( f (x, y) = k = ct., ∀(x, y) ∈ D), then: ¨ Ix = k

¨ y dx dy; I y = k 2

D

¨ =k D

(

) x + y dx dy. 2

2

x 2 dx dy; I O D

184

6 Double and Triple Integrals

y

Fig. 6.17 The quarter disk of radius R centered at the origin

D

R x

O

Example 6.5.2 Find the moments of inertia with respect to the axis O x, respectively with respect to the origin O of the non-homogeneous lamina having density f (x , y){ = x, where: } D = (x, y) ∈ R2 ; x 2 + y 2 ≤ R 2 , x ≥ 0, y ≥ 0 (Fig. 6.17). The plane domain D is the quarter disk of radius R centered at the origin that lies in the first quadrant of the axis system. By passing to polar coordinates (Fig. 6.18): {

π x = ρ cos θ , 0 ≤ ρ ≤ R, 0 ≤ θ ≤ , y = ρ sin θ 2

we get: π

¨ Ix =

∫2 y 2 x dx dy = 0

D π

=

R5 5

∫2

π ⎛ R ⎞ 2 ∫ 5 ∫ R ⎝ ρ 4 sin2 θ cos θ dρ ⎠dθ = sin2 θ cos θ dθ 5

0

0

| R 5 sin3 θ || π R5 2 ' 2 = · sin θ (sin θ ) dθ = 5 3 |0 15

0

¨ IO =

π

(

) 2

∫2

x 2 + y x dx dy =

D

Fig. 6.18 The variation domain of the polar coordinates θ and ρ

0

⎛ ⎝

∫R 0

⎞ R5 ρ 4 cos θ dρ ⎠dθ = 5

π

∫2

cos θ dθ =

R5 . 5

0

ρ R Ω O

π 2

θ

6.6 Riemann–Green Formula

185

6.6 Riemann–Green Formula The Riemann–Green formula makes the connection between the double integral and the line integral of the second kind, being used mainly for the processing of line integrals on closed curves. Let D ⊂ R2 be a bounded plane domain whose boundary C = ∂ D is a piecewise smooth curve formed by a finite union of simple closed curves. Let P, Q : D → R be two real continuous functions with the property that there are the first order partial derivatives ∂∂Py and ∂∂Qx continuous on D. With these details, the Riemann–Green formula is expressed as follows: ¨ ( D

) ∫ ∂Q ∂P (x, y) − (x, y) dx dy = ↺ P(x, y)dx + Q(x, y)dy ∂x ∂y

(6.17)

C

In this formula, the orientation of the curve C = ∂ D is chosen so that when we traverse the curve, the domain D must always be on the left. In Fig. 6.19 we have exemplified the orientation of the curve for a domain whose boundary consists of a single closed curve and in Fig. 6.20 for a domain whose boundary consists of a finite union of closed curves. Definition 6.6.1 By elementary domain of Green type (G− elementary domain) we mean any of the four domains represented in Fig. 6.21. We will prove the Riemann–Green formula (6.17) in stages. Stage 1. Fig. 6.19 The orientation of the curve for a domain whose boundary consists of a single closed curve

Fig. 6.20 The orientation of the curve for a domain whose boundary consists of a finite union of closed curves

186

6 Double and Triple Integrals

Fig. 6.21 Elementary domains of Green-type

y

Fig. 6.22 Elementary domain of Green-type

E

d

f

y

Δ c

O

A

B

a

x

b

x

Lemma 6.6.1 The Riemann–Green formula is established for any G—elementary domain. Proof First, we consider an elementary domain of Green type Δ as in Fig. 6.22. More specifically, such a domain is defined as follows: } { Δ = (x, y) ∈ R2 ; a < x < b, c < y < f (x) where f : [a, b] → [c, d] is a continuous, strictly increasing, surjective function (Fig. 6.22). ͡

Its boundary consists of the segments AB, B E and the arc E A. According to the computation method of the double integral for simple domains with respect to O y axis (formula 6.2), we have: ¨ ∂P − (x, y)dx dy ∂y Δ

∫b =− a

⎛ ⎝

∫f (x) c

⎞ ∂P (x, y)dy ⎠dx = − ∂y

∫b

∫b P(x, f (x))dx +

a

P(x, c)dx. a

(6.18)

6.6 Riemann–Green Formula

187 ͡

Considering the following parametric representations of the arc AE and the segments AB and B E: ͡

AE : x = t, y = f (t), t ∈ [a, b] AB : x = t, y = c, t ∈ [a, b] B E : x = b, y = t, t ∈ [c, d] we deduce: ∫

∫b P(x, y)dx =

P(t, f (t))dt

͡

a

∫

∫b

AE

P(x, y)dx =

P(t, c)dt a

AB

∫

P(x, y)dx = 0.

(6.19)

BE

From (6.18) and (6.19), it follows that: ∫ ¨ ∫ ∫ ∫ ∂P dx dy = − P dx + P dx + P dx = ↺ P dx. ∂y Δ

AB

͡

BE

(6.20)

∂Δ

AE

On the other hand, we have: ¨ Δ

∂Q (x, y)dx dy = ∂x

∫d c

⎛ ⎜ ⎝

∫b

f −1 (y)

⎞ ∂Q ⎟ (x, y)dx ⎠dy ∂x

∫d =

∫d Q(b, y)dy −

c

( ) Q f −1 (y), y dy.

c

͡

If we consider for arc AE the parametric representation: ͡

AE : x = f −1 (t), y = t, t ∈ [c, d] we obtain:

(6.21)

188

6 Double and Triple Integrals

∫

∫d Q(x, y)dy =

͡

( ) Q f −1 (t), t dt.

(6.22)

c

AE

For the segments AB and B E, we have: ∫

∫ Q(x, y)dy = 0 and AB

∫d Q(x, y)dy =

Q(b, t)dt.

(6.23)

c

BE

From (6.21), (6.22) and (6.23), it results that: ∫ ∫ ∫ ∫ ¨ ∂Q dx dy = Qdy + Q dy + Q dy = ↺ Q dy. ∂x Δ

AB

͡

BE

EA

(6.24)

∂Δ

Adding formulas (6.20) and (6.24), we obtain the Riemann–Green formula for the G-elementary domain Δ considered in Fig. 6.22. It is obvious that the proofs of the Riemann–Green formula for the other G− elementary domains in Fig. 6.21 are absolutely analogous. Stage 2. Lemma 6.6.2 The Riemann–Green formula is valid for any triangular domain. Proof Let Δ be an arbitrary triangular domain having as boundary the triangle ABC (Fig. 6.23). Let E be the point of intersection between the straight line parallel to the axis O y and passing through A and the straight line parallel to the axis O x and passing through B. Let F be the intersection point between the straight lines AE and BC. Fig. 6.23 Triangular domain

A

y

Δ1 B

Δ3 E Δ2 F G

O

C x

6.6 Riemann–Green Formula

189

We will denote also by G the intersection point between the straight lines AE and the straight line parallel to the axis O x and passing through C. The domain Δ is the union of domains Δ1 , Δ2 , Δ3 , where Δ1 has the boundary AB E, Δ2 has the boundary B E F and Δ3 has the boundary AFC. We notice that Δ1 and Δ2 are G—elementary domains, while Δ3 does not have this property. On the other hand, it is clear that Δ3 can be described as the difference of two G—elementary domains. Indeed, if we denote by Δ4 the domain whose boundary is AGC and by Δ5 the domain whose boundary is F GC, then Δ4 , Δ5 are G—elementary domains and Δ3 = Δ4 \Δ5 . Applying the Riemann–Green formula for each of the G—elementary domains Δ1 , Δ2 , Δ4 , Δ5 , we obtain: ) ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ( ∂P ∂Q − dx dy = + + = + + − ∂x ∂y Δ Δ1 Δ2 Δ3 Δ1 Δ2 Δ4 Δ5 ⎞ ⎛ ⎞⎛ ⎞ ⎛ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ⎟ ⎜ ⎟⎜ ⎟ ⎜ + ⎠ = ⎝ + + ⎠ + ⎝ + + ⎠⎝ + AB

⎛

⎜ −⎝

BE

∫

∫

CA

CF

∫ − FG

⎟ ⎠=

+

GC

∫

⎞

∫

+

FG

+

EA

GC

∫

+

= AB

AG

∫ +

EA

∫

FC

EB

∫ +

∫

∫ = ↺ P dx + Q dy.

FE

AB

∫ −

BF

⎛

⎜ +⎝

∫ +

BF

∫

BF

+

CA

∫ +

FE

∫

GC

⎞

AE

⎟ ⎠+

FC

∫

∫

+ EF

∫

CA

∫

+

+

FG

∫ =

∫ +

AB

GC

BC

∫ + CA

∂Δ

Theorem 6.6.1 The Riemann–Green formula is valid for any polygonal domain. Proof Clearly, any polygonal domain is a finite union of triangular domains which have no common interior points. For simplicity, let us consider the domain Δ represented in Fig. 6.24, whose boundary is the quadrilateral ABC D. Obviously, Δ = Δ1 ∪ Δ2 , where Δ1 is the triangular domain AB D and Δ2 is the triangular domain BC D. From Lemma 6.6.2, we have: ) ) ) ¨ ( ¨ ( ¨ ( ∂P ∂P ∂P ∂Q ∂Q ∂Q − − − dx dy = dx dy + dx dy ∂x ∂y ∂x ∂y ∂x ∂y Δ1 Δ2 Δ ∫ ∫ = ↺ P dx + Q dy + ↺ P dx + Q dy AB D

BC D

190

6 Double and Triple Integrals

B

Fig. 6.24 Polygonal domain

A

Δ1

Δ2 C

D

⎛ ⎜ =⎝

⎜ =⎝

+

+

AD

⎛

DB

+ DC

⎛

⎞

⎟ ⎜ ⎠+⎝

BA

∫

∫

AD

∫

∫

∫

DC

∫ +

∫ +

CB

BA

∫

∫

⎞

+ CB

∫ +

⎞ ⎟ ⎠

BD

∫ ⎟ = ↺ P dx + Q dy. ⎠ ∂Δ

Stage 3. Theorem 6.6.2 The Riemann–Green formula is valid for any domain whose boundary is a simple closed piecewise smooth curve. Proof Indeed, it can be shown that there is a sequence of polygonal lines {Cn }, inscribed in C = ∂ D such that: ∫ ∫ lim P(x, y)dx + Q(x, y)dy = P(x, y)dx + Q(x, y)dy. n→∞

Cn

C

If we denote by Dn the bounded domain whose boundary is the curve Cn , then from Lemma 6.6.2, we deduce: ) ¨ ( ∂Q ∂P (x, y) − (x, y) dx dy ∂x ∂y D ( ) ¨ ∂Q ∂P = lim (x, y) dx dy (x, y) − n→∞ ∂y ∂x Dn ∫ ∫ = lim P(x, y)dx + Q(x, y)dy = P(x, y)dx + Q(x, y)dy. n→∞

Cn

C

∫ Example 6.6.1 Compute ↺ (y − x y)dx + (x y − x)dy, where: ∂D

6.6 Riemann–Green Formula

191

y

Fig. 6.25 Ellipse centered in the origin and of semi-axis a and b

b

D O

a x

} { 2 2 D = (x, y) ∈ R2 ; ax 2 + by2 ≤ 1 (Fig. 6.25). If we denote by P(x, y) = y − x y and by Q(x, y) = x y − x, then ∂Q ∂P = y − 1. = 1 − x and ∂x ∂y From the formula Riemann–Green (6.17), we have: ∫ ↺ (y − x y)dx + (x y − x)dy ∂D

¨ (x + y − 2)dx dy.

= D

According to Example 6.4.4, we obtain: ¨ (x + y − 2)dx dy = −2 π a b. D

− → − → − → − → Remark 6.6.1 If V : D ⊂ R2 → R2 , V (x, y) = P(x, y) i + Q(x, y) j is a 1 vector field of C -class on the squareable bounded domain D and C = ∂ D, then. ∫

∫ − → − V ·→ τ = ↺ P(x, y)dx + Q(x, y)dy

C+

C

¨ ( = D

) ∂Q ∂P (x, y) dx dy. (x, y) − ∂y ∂x

Remark 6.6.2 If D ⊂ R2 is a squareable domain for which the Riemann–Green formula is valid, then: ∮ 1 area(D) = x dy − y dx. (6.25) 2 ∂D

192

6 Double and Triple Integrals

Proof Let us denote by P(x, y) = − 2y and by Q(x, y) = 2x . Then, using the Riemann–Green formula and Proposition 6.2.1, it results that: 1 2

∮ ∂D

¨ (

) ∂Q ∂P x dy − y dx = (x, y) − (x, y) dx dy ∂x ∂y D ) ¨ ( ¨ 1 1 = dx dy = + 1 dx dy = area(D). 2 2 D

D

Example 6.6.2 Compute the area of the elliptical domain D : Because the boundary of the domain D is the ellipse parametric equations of the ellipse: {

x2 a2

+

y2 b2

x2 a2

+

y2 b2

≤ 1.

= 1, then, using the

x = a cos t , t ∈ [0, 2 π ], a, b > 0 y = b sin t

and the formula (6.25), we obtain: 1 area(D) = 2 =

1 2

∮ ∂D

∫2 π

1 x dy − y dx = 2

∫2 π (a cos t b cos t + b sin t a sin t) dt 0

( ) 1 ab cos2 t + ab sin2 t dt = 2

0

∫2 π a b dt = π a b. 0

6.7 Improper Double Integrals In this paragraph we present the notion of improper double integral in case where the integrated domain is unbounded. Let D ⊂ R2 be an unbounded domain and let f : D → R. We will assume that f is integrable on any squareable bounded subdomain of D. ˜ Definition 6.7.1 We say that the double integral f (x, y)dx dy is convergent if D

for any sequence of squareable bounded domains {Dn } with the properties: (i) D 1 ⊂ D2 ⊂ · · · ⊂ Dn ⊂ · · · (ii) D n ⊂∪Dn+1 , ∀ n ∈ N∗ (iii) D = ∞ n=1 Dn ˜ the limit lim f (x, y)dx dy exists, it is finite and does not depend on the choice n→∞

Dn

of the sequence {Dn }. In case of convergence, we will write

6.7 Improper Double Integrals

193

¨

¨ f (x, y)dx dy = lim

f (x, y)dx dy.

n→∞

D

Dn

˜ Otherwise, if the limit lim f (x, y)dx dy does not exist or it is infinite, we say n→∞ Dn ˜ f (x, y)dx dy is divergent. that the integral D

˜ Example 6.7.1 Establish the nature of the improper double integral x y dx dy. R2 } { Let us denote by Dn = (x, y) ∈ R2 ; x 2 + y 2 < n 2 , n ∈ N∗ . It is obvious that the sequence {Dn } of bounded domains has the properties (i)–(iii) of Definition 6.7.1. Since Dn is the disk of radius n centered at the origin, passing to polar coordinates we have: ⎞ ⎛ ¨ ∫2 π ∫n ∫2 π n4 3 ⎠ ⎝ x y dx dy = ρ cos θ sin θ dρ dθ = cos θ sin θ dθ = 0. 4 0

Dn

0

0

On the other hand, we consider the domains: { } Dn' = (x, y) ∈ R2 ; −n < x < 2 n, −n < y < 2 n , n ∈ N∗ . { } Clearly, the sequence Dn' fulfills also the properties (i)–(iii) of Definition 6.7.1. Furthermore, we have: ¨

∫2 n x y dx dy = −n

Dn'

⎞ ⎛ 2n ⎞ ⎛ 2n ⎞ ⎛ 2n ∫ ∫ ∫ 4 ⎝ x y dx ⎠dy = ⎝ x dx ⎠ · ⎝ y dy ⎠ = 9n . 4 −n

−n

−n

˜ 4 x y dx dy = 0 and lim x y dx dy = lim 9n4 = ∞, we deduce that n→∞ ' n→∞ Dn Dn ˜ the improper double integral x y dx dy is divergent. Since

