integral calculus problems -...

INTEGRAL CALCULUS

PROBLEMS

CLAUDIA TIMOFTE

INTEGRAL CALCULUS

PROBLEMS

To my students

Preface

The present book is a collection of problems in integral calculus. The book is

based on some lectures I delivered for a number of years at the Faculty of Physics of

the University of Bucharest and covers the curriculum on integral calculus for the

students of the first year of this faculty.

Each chapter contains a brief review of the corresponding theoretical results,

worked out examples and proposed problems. Since the ”learning-by-doing” method

is a successful one, the student is encouraged to solve as many exercises as possible.

The basic prerequisites for studying integral calculus using this book are undergrad-

uate courses in linear algebra and differential calculus.

It is my hope that this book will serve as an useful outlook for the students of

the first year of the Faculty of Physics of the University of Bucharest.

I would like to thank to my students for their continuous questions, comments

and suggestions, which helped me to improve the content of these notes.

Claudia Timofte

7

Contents

1 Definite Integrals. Improper Integrals 11

1.1 Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2 Binomial Differentials. Chebyshev’s Substitutions . . . . . . . . . . . 17

1.3 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.4 Integrals depending on a parameter . . . . . . . . . . . . . . . . . . . 29

2 Line Integrals 37

3 Multiple Integrals 49

3.1 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.2 Applications of double integrals to problems of mechanics . . . . . . 55

3.3 Triple and n-fold multiple integrals . . . . . . . . . . . . . . . . . . . 58

4 Surface Integrals 67

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

9

Chapter 1

Definite Integrals. ImproperIntegrals

1.1 Definite Integrals

Definition 1.1 Let I ⊆ R be an interval and f : I → R. The function f possesses

an antiderivative on I if there exists F : I → R such that F is derivable and F ′(x) =

f(x), ∀x ∈ I. The function F is called a primitive function or an antiderivative

for f .

Remark 1.2 If F is a primitive of the function f , then there exists a real constant

C such that the indefinite integral of f can be written as∫f(x) dx = F (x) + C. (1.1)

Remark 1.3 Every continuous function possesses an infinite number of antideriva-

tives.

Remark 1.4 If F1 and F2 are antiderivatives of the function f , then there exists

C ∈ R such that:

F1(x) = F2(x) + C, ∀x ∈ I. (1.2)

11

12 INTEGRAL CALCULUS

Theorem 1.5 Let I ⊆ R be and interval. If the functions f, g : I → R have an-

tiderivatives on I and α, β ∈ R, then the function αf+βg possesses an antiderivative

on I and ∫(αf(x) + βg(x)) dx = α

∫f(x) dx+ β

∫g(x) dx.

Theorem 1.6 If the function f : I → R has an antiderivative on I, then f has the

Darboux property on I.

Remark 1.7 If f : J → R possesses an antiderivative F and φ : I → J is derivable,

then (f φ)φ ′ possesses an antiderivative and∫f(φ(t))φ ′(t) dt = F φ+ C, C ∈ R. (1.3)

The following integration by change of variable (substitution) formula holds true:∫f(x) dx =

∫f(φ(t))φ ′(t)dt = F φ+ C, C ∈ R

Also, we recall the integration by parts formula∫f(x)g ′(x) dx = f(x)g(x)−

∫f ′(x)g(x) dx.

The table of antiderivatives for some elementary functions!

Integration of rational and irrational expressions. Euler’s substitutions! Inte-

grating trigonometric expressions. Trigonometric substitutions!

Example 1.8 The value of the indefinite integral

I =

∫ (2x+ x4

)dx

is

I = x2 +x5

5+ C, C ∈ R.

IMPROPER INTEGRALS 13

Example 1.9 Evaluate the indefinite integral

I =

∫ (3x2 +

√x− x3

)dx.

We have

I = x3 +2

3x3/2 − x4

4+ C, C ∈ R.

Exercise 1.10 Evaluate the following integral:

I =

∫e√x

√xdx.