˜

R2

The improper double integral as defined above, retains the main properties of the double integral on a bounded domain. Thus, linearity, monotony, additivity with respect to the integrated domain and the change of variables formula are properties that remain true even in the case of double integral on an unbounded domain. For positive functions defined on an unbounded domain, the following theorem can be proved. Theorem 6.7.1 Let D ⊂ R2 be an unbounded domain and let f : D → R+ be a positive function,˜integrable on any squareable subdomain of D. Then the improper f (x, y)dx dy is convergent if and only if there is at least one double integral D

194

6 Double and Triple Integrals

sequence of bounded squareable domains, ˜ with the properties (i)–(iii) for which the f (x, y)dx dy. sequence {an } is bounded, where an = Dn

Moreover, we have: ¨ ¨ f (x, y)dx dy = lim an = lim f (x, y)dx dy. n→∞

n→∞

D

Dn

˜

e−x −y dx dy is R2 ∫∞ 2 convergent, and then find that the value of the Euler–Poisson integral is −∞ e−x dx = √ π. 2 2 Since the function f (x, y) = e−x −y ≥ 0, ∀(x, y) ∈ R2 is positive, it is enough to find a sequence {Dn } of bounded squareable domains for which the sequence of positive numbers {an } is bounded, where: Example 6.7.2 Prove that the following improper double integral

2

2

¨ an =

f (x, y)dx dy. Dn

} { If we choose Dn = (x, y) ∈ R2 ; x 2 + y 2 < n 2 , n ∈ N∗ , then the sequence {Dn } fulfills the properties (i)–(iii) from Definition 6.7.1. Because Dn is the disk of radius n centered at the origin, passing to polar coordinates we obtain: ⎛ ⎞ ∫2 π ∫n ¨ 2 2 2 an = e−x −y dx dy = ⎝ e−ρ · ρ dρ ⎠dθ 0

Dn

0

| ) ∫2 π( ∫2 π ) ( ) 1 −ρ 2 ||n 1 ( 2 2 − e | dθ = 1 − e−n dθ = 1 − e−n · π. = 0 2 2 0

0

Because lim an = π , from Theorem 6.7.1, it follows that n→∞

R2

convergent and its value is: ¨

˜

e−x

2

−y 2

dx dy = lim an = π. n→∞

R2

On the other hand, let us consider the domains: { } Dn' = (x, y) ∈ R2 ; |x| < n, |y| < n , n ∈ N∗ .

e−x

2

−y 2

dx dy is

6.8 Volume of a Space Figure

195

{ } Obviously, the sequence of domains Dn' fulfills also the properties (i)–(iii) from Definition 6.7.1. Further, we have: ⎞ ⎛ ¨ ∫n ∫n 2 2 2 2 e−x −y dx dy = ⎝ e−x e−y dx ⎠dy −n

Dn'

⎛

=⎝

−n

∫n

⎞ ⎛

e−x dx ⎠ · ⎝ 2

−n

∫n

⎞

⎛

e−y dy ⎠ = ⎝ 2

−n

∫n

⎞2 e−x dx ⎠ . 2

−n

From Theorem 6.7.1, it follows that: ¨ ¨ 2 2 2 2 π= e−x −y dx dy = lim e−x −y dx dy n→∞

Dn'

D

⎛ n ⎞2 ⎛ ∞ ⎞2 ∫ ∫ 2 2 = lim ⎝ e−x dx ⎠ = ⎝ e−x dx ⎠ n→∞

−n

−∞

∫∞ 2 (here we used the fact that −∞ e−x dx is convergent, see Example 3.2.4). We thus calculated the Euler–Poisson integral, namely: ∫∞

e−x dx = 2

√ π.

∞

The most remarkable result is another property of improper double integral that has no analogue in the one-dimensional case, that is, the convergence of a improper double integral implies its absolute convergence. The following result can be proved: Theorem for the improper double ˜ 6.7.2 The necessary and sufficient condition ˜ f (x, y)dx dy to be convergent is that | f (x, y)|dx dy to be convergent. integral D

D

In other words, for improper double integral the concepts of convergence and absolute convergence are equivalent.

6.8 Volume of a Space Figure As we have seen in this chapter, the transition from the simple integral to the double integral, in addition to many analogies, also implies some substantial changes, both in terms of concepts and in terms of reasoning. These changes have their origin mainly in the theory of measurable plane sets (which have area).

196

6 Double and Triple Integrals

In contrast to this situation, the transition from the double integral to the triple integral does not involve any complication. To begin with, it is necessary to introduce the notion of volume. From elementary geometry, it is known that the volume of a rectangular parallelepiped is equal to the product of the lengths of its edges. In particular, if T is a rectangular parallelepiped with edges parallel to the coordinate axes, this means that: T = [a1 , a2 ] × [b1 , b2 ] × [c1 , c2 ], then: def

Vol(T ) = (a2 − a1 )(b2 − b1 )(c2 − c1 ). The same formula of volume remains true even if the rectangular parallelepiped does not contain one or more faces. Definition 6.8.1 By an elementary space subset E ⊂ R3 we mean any finite union of rectangular parallelepipeds with edges parallel to the coordinate axes, which may or may not contain one or more faces and which have in common two by two only at most one face. More precisely, E ⊂ R3 is an elementary set if there is a finite number of rectangular parallelepipeds Ti , i = 1, p, with edges parallel to the coordinate axes, such that: E=

p ∪

◦

◦

Ti and Ti ∩ T j = ∅, for i /= j.

i=1

The volume of such elementary set is by definition the sum of the volumes of the parallelepipeds that compose it: p def Σ

Vol(E) =

Vol(Ti ).

i=1

Next, we note by T the family of all elementary space sets. We recall that a set T ⊂ R3 is bounded if there is r > 0 such that T is included in the ball with the center at origin and radius r (i.e. T ⊆ B(O, r )). If T ⊂ R3 is a bounded set, then we will define: V∗ (T ) = sup{Vol(E); E ∈ T , E ⊂ T } the interior volume of T . V ∗ (T ) = inf{Vol(F); F ∈ F , F ⊃ T } the exterior volume of T . If the set T does not contain any elementary set, we will define V∗ (T ) = 0. Clearly we have: V∗ (T ) ≤ V ∗ (T ).

6.8 Volume of a Space Figure

197

Definition 6.8.2 We say that the bounded set T ⊂ R3 is cubable (has volume) if: V∗ (T ) = V ∗ (T ) = V (T ). The common value V (T ) is called the volume of T and it is denoted by Vol(T ). Remark 6.8.1 Any elementary space set is cubable and its volume coincides with the volume defined in Definition 6.8.1. Example 6.8.1 Let D ⊂ R2 be a squareable bounded domain and let h > 0 be a positive number. By a generalized right cylinder with base D and height h we understand the following solid: { } T = (x, y, z) ∈ R3 ; (x, y) ∈ D, 0 ≤ z ≤ h . We will show that the generalized right cylinder T is cubable and its volume is: Vol(T ) = h · area(D). Indeed, we recall that E denotes the family of all elementary plane sets. Since D is a squareable bounded domain, from Theorem 2.7.1, it follows that for any ε > 0, there are E ε , Fε ∈ E with the properties: E ε ⊂ D ⊂ Fε and area(Fε ) − area(E ε ) < ε. We remark that if E =

∪p i=1

PE =

p ∪

[ai , bi ] × [ci , di ] ∈ T , then

[ai , bi ] × [ci , di ] × [0, h] ∈ E .

i=1

Further, we have: h · area(E ε ) ≤ h · area(D) = h · sup{area(E); E ⊂ D, E ∈ E } ≤ V∗ (T ) ≤ V ∗ (T ) ≤ h · sup{area(F); F ⊃ D, F ∈ E } = h · area(D) ≤ h · area(Fε ). Taking into account that h · (area(Fε ) − area(E ε )) < h · ε and ε > 0 is arbitrary, it follows that: V∗ (T ) = V ∗ (T ) = h · area(D). Next, we will show that any curvilinear cylinder (cylindroid) is cubable. By a curvilinear cylinder with base D ⊂ R2 , we understand a solid T bounded by the base D, a surface z = f (x, y), (x, y) ∈ D, and the lateral cylindrical surface (Fig. 6.26).

198

6 Double and Triple Integrals

Fig. 6.26 Curvilinear cylinder

z = f ( x, y)

z

T

y

O x

D ∂D

More precisely, if D ⊂ R2 is a squareable bounded domain and f : D → R+ is a positive continuous function, then the corresponding curvilinear cylinder is: } { T = (x, y, z) ∈ R3 ; (x, y) ∈ D, 0 ≤ z ≤ f (x, y) . Theorem 6.8.1 Any curvilinear cylinder is cubable and its volume is given by the formula: ¨ Vol(T ) =

f (x, y)dx dy.

(6.26)

D

Proof If we consider that ρ : D1 , D2 , . . . , D p is an arbitrary partition of D and (ξi , ηi ) ∈ Di , i = 1, p, are arbitrary points, then we shall denote by: { } Ti = (x, y, z) ∈ R3 ; (x, y) ∈ Di , 0 ≤ z ≤ f (ξi , ηi ) , 1 ≤ i ≤ p. Taking into account to Example 6.8.1, we have: Vol(Ti ) = f (ξi , ηi ) · area(Di ). Therefore, the volume of the cylindrical solid T can be approximated by the sum: Vol(T ) ≈

p Σ i=1

Vol(Ti ) =

p Σ

f (ξi , ηi ) · area(Di ).

i=1

But we notice that the last sum is the Riemann sum of the function f , associated to the partition ρ and the intermediate points (ξi , ηi ). The exact value of the volume of the curvilinear cylinder T is: Vol(T ) = lim

∥ρ∥→0

p Σ i=1

Vol(Ti ) = lim

∥ρ∥→0

p Σ i=1

f (ξi , ηi ) · area(Di ).

6.9 Triple Integrals. Definition and Basic Properties

199

Here it is taken into account that f is continuous on D, hence integrable on D and we obtained ¨ Vol(T ) = f (x, y)dx dy. D

Corollary 6.8.1 Let D ⊂ R2 be a squareable bounded domain and let f, g: D → R be two continuous functions such that f (x, y) ≤ g(x, y), ∀(x, y) ∈ D. Then the space domain: { } T = (x, y, z) ∈ R3 ; (x, y) ∈ D, f (x, y) ≤ z ≤ g(x, y) is cubable and its volume is determined by the formula: ¨ Vol(T ) =

(g(x, y) − f (x, y))dx dy. D

The following theorem results immediately from Definition 6.8.2; its proof is completely analogous to the case of plane sets (Theorem 2.7.1). Theorem 6.8.2 A bounded subset T ⊂ R3 is cubable if and only if for any ε > 0, there are two elementary space subsets Pε and Q ε with the properties: Pε ⊂ T ⊂ Q ε and Vol(Q ε ) − Vol(Pε ) < ε. Definition 6.8.3 A subset S ⊂ R3 is said to be of volume zero if for any ε > 0, there is an elementary space set Q ε ∈ T such that S ⊂ Q ε and Vol(Q ε ) < ε. The following theorem can be proved. Theorem 6.8.3 A subset T ⊂ R3 is cubable if and only if its boundary is of volume zero.

6.9 Triple Integrals. Definition and Basic Properties In the following we will denote by T ⊂ R3 a bounded cubable subset. Definition 6.9.1 It is called a partition of T any finite family of cubable subdomains ∪p ρ : T1 , T2 , . . . , T p which have no pairwise common interior points and T = i=1 Ti . By the norm of the partition ρ we mean the largest of the diameters of the domains Ti , i = 1, p, i.e.:

200

6 Double and Triple Integrals

∥ρ∥ = max{diam(Ti ); 1 ≤ i ≤ p} ) } { ( where diam(Ti ) = sup dist M ' , M '' ; M ' , M '' ∈ Ti . If ρ : T1 , T2 , . . . , T p is a partition of T and f : T → R is a bounded function, then we will denote by: m = inf{ f (x, y, z); (x, y, z) ∈ T }, M = sup{ f (x, y, z); (x, y, z) ∈ T } m i = inf{ f (x, y, z); (x, y, z) ∈ Ti }, 1 ≤ i ≤ p Mi = sup{ f (x, y, z); (x, y, z) ∈ Ti }, 1 ≤ i ≤ p. Definition 6.9.2 The lower Darboux sum, respectively the upper Darboux sum associated to the bounded function f and the partition ρ are defined as follows: sρ =

p Σ

m i · Vol(Ti ), Sρ =

i=1

p Σ

Mi · Vol(Ti ).

i=1

Because m ≤ m i ≤ Mi ≤ M, ∀ 1 ≤ i ≤ p and Vol(T ) =

p Σ

Vol(Ti ), it results:

i=1

m · Vol(T ) ≤ sρ ≤ Sρ ≤ M · Vol(T ).

(6.27)

The properties of Darboux sums for three-variable functions are similar to the properties of Darboux sums for two-variable functions, so we will not recall them. If we denote by P the family of all partition of T , then we will define the lower (upper) Darboux triple integral: { } { } I∗ = sup sρ ; ρ ∈ P and I ∗ = inf Sρ ; ρ ∈ P . The existence of I∗ and I ∗ results from inequalities (6.27). From the properties of the Darboux sums, it follows that: I∗ ≤ I ∗ . Definition 6.9.3 Let f : T → R be a real bounded function defined on the cubable bounded domain T ⊂ R3 . We say that the function f is integrable in the Darboux sense on T if. I∗ = I ∗ = I.

6.9 Triple Integrals. Definition and Basic Properties

The common value I is denoted by I =

201

˝

f (x, y, z)dx dy dz and it is called

T

the triple integral of the function f on the domain T ; the real number I is uniquely determined by the function f and T . Example 6.9.1 Prove that any constant function on a bounded cubable set T ⊂ R3 is integrable. If f (x, y, z) = k, ∀(x, y, z) ∈ T , then ˚ k dx dy dz = k · Vol(T ). T

Indeed, if ρ : T1 , T2 , . . . , T p is an arbitrary partition of T , then: m i = Mi = k, ∀ i = 1, p and sρ = Sρ =

p Σ

k · Vol(Ti ) = k · Vol(T ),

i=1

whence, it results that: ˚

∗

I∗ = I = k · Vol(T ) =

k dx dy dz. T

In particular, we have: ˚ 1 dx dy dz = Vol(T ). T

Next, we define the Riemann sums. Definition 6.9.4 Let T ⊂ R3 be a bounded cubable subset, f : T → R a bounded function, ρ : T1 , T2 , . . . , T p an arbitrary partition of T and Pi (ξi , ηi , ζi ) ∈ Ti an arbitrary intermediate point, ∀ i = 1, p. The Riemann sum of the function f , associated to the partition ρ and the intermediate points Pi is given by: σρ ( f ; Pi ) =

p Σ i=1

f (Pi ) · Vol(Ti ) =

p Σ

f (ξi , ηi , ζi ) · Vol(Ti )

i=1

Since m i ≤ f (Pi ) ≤ Mi , ∀ i = 1, p, it results that: sρ ≤ σρ ( f ; Pi ) ≤ Sρ for any system of intermediate points Pi ∈ Ti .

202

6 Double and Triple Integrals

Definition 6.9.5 We say that the function f : T → R is integrable in the Riemann sense on the bounded cubable domain T if there is a finite number I , with the property that for any ε > 0, there is δε > 0 such that for any partition ρ of T , with ∥ρ∥ < δε and any intermediate points Pi ∈ Ti , we have: | | |σρ ( f ; Pi ) − I | < ε. The number I is called the triple integral of function f on the domain T and it is denoted by: ˚ I =

f (x, y, z)dx dy dz. T

We will also write: ˚ f (x, y, z)dx dy dz = lim σρ ( f ; Pi ), ∥ρ∥→0

T

the exact meaning being that of Definition 6.9.5. Remark 6.9.1 The properties of the triple integral are completely analogous to the properties of the double integral. In particular, it can be shown that any continuous function on a compact cubable domain is integrable. Moreover, it can be shown that both definitions of the triple integral, with Riemann sums and Darboux sums, coincide.