Solution. We have:

I = 2

∫e√x 1

2√xdx = 2

∫e√x(√x) ′

dx = e√x + C, C ∈ R.

Exercise 1.11 Evaluate the following indefinite integrals:

a)

∫x3 + x+ 1

x(x2 + 1)dx;

b)

∫x√1− x2 dx;

c)

∫1

(x2 + 1)3/2dx.

Exercise 1.12 Evaluate the following indefinite integrals:

a)

∫x+ 1

exdx;

b)

∫x(x2 − 1

)5dx;

c)

∫1

x2x+ 2dx.

Let us revisit now the definition of a definite integral.

Definition 1.13 Let f : [a, b] → R and let us divide the interval [a, b] into n parts

with the aid of the division points a = x0 < x1 < · · · < xn = b. Then, ∆ =


(x0, . . . , xn) is called a partition of the interval [a, b]. Also, let us choose an arbitrary

point ξi in each subinterval [xi−1, xi], i = 1, n and let

σ∆(f, ξ) =

n∑i=1

f(ξi)(xi − xi−1)

be the Riemann sum associated to the partition ∆.

The function f is called integrable on the interval [a, b] if there exists I ∈ R such

that for any ε > 0 there exists ηε > 0 such that for any partition ∆ = (x0, . . . , xn)

with the norm max1≤i≤n

(xi − xi−1) < ηε and for any choice of the points ξi we have:

| σ∆(f, ξ)− I |< ε.

The number I is called the definite integral of the function f on [a, b].

We shall use the following notation:

I =

∫ b

af(x) dx.

Remark 1.14 If f is integrable on [a, b], then the number I is unique.

Proposition 1.15 If f is integrable on [a, b], then

∫ b

af(x) dx = −

∫ a

bf(x) dx

and ∫ a

af(x) dx = 0.

Proposition 1.16 If f, g : [a, b] → R are integrable functions and α, β ∈ C, then

αf + βg is integrable and

∫ b

a(αf(x) + βg(x)) dx = α

∫ b

af(x) dx+ β

∫ b

ag(x) dx.


Proposition 1.17 If F is a primitive function for f : [a, b] → R, then

∫ b

af(x) dx = F (b)− F (a) (Leibniz −Newton).

Proposition 1.18 If f : [a, b] → R is integrable and f(x) ≥ 0 for any x ∈ [a, b],

then

∫ b

af(x) dx ≥ 0.

Proposition 1.19 If f is integrable and there exit m,M ∈ R such that m ≤ f(x) ≤M , for any x ∈ [a, b], then

m(b− a) ≤∫ b

af(x) dx ≤M(b− a).

Proposition 1.20 If f : [a, b] → R is integrable, then |f | is integrable, as well.

Proposition 1.21 If f and |f | are integrable, then

∣∣∣ ∫ b

af(x) dx

∣∣∣ ≤ ∫ b

a|f(x)|dx.

Proposition 1.22

∫ b

af(x) dx =

∫ c

af(x) dx+

∫ b

cf(x) dx.

Proposition 1.23 If the function f : [a, b] → R is continuous on [a, b], then the

function F : [a, b] → R, F (x) =∫ x

af(t) dt is a primitive function for f .

Proposition 1.24 If the function f : [a, b] → R is continuous on [a, b], then there

exists at least one point ξ ∈ [a, b] such that

∫ b

af(x) dx = f(ξ)(b− a).

Theorem 1.25 Any continuous function f : [a, b] → R is integrable on [a, b].

Theorem 1.26 Any integrable function f : [a, b] → R is bounded on [a, b].


Theorem 1.27 (Lebesgue) f : [a, b] → R is integrable on [a, b] if and only if f

is bounded on [a, b] and continuous on [a, b], except possibly on a set of Lebesgue

measure zero.

Theorem 1.28 Any integrable function f : [a, b] → R is bounded on [a, b].

Theorem 1.29 Any monotone bounded function f : [a, b] → R is integrable on

[a, b].