6.10 Computing Triple Integral. Change of Variables in Triple Integral Definition 6.10.1 A domain T ⊂ R3 is called simple with respect to Ozaxis if there are a bounded squareable domain D ⊂ R2 and two functions ϕ, ψ : D → R ◦

with the property ϕ(x, y) < ψ(x, y), ∀(x, y) ∈ D , such that: { } T = (x, y, z) ∈ R3 ; (x, y) ∈ D, ϕ(x, y) ≤ z ≤ ψ(x, y) . From Corollary 6.8.1, it results that such a domain has volume and: ¨ Vol(T ) =

¨ ψ(x, y)dx dy −

D

ϕ(x, y)dx dy. D

6.10 Computing Triple Integral. Change of Variables in Triple Integral

203

Remark 6.10.1 In the case of a simple domain with respect to Oz axis, any straight ◦

line parallel to the Oz axis through an inner point (x, y) ∈ D , cuts the boundary of the domain in at most two points. The following theorem shows us how to compute the triple integral on a simple domain with respect to Oz axis. Theorem 6.10.1 Let T ⊂ R3 be a simple domain with respect to Oz axis and let f : T → R be a continuous function. Then: ¨

˚ f (x, y, z)dx dy dz = T

⎛ ⎜ ⎝

ψ(x,y) ∫

⎞ ⎟ f (x, y, z)dz ⎠dx dy.

(6.28)

ϕ(x,y)

D

Similarly, simple domains with respect to O x, respectively O y axis can be considered, and formula (6.28) can be adapted for such simple domains. Remark 6.10.1 If T is a rectangular parallelepiped with edges parallel to the coordinate axes, i.e.: } { T = (x, y, z) ∈ R3 ; a1 ≤ x ≤ a2 , b1 ≤ y ≤ b2 , c1 ≤ z ≤ c2 ⊂ R3 then, for any continuous function f : T → R the triple integral is calculated by the formula: ⎞ ⎞ ⎛ ⎛ ˚ ∫a2 ∫b2 ∫c2 f (x, y, z)dx dy dz = ⎝ ⎝ f (x, y, z)dz ⎠dy ⎠dx. (6.29) a1

T

c1

b1

In the right member of the formula (6.29) the order of integration can be changed and other 5 computation formulas analogous to (6.29) are obtained (Fig. 6.27). Remark 6.10.2 If f : [a1 , a2 ] → R, g: [b1 , b2 ] → R, h : [c1 , c2 ] → R are three continuous functions and T = [a1 , a2 ] × [b1 , b2 ] × [c1 , c2 ], then: ˚

∫a2 f (x) · g(y) · h(z)dx dy dz =

∫b2 f (x) dx ·

a1

T

Example 6.10.1 Compute

˝

∫c2 g(y) dy ·

b1

h(z) dz. c1

x y z dx dy dz, where:

T

{ } T = (x, y, z) ∈ R3 ; 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c From Remark 6.10.2, it results that:

204

6 Double and Triple Integrals

z

Fig. 6.27 Rectangular parallelepiped

c

O x

˚

∫a x y z dx dy dz = 0

T

⎛ ⎝

∫b

y

a

⎞ ⎞ ⎛ c ∫ ⎝ x y z dz ⎠dy ⎠dx

0

∫a

0

∫b x dx ·

=

b

0

∫c y dy ·

0

z dz =

a 2 b2 c2 . 8

0

˝

Example 6.10.2 Compute T z dx dy dz, where T is the conical domain bounded √ by the surfaces: z = 0, z = 1, z = x 2 + y 2 (Fig. 6.28). Geometrically, the equation x 2 + y 2 = z 2 represents a cone with the vertex at the origin and symmetrical around the Oz axis. We notice that if we denote by D the disk x 2 + y 2 < 1, then: } { √ T = (x, y, z) ∈ R3 ; (x, y) ∈ D, x 2 + y 2 < z < 1 According to formula (6.28), we obtain: z

Fig. 6.28 Cone with the vertex at the origin and symmetrical around the Oz axis

1

O x

y

6.10 Computing Triple Integral. Change of Variables in Triple Integral

⎛

¨

˚

∫1

205

⎞

⎜ ⎟ ⎜ z dz ⎟ ⎝ ⎠dx dy √ D x 2 +y 2 ¨ ( ( )) 1 1 − x 2 + y 2 dx dy = 2 D ⎞ ⎛ ∫2 π ∫2 π ∫1 ) π 1 1 1 ⎝ ( 2 ⎠ dθ = . 1 − ρ ρ dρ dθ = = 2 4 4 2

z dx dy dz = T

0

0

0

Remark 6.10.3 If the integrated domain T is not simple with respect to any of the coordinate axes, then the domain T can be divided, by planes parallel to one of the coordinate planes, into a finite number of simple∪ subdomains T1 , T2 , . . . , Tn , which n Ti , we have: do not have common interior points. Since T = i=1 ˚ f (x, y, z)dx dy dz =

n ˚ Σ

f (x, y, z)dx dy dz.

i=1 T i

T

Next, we present the change of variables theorem in triple integral. Let Ω ⊂ R3 be an open, cubable, connected, bounded subset and let F: Ω → R3 , let F(u, v, w) = (x(u, v, w), y(u, v, w), z(u, v, w)), ∀(u, v, w) ∈ Ω be a vector function with the properties: (i) F is of C 1 -class on Ω (ii) F: Ω → T = F(Ω) is bijective (iii) F is a regular transformation on Ω, i.e. det JF (u, v, w) =

D(x, y, z) (u, v, w) /= 0, ∀(u, v, w) ∈ Ω. D(u, v, w)

Such a vector function is called a change of variables. As in the two-dimensional case, ( ) ( )it can be shown that. T \T = F Ω\Ω and T = F Ω . Theorem 6.10.2 (The change of variables formula in triple integral) Let F: Ω ⊂ R3 → T ⊂ R3 be a change of variables defined by F(u, v, w) = (x(u, v, w), y(u, v, w), z(u, v, w)), ∀(u, v, w) ∈ Ω and let f : T → R be a continuous function. Then:

206

6 Double and Triple Integrals

˚ f (x, y, z) dx dy dz T

˚

= Ω

| | | | D(x, y, z) | f (x(u, v, w), y(u, v, w), z(u, v, w))| (u, v, w)||du dv dw. D(u, v, w) (6.30)

Formula (6.30) is used in practice in order to transform the initial integral into an easier to compute integral, either by simplifying the integrated domain Ω, either by simplifying the integrated function f . Mapping F transforms the domain Ω into the domain T . Consequently, the specification of a point (u, v, w) belonging to Ω uniquely determines the corresponding point (x, y, z) of T . In other words, the numbers u, v, w can be regarded as coordinates (different from Cartesian coordinates) of the points of the domain T . They are called curvilinear coordinates. In the following we will present three space coordinate systems that are most frequently used in applications, namely: spherical coordinates, cylindrical coordinates and generalized spherical coordinates. Case 1. Spherical coordinates. The spherical coordinates are expressed by the formulas: ⎧ ⎨ x = ρ sin θ cos ϕ y = ρ sin θ sin ϕ , 0 < ρ < ∞, 0 < θ < π, 0 < ϕ < 2π. ⎩ z = ρ cos θ

(6.31)

The meaning of the notations is shown in Fig. 6.29, namely: ρ is the distance from the origin O to the point M(x, y, z), θ is the angle between the positive direction of the Oz axis and the straight line segment O M and ϕ is the angle formed by the positive direction of the O x axis and the projection O N of the segment O M to O x y plane. Obviously we have: x = ∥O A∥ = ∥O N ∥ cos ϕ = ρ sin θ cos ϕ y = ∥O B∥ = ∥O N ∥ sin ϕ = ρ sin θ sin ϕ z = ∥OC∥ = ρ cos θ The Jacobian of the transformation is given by:

6.10 Computing Triple Integral. Change of Variables in Triple Integral

207

z

Fig. 6.29 The spherical coordinates

C

θ

M ( x, y , z )

ρ

B

O

y

ϕ

A

N ( x, y , 0 )

x

| | | sin θ cos ϕ ρ cos θ cos ϕ −ρ sin θ sin ϕ | | | D(x, y, z) = | sin θ sin ϕ ρ cos θ sin ϕ ρ sin θ cos ϕ || = ρ 2 sin θ > 0. D(ρ, θ, ϕ) || | cos θ −ρ sin θ 0 Example 6.10.3 Compute

˝

x y z dx dy dz, where T is the solid bounded by the

T

surfaces: x = 0, y = 0, z = 0, x 2 + y 2 + z 2 = 1. From a geometric point of view, the domain T is the first octant of the ball (solid sphere) of radius R = 1 centered at the origin, having the equation x 2 + y 2 + z 2 ≤ 1 (Fig. 6.30). In spherical coordinates, the integrated domain T becomes} Ω, where: { Ω = (ρ, θ, ϕ) ∈ R3 ; 0 < ρ < 1, 0 < θ < π2 , 0 < ϕ < π2 (Fig. 6.31). From Theorem 6.10.2, we deduce: z

Fig. 6.30 The first octant of the solid sphere of radius R = 1 centered at the origin

T O

x

y

208

6 Double and Triple Integrals

Fig. 6.31 The variation domain of the spherical coordinates ρ, θ, ϕ

π

ϕ

2 Ω

ρ

π

O

2

θ

˚

˚ x y z dx dy dz = T

1

ρ 3 sin2 θ cos θ sin ϕ cos ϕ · ρ 2 sin θ dρ dθ dϕ Ω

⎛ π ⎞ ⎛ π ⎞ ⎛ ⎞ ∫2 ∫2 ∫1 ⎜ ⎟ ⎜ ⎟ = ⎝ sin ϕ cos ϕ dϕ ⎠ · ⎝ sin3 θ cos θ dθ ⎠ · ⎝ ρ 5 dρ ⎠ ( =

0

0

| ) ( 4 | ) ( 6| ) 1 sin θ || π ρ ||1 sin ϕ || π 2 2 · · = . 48 2 |0 4 |0 6 |0

0

2

Case 2. Cylindrical coordinates. The cylindrical coordinates are expressed by the formulas: ⎧ ⎨ x = ρ cos θ y = ρ sin θ , 0 < ρ < ∞, 0 < θ < 2 π, z ∈ R. ⎩ z =z

(6.32)

The meaning of the notations is shown in Fig. 6.32. The Jacobian is: | | | cos θ −ρ sin θ 0 | | | D(x, y, z) | = | sin θ ρ cos θ 0 || = ρ > 0. D(ρ, θ, z) | 0 0 1| Let us return to Example 6.10.2 and compute the triple integral using the change of variables in cylindrical coordinates. ˝ Example 6.10.4 Compute z dx dy dz, where T is the domain bounded by the T √ surfaces: z = 0, z = 1, z = x 2 + y 2 . In cylindrical coordinates, the integral over the conical domain T is the integral over Ω, where: { } Ω = (ρ, θ, z) ∈ R3 ; 0 < ρ < z, 0 < θ < 2 π, 0 < z < 1 .

6.10 Computing Triple Integral. Change of Variables in Triple Integral

209

z

Fig. 6.32 Cylindrical coordinates

M ( x, y , z )

z

O

θ

ρ

y N ( x, y , 0 )

x

From Theorem 6.10.2, it results that: ˚

˚ z dx dy dz =

T

⎞ ⎞ ⎛ ⎛ ∫2 π ∫1 ∫ z z · ρ dρ dθ dz = ⎝ ⎝ z · ρ dρ ⎠dz ⎠dθ

Ω

0

0

0

⎛ 2 π⎛ 1 ⎞ ⎛ 1 ⎞ ⎞ ⎛ 2π ⎞ ∫ ∫ 3 ∫ ∫ 3 z z 1 π =⎝ ⎝ dz ⎠dθ ⎠ = ⎝ 1 dθ ⎠ · ⎝ dz ⎠ = 2π · = . 2 2 8 4 0

0

0

0

Case 3. Generalized spherical coordinates. The generalized spherical coordinates are expressed by the formulas: ⎧ ⎨ x = a ρ sin θ cos ϕ y = b ρ sin θ sin ϕ , 0 < ρ < ∞, 0 < θ < π, 0 < ϕ < 2 π. ⎩ z = c ρ cos θ The Jacobian of transformation is Example 6.10.5 Compute

˝ ( x2 T

9

D(x,y,z) D(ρ,θ,ϕ)

+ y2 +

z2 25

= a b c ρ 2 sin θ > 0. ) dx dy dz, where:

} { z2 x2 + y2 + ≤4 . T = (x, y, z) ∈ R3 ; 1 ≤ 9 25 In this case, the generalized spherical coordinates are:

(6.33)

210

6 Double and Triple Integrals

⎧ ⎨ x = 3 ρ sin θ cos ϕ y = ρ sin θ sin ϕ ⎩ z = 5 ρ cos θ and the domain T is described by: { } Ω = (ρ, θ, ϕ) ∈ R3 ; 1 ≤ ρ ≤ 2, 0 ≤ θ ≤ π, 0 ≤ ϕ ≤ 2 π . According to Theorem 6.10.2, we have: ˚ ( T

) ˚ x2 z2 2 +y + dx dy dz = ρ 2 · 15 ρ 2 sin θ dρ dθ dϕ 9 25 Ω ⎞ ⎞ ⎛ ⎛ ∫2 π ∫π ∫2 = 15 ⎝ ⎝ ρ 4 sin θ dρ ⎠dθ ⎠dϕ 0

0

1

⎛ 2π ⎞ ⎛ π ⎞ ⎛ 2 ⎞ ∫ ∫ ∫ = 15⎝ 1 dϕ ⎠ · ⎝ sin θ dθ ⎠ · ⎝ ρ 4 dρ ⎠ 0

0

( | ) ( | = 15 ϕ ||2 π · − cos 0

1

| ) ( 5| ) | ρ ||2 θ ||π · = 372π. 0 5 |1

6.11 Applications of the Triple Integral in Geometry and Mechanics 6.11.1 Volume of a Space Domain If T ⊂ R3 is a cubable bounded domain, then the volume of T is given by the formula: ˚ Vol(T ) = 1 dx dy dz. (6.34) T

6.11 Applications of the Triple Integral in Geometry and Mechanics

211

6.11.2 Mass of a Solid By solid we mean a material set in the shape of a cubable bounded domain T ⊂ R3 , together with a positive and continuous function f : T → R+ , called density. The solid is generally considered non-homogeneous, its density being variable. Proceeding similarly as in the case of the mass of a non-homogeneous lamina, we find that the mass of the solid is expressed through the triple integral as: ˚ Mass(T ) =

f (x, y, z) dx dy dz.

(6.35)

T

6.11.3 Coordinates of the Center of Mass of a Solid The coordinates of the center of mass of a non-homogeneous solid T with the density function f are described by the expressions: ˝

x f (x, y, z) dx dy dz

xG = ˝ T

f (x, y, z) dx dy dz

T

˝

y f (x, y, z)dx dy dz

yG = ˝ T

f (x, y, z)dx dy dz

T

˝

z f (x, y, z)dx dy dz

zG = ˝ T

f (x, y, z)dx dy dz

.

(6.36)

T

Remark 6.11.1 In particular, if the solid is homogeneous ( f (x, y, z) = k = ct., ∀(x, y, z) ∈ T ), the calculation formulas for the mass and the coordinates of the center of mass become: ˚ Mass(T ) = k dx dy dz = k · Vol(T ) ˝ xG =

T

x dx dy dz

T

Vol(T )

, yG =

˝

˝

y dx dy dz

T

Vol(T )

, zG =

z dx dy dz

T

Vol(T )

.

212

6 Double and Triple Integrals

6.11.4 Moment of Inertia of a Solid The moments of inertia of a non-homogeneous solid T having the density f with respect to Oz axis (respectively, to O x y plane or to the origin O) are expressed by the formulas: ˚ ( 2 ) x + y 2 f (x, y, z)dx dy dz I Oz = T

˚ IO x y =

z 2 f (x, y, z)dx dy dz T

˚ IO =

(

) x 2 + y 2 + z 2 f (x, y, z)dx dy dz.

(6.37)

T

Formulas (6.37) take equivalent forms in the event of a change in the coordinate axis or the coordinate plane for which the moment of inertia is computed. Remark 6.11.2 For a homogeneous solid of constant density f (x, y, z) = k, the formulas (6.37) are given by: ˚ I Oz = k ·

(

) x 2 + y 2 dx dy dz

T

˚ IO x y = k ·

z 2 dx dy dz T

˚ IO = k ·

(

) x 2 + y 2 + z 2 dx dy dz.