Proposition 1.30 (Integration by parts) If f, g : [a, b] → R are of class C1 on [a, b],

then ∫ b

af(x)g ′(x) dx = f(b)g(b)− f(a)g(a)−

∫ b

af ′(x)g(x) dx.

Proposition 1.31 If φ : [a, b] → R, f : φ([a, b]) → R, f continuous, φ ∈ C1, then

∫ b

af(φ(t))φ ′(t) dt =

∫ φ(b)

φ(a)f(x) dx.

Proposition 1.32 If f(x) ≤ g(x) on [a, b], then

∫ b

af(x) dx ≤

∫ b

ag(x) dx.

Proposition 1.33 Let f, g : [a, b] → R be Riemann integrable functions over [a, b].

Then, the product fg : [a, b] → R is Riemann integrable over [a, b].

Proposition 1.34 Let f, g : [a, b] → R be Riemann integrable functions over [a, b].

Then, minf, g and maxf, g are Riemann integrable over [a, b].

Exercise 1.35 Compute the following integral:

I =

∫ π

0x sinxdx.

Solution. Integrating by parts, we get

I = −∫ π

0x (cosx) ′ dx = −x (cosx)

∣∣π0+

∫ π

0cosx dx = π.


1.2 Binomial Differentials. Chebyshev’s Substitutions

Let us consider integrals of the following form:

I =

∫xα(axβ + b)γ dx, (1.4)

where a, b ∈ R and α, β, γ ∈ Q.

Remark 1.36 An expression of the form xα(axβ + b)γ dx is called a binomial dif-

ferential. Chebyshev proved a remarkable theorem, stating that there are only three

cases for which such an expression can be integrated in terms of elementary functions

(the corresponding integrals can be rationalized by means of proper substitutions).

Let

α =m

′

m, β =

n′

n, γ =

p′

p, m, n, p ∈ N∗. (1.5)

We have the following three cases.

Case 1. γ ∈ Z. If we denote by z the least common multiple of the denominators

of the rational numbers α and β, then the change of variables

z√x = t

leads us to a rational integral.

Case 2.α+ 1

β∈ Z. If x = 0, then the change of variables

axβ + b = tp

leads us to an expression that can be integrated in terms of elementary functions.

Case 3.α+ 1

β+ γ ∈ Z. If x = 0, then the change of variables

axβ + b = tpxβ

leads us to a rational integral.


Exercise 1.37 Compute the following integrals:

1) I =

∫ 4

1

√x(x2 + 1)2 dx.

2) I =

∫ √x (2 3

√x+ 1)2 dx.

3) I =

∫ 2

1

√x√

3 4√x+ 1

dx.

4) I =

∫ 4

1

√1 +

√xdx.

5) I =

∫x−1/2 (1− x−4/3)−5/8 dx.


I =

∫ ∞

0xn e−x

2dx.


I =

∫ 1

0

√x− x2 dx.


I =

∫1

x2√x2 − 1

dx.

1.3 Improper Integrals

Up to now, when speaking of definite integrals we assumed that the interval of inte-

gration [a, b] was compact and the integrand was bounded on [a, b]. However, it often

becomes necessary to consider integrals over noncompact sets and also integrals for

which the integrand is unbounded. Such integrals are called improper or generalized

integrals.


Let us consider now integrals of the form

I =

∫ b

af(x) dx

I. Improper integrals of the first kind. In this case, we have

b− a = ∞, f is bounded.

We include here integrals of the form

I =

∫ ∞

af(x) dx,

I =

∫ b

−∞f(x) dx

or

I =

∫ ∞

−∞f(x) dx,

with a, b ∈ R.

II. Improper integrals of the second kind. In this case,

b− a <∞, f is unbounded on the interval of integration.

III. Improper integrals of the third kind. In this situation,

b− a = ∞, f is not bounded on the entire interval of integration.

Let us recall here some results about improper integrals of the first kind.

Definition 1.41 Let f : [a,∞) → R such that f is integrable on any closed interval

[a, b], a < b. If there exists the limit

limb→∞

∫ b

af(x) dx <∞,


then the integral I =

∫ ∞

af(x) dx is called convergent and

∫ ∞

af(x) dx = lim

b→∞

∫ b

af(x) dx.