T

Example 6.11.1 Determine the coordinates of the center of mass and the moment of inertia with respect to O y axis of the non-homogeneous solid T with the density f , where: { } T = (x, y, z) ∈ R3 ; x 2 + z 2 ≤ 9, 1 ≤ y ≤ 3 and f (x, y, z) = 2 y. From a geometric point of view, the space domain T is the region that lies inside the cylinder given by x 2 + z 2 = 9 and between the planes y = 1 and y = 3. Let us denote by D the projection of the domain T to the coordinate plane O x z. Then D is the disk of radius 3 centered at the origin: x 2 + z 2 ≤ 9. Therefore:

6.11 Applications of the Triple Integral in Geometry and Mechanics

213

z

Fig. 6.33 Cylinder symmetrical around the Oy axis

T

D O

1

3

y

x

{ } T = (x, y, z) ∈ R3 ; (x, z) ∈ D, 1 ≤ y ≤ 3 . Obviously, because the cylinder T is symmetrical around the O y axis, then its center of mass is situated on the axis O y, hence x G = z G = 0 (Fig. 6.33). It remains to compute the coordinate yG . According to formulas (6.36), we have: ˝ yG = ˝

˝

y f (x, y, z)dx dy dz

= ˝

T

= ˝

2 y dx dy dz

T

T

y 2 dx dy dz

T

T

f (x, y, z) dx dy dz

˝

2 y 2 dx dy dz

y dx dy dz

T

whence: ˚

∫3 y dx dy dz = 1

T

⎛ ⎝

¨

⎞

∫3

y dx dz ⎠dy =

1

D

⎛ ⎞ ¨ y⎝ 1 dx dz ⎠dy D

∫3 y dy = 9 π · 4 = 36 π

= Area(D) · 1

∫3

˚ y 2 dx dy dz = T

∫3 = 1

⎛ ⎝

¨

1

D

⎞ y 2 dx dz ⎠dy

⎛ ⎞ ¨ y2⎝ 1 dx dz ⎠dy = 78 π. D

Thus, the coordinate yG of the center of mass is: yG =

13 78 π = . 36 π 6

The moment of inertia with respect to O y axis is:

214

6 Double and Triple Integrals

˚ IO y =

(

) x 2 + z 2 f (x, y, z)dx dy dz

T

˚ = T

∫3 =2 1

∫3 =2 1

∫3 =2 1

(

x2 + z ⎛

y⎝ ⎛

¨

) 2

(

∫3 · 2 y dx dy dz = 2 1

⎞

⎛ ⎝

¨

(

D

) x 2 + z 2 dx dz ⎠dy

D

y⎝

0

∫2π 0

0

⎞ ∫3 81 ⎠ dθ dy = 81 π y dy = 324 π. 4 1

⎞

x 2 + z y dx dz ⎠dy

⎞ ⎞ ⎛ ∫2π ∫3 y ⎝ ⎝ ρ 3 dρ ⎠dθ ⎠dy ⎛

) 2

Chapter 7

Surface Integrals

7.1 Parameterized Surface Canvases. Definition of a Surface The notion of parameterized surface canvas is a natural generalization of the notion of parameterized path. Definition 7.1.1 Let Δ ⊂ R2 be an open connected subset (domain). Any vector function r : Δ → R3 of C 1 − class is called a parameterized surface canvas. If we denote by x, y, z the scalar components of the vector function r , then: r (u, v) = (x(u, v), y(u, v), z(u, v)), ∀(u, v) ∈ Δ. ⎧ ⎨ x = x(u, v) The equations y = y(u, v), (u, v) ∈ Δ , are called the parametric equations ⎩ z = z(u, v) of the canvas or a parametric representation of the canvas, and u and v are called parameters. We will denote by (Δ, r ) a parameterized surface canvas. The direct image r (Δ) of the domain Δ through the vector function r , i.e. the subset: r (Δ) = {r (u, v); (u, v) ∈ Δ} = {(x(u, v), y(u, v), z(u, v)); (u, v) ∈ Δ} ⊂ R3 is called the support of parameterized surface canvas (Δ, r ). The fact that the vector function r : Δ → R3 is of C 1 − class is equivalent to the fact that the scalar functions x, y, z : Δ → R are of C 1 − class on Δ, that is, they are continuous and their first order partial derivatives are continuous on Δ. Next we will use some notations specific to differential geometry.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Integral Calculus for Engineers, https://doi.org/10.1007/978-981-19-4793-3_7

215

216

7 Surface Integrals

{ − → − → − →{ With respect to the Cartesian coordinate system O, i , j , k , the vector equation of a surface canvas (Δ, r ) is: − → − → − → → r (u, v) = − r (u, v) = x(u, v) i + y(u, v) j + z(u, v) k , ∀(u, v) ∈ Δ. Partial derivatives with respect to the parameters u and v are denoted by: ∂x ∂y ∂y ∂z ∂z ∂x ; yu = ; yv = ; zu = ; zv = ; xv = ∂v ∂u ∂v ∂u ∂v ∂u ∂ r → ∂ r→ → r→v = → = xu →i + yu →j + z u k; = xv →i + yv →j + z v k. r→u = ∂u ∂v xu =

We also note with: | | | | | xu z u | | yu z u | | | |; | ; B = B(u, v) = −| A = A(u, v) = | yv z v | xv z v | | | |x y | C = C(u, v) = || u u || xv yv ∥− ∥2 r u ∥ = xu2 + yu2 + z u2 E = E(u, v) = ∥→ F = F(u, v) = r→u · r→v = xu xv + yu yv + z u z v ∥→ ∥2 r ∥ = x 2 + y2 + z2. G = G(u, v) = ∥− v

v

v

v

We remark that: ∥→ ∥ √ − → − → − → → → − → r u×− r v = A i + B j + C k and ∥ − r v ∥ = A2 + B 2 + C 2 . r u ×− → → If we denote by θ the angle between the vectors − r u and − r v , then: rv ∥2 − (→ E · G − F 2 = ∥→ ru ∥2 ∥→ ru · r→v )2 = ∥ 2∥ rv ∥2 − ∥→ rv ∥2 cos2 θ = ru ∥2 ∥→ = ∥r→u ∥∥→ rv ∥2 sin2 θ = ∥→ ru × r→v ∥2 = A2 + B 2 + C 2 = ∥→ ru ∥2 ∥→ Thus, we obtained the following identity, called Lagrange identity: E · G − F 2 = A2 + B 2 + C 2 .

(7.1)

Definition 7.1.2 A parameterized surface canvas r : Δ → R3 is called simple if the vector function r is injective, i.e. ∀(u 1 , v1 ), (u 2 , v2 ) ∈ Δ, (u 1 , v1 ) /= (u 2 , v2 ) implies r (u 1 , v1 ) /= r (u 2 , v2 ). Definition 7.1.3 A parameterized surface canvas r : Δ → R3 is called smooth if: ∥− ∥2 → ∥→ r u ×− r v ∥ = A2 + B 2 + C 2 > 0, ∀(u, v) ∈ Δ.

7.1 Parameterized Surface Canvases. Definition of a Surface

217

Example 7.1.1 Let r : Δ → R3 be the parameterized surface canvas defined by: − → − → − → − → r (u, v) = R sin u cos v i + R sin u sin v j + R cos u k , ∀(u, v) ∈ Δ where { π{ π Δ = (u, v) ∈ R2 ; 0 < u < , 0 < v < 2 2 The parametric equations are: ⎧ ⎨ x = R sin u cos v y = R sin u sin v, (u, v) ∈ Δ . ⎩ z = R cos u We observe that for any (u , v) ∈ Δ, the point (x(u, v), y(u, v), z(u, v)) verifies the equation x 2 + y 2 + z 2 = R 2 , whence we deduce that the support of this canvas is the portion of the sphere of radius R centered at the origin that lies in the first octant (x > 0 , y > 0, z > 0) (Fig. 7.1). Using the above notations, we have: xu = R cos u cos v, xv = −R sin u sin v yu = R cos u sin v, yv = R sin u cos v z u = −R sin u, z v = 0 z

Fig. 7.1 The portion of the sphere of radius R centered at the origin that lies in the first octant

y

O

x

Fig. 7.2 The diagramme of the parameter change

Φ = (ϕ 1 , ϕ 2 )

Δ1 ⊂ R 2

Δ2 ⊂ R2

Φ −1

r1

r2 S ⊂R3

218

7 Surface Integrals

and further: A = R 2 sin2 u cos v, B = R 2 sin2 u sin v C = R 2 sin u cos u E = R 2 , F = 0, G = R 2 sin2 u. Since E · G − F 2 = A2 + B 2 + C 2 = R 4 sin2 u > 0, ∀(u, v) ∈ Δ, it results that this surface canvas is smooth. We also notice that the considered canvas is simple. Definition 7.1.4 Two parameterized surface canvases (Δ1 , r1 ) and (Δ2 , r2 ) are called equivalent with the same orientation if there is a diffeomorphism. ϕ : Δ1 ⊂ R2 → Δ2 ⊂ R2 with the properties: det Jϕ (u 1 , v1 ) > 0, ∀ (u 1 , v1 ) ∈ Δ1 and r1 = r2 ◦ ϕ. Recall that ϕ : Δ1 → Δ2 is diffeomorphism if ϕ is bijective, of C 1 − class on Δ1 and its inverse ϕ−1 : Δ2 → Δ1 is of C 1 − class on Δ2 . Moreover, we recall that if ϕ1 and ϕ2 are the scalar components of ϕ, then: | | ∂ϕ1 (u , v ) D(ϕ1 , ϕ2 ) | ∂u 1 1 1 det Jϕ (u 1 , v1 ) = (u 1 , v1 ) = | ∂ϕ | ∂u 12 (u 1 , v1 ) D(u 1 , v1 )

|

∂ϕ1 ,v | ∂v1 (u 1 1 ) | |. ∂ϕ2 ,v | ∂v1 (u 1 1 )

The equality r1 = r2 ◦ ϕ means: r1 (u 1 , v1 ) = r2 (ϕ1 (u 1 , v1 ), ϕ2 (u 1 , v1 )), ∀ (u 1 , v1 ) ∈ Δ1 If det Jϕ < 0 on Δ1 , we say that the two surface canvases are equivalent to opposite orientation. The function ϕ is also called parameter change. We will use the notation (Δ1 , r1 ) ∼ (Δ2 , r2 ). Remark 7.1.1 Any two equivalent parameterized surface canvases have the same support. Remark 7.1.2 Any parameterized surface canvas equivalent to a simple (smooth) parameterized surface canvas is also simple (smooth). Example 7.1.2 Show that the following parameterized surface canvases (Δ1 , r1 ) and (Δ2 , r2 ) are equivalent, where: − → − → − → − → r 1 (u 1 , v1 ) = R sin u 1 cos v1 i + R sin u 1 sin v1 j + R cos u 1 k { π{ π ; Δ1 = (u 1 , v1 ) ∈ R2 ; 0 < u 1 < , 0 < v1 < 2 2 √ − → − → − → − → r 2 (u 2 , v2 ) = u 2 i + v2 j + R 2 − u 22 − v22 k

7.1 Parameterized Surface Canvases. Definition of a Surface

219

{ } Δ2 = (u 2 , v2 ) ∈ R2 ; u 22 + v22 < R 2 , u 2 > 0, v2 > 0 . Indeed, it is verified by direct computation that the application ϕ : Δ1 → Δ2 , defined by ϕ(u 1 , v1 ) = (R sin u 1 cos v1 , R sin u 1 sin v1 ) , (u 1 , v1 ) ∈ Δ1 satisfies the conditions of Definition 7.1.4, thus (Δ1 , r1 ) ∼ (Δ2 , r2 ). Definition 7.1.5 A subset S ⊂ R3 is called elementary surface if there is a simple and smooth parameterized surface canvas (Δ , r ) such that S = r (Δ) and, in addition, the function r −1 : S → Δ to be continuous. The pair (Δ , r ) is by definition a parametric representation of the surface S. Therefore, a subset S ⊂ R3 is an elementary surface if there is an open connected domain Δ ⊂ R2 and a vector function r : Δ → S of C 1 − class, with the properties: (i) r is bijective (ii) r −1 : S → Δ is continuous (iii) A2 + B 2 + C 2 > 0, ∀(u, v) ∈ Δ. Remark 7.1.3 It can be shown that if (Δ, r ) is a parametric representation of the elementary surface S and Δ' ⊂ R2 (is another )domain such that the mapping ϕ : Δ' → Δ is a diffeomorphism, then Δ' , r ◦ ϕ is also a parametric representation of S. Remark 7.1.4 The most convenient particular case of elementary surface is the graph of a function f : D ⊂ R2 → R, also known as an explicit surface. Let D ⊂ R2 be a domain and let f : D → R be a function of C 1 − class (Fig. 7.2). We associate this function the parameterized surface canvas (D , r ): → ∀(x, y) ∈ D. r (x, y) = r→(x, y) = x →i + y →j + f (x, y)k, Next, we will show that the support of this canvas, i.e., not

S = r (D) = {(x, y, f (x, y)); (x, y) ∈ D} = G f ⊂ R3 that represents the graph of the function f , is an elementary surface (Fig. 7.3). If we denote by p = ∂∂ xf and q = ∂∂ yf , then: | | | | | |1 |1 p| |0 p| | | | | = −q; C = || = − p; B = −| A=| | | 0 0q 1q and A2 + B 2 + C 2 = 1 + p 2 + q 2 > 0, ∀(x, y) ∈ D,

| 0 || =1 1|

220

7 Surface Integrals

z

Fig. 7.3 The elementary surface

S

O y

x

D

whence we deduce that the canvas (D, r ) is smooth. On the other hand, it is obvious that r : D → S is bijective. Finally, it is clear that the function r −1 : S → D, defined by r −1 (x, y, z) = (x, y), ∀(x, y, z) ∈ S is continuous. Definition 7.1.6 A set S ⊂ R3 is called a surface if any point of it belongs to an open subset V of R3 such that V ∩ S is an elementary surface. It is obvious that any elementary surface is a surface within the meaning of Definition 7.1.6. In particular, any explicit surface is a surface (Remark 7.1.4). Remark 7.1.5 The support of a smooth parameterized surface canvas is a surface. Indeed, let (Δ, r ) be a smooth parameterized surface canvas, defined by: → ∀(u, v) ∈ Δ. r (u, v) = r→(u, v) = x(u, v)→i + y(u, v) →j + z(u, v)k, Let S = r (Δ) be its support and let M0 (x(u 0 , v0 ), y(u 0 , v0 ), z(u 0 , v0 )) ∈ S be an arbitrary point. Since at least one of the parameters A , B , C is nonzero, we can assume that: | | | xu yu | |(u 0 , v0 ) = D(x, y) (u 0 , v0 ) /= 0. | C(u 0 , v0 ) = | xv yv | D(u, v) Consider the vector function ϕ : Δ → R2 defined by: ϕ(u, v) = (x(u, v), y(u, v)), ∀(u, v) ∈ Δ. , v /= 0, from the local inversion theorem, it results that there are Since D(x,y) D(u,v) (u 0 0 ) an open neighborhood U0 ⊂ Δ of the point (u 0 , v0 ) and an open neighborhood D of the point (x0 , y0 ) = ϕ(u 0 , v0 ) such that the map ϕ : U0 → D is diffeomorphism and D(x,y) v) /= 0, ∀(u, v) ∈ U0 . D(u,v) (u,

7.1 Parameterized Surface Canvases. Definition of a Surface

221

If we denote by u = u(x, y), v = v(x, y) the scalar components of the inverse function ϕ−1 : D → U0 and by f (x, y) = z(u(x, y), v(x, y)), ∀(x, y) ∈ D, then f is function of C 1 − class on D. On the other hand, since: x(u 0 , v0 ) = x0 ; y(u 0 , v0 ) = y0 ; z(u 0 , v0 ) = z(u(x0 , y0 ), v(x0 , y0 )) = f (x0 , y0 ) we deduce that M0 ∈ G f = {(x, y, f (x, y)); (x, y) ∈ D}. As G f ⊂ R3 is an elementary surface (Remark 7.1.4), it follows that the support S = r (Δ) is a surface within the meaning of Definition 7.1.6. Another way of analytic representation of a surface is the implicit representation. Remark 7.1.6 Let D ⊂ R3 be an open subset and let F : D → R be a function of ( )2 ( )2 ( )2 C 1 − class on D. Moreover, if we assume that ∂∂ Fx + ∂∂Fy + ∂∂zF > 0 on D, then S = {(x, y, z) ∈ D; F(x, y, z) = 0} is a surface called implicit surface. Indeed, let M0 (x0 , y0 , z 0 ) ∈ S. Since the first-order partial derivatives of F are not simultaneously zero at M0 , we can assume that ∂∂zF (M0 ) /= 0. From the implicit function theorem, it follows that there are an open neighborhood U of x0 , an open neighborhood V of y0 , an open neighbourhood W of z 0 and an unique implicit function (x, y) → f (x, y) : U × V → W of C 1 − class with the properties: (a) f (x0 , y0 ) = z 0 . (b) F(x, y, f (x, y)) = 0, ∀(x, y) ∈ U × V . If the set G f = {(x, y, f (x, y)); (x, y) ∈ U × V } represents the graph of the implicit function f : U × V → W , from the property (a), it results that M0 ∈ G f , and from (b), it follows that G f ⊂ S. Since G f is an elementary surface, we deduce that the set S = {(x, y, z) ∈ D ; F(x, y, z) = 0} is a surface within the meaning of Definition 7.1.6. In conclusion, a surface can be represented analytically in three ways: parametric, explicit, or implicit. These three analytical representations are locally equivalent, in the sense that any point on a surface belongs to an elementary surface that can be represented in all three possible ways. In practice, all three analytical representations are used, but the parametric representation proves to be more useful in theoretical problems. It will be used mainly in the following. Definition 7.1.7 A parameterized surface is any equivalence class of parameterized surface canvases. Therefore, ~ S is a parameterized surface if there is a parameterized surface canvas r : Δ → R3 such that: { } ~ S = r1 : Δ1 → R3 parametrized surface canvas; (Δ1 , r1 ) ∼ (Δ, r ) .