In the opposite case, we shall say that I =

∫ ∞

af(x) dx is divergent.

Theorem 1.42 (Cauchy) Let f : [a,∞) → R be a continuous function. The

integral I =

∫ ∞

af(x) dx is convergent if and only if ∀ ε > 0∃Mε > a such that

∣∣∣ ∫ β

αf(x) dx

∣∣∣ < ε, ∀α, β, β > α > Mε.

Definition 1.43 The improper integral I =

∫ ∞

af(x) dx is called absolutely conver-

gent if the integral I =

∫ ∞

a| f(x) | dx is convergent.

Theorem 1.44 Let f : [a,∞) → R be a continuous function. If the improper

integral I =

∫ ∞

af(x) dx is absolutely convergent, then I is convergent.

Theorem 1.45 (Comparison Test) Let f, g : [a,∞) → R be continuous functions.

If | f(x) |≤ g(x), ∀x ≥ a and if the improper integral I =

∫ ∞

ag(x) dx is convergent,

then I =

∫ ∞

af(x) dx is absolutely convergent.

Remark 1.46 If f, g are continuous, 0 ≤ f(x) ≤ g(x), ∀x ≥ a and the integral∫ ∞

af(x) dx is divergent, then

∫ ∞

ag(x) dx is divergent, too.

Remark 1.47 If f, g > 0 are integrable on [a, x], ∀x ∈ [a,∞) and there exists

∃L = limx→∞

f(x)

g(x)∈ (0,∞),

then the integrals

∫ ∞

af(x) dx and

∫ ∞

ag(x) dx have the same nature.


Proposition 1.48 If ∃ limt→∞

| f(t) |1/t<∞, then if L < 1, the integral

∫ ∞

af(x) dx

is absolutely convergent and if L > 1, the integral

∫ ∞


Proposition 1.49 If f : [a,∞) → R+ is nonincreasing and n0 = minn ∈ N | a <

n, then the integral

∫ ∞

af(x) dx and the sum

∑n≥n0

f(n) have the same nature.

Proposition 1.50 (P.G. Dirichlet 1805-1859) If f, φ : [a,∞) → R and f is con-

tinuous, φ is nonincreasing, limx→∞

φ(x) = 0 and ∃M > 0 such that

∣∣∣ ∫ x

af(t)dt

∣∣∣ ≤M, ∀x ∈ [a,∞),

then the integral

∫ ∞

af(x)φ(x) dx is convergent.

Proposition 1.51 (Abel) If f, φ : [a,∞) → R and f is continuous, φ is monotone,

limx→∞

φ(x) ∈ R and the integral

∫ ∞

af(x) dx is convergent, then

∫ ∞

af(x)φ(x) dx is

convergent.

Lemma 1.52 If a > 0, then the integral

∫ ∞

a

dx

xαis convergent for α > 1 and

divergent for α ≤ 1.

Theorem 1.53 Let f : [a,∞) → R be a continuous function and F : [a,∞) → R be

a primitive function for f . Then, the improper integral

∫ ∞

af(x) dx is convergent if

and only if ∃ limx→∞

F (x) dx and in this case we have:

∫ ∞

af(x) dx = lim

b→∞F (b)− F (a).

Let us recall now some useful results about improper integrals of the second

kind. If b− a ∈ R, but f is unbounded in the neighbourhood of the points a < x1 <


x2 < · · · < xq < b, then the integral

∫ b

af(x) dx is called an improper integral of the

second kind. Such an integral will be convergent if

∃ limεj ,ε

′j

(

∫ x1−ε1

af(x) dx+

∫ x2−ε2

x1+ε′1

f(x) dx+ · · ·+∫ b

xq+ε′q

f(x) dx).

It suffices then to consider only improper integrals of the form

∫ b

af(x) dx, where

the function f is unbounded in the neighbourhood of the point b.