222

7 Surface Integrals

Since (Δ, r ) ∼ (Δ, r ), we deduce that (Δ , r ) ∈ ~ S. A parameterized surface S˜ is called simple (smooth) if the surface canvas that determines it is simple (smooth). The support of ~ S is the support S of any canvas from ~ S. Usually, we will identify the surface ~ S with its support S. Let S = r (Δ) be a smooth parameterized surface and let (u 0 , v0 ) ∈ Δ. Since ∥→ ru (u 0 , v0 ) × r→v (u 0 , v0 )∥2 = A2 (u 0 , v0 ) + B 2 (u 0 , v0 ) + C 2 (u 0 , v0 ) > 0 → → r u (u 0 , v0 ) and − r v (u 0 , v0 ) are non-collinear, thus it results that the vectors − determine a plane. → → Definition 7.1.7 The plane determined by the vectors − r u (u 0 , v0 ) and − r v (u 0 , v0 ) passing through the point M0 (x(u 0 , v0 ), y(u 0 , v0 ), z(u 0 , v0 )) of the surface S is called the tangent plane to the surface S at M0 . The straight line perpendicular to the tangent plane at the point M0 is called the normal line to the surface S at the point M0 . This plane does not depend on the local representation of S. ϕ

Indeed, if we consider (s, t) ∈ Δ1 ⊂ R2 −→(u, v) ∈ Δ ⊂ R2 a diffeomorphism, then r1 = r ◦ ϕ is another local representation of S. At the point (x0 , y0 , z 0 ), we have: ∂v ∂u + r→v (u 0 , v0 ) · ∂s ∂s ∂u ∂v + r→v (u 0 , v0 ) · r1 )t (s0 , t0 ) = r→u (u 0 , v0 ) · (→ ∂t ∂t

r1 )s (s0 , t0 ) = r→u (u 0 , v0 ) · (→

and ( r1 )t = (→ ru × r→v ) · r1 )s × (→ (→

∂v ∂u ∂u ∂v · − · ∂s ∂t ∂s ∂t

) = (→ ru × r→v ) ·

D(u, v) , D(s, t)

whence, we deduce that the normal line to the surface S at the point (x0 , y0 , z 0 ) is uniquely determined. − → − → − → → → r u ×− Since the vector − r v = A i + B j +C k is perpendicular to the tangent plane, means it is collinear with the normal line to the surface, it follows that A, B, C are the direction parameters of the normal line. Further, we denote by: − → − → − → → − → A i + B j +C k r v r u ×− − → ∥ = n = ∥− √ → ∥→ r u ×− r v∥ A2 + B 2 + C 2 the unit normal vector to surface S. In the case of an explicit surface, the direction parameters of the normal line to surface are: − p, −q, 1 (Remark 7.1.4).

7.2 The Area of a Smooth Surface

223

7.2 The Area of a Smooth Surface To begin with, we approach the problem of the area of an explicitly defined smooth surface. Let D ⊂ R2 be a squareable bounded domain and let f : D → R be a function of C 1 − class on D. If we denote by p = ∂∂ xf and q = ∂∂ yf , then p, q are continuous on D. Let S (respectively, S) be the graph of the function f : D → R (respectively, the graph of the function f : D → R). Thus: { } S = {(x, y, f (x, y)); (x, y) ∈ D} and S = (x, y, f (x, y)); (x, y) ∈ D . If we denote by ┌ = S \ S the boundary of the surface S and by C the boundary of the domain D, then: ┌ = {(x, y, f (x, y)); (x, y) ∈ C}. The sets D and C are the orthogonal projections to the plane O x y of the sets S and ┌, respectively. Let ρ : D1 , D2 , .., D p be an arbitrary partition of D and let Mi (xi , yi ) ∈ Di , 1 ≤ i ≤ n, be some arbitrary points. We will denote by Pi ((xi , yi , f (xi , yi ))) the corresponding points on the surface S (Fig. 7.4). → n i the unit Let us denote by πi the tangent plane to S at the point Pi and by − normal vector to S at Pi , oriented upward. → n i with Oz axis, then: If we denote by γi the angle formed by the unit vector − 1 cos γi = √ 1 + pi2 + qi2 where pi = ∂∂ xf (xi , yi ) and qi = ∂∂ yf (xi , yi ). Let Si (respectively, Ti ) be the portion from the surface S (respectively, from the tangent plane πi ) cuts by the cylinder whose generatrix is parallel to Oz and whose directrix curve is the boundary ∂ Di . We will approximate the area of the surface portion Si by the area of the plane n Σ area(Ti ). Therefore: portion Ti and the area of S by i=1

area(Si ) ≈ area(Ti ) and area(S) =

n Σ i =1

area(Si ) ≈

n Σ i =1

area(Ti ).

224

7 Surface Integrals

Fig. 7.4 The area of an elementary surface

γi

ni

Pi S

Γ z

O

y

Mi

x

Di D

C

Because γi is the angle between the plane πi and the coordinate plane O x y, from the elementary geometry, it follows that: area(Di ) = area(Ti ) · cos γi and further: area(Ti ) =

√ 1 + pi2 + qi2 · area(Di ).

(7.2)

From (7.2), we have: n Σ

area(Ti ) =

i =1

n √ Σ

1 + pi2 + qi2 · area(Di ).

(7.3)

i =1

Definition 7.2.1 It is called the area of the smooth surface S the following real number: not

area(S) = area (S) = A = lim

∥ρ∥→0

n √ Σ i=1

1 + pi2 + qi2 · area(Di ).

(7.4)

7.2 The Area of a Smooth Surface

225

The exact meaning is this: There is a finite real number A ∈ R+ with the property that for any ε > 0, there is δε > 0 such that for any partition ρ : D1 , D2 , ... , D p of D, with ∥ρ∥ < δε , and for any points Mi (xi , yi ) ∈ Di , we have: | | n √ | | Σ | | 2 2 1 + pi + qi · area(Di )| < ε. |A − | | i =1

Notice that the sum from the √ right member of the relation (7.3) is the Riemann sum of the function g(x, y) = 1 + p 2 (x, y) + q 2 (x, y), associated to the partition ρ and the intermediate points Mi (xi , yi ) ∈ Di . Since g is continuous on D, thus integrable on D, according to (7.4), it follows that: area(S) = lim

∥ρ∥→0

¨

n √ Σ 1 + pi2 + qi2 · area(Di ) = lim σρ (g; Mi ) = i=1

g(x, y)dx dy =

= D

¨ √

∥ρ∥→0

1 + p 2 (x, y) + q 2 (x, y)dx dy.

D

Therefore, we proved the following result. Theorem 7.2.1 Any smooth explicit surface S : z = f (x, y), (x, y) ∈ D, where D ⊂ R2 is a bounded squareable domain and f : D → R is a function of C 1 − class on D, has area given by the formula: area(S) =

¨ √

1 + p 2 (x, y) + q 2 (x, y)dxdy.

(7.5)

D

Example 7.2.1 Compute the area of the surface S : x 2 + y 2 =

R2 2 z , h2

0 ≤ z ≤ h.

Geometrically, the surface S represents a cone with the vertex at the origin, the radius R and the height h (Fig. 7.5). Notice also that the surface S is the graph of the function: h √ 2 x + y 2 , (x, y) ∈ D R } { where D = (x, y) ∈ R2 ; x 2 + y 2 < R 2 . Next, we have: f (x, y) =

p=

∂f ∂f x h y h ; q= = · √ = · √ 2 2 2 R ∂y R ∂x x +y x + y2

and 1 + p2 + q 2 =

R2 + h2 . R2

226

7 Surface Integrals

z

Fig. 7.5 A cone with the vertex at the origin, the radius R and the height h

h

x

O

y

Taking into account to formula (7.5), the area of the surface S is expressed by: ¨ √

R2 + h2 dx dy = R2

√

R2 + h2 · R

¨

1 dx dy = D √D √ R2 + h2 R2 + h2 = · area(D) = · π R2 = π · R · G R R √ where we denoted by G = R 2 + h 2 the generatrix of the cone. Next, we study the area of a simple smooth parameterized surface. If S is a simple smooth parameterized surface, then S is the support of a simple smooth parameterized surface canvas. Therefore, there is a vector function r : Δ ⊂ R2 → R3 , area(S) =

r (u, v) = (x(u, v), y(u, v), z(u, v)), ∀ (u, v) ∈ Δ such that: S = r (Δ) = {(x(u, v), y(u, v), z(u, v)); (u, v) ∈ Δ}. ∥→ ∥2 → r u ×− Obviously, r is of C 1 − class on Δ, ∥− r v ∥ = A2 +B 2 +C 2 > 0, ∀(u, v) ∈ Δ, and r : Δ → S is bijective. Also, if the vector function r is extendable by continuity to Δ, we will denote by: S = r (Δ) = {(x(u, v), y(u, v), z(u, v)); (u, v) ∈ Δ}. If we denote by ┌ = S \ S the boundary of S and by C the boundary of Δ, then: ┌ = r (C) = {(x(u, v), y(u, v), z(u, v)), (u, v) ∈ C}.

7.2 The Area of a Smooth Surface

227

We mention that, in general, the correspondence between ┌ and C is not bijective. The surface S is called closed if S = S. A closed parameterized surface has no boundary. Example 7.2.2. Let S be the parameterized surface having the representation: r (u, v) = (R sin u cos v, R sin u sin v, R cos u), (u, v) ∈ Δ = (0, π ) × (0, 2π ). The parametric equations are: ⎧ ⎨ x = R sin u cos v y = R sin u sin v, u ∈ (0, π ), v ∈ (0, 2π ) . ⎩ z = R cos u We notice that r (0 , v) = (0 , 0 , R) , ∀ v ∈ [0 , 2 π]. Therefore, the image of any point on the segment AE through the vector function r is the point P(0, 0, R) (Figs. 7.6 and 7.7). Similarly, the image of any point on the segment B F is the point P ' (0, 0, −R). On the other hand, the image of any point M ∈ AB ∪ E F will be a point of coordinates: x = R sin u, y = 0, z = R cos u , u ∈ [0 , π]. Fig. 7.6 The variation domain of parameters u and v

v

E ( 0, 2 π )

F (π , 2 π ) Δ

B (π , 0 ) u

A z

Fig. 7.7 The sphere of radius R centered at the origin, without the meridian P Q P'

P ( 0, 0, R )

O

Q ( R , 0, 0 ) x

P ' ( 0, 0, − R )

y

228

7 Surface Integrals

Since x 2 + y 2 + z 2 = R 2 and x ≥ 0, it follows that the image of the boundary of the domain Δ through the vector function r is the meridian P Q P ' on the sphere of radius R centered at the origin. Therefore: S = r (Δ) is the sphere of radius R centered at the origin, without the meridian P Q P '. ( ) S = r Δ is the sphere of radius R centered at the origin. The boundary of S is ┌ = S\S = P Q P ' . Definition 7.2.2 Let S be a simple and smooth parameterized surface and let r (u, v) = (x(u, v), y(u, v), z(u, v)), ∀(u, v) ∈ Δ s be a parametric representation of S. Then the surface S is squareable and its area is described by the expressions: ¨ √

E · G − F 2 du dv = area(S) = area(S) = Δ ¨ √ ¨ ∥→ = ru × r→v ∥du dv. A2 + B 2 + C 2 du dv =

(7.6)

Δ

Δ

Remark 7.2.1 It can be shown that the area of a smooth parameterized surface does not depend on the chosen parameterization. ϕ

Indeed, if we consider (s , t) ∈ Δ1 ⊂ R2 −→ (u , v) ∈ Δ ⊂ R2 a diffeomorphism, then r1 = r ◦ ϕ is another local representation of S. From Definition 7.1.7, we have: | | | D(u, v) | |. | ∥(→ ru × r→v ∥ · | r1 )t ∥ = ∥→ r1 )s × (→ D(s, t) | The assertion follows using the change of variable ϕ in the last term of (7.6). Remark 7.2.2 Formula (7.6) generalizes formula (7.5) which expresses the area of an explicit surface. let the explicit surface S : z = f (x, y), (x, y) ∈ D ⊂ R2 with f ∈ ) (Indeed, C1 D . From Remark 7.1.4, we have A = − p, B = −q, C = 1, whence, it results that: ¨ √ 1 + p 2 + q 2 dxdy. area(S) = D

Example 7.2.3 Compute the area of the sphere S : x 2 + y 2 + z 2 = R 2 . A parametric representation of the sphere is:

7.2 The Area of a Smooth Surface

229

⎧ ⎨ x = R sin u cos v y = R sin u sin v, (u, v) ∈ Δ = (0, π ) × (0, 2π ) . ⎩ z = R cos u As shown in Example 7.1.1, in this case we have: E · G − F 2 = A2 + B 2 + C 2 = R 4 sin2 u hence, applying formula (7.6), we obtain: ¨

∫ 2π ∫ π R 2 sin u dudv = R 2 dv sin u du 0 0 Ʌ | π = − 2π R 2 cos u |0 = 4π R 2 .

area(S) =

Example 7.2.4 Find the area of the torus. Let us consider a circle of radius a centered at the point (b, 0), where 0 < a < b, lying in the plane O x y. The torus is the surface T obtained by revolving, in space, this circle around the axis O y. If θ is the angle in Fig. 7.8 and ϕ is the angle of rotation of the circle around the axis O y, then the parametric equations of the torus are (Fig. 7.9): ⎧ ⎨ x = (b + a cos θ ) cos ϕ y = a sin θ, (θ, ϕ) ∈ Δ = (0, 2π ) × (0, 2π ), a, b > 0 . ⎩ z = (b + a cos θ ) sin ϕ Further, we have: xθ = −a sin θ cos ϕ, xϕ = −(b + a cos θ ) sin ϕ yθ = a cos θ, yϕ = 0 z θ = −a sin θ sin ϕ, z ϕ = (b + a cos θ ) cos ϕ E = a 2 , F = 0, G = (b + a cos θ )2 E · G − F 2 = a 2 (b + a cos θ )2 .

Fig. 7.8 The circle centered in (b, 0) and of radius a

y a

O

b

θ x

230

7 Surface Integrals

¨ area(T ) =

∫2π a(b + a cos θ )dθ dϕ = a

Δ

∫2π (b + a cos θ )dθ =

dϕ 0

0

2 = 2π a(bθ + a sin θ )|2π 0 = 4π ab.

7.3 Surface Integral of the First Kind Let S be a simple smooth parameterized surface that is given by the parameterization r (u, v) = (x(u, v), y(u, v), z(u, v)), ∀(u, v) ∈ Δ ( ) and let F : S = r Δ → R be a continuous function. Of course, we suppose that Δ ⊂ R2 is a squareable bounded domain. Definition 7.3.1 The surface ˜ integral of the first kind of the function F over the F(x, y, z)dσ and is defined by the following double surface S is denoted by S

integral: ¨ F(x, y, z)dσ = S

¨

=

√ F(x(u, v), y(u, v), z(u, v)) E(u, v) · G(u, v) − F 2 (u, v)du dv.

(7.7)

Δ

Remark 7.3.1 The surface integral of the first kind

˜

F(x, y, z)dσ is independent

S

of its parameterization, in the sense that for two equivalent parameterizations, the value of the integral is the same, as a result of Theorem 6.4.1 of changing variables in the double integral. Remark 7.3.1 Under the above conditions, if F(x, y, z) = 1, (x, y, z) ∈ S, then, from (7.6), it results that: ¨ √ ¨ 1 dσ = E(u, v) · G(u, v) − F 2 (u, v)dudv = area(S). S

Δ

Example 7.3.1 Compute

˜ ( S

) x − y + z 2 dσ , where the surface S is given by:

S : x 2 + y 2 + z 2 = R 2 , z > 0.

7.3 Surface Integral of the First Kind

231

The surface S represents the upper half of the sphere of radius R centered at the origin. A parametric representation of this surface is: ⎧ ⎨ x = R sin u cos v ) ( y = R sin u sin v, (u , v) ∈ 0, π2 × (0, 2 π ) ⎩ z = R cos u (see Example 7.1.1). Taking into account that in this case E · G − F 2 = R 4 sin2 u, from formula (7.7), we get: ¨

(

) x − y + z 2 dσ = S ¨ ( ) R sin u cos v − R sin u sin v + R 2 cos2 u R 2 sin u du dv = = Δ ) ∫ π (∫ 2π 2 ) ( 2 2 2 3 sin u cos v − sin u sin v + R sin u cos u dv du = =R 0

= 2π R

∫

4

0 π 2

sin u cos u du = − 2π R

0

2

|π u || 2 2π R 4 . = | 3 0 3

4 cos

3

Remark 7.3.2 If S : z = f (x , y) , (x , y) ∈ D is an explicit smooth surface, where D ⊂ R2 is a squareable bounded domain, f : D → R is of C 1 − class on D, and F : S → R is continuous, then: ¨ F(x, y, z)dσ = S

¨ =

√ F(x, y, f (x, y)) 1 + p 2 (x, y) + q 2 (x, y)dx dy.

(7.8)

D

˜ Example 7.3.2 Compute S (x + z)dσ , where the surface S is the portion of the √ cone z = x 2 + y 2 that lies inside the circular cylinder given by x 2 + y 2 = 2 y. We notice that the projection of the surface S to the plane O x y is the domain: { } { } D = (x, y) ∈ R2 ; x 2 + y 2 − 2y ≤ 0 = (x, y) ∈ R2 ; x 2 + (y − 1)2 ≤ 1 (i.e., the disk of radius √ 1 centered at the point (0 , 1); Fig. 7.10). Therefore, S : z = x 2 + y 2 , (x, y) ∈ D. Further, we have: p=

∂z ∂z x y ; q= ; 1 + p 2 + q 2 = 2. = √ = √ 2 2 2 2 ∂x ∂ y x +y x +y

232

7 Surface Integrals

z

Fig. 7.9 The portion of a cone that lies inside a circular cylinder

O D

x y

Fig. 7.10 The disk of radius 1 centered at the point (0,1)

2

D

1

O

According to formula (7.8), we have: ¨ I =

(x + z)dσ = S

¨ (

x+

√

) √ x 2 + y 2 · 2dx dy.

D

Since we are integrating over a disk, it makes sense to use polar coordinates: ∫

x = ρ cos θ , 0 ≤ ρ ≤ 2 sin θ, 0 ≤ θ ≤ π y = ρ sin θ

and therefore, we get: ⎛ 2 sin θ ⎞ π ∫ √ ∫ I = 2 ⎝ (ρ cos θ + ρ)ρ dρ ⎠dθ = 0

0

π | √ ∫ ρ 3 ||2 sin θ dθ = 2 (cos θ + 1) · | 3 0 0

√ ∫π ) 8 2 ( 3 sin θ cos θ + sin3 θ dθ = = 3 0

y

x

7.3 Surface Integral of the First Kind

233

√ ∫π √ ∫π ) 8 2 ( 8 2 3 1 − cos2 θ sin θ dθ = = sin θ dθ = 3 3 0 0 √ √ ( 3 )| | cos θ 8 2 32 2 | π · − cos θ | = . = 0 3 3 9 Remark 7.3.2 If the surface S is a piecewise smooth surface, i.e., it is a finite union of simple smooth surfaces Si , i = 1 , n, such that two by two have no common interior points, then: area(S) =

n Σ

¨ F(x, y, z)dσ =

area(Si ) and

i=1

n ¨ Σ

F(x, y, z)dσ .

i=1 S i

S

Remark 7.3.3 The surface integral of the first kind is frequently used in engineering applications. For example, having a thin shell which is a material surface S, smooth or piecewise smooth, whose variable density is given by the continuous positive function F : S → R+ , then the mass, the center of mass and the moments of inertia of the surface are expressed through the surface integrals of the first kind as follows. 1. The Mass of the Shell S is Given by: ¨ Mass(S) =

F(x, y, z)dσ . S

2. The Coordinates of the Center of Mass of the Shell S Are Defined by: ˜ xG = ˜ S

,

F(x, y, z)dσ

S

˜

S zG = ˜

˜

x F(x, y, z)dσ

yG = ˜

y F(x, y, z)dσ

S

F(x, y, z)dσ

S

z F(x, y, z)dσ F(x, y, z)dσ

.

S

3. The moments of inertia of the thin shell S with respect to the origin O(0, 0, 0), with respect to the coordinate axes O x, O y, Oz and with respect to the coordinate planes O x y, O x z, O yz are given by the formulas: ¨ IO = S

(

) x 2 + y 2 + z 2 F(x, y, z)dσ

234

7 Surface Integrals

¨ IOx = S

¨ IOy = S

¨ IOz =

(

) y 2 + z 2 F(x, y, z)dσ ,

(

) x 2 + z 2 F(x, y, z)dσ

(

) x 2 + y 2 F(x, y, z)dσ

S

¨

IOx y =

z 2 F(x, y, z)dσ , S

¨ IOx z =

y 2 F(x, y, z)dσ S

¨ IO yz =

x 2 F(x, y, z)dσ . S

7.4 Surface Integral of the Second Kind Before we really define the surface integral of the second kind, we first need to introduce the idea of an oriented surface, similar to an oriented curve. Let S be a smooth surface and let − → − → − → − → r (u, v) = x(u, v) i + y(u, v) j + z(u, v) k , ∀(u, v) ∈ Δ be one of its parametric representation. − → → → r u ×− Because the surface is smooth, it follows that − r v /= 0 , for any (u, v) ∈ Δ. We notice that at each interior point M(x(u, v), y(u, v), z(u, v)) ∈ S, there are → → n (M), where: n (M) and −− two unit normal vectors to S, namely − → − → r v r u ×− − → ∥. n (M) = ∥− → − → ∥ r u × r v∥ Definition 7.4.1 The smooth surface S is said to be orientable (two-sided) if the → n (M) : S → R3 is continuous. map M → − → It is obvious that if the application M → − n (M) : S → R3 is continuous, then − → 3 also the map M → − n (M) : S → R is continuous. If a surface is orientable, then choosing a side of this surface returns to choose one of the two continuous → n (M). applications M → ±− Therefore, we have two possible orientations (or two sides) of the surface, namely:

7.4 Surface Integral of the Second Kind

235

( →) ( ) → S+ = S, − n and S− = S, −− n → Remark 7.4.1 The property of the mapping M → − n (M) : S → R3 to be continuous on the orientable surface S is a global property and it covers the entire surface. This implies the following property: let M0 ∈ S be an arbitrary interior point and let C be an arbitrary closed curve lying on the surface S, which passes through the point M0 and has no common points with the boundary of S; let us suppose that we choose an orientation to the normal line to S at M0 , and that is the orientation → n (M0 ). Imagine that we are moving the unit normal given by the unit normal vector − − → vector n (M) on the curve C, starting from the point M0 , in such a way the vector always remains perpendicular to the surface, we return to the point M0 with the same orientation of the normal, that is: − → → lim n (M) = − n (M0 ). M → M0 M ∈C If there is a closed curve lying on the surface S such that after it has been traversed, the unit normal vector changes its orientation to the opposite, the surface is called one-sided. Example 7.4.1 The simplest orientable surface is the plane. Also, quadric surfaces (sphere, ellipsoid, hyperboloids, paraboloids, cone, cylinders, pairs of planes) are orientable surfaces. When the surface is closed, so when it is the boundary of a bounded spatial domain, then it is orientable and its two faces are called: the outer face and the inner face. Example 7.4.2 Any elementary surface is an orientable surface. Indeed, let − → − → − → − → r (u, v) = x(u, v) i + y(u, v) j + z(u, v) k , ∀(u, v) ∈ Δ be a parametric representation of an elementary surface S. → We notice that the application M → − n (M) : S → R3 is continuous on S, because it is the composition of continuous functions r − 1 : S → Δ and (u, v) → → − → r v r u ×− : Δ → R3 . → → r v∥ r u ×− ∥− In particular, any explicit smooth surface is orientable having the sides: the upper side which corresponds to the normal oriented upward (which forms an acute angle with the positive half-axis Oz) (Fig. 7.11) and the lower side which corresponds to the normal oriented downward (which forms an obtuse angle with the positive half-axis Oz) (Fig. 7.12).

236 Fig. 7.11 The orientation of the upper side of an explicit smooth surface

7 Surface Integrals

γ

z

y

O x

Fig. 7.12 The orientation of the lower side of an explicit smooth surface

z

γ

O

y

x

} { Example 7.4.3 The sphere S = (x, y, z) ∈ R3 ; x 2 + y 2 + z 2 = R 2 is orientable closed surface; thus, it has two faces: the outer face corresponding to the unit normal vector that point outward and the inner face corresponding to the unit normal vector facing inward. Indeed, for any point M(x , y , z) on the sphere, the unit vector of the outer −−→ → → normal is − n (M) = R1 O M. Clearly, the map M → − n (M) : S → R3 is continuous on S. Example 7.4.4 The classic example of a non-orientable (one-sided) surface is the Möbius strip. A model of this surface is obtained by twisting a rectangular strip of paper ABC D such that the point A coincides with the point C and B with D (Fig. 7.13). It is easily seen that after we traverse the middle line E F of the M˝obius strip, the orientation of the normal to the surface is reversed.

Fig. 7.13 The Möbius strip

7.4 Surface Integral of the Second Kind

237

z

Fig. 7.14 The outer face of the sphere centered in the origin and of radius R

γ u

M

u'

O

y

γ' M'

x

n

n

z

Fig. 7.15 The lower side of the cone with vertex in the origin

h

γ n

y

O x

z

Fig. 7.16 A simple domain with respect to the three coordinate axes

γ

S2

n

S3

γ n

γ

S1 O

x

n

D

∂D

y

238

7 Surface Integrals

Let S be a simple smooth orientable parameterized surface and let − → − → − → − → r (u, v) = x(u, v) i + y(u, v) j + z(u, v) k , ∀(u, v) ∈ Δ be one of its parametric representation, where Δ ⊂ R2 is a squareable bounded ( →) n domain and x, y, z are functions of C 1 − class on Δ. We will denote by S+ = S, − − → − → → r u× r v . the face of the surface corresponding to the unit normal vector − n = ∥− → → r v∥ r u ×− − → 3 We also consider a continuous vector field F = (P, Q, R) : Ω ⊂ R → R3 , where Ω ⊂ R3 is a space domain that contains the surface S. Definition 7.4.2 The surface integral of the second kind of the vector field ( →) − → n of the surface S, means the following F = (P, Q, R) over the face S+ = S, − expression: ¨ P(x, y, z)dy ∧ dz + Q(x, y, z)dz ∧ dx + R(x, y, z)dx ∧ dy = S+

¨

=

F→ · n→dσ =

S

¨ (P(x, y, z) cos α + Q(x, y, z) cos β + R(x, y, z) cos γ )dσ

=

(7.9)

S

→ n to the surface S where α, β, γ are the angles formed by the unit normal vector − with the positive directions of the coordinate axes O x, O y, Oz, respectively, i.e.: → (x, y, z) ∈ S. n→(x, y, z) = cos α(x, y, z)→i + cos β(x, y, z) →j + cos γ (x, y, z)k, If we use the usual notations presented in paragraph 7.1, then: | | | | | | | xu z u | | xu yu | | yu z u | | | | | | | ; B = −| ; C =| A=| yv z v | xv z v | xv yv | cos α = ± √ cos β = ± √ cos γ = ± √

A A2

+ B2 + C 2 B

A2 + B 2 + C 2 C

A2

+ B2 + C 2

.

Note that we must have to choose the +" or "−" sign in front of the square root to match the orientation of the normal to the surface.

7.4 Surface Integral of the Second Kind

239

Taking into account the computation of the surface integral of the first kind, the formula (7.9) becomes: ¨ P(x, y, z)dy ∧ dz + Q(x, y, z)dz ∧ dx + R(x, y, z)dx ∧ dy = S+

¨ [P(x(u, v), y(u, v), z(u, v)) · A(u, v)+

=± Δ

+ Q(x(u, v), y(u, v), z(u, v)) · B(u, v)+ +R(x(u, v), y(u, v), z(u, v)) · C(u, v)]du dv.

(7.10)

Remark 7.4.2 The surface integral of the second kind depends on the surface orientation. ( ) → Indeed, if S− = S, −− n is the other face of the surface S, then: ¨ P(x, y, z)dy ∧ dz + Q(x, y, z)dz ∧ dx + R(x, y, z)dx ∧ dy = S−

¨

= S

− → F · (−→ n )dσ = −

¨

− → F · n→dσ =

S

¨

P(x, y, z)dy ∧ dz + Q(x, y, z)dz ∧ dx + R(x, y, z)dx ∧ dy.

=− S+

Remark 7.4.3 In physical applications, the surface integral of the second kind ˜ − − → → →dσ is often referred to as the total flux ϕ of the vector field F through the Ss F ·n − → face S+ (respectively, S− ) of the surface S. Specifically, let us suppose that F is the particle velocities field of a fluid in stationary flow (the velocities of fluid particles ˜ − → depend on space positions but not on the time). Then, ϕ = S F · n→dσ represents the volume of fluid flowing through the surface S per time unit, in the direction of the unit normal vector that defines the selected face. ˜ Example 7.4.5 Compute x 2 dy ∧ dz + y 2 dz ∧ dx + zdx ∧ dy,, S+

where S+ is the outer face of the sphere x 2 + y 2 + z 2 = R 2 . The ⎧ parametric equations of the sphere are: ⎨ x = R sin u cos v y = R sin u sin v , u ∈ [0, π ], v ∈ [0, 2π ]. ⎩ z = R cos u From Example 7.1.1, we have: A = R 2 sin2 u cos v, B = R 2 sin2 u sin v, C = R 2 sin u cos u

240

7 Surface Integrals

A2 + B 2 + C 2 = R 4 sin2 u, hence: cos α = ± sin u cos v, cos β = ± sin u sin v, cos γ = ± cos u.