Definition 1.54 Let f : [a, b) → R such that f is continuous on [a, b) (f is un-

bounded in the neighbourhood of b). If there exists the limit

limc→b

∫ c

af(x) dx <∞,

then the improper integral I =

∫ b

af(x) dx is called convergent and

∫ b

af(x) dx = lim

c→b

∫ c

af(x) dx.

In the opposite case, we shall say that I =

∫ b


Theorem 1.55 (Cauchy) Let f : [a, b) → R be a continuous function. The integral

I =

∫ b

af(x) dx is convergent if and only if ∀ ε > 0∃δε > 0 such that

∣∣∣ ∫ β

αf(x) dx

∣∣∣ < ε, ∀α, β such that b− δε < α < β < b.


∫ b

af(x) dx is called absolutely conver-

gent if the integral I =

∫ b

a| f(x) | dx is convergent.


Theorem 1.57 If the improper integral I =

∫ b

af(x) dx is absolutely convergent,

then I is convergent.

Theorem 1.58 (Comparison Test) Let f, g : [a, b) → R be continuous functions. If

| f(x) |≤ g(x), ∀x ∈ [a, b) and if the improper integral I =

∫ b

ag(x) dx is convergent,

then I =

∫ b

af(x) dx is absolutely convergent.

Remark 1.59 If f, g are continuous, 0 ≤ f(x) ≤ g(x), ∀x ≥ a and

∫ ∞

af(x) dx is

divergent, then

∫ ∞

ag(x) dx is divergent, too.

Lemma 1.60 If b > 0, then the integral

∫ b

0

dx

xβis convergent for β < 1 and diver-

gent for β ≥ 1.

Remark 1.61 Improper integrals of the third kind, i.e. integrals for which b−a = ∞and f is not bounded on the entire interval of integration, can be studied by reducing

them to the previous cases. Indeed, if we consider the integral

I =

∫ ∞

af(x) dx,

where f is continuous on [a, c) and (c,∞), then we can write

I =

∫ ∞

af(x) dx =

∫ b

af(x) dx+

∫ ∞

bf(x) dx, with c < b <∞

and we can use our previous results for deciding upon the nature of the integral I.

Exercise 1.62 Evaluate the integral

I =

∫ ∞

1

1

x2dx.


Solution. We have

I = lima→∞

∫ a

1

1

x2dx = lim

a→∞

(−1

a+

1

1

)= 1.

Thus, the integral is convergent.


I =

∫ ∞

1

1

x2 (1 + ex)dx.

Solution. Since1

x2(1 + ex)≤ 1

x2

and the integral ∫ ∞

1

1

x2dx

is convergent, it follows that the given integral is convergent, as well.


I =

∫ ∞

1

sinx

x3dx.

Exercise 1.65 Prove that the integral

I =

∫ ∞

1

sinx

xdx.

is convergent.

Hint. We use Proposition 1.46, with

φ(x) =1

x

and

f(x) = sinx.



I =

∫ ∞

1

sinx

xdx

is semi-convergent.


I =

∫ ∞

1sin(x2)dx

is convergent.

Hint. Use the change of variables x2 = t.


I =

∫ 1

0

1√xdx.

Solution. We get

I = limb→0+

∫ 1

b

1√x

dx = limb→0+

(2− 2√b) = 2.

Thus, the given improper integral is convergent.


I =

∫ ∞

0xe−x dx.

Solution. We get

I = lima→∞

∫ a

0xe−x dx = lim

a→∞

(−xe−x − e−x

) ∣∣∣a0= −ae−a − e−a + 1 = 1.

So, the improper integral∫∞0 xe−x dx is convergent to 1.



I =

∫ ∞

−∞

1

1 + x2dx.

Solution.

I =

∫ 0

−∞

1

1 + x2dx+

∫ ∞

0

1

1 + x2dx.

Since the integrand is an even function, we have

I = 2

∫ ∞

0

1

1 + x2dx = lim

a→∞2

∫ a

0

1

1 + x2dx = lim

a→∞arctanx

∣∣∣a0= π.