(7.11)

Since S+ is the outer face of the sphere, we notice that for the normal to the outward-facing sphere ) must choose the sign “+” in the formulas (7.11). ( we Indeed, if u ∈ 0, π2 , then the corresponding point M on the sphere is on the upper hemisphere and the outer normal will describe a acute angle γ with the axis Oz (cos γ = ( cos)u > 0) (Fig. 7.14). If u ' ∈ π2 , π , then the corresponding point M ' on the sphere is on the lower hemisphere normal will make an obtuse angle γ ' with the ( and' the outward-facing ) ' axis Oz cos γ = cos u < 0 . From the formula (7.10), it results that: ¨ x 2 dy ∧ dz + y 2 dz ∧ dx + zdx ∧ dy = S+

⎞ ⎛ ∫2π ∫π ( 4 4 ) = ⎝ R sin u cos3 v + R 4 sin4 u sin3 v + R 3 sin u cos2 u du ⎠dv = 0

0

⎞ ⎛ π ⎞ ⎛ 2π ∫ ∫ = R 4 ⎝ cos3 vdv ⎠ · ⎝ sin4 udu ⎠ 0

0

⎞ ⎛ π ⎞ ⎛ 2π ∫ ∫ + R 4 ⎝ sin3 v dv ⎠ · ⎝ sin4 u du ⎠+ 0

∫2π + R3

⎛ ⎝

0

0

∫π

⎞

sin u cos2 u du ⎠dv.

0

∫ 2π

( ∫ 2π ( ) cos vdv = 0 1 − sin2 v cos vdv = sin v − 3

Since we have 0 ∫ 2π 0, and similarly, 0 sin3 vdv = 0, then it follows that:

sin3 v 3

¨ x 2 dy ∧ dz + y 2 dz ∧ dx + zdx ∧ dy = S+

⎛ ⎞ | ∫2π ∫π cos3 u ||π 4π R 3 . = R 3 ⎝ sin u cos2 u du ⎠dv = −2π R 3 · = 3 |0 3 0

0

)|| |2π = |0

7.4 Surface Integral of the Second Kind

241

Remark 7.4.1 If the smooth surface is explicitly given by S : z = f (x, y), (x, y) ∈ D where D ⊂ R2 is a squareable bounded domain and f : D → R is a function of C 1 − class on D, then we obtain cos α =

−p −q 1 √ ; cos β = √ ; cos γ = √ . 2 2 2 2 ± 1+ p +q ± 1+ p +q ± 1 + p2 + q 2

If S+ is the upper side of the surface, corresponding to the normal oriented upward, then cos γ > 0 and we choose the sign “+” in front of the square root from the denominator. For the lower side S− , the unit normal vector will point downward, hence cos γ < 0 and we will choose the sign “−” in front of the square root. In this case, formula (7.9) becomes: ¨ P(x, y, z)dy ∧ dz + Q(x, y, z)dz ∧ dx + R(x, y, z)dx ∧ dy = S+

¨

=

(− p · P(x, y, f (x, y)) − q · Q(x, y, f (x, y))+ D

+ R(x, y, f (x, y)))dxdy.

(7.12)

Example 7.4.6 Compute the total flux ϕ of the vector field: − → − → − → − → V (x, y, z) = (y + z) i − (z + x) j − (x + y) k through the lower side of the cone S : x 2 + y 2 = z 2 , 0 ≤ z ≤ h. − → The total flux of the vector field V through the lower face S− of the cone is given by (Fig. 7.15): ¨ ϕ=

(y + z)dy ∧ dz − (z + x)dz ∧ dx − (x + y)dx ∧ dy. S−

The surface S is explicitly defined as: S:z=

√

x 2 + y 2 , (x, y) ∈ D,

where } { D = (x, y) ∈ R2 ; x 2 + y 2 ≤ h 2 .

242

7 Surface Integrals

Also, we have: p=

√ √ y x ∂z ∂z = √ = √ ; q= ; 1 + p 2 + q 2 = 2. ∂y ∂x x 2 + y2 x 2 + y2

Since S− is the lower side of the cone, then cos γ < 0 and further, we have: 1 cos γ = −√ ; cos α = √ √x 2 2 ; cos β = √ √y 2 2 . 2 2·

x +y

The total flux becomes: ¨ ( ϕ= S

x (y + z) √ √ 2 · x 2 + y2

2·

x +y

)

) y 1 −(z + x) √ √ − (x + y) √ dσ = − 2 2 · x 2 + y2 ) ⎛( √ x y + x 2 + y 2 √ √ ¨ ⎜ 2 · x 2 + y2 ⎜ (√ = ) ⎝ y + − x 2 + y 2 + x √ √ D 2 · x 2 + y2 ⎞ ⎛ ) ¨ ∫2π ∫h x+y √ + √ 2dxdy = 2 xdxdy = 2 ⎝ ρ 2 cos θ dρ ⎠dθ = 2 D

=

2h 3 3

0

0

∫2π cos θ dθ = 0. 0

7.5 Integral Formulas Integral formulas are formulas for linking different types of integrals. Such a formula has already been presented in Chap. 6, namely the Riemann–Green formula, which establishes the connection between the double integral on a domain and the line integral of the second kind on the boundary of the domain. The analogous formula for three-dimensional space is the Gauss–Ostrogradsky formula, which establishes an important link between the triple integral and the surface integral of the second kind and is used, in particular, for the computation of integrals on closed surfaces. Theorem 7.5.1 (Gauss–Ostrogradsky formula). Let T ⊂ R3 be a domain simple with respect to the three coordinate axes and let P , Q, R : T → R be three continuous functions with the property that there

7.5 Integral Formulas

243

are partial derivatives ∂∂ Px , ∂∂Qy , ∂∂Rz continuous on T . We also assume that the closed surface S = T \T that bounded the domain T is a piecewise smooth orientable surface and S+ is its outer face. Then we have the formula: ¨ P(x, y, z)dy ∧ dz + Q(x, y, z)dz ∧ dx + R(x, y, z)dx ∧ dy = S+

˚ (

= T

) ∂P ∂Q ∂R (x, y, z) + (x, y, z) + (x, y, z) dxdydz. ∂x ∂y ∂z

(7.13)

Proof Since the domain T ⊂ R3 is simple with respect to O z axis, it follows that there are a squareable bounded domain D ⊂ R2 and two continuous functions ϕ, ψ : D → R, with the property ϕ(x) < ψ(x), ∀(x, y) ∈ D such that: { } T = (x, y, z) ∈ R3 ; ϕ(x, y) < z < ψ(x, y), ∀(x, y) ∈ D . We denote by S1 the graph of the function z = ϕ(x, y), (x, y) ∈ D, respectively by S2 the graph of the function z = ψ(x, y), (x, y) ∈ D, and by S3 the lateral cylindrical surface with the generators parallel to O z axis and the directrix curve ∂ D. We notice that the surface S = S1 ∪ S2 ∪ S3 is the boundary of the domain T . The assumption that S is piecewise smooth means that ϕ and ψ are of C 1 − class on D. The outer face of the surface S means the face corresponding to the normal facing outward (Fig. 7.16). This means for S1 the lower face surface, and for S2 the upper face surface. Because, for the lower face S1 of the surface, the angle formed by O z axis with the downward-facing normal is obtuse, it results cos γ < 0, hence: cos γ = − √ 1+

(

∂ϕ ∂x

1 )2

(

+

∂ϕ ∂y

Furthermore, we have: ¨ ¨ R(x, y, z)dx ∧ dy = R(x, y, z) · √ (S1 )−

1+

S1

¨ =− D

¨

R(x, y, ϕ(x, y)) √ ( )2 ( )2 · 1 + ∂ϕ + ∂ϕ ∂x ∂y R(x, y, ϕ(x, y))dxdy.

=− D

√ 1+

(

∂ϕ ∂x

)2 .

(

−1 )2

∂ϕ ∂x

)2

( +

+

∂ϕ ∂y

(

∂ϕ ∂y

)2 dσ =

)2 dxdy =

(7.14)

244

7 Surface Integrals

Similarly, for the upper face of the surface S2 , cos γ > 0, hence: ¨

¨ R(x, y, z)dx ∧ dy = (S2 )+

R(x, y, z) · √ 1+

S2

¨ =

√ D

¨ =

R(x, y, ψ(x, y)) ( )2 ( )2 · 1 + ∂ψ + ∂ψ ∂y ∂x

√

(

1+

∂ψ ∂x

(

)2

∂ψ ∂x

( +

1 )2

∂ψ ∂y

+

(

∂ψ ∂y

)2 dσ =

)2 dxdy =

R(x, y, ψ(x, y))dxdy.

(7.15)

D

For the outer face of the lateral cylindrical surface S3 , cos γ = 0, because γ = π2 . It turns out that: ¨ ¨ R(x, y, z)dx ∧ dy = R(x, y, z) cos γ dσ = 0. (7.16) (S3 )+

S3

Therefore, we have: ¨ ¨ R(x, y, z)dx ∧ dy = R(x, y, z)dx ∧ dy+ (S1 )−

S+

¨ R(x, y, z)dx ∧ dy

+ (S2 )+

¨

R(x, y, z)dx ∧ dy =

+ (S3 )+

¨

R(x, y, ψ(x, y))dxdy

= D

¨

−

R(x, y, ϕ(x, y))dxdy. D

On the other hand, from Theorem 6.10.1, it results that: ⎛ ⎞ ψ(x,y) ˚ ∫ ¨ ∂R ∂R ⎜ ⎟ (x, y, z)dz ⎠dxdy = (x, y, z)dxdydz = ⎝ ∂z ∂z D

T

¨ =

ϕ(x,y)

| | R(x, y, z)||ψ(x,y) dxdy = ϕ(x,y)

D

(7.17)

7.5 Integral Formulas

245

¨ =

R(x, y, ψ(x, y))dxdy D

¨

−

R(x, y, ϕ(x, y))dxdy.

(7.18)

D

From Eqs. (7.17) and (7.18), we deduce: ˚ T

∂R (x, y, z)dxdydz = ∂z

¨ R(x, y, z)dx ∧ dy.

(7.19)

S+

Similarly, using the fact that the domain T is simple also with respect to O y axes, respectively O x, we get: ˚ T

∂Q (x, y, z)dxdydz = ∂y

¨ Q(x, y, z)dz ∧ dx

(7.20)

P(x, y, z)dy ∧ dz.

(7.21)

S+

and, respectively: ˚ T

∂P (xs, y, z)dxdydz = ∂x

¨ S+

Finally, adding Eq. (7.19), (7.20) and Eq. (7.21), we obtain the Gauss–Ostrogradsky formula: ˚ (

) ∂P ∂Q ∂R (x, y, z) + (x, y, z) + (x, y, z) dxdydz = ∂x ∂y ∂z T ¨ = P(x, y, z)dy ∧ dz + Q(x, y, z)dz ∧ dx + R(x, y, z)dx ∧ dy. S+

Remark 7.5.1 Among the examples of simple domains with respect to the three coordinate axes we mention: the sphere, the ellipsoid, the rectangular parallelepiped with the edges parallel to the axes, and so on. Without going into details, we mention that the Gauss–Ostrogradsky formula also remains valid for domains T that are finite unions of simple domains with respect to the three coordinate axes, two by two having in common at most piecewise smooth surfaces. By writing the Gauss– Ostrogradsky formula for each of the simple domains Ti that make up the domain T , adding these formulas and using the additivity property of the triple integral and the surface integral, the Gauss–Ostrogradsky formula for the domain T is obtained. This is explained by the fact that the surface integral oven an intersection surface of two neighboring simple domains appears in the sum of the right side twice, once on

246

7 Surface Integrals

the upper face and once on the lower face, so its contribution to the right side is zero. Thus, only the integral on the outer face of the domain T remains in the right side. Remark 7.5.2 Taking into account the connection between the surface integral of the second kind and the surface integral of the first kind, the Gauss–Ostrogradsky formula is also written: ) ¨ ˚ ( ∂P ∂Q ∂R + + dxdydz (P cos α + Q cos β + R cos γ )dσ = ∂x ∂y ∂z S

T

(7.22) where α, β, γ are the angles that the outer normal to the surface makes with the positive directions of the coordinate axes O x, O y and Oz. − → Remark 7.5.3 If we consider the vector field V : T ⊂ R3 → R3 of C 1 − class on T having the components P, Q, R, then: ∂R ∂Q − → − → − → − → − → ∂P + . + V = P · i + Q · j + R · k and div V = ∂y ∂z ∂x − → − → − → → Let also − n = cos α · i + cos β · j + cos γ · k be the unit normal vector of the outer normal to the surface S. With these notations, the Gauss–Ostrogradsky formula becomes: ¨

− → − V ·→ n dσ =

S

˚

− → div V dxdydz

(7.23)

T

and in this form it is also called the flux—divergence formula, because it shows that − → the flux of the vector field V through the surface S, according to the outer normal, is equal to the triple integral of its divergence on the domain T . Example 7.5.1 Using the Gauss–Ostrogradsky formula compute the integral. ¨ x 3 dy ∧ dz + y 3 dz ∧ dx + z 3 dx ∧ dy, S+

where S+ is the outer face of the cube: { } T = (x, y, z) ∈ R3 ; 0 ≤ x ≤ a, 0 ≤ y ≤ a, 0 ≤ z ≤ a . First, we denote by P(x, y, z) = x 3 , Q(x, y, z) = y 3 , R(x, y, z) = z 3 and from the Gauss–Ostrogradsky formula (7.13), we deduce:

7.5 Integral Formulas

247

¨ x 3 dy ∧ dz + y 3 dz ∧ dx + z 3 dx ∧ dy = S+

˚ =

( 2 ) 3x + 3y 2 + 3z 2 dxdydz

T

∫a =3

∫a dx

∫a dy

0

0

∫a

∫a (

=3 0

∫a =3 0

) x 2 + y 2 + z 2 dz =

0

)| z 3 ||a x z+y z+ dy 3 |0 2

dx

(

2

0

) ∫a ( a3 2 2 ax + ay + dy = dx 3 0

)| ∫a ( y3 a 3 ||a 2 =3 ax y + a + y | dx 0 3 3 0

) ∫a ( a4 a4 2 2 a x + + dx = =3 3 3 0 ( )| 3 2a 4 ||a 2x =3 a x | = 3a 5 . + 0 3 3 The Stokes formula expresses a relationship between surface integrals and line integrals. It generalizes Riemann–Green formula, the latter being a special case of the former when the surface in question is a part of the plane O x y. We will present this formula for the particular case of explicit surfaces. Theorem 7.5.2 (Stokes formula) Let S : z = f (x, y), (x, y) ∈ D be an explicit smooth surface, where D ⊂ R2 is a bounded domain whose boundary γ is a smooth curve and f : D → R is a function of C 2 − class on D. Let Ω ⊂ R3 be a space domain that includes the surface S and let P, Q, R : Ω → R be three functions of C 1 − class on Ω. If we denote by ┌ = S\S = {(x, y, f (x, y)); (x, y) ∈ γ } the boundary of the surface S, then we have the formula: ∫  P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz = ┌

¨ ( = S+

) ( ) ( ) ∂R ∂Q ∂P ∂R ∂Q ∂P − dy ∧ dz + − dz ∧ dx + − dx ∧ dy . ∂y ∂z ∂z ∂x ∂x ∂y (7.24)

248

7 Surface Integrals

z

Fig. 7.17 An explicit smooth surface

n

S

Γ

O

D

x

Fig. 7.18 A finite union of explicit surfaces

y

γ

Sp

S1 S2

There is the following compatibility link (* ) between the direction of travel of the curve ┌ and the face of the surface S on which the integral is made in the right member of the formula (7.24): if the curve ┌ is traversed counter-clockwise (respectively, clockwise), then the surface integral in the right member is made on the upper (respectively, lower) face of the surface (Fig. 7.17) (*) Assuming that the coordinate system is rectangular). Proof If x = ϕ(t), y = ψ(t), t ∈ [a, b] is a parametric representation of the plane curve γ , then the spatial curve ┌ = {(x, y, f (x, y)); (x, y) ∈ γ } is described by the following parametric equations: ⎧ ⎨ x = ϕ(t) , t ∈ [a, b]. y = ψ(t) ⎩ z = f (ϕ(t), ψ(t)) Taking into account the computation mode for the line integral of the second kind, we have: ∫ ∫b P(ϕ(t), ψ(t), f (ϕ(t), ψ(t))) · ϕ'(t)dt =  P(x, y, z)dx = ┌

a

∫ =  P(x, y, f (x, y))dx. γ

Further, from Riemann–Green formula, it results that:

(7.25)

7.5 Integral Formulas

249

) ¨ ( ∫ ∂P ∂P ∂f + · dxdy.  P(x, y, f (x, y))dx = − ∂y ∂z ∂ y γ

(7.26)

D

If we note by p =

∂f ∂x

and q =

∂f , ∂y

further we have:

¨

¨ √ ∂P ∂P 1 · 1 + p 2 + q 2 dxdy = · √ dxdy = − 2 2 ∂y ∂y 1+ p +q D D ¨ ¨ ∂P ∂P dx ∧ dy. (7.27) cos γ dσ = − =− ∂y ∂y

−

S+

S

Similarly, we have: ¨ √ −q ∂P ∂P ∂f · dxdy = · √ · 1 + p 2 + q 2 dxdy = 2 2 ∂z ∂y ∂z 1+ p +q D D ¨ ¨ ∂P ∂P dz ∧ dx. (7.28) cos βdσ = = ∂z ∂z ¨

−

S

S+

From Eqs. (7.25), (7.26), and Eq. (7.28), we deduce: ¨ ¨ ∫ ∂P ∂P dx ∧ dy. dz ∧ dx −  P(x, y, z)dx = ∂y ∂z ┌

S+

(7.29)

S+

Similarly, it is shown that: ¨ ¨ ∫ ∂Q ∂Q dx ∧ dy − dy ∧ dz  Q(x, y, z)dy = ∂x ∂z

(7.30)

¨ ∫ ¨ ∂R ∂R  R(x, y, z)dz = dz ∧ dx. dy ∧ dz − ∂x ∂y

(7.31)

┌

┌

S+

S+

S+

S+

Adding the relations (7.29), (7.30), and (7.31) we obtain Stokes formula (7.24) from the theorem statement. Remark 7.5.4 Stokes formula also holds for surfaces that are finite unions of explicit surfaces such as those in Theorem 7.5.2, two by two having in common at most oriented curve arcs that are portions of the oriented boundaries of these surfaces. Indeed, writing Stokes formula for each of the surfaces Si and adding up the obtained p ∪ Si (Fig. 7.18). formulas we obtain Stokes formula for the surface S = i=1

250

7 Surface Integrals

The explanation consists in the fact that the line integral along an intersection curve of two neighboring surfaces intervenes twice in the sum of the left member of formula (7.24), with opposite orientations, hence its contribution in this sum is zero. In this way, on the left side of the formula appears only the line integral on the boundary of surface. On the other hand, it is obvious that: ¨ =

p ¨ Σ

.

i=1 S i

S

Remark 7.5.5 Taking into account the connection between the surface integral of the second kind and the surface integral of the first kind, Stokes formula is also written: ∫  P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz = ┌

¨ ((

= S

∂R ∂Q − ∂y ∂z

)

( cos α +

∂P ∂R − ∂z ∂x

)

( cos β +

∂Q ∂P − ∂x ∂y

)

) cos γ dσ . (7.32)

Remark 7.5.6. Under the conditions of Theorem 7.5.2, if we denote by: − → − → − → − → V = P· i +Q· j +R· k − → Curl V =

(

∂Q ∂R − ∂y ∂z

)

− → · i +

(

∂P ∂R − ∂z ∂x

)

− → · j +

(

∂Q ∂P − ∂x ∂y

)

− → · k

− → − → − → − → n = cos α · i + cos β · j + cos γ · k − → − → − → → d− r = dx · i + dy · j + dz · k then Stokes formula (7.24) is also written in vector form: ¨ ∫ − → − − → → →  V ·d r = Curl V · − n dσ . ┌

(7.33)

S

→ n is the unit normal vector to the surface oriented upward. Of course, here − The Stokes formula, in the form (7.33), shows that the circulation of the vector − → field V along the oriented curve ┌ = ∂ S is equal to the flux of the curl of vector − → field V through that surface in the direction of normal oriented upward.

7.5 Integral Formulas

251

− → Remark 7.5.7. If the vector field V = (P, Q, R) is irrotational on Ω, that is − → − → curl V = 0 on Ω, from (7.33), we deduce that the line integral of the second − → kind of the vector field V along any closed curve included in Ω is zero, hence the differential form: ω = P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz is exact on Ω. − → Moreover, in this case, the line integral of the second kind of the vector field V is independent on the path in Ω (Remark 5.6.1). The Stokes formula is used to compute the line integral of the second kind of the − → − → vector field V along a closed curve ┌ when the flux of curl V through the surface bounded by the curve ┌ is easier to compute. Example 7.5.2 Using Stokes formula, compute the line integral of the second kind: ∫  (z − y)dx + (x − z)dy + (y − x)dz ABC

where A(a, 0, 0), B(0, b, 0), C(0, 0, c), a, b, c > 0, are the vertices of the triangle ABC and the direction of travel is A → B → C → A. The plan determined by the points A, B, C has the equation: ax + by + cz = 1. Notice that the triangle ABC is the boundary of the explicit surface (Fig. 7.19): ( y) x , (x, y) ∈ D S : z =c 1− − a b where D is the right triangle (full) O AB, namely (Fig. 7.20): { ( x ){ . D = (x, y) ∈ R2 ; 0 ≤ x ≤ a, 0 ≤ y ≤ b 1 − a

z c C

Fig. 7.19 The triangle ABC

γ

n

S O a x

A

b B

y

252

7 Surface Integrals

y

Fig. 7.20 projection of the triangle ABC in the plane xOy

b

D O

a

x

If we denote by P(x, y, z) = z − y, Q(x, y, z) = x − z, R(x, y, z) = y − x, then we have: ∂P ∂Q ∂Q ∂R ∂R ∂P = −1; = 1; = 1; = −1; = −1; = 1. ∂y ∂z ∂x ∂z ∂x ∂y From Stokes formula (7.32), it results that: ∫ ¨  (z − y)dx + (x − z)dy + (y − x)dz = 2(cos α + cos β + cos γ )dσ ABC

S

where α, β, γ are the angles formed by the normal to the surface S, oriented upward, with the axes O x, O y, Oz. On the other hand, we have: p=

c ∂z c ∂z = − ,q = =− ∂x a ∂y b

and: 1 + p2 + q 2 =

a 2 b2 + b2 c2 + c2 a 2 . a 2 b2

Since cos γ > 0, we deduce: −p bc cos α = √ =√ 2 2 2 2 a b + b2 c2 + c2 a 2 1+ p +q −q ca cos β = √ =√ 2 2 2 2 a b + b2 c2 + c2 a 2 1+ p +q ab 1 . =√ cos γ = √ 2 2 2 2 a b + b2 c2 + c2 a 2 1+ p +q Therefore, we have:

7.5 Integral Formulas

253

∫  (z − y)dx + (x − z)dy + (y − x)dz ABC

¨ ( ) c c =2 + + 1 dxdy = a b D ) ¨ (c c 2(ab + bc + ca) + +1 · · area(D) = 1dxdy = =2 a b ab D

2(ab + bc + ca) ab = · = ab + bc + ca. ab 2 Remark 7.5.8 We recall that the computation of the integral in Example 7.5.2 as a line integral of the second kind was performed in Example 5.5.5.

References

1. Bucur, I., Analyse mathématique. Calcul intégral, Conspress, Bucarest, 2014. 2. Ilyn, V.A., Poznyak, E.G., Fundamentals of mathematical analysis, Mir Publishers, Moscow, Part I and Part II, 1982. 3. Krasnov, M., Kiselev, A., Makarenko, G., Shikin, E., Mathematical analysis for engineers, Mir Publishers, Moscow, Part I, 1989 and Part II, 1990. 4. Lange, S., Calculus of several variables, Third Edition, Springer, New York Inc., 1987. 5. Nikolsky, S.M., A course of mathematical analysis, Mir Publishers, Moscow, Part I and Part II, 1977. 6. Paltineanu, G., Bucur, I., Zamfir, M., Differential Calculus for Engineers, Springer, 2022, ISBN: 978-981-19-2553-5 7. P˘altineanu, G., Zamfir, M., Higher Mathematics 2 (Integral Calculus), Conspress, Bucharest, 2016 (In Romanian). 8. P˘altineanu, G., Matei, P., Mateescu, G.D., Numerical analysis, Conspress, Bucharest, 2010 (In Romanian). 9. Popescu, S.A., Mathematical analysis II. Integral calculus, Conspress, Bucharest, 2016. 10. Rudin, W., Principles of mathematical analysis, McGraw-Hill, Inc., 1964. 11. Rudin, W., Real and complex analysis, Third Edition, McGraw-Hill, Inc., 1987. 12. St˘an˘as¸il˘a, O., Mathematical analysis, E.D.P., Bucharest, 1981 (In Romanian). 13. Trench, W.F., Introduction to real analysis, Library of Congress Cataloging-in-Publication Data, 2003.

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Integral Calculus for Engineers, https://doi.org/10.1007/978-981-19-4793-3

255

Index

A Abel-Dirichlet criterion for improper integral, 82 Absolutely convergent improper integral, 77 Antiderivative, 1, 2 Approximation error at a point for Lagrange polynomial, 62, 63 Area of a circular sector, 55 Area of a curvilinear sector, 58 Area of a curvilinear trapezoid, 25, 52 Area of a plane figure, 49 Area of a simple smooth parameterized surface, 226 Area of a smooth explicit surface, 225 Area of a torus, 229 Area of cardioid, 60 Area of ellipse, 54 Area of sphere, 228

B Binomial integral, 21

Closed curve, 118 Closed parameterized path, 113 Computation of Dirichlet integral, 100 Computation of Euler-Poisson integral, 103 Computing triple integral on simple domain, 202, 203 Convergent and divergent for improper double integral, 192, 193 Convergent and divergent for improper integral, 71 Coordinates of center of mass of a lamina, 181 Coordinates of center of mass of a shell, 233 Coordinates of center of mass of a solid, 211 Coordinates of center of mass of wire, 134 Cubable bounded space set, 197 Curvilinear cylinder (cylindroid), 197 Curvilinear sector, 58 Curvilinear trapezoid, 25 Cylindrical coordinates, 208 Cylindrical helix, 115

C Cauchy integral criterion, 82 Change of variable for definite integral, 48 Change of variable for improper integral, 84 Change of variable for indefinite integral. First method, 6 Change of variable for indefinite integral. Second method, 8 Change of variables in double integral, 175 Change of variables in triple integral, 205 Circular sector, 55

D Darboux criterion for integrability for function of one variable, 32 Darboux criterion for integrability for function of two variables, 163 Darboux integrals for function of one variable, 30 Darboux integrals for function of two variables, 162 Darboux sums for function of one variable, 28, 161

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Integral Calculus for Engineers, https://doi.org/10.1007/978-981-19-4793-3

257

258 Darboux sums for function of three variables, 200 Darboux sums for function of two variable, 200 Definite integral, 30, 35 Diameter of a plane set, 160 Diameter of a space domain, 199 Diffeomorphism, 218 Differential closed form, 154 Differential exact form, 150 Differential form, 150 Dirichlet integral, 84 Domain in, 202 Domain in ϒ 2 simple with respect to coordinate axis, 166 Double integral, 163, 164

E Elementary domain of Green type, 185 Elementary plane set, 50 Elementary space set, 196 Elementary surface, 219 Elliptic integrals, 125 Equivalent paths, 116 Equivalent surface canvases, 218 Euler integral of the first kind (beta-function), 104 Euler integral of the second kind (gamma-function), 104 Euler-Poisson integral, 80 Explicit surface, 219

F First Comparison test for improper integral, 71 Flux-divergence formula, 246 Fresnel integrals, 85 Fubini Theorem, 92

G Gauss-Ostrogradsky formula, 242 Generalized polar coordinates, 179 Generalized right cylinder, 197 Generalized spherical coordinates, 209

I Implicit surface, 221 Improper double integral, 192 Improper integral depending on a parameter pointwise convergent, 93

Index Improper integral depending on a parameter uniform convergent, 94 Improper integral of the first kind, 71 Improper integral of the second kind, 71 Indefinite integral, 2 Integrable function of one variable in the Darboux sense, 30 Integrable function of one variable in the Riemann sense, 35 Integrable function of three variables in the Darboux sense, 200 Integrable function of three variables in the Riemann sense, 202 Integrable function of two variables in the Darboux sense, 162 Integrable function of two variables in the Riemann sense, 164 Integral depending on a parameter, 88 Integration by parts for definite integral, 48 Integration by parts for indefinite integral, 9 L Lagrange interpolation polynomial, 61 Lagrange interpolation polynomial for equidistant nodes, 62 Lamina, 180 Lebesque criterion for integrability for function of one variable, 42 Lebesque criterion for integrability for function of two variables, 163 Leibniz formula for differentation of the integral depending on a parameter, 91, 92 Length of a rectifiable curve (path), 119, 123 Length of the cardioid, 127 Length of the circle, 124 Length of the cylindrical helix, 123 Line integral independent on the path, 150 Line integral of the first kind, 132 Line integral of the second kind, 140, 141 Locally integrable function, 71 M Mass of a lamina, 181 Mass of a shell, 233 Mass of a solid, 211 Mass of a wire, 134 Mean value formula for double integral, 166 Mean value formula for line integral, 140 Mean value formula for simple integral, 45

Index Mechanical work, 143 M˝obius strip, 236 Moments of inertia of a lamina, 183, 184 Moments of inertia of a shell, 233 Moments of inertia of a solid, 212 Moments of inertia of a wire, 135

N Natural parameterization of a path, 129 Newton-Côtes coefficients, 64 Newton-Côtes quadrature formula, 64 Newton-Leibniz formula, 47 Normal line, 222 Norm of a partition of a compact interval, 28 Norm of a partition of a plane bounded domain, 161 Norm of a partition of a space bounded domain, 199, 200 Null subset, 41

O Orientable (two-sider) surface, 234 Orientation of a curve, 118

P Parameter change, 116, 218 Parameterized curve, 117 Parameterized path, 113 Parameterized simple surface canvas, 216 Parameterized smooth surface canvas, 216 Parameterized surface, 221 Parameterized surface canvas, 215 Parametric representation of a path, 113 Parametric representation of a surface canvas, 215 Partition of a bounded plane domain, 161 Partition of a bounded space domain, 199 Partition of a compact interval, 28 Piecewise smooth curve, 118 Polar coordinates, 176 Polygonal plane set, 49 Primitive function, 1 Primitives of rational functions, 11 Primitives of trigonometric functions, 18 Principal value of improper integral in Cauchy sense, 74

R Rectifiable curve (path), 118, 123

259 Reducing a double integral to an iterated simple integral, 168, 170 Refinement of a partition of a bounded plane domain, 161 Refinement of a partition of a compact interval, 28 Repeated Simpson formula, 68 Repeated trapezoidal formula, 66 Riemann criterion for integrability for function of one variable, 38 Riemann criterion for integrability for function of two variables, 165 Riemann-Green formula, 185 Riemann sum for function of one variable, 34 Riemann sums for function of three variables, 201 Riemann sums for function of two variables, 163 S Second Comparison test for improper integral, 78 Set of area zero, 56 Set of volume zero, 199 Shell, 233 Simple curve, 118 Simple path, 115 Simpson formula (rule), 67 Singular point of a path, 115 Smooth curve, 118 Smooth path, 116 Solid, 211 Spherical coordinates, 206 Squareable bounded plane set, 51 Star domain, 154 Stokes formula, 247 Support of a curve, 118 Support of a parametrized path, 113 Support of a parametrized surface canvas, 221 Surface, 220 Surface integral of the first kind, 230 Surface integral of the second kind, 238 T Table of antiderivatives of the basic elementary functions, 3 Tangent plane, 222 Total flux of a vector field through a surface, 239 Trapezoidal formula (rule), 64

260 Triple integral, 201, 202 U Union of two paths, 118 Unit normal vector, 222 V Volume of a curvilinear cylinder, 198

Index Volume of a space domain, 210 Volume of a space figure, 195

W Weierstrass criterion for uniform convergence of an improper integral, 94