Exercise 1.71 Study the nature of the integral

I =

∫ ∞

1e−x

2dx.

Solution. We note that for x ≥ 1, we have

e−x2 ≤ e−x.

Since the integral

∫ ∞

1e−x dx is convergent, using the comparison test, it follows

that the given integral is convergent.


I =

∫ ∞

e

1

x ln(x)dx.

Solution. We have ∫1

x ln(x)dx = ln(ln(x)) + C.

Then, ∫ ∞

e

1

x ln(x)dx = lim

a→∞ln(ln(x))

∣∣∣ae= ∞,

which implies that the given improper integral is divergent.


Exercise 1.73 Show that the improper integral

I =

∫ 1

0

1√1− x2

dx

is convergent.


I =

∫ 1

0

1√x(1 + ex)

dx.

Hint. Compare with the integral ∫ 1

0

1√x

dx.

Exercise 1.75 Show that the integral

∫ ∞

0

1

xαdx is divergent for all α ∈ R.

Hint. Use Lemma 1.46 and Lemma 1.54.


I =

∫ ∞

1

cos2 x

x2dx

is convergent.

Hint. Compare the given improper integral with the integral

J =

∫ ∞

1

1

x2dx,

which is convergent.

Exercise 1.77 Show that the integral

I =

∫ ∞

0e−x

2/2 dx

is convergent.


Exercise 1.78 Determine whether the improper integral

I =

∫ π/2

0tan(x) dx

is convergent or divergent.

Exercise 1.79 Evaluate the following improper integrals:

a) I =

∫ 5

1

1

(x− 1)1/2dx; b) I =

∫ 5

1

1

(x− 1)3/2dx.

Exercise 1.80 Evaluate the following improper integrals:

a) I =

∫ ∞

1

sin2(x)

1 + x2dx; b) I =

∫ 1

−3

1

(x− 1)(x+ 3)dx.

Let us briefly discuss now the notion of principal value for divergent improper

integrals. We consider the improper integral of the first kind

I =

∫ ∞

−∞f(x) dx, (1.6)

where f is continuous on R.


∫ ∞

−∞f(x) dx is called convergent in the

sense of the principal value (Cauchy) if

∃ lima→∞

∫ a

−af(x) dx = v.p.

∫ ∞

−∞f(x) dx.

Remark 1.82 If the integral I is convergent in the classical sense, then I is con-

vergent in the sense of the principal value. The converse implication is false.

Example 1.83 The integral I =

∫ ∞

−∞xdx is divergent in the classical sense of

convergence, but it is convergent in the sense of principal value and we have

v.p.

∫ ∞

−∞x dx = 0.



∫ b

af(x) dx, where f is continuous on

[a, c) ∪ (c, b), a < c < b is called convergent in the sense of the principal value

(Cauchy) if

∃ limδ→0

(∫ c−δ

af(x) dx+

∫ b

c+δf(x) dx

)= v.p.

∫ b

af(x) dx.

Example 1.85 The integral I =

∫ b

a

1

xdx, a < 0 < b is divergent in the classical

sense of convergence, but it is convergent in the sense of principal value and we have

v.p.

∫ b

a

1

xdx = ln

∣∣∣ ba

∣∣∣.1.4 Integrals depending on a parameter

Definition 1.86 Let f : [a, b]× [c, d] → R be a continuous function and let

g1(x) =

∫ d

cf(x, y) dy

g2(y) =

∫ b

af(x, y) dx

g1 and g2 are called functions defined by integrals which depend on a parameter.

Remark 1.87 If f ∈ C([a, b]× [c, d]), then g1 ∈ C([a, b]), g2 ∈ C([c, d]) and

limx→x0

d∫c

f(x, y) dy =

d∫c

(limx→x0

f(x, y)

)dy ∀x0 ∈ [a, b]

limy→y0

b∫a

f(x, y) dx =

b∫a

(limy→y0

f(x, y)

)dx, ∀y0 ∈ [c, d].

These equalities are no longer true if the above integrals are improper.


Definition 1.88 Let f : [a, b] × [c, d) → R be a continuous function, which is un-

bounded in the neighbourhood of y = d. The improper integral∫ d

cf(x, y) dy = lim

d1→d

∫ d1

cf(x, y) dy

is called uniformly convergent if ∀ε > 0, ∃δε > 0 such that∣∣∣ ∫ β

αf(x, y) dy

∣∣∣ < ε,

for all α < β, α, β ∈ (d− δε, d).

Definition 1.89 Let f : [a, b]× [c,∞) → R be a continuous function. The improper

integral ∫ ∞

cf(x, y) dy = lim

d→∞

∫ d

cf(x, y) dy

is called uniformly convergent if ∀ε > 0, ∃Mε > c such that∣∣∣ ∫ β

αf(x, y) dy

∣∣∣ < ε,

for all β > α > Mε, ∀x ∈ [a, b].

Theorem 1.90 a) Let f : [a, b]×[c, d) → R be a continuous function. If | f(x, y) |≤g(y), ∀(x, y) ∈ [a, b] × [c, d), where g : [c, d) → R+ is a continuous function and∫ d

cg(y) dy is convergent, then

∫ d

cf(x, y) dy is uniformly convergent.

b) Let f : [a, b] × [c,∞) → R be a continuous function. If | f(x, y) |≤g(y), ∀(x, y) ∈ [a, b] × [c,∞), where g : [c,∞) → R+ is a continuous function and∫ ∞

cg(y) dy is convergent, then

∫ ∞

cf(x, y) dy is uniformly convergent.

Theorem 1.91 Let f : [a, b]×[c,∞) → R be a continuous function. If

∫ ∞

cf(x, y) dy

is uniformly convergent, then the function g : [a, b] → R, g(x) =

∫ ∞

cf(x, y) dy is

continuous and

limx→x0

(∫ ∞

cf(x, y) dy

)=

∫ ∞

c

(limx→x0

f(x, y)

)dy, ∀x0 ∈ [a, b].


Theorem 1.92 (Leibniz) Let f : [a, b] × [c, d] → R be a continuous function. If

there exists∂f

∂xand it is continuous, then the function g : [a, b] → R,

g(x) =

∫ d

cf(x, y) dy

is derivable and

g ′(x) =

∫ d

c

∂f

∂x(x, y) dy.

d

dx

(∫ d

cf(x, y) dy

)=

∫ d

c

∂f

∂x(x, y) dy.

Theorem 1.93 (Leibniz) Let f : [a, b]×[c, d] → R be a continuous function. If there

exists∂f

∂xand it is continuous and if φ,ψ : [a, b] → [c, d] are derivable on (a, b), then

d

dx

(∫ ψ(x)

φ(x)f(x, y) dy

)=

∫ ψ(x)

φ(x)

∂f

∂x(x, y) dy + f(x, ψ(x))ψ ′(x)− f(x, φ(x))φ ′(x),

∀x ∈ (a, b).

Theorem 1.94 (Leibniz) Let f : [a, b] × [c,∞) → R be a continuous function. If

∃ ∂f∂x

and it is continuous, if

∫ ∞

cf(x, y) dy is convergent and if

∫ ∞

c

∂f

∂x(x, y) dy is

uniformly convergent, then the function g : [a, b] → R,

g(x) =

∫ ∞

cf(x, y) dy

is derivable and

d

dx

(∫ ∞

cf(x, y) dy

)=

∫ ∞

c

∂f

∂x(x, y) dy.

Remark 1.95 A similar result holds true for the case of improper integrals of the

second kind.


Exercise 1.96 Let

g(x) =

∫ x2

0arctan

( yx2

)dy.

Compute g ′(x).


I(a) =

∫ π/2

0

1

sinxln

1 + a sinx

1− a sinxdx,

for |a| < 1.


I(a) =

∫ π/2

0

ln (1 + a cosx)

cosxdx,

for |a| < 1.


I(a) =

∫ π/2

0

arctan (a tanx)

tanxdx,

for a > 0.


I(a) =

∫ π/2

0ln(a2 − sin2 x

)dx,

for a > 1.


I(a) =

∫ 1

0

arctan (a x)

x√1− x2

dx,

for a ≥ 0.


Euler’s Functions

Γ(p) =

∫ ∞

0e−ttp−1dt, p ∈ R (Gamma function)

B(p, q) =

∫ 1

0tp−1(1− t)q−1dt, p, q ∈ R (Beta function)

B(p, q) =Γ(p)Γ(q)

Γ(p+ q)

Proposition 1.102 The following properties hold true:

1) Γ is defined and continuous for p > 0.

2) B is defined and continuous for p, q > 0.

3) Γ(p) > 0, p > 0.

4) Γ(1) = 1.

5) Γ(p+ 1) = pΓ(p), p > 0.

6) Γ(n+ 1) = n!, n ∈ N.

7) Γ(p)Γ(1− p) =π

sin(πp), 0 < p < 1.

8) Γ(1

2) =

√π.

Remark 1.103 We can extend the notion of a ”factorial” if we put

x! = Γ(x+ 1), ∀x > −1.

Also, the function Γ can be extended for p ∈ R \ 0,−1,−2, · · · if we put

Γ(x) =Γ(x+ n+ 1)

x(x+ 1) · · · (x+ n), ∀n ∈ N.


Proposition 1.104 The following properties for Euler’s Beta function hold true:

1) B(p, q) = B(q, p).

2) B(p, q) =q − 1

p+ q − 1B(p, q − 1), p > 0, q > 1.

3) B(p, q) =(p− 1)(q − 1)

(p+ q − 1)(p+ q − 2)B(p− 1, q − 1), p, q > 1.

4) B(p, q) =

∞∫0

xp−1

(1 + x)p+qdx, p, q > 0.

(Hint: use the change of variables t =x

x+ 1in B(p, q)).

5) B(p, q) =1

(b− a)p+q−1

∫ b

a(x− a)p−1(b− x)q−1 dx, a < b.

(Hint: use the change of variables x = a+ (b− a)t in B(p, q)).

6) B(p, q) = 2

∫ π/2

0cos2p−1 x sin2q−1 xdx.

(Hint: use the change of variables t = cos2 x in B(p, q)).

Exercise 1.105 Test the following improper integrals for convergence:

1) I =

∫ ∞

0

1√x+ 1

dx;

2) I =

∫ ∞

03 ex dx;

3) I =

∫ ∞

1

dx

x5;

4) I =

∫ ∞

1

dx

x4(1 + ex).

Exercise 1.106 Discuss the nature of the following improper integrals:

1) I =

∫ ∞

1

x+ 1√x3

dx;


2) I =

∫ ∞

1

sinx

x5dx;

3) I =

∫ ∞

1

cosx

xdx.

Exercise 1.107 Decide whether the following integral is convergent or divergent:

I =

∫ ∞

−∞

1

x2 + 4dx.

Exercise 1.108 Test the following integrals for convergence:

1) I =

∫ ∞

0sinxdx;

2) I =

∫ 1

0

2√1− x2

dx.

3) I =

∫ ∞

0e−x

2dx.

Exercise 1.109 Discuss the nature of the following improper integrals:

1) I =

∫ 1

0

1

x√xdx;

2) I =

∫ ∞

1

sin y

x2 + y2dy, x ∈ R;

3) I =

∫ 1

0

cosxy√y

dy, x ∈ R.

Exercise 1.110 Using Euler’s functions, compute I =

∫ 1

0

13√1− x3

dx.

Exercise 1.111 Test the following integral for convergence:

I =

∫ ∞

−∞

2x

(1 + x2)2dx.

Exercise 1.112 Using Euler’s functions, evaluate the integral

I =

∫ π/2

0

√tanxdx.


Exercise 1.113 Using Euler’s functions, compute

I =

∫ ∞

0

4√x

(1 + x)2dx.


I =

∫ 1

0

13√1− x3

dx.


I =

∫ 1

−1

(1 + x

1− x

)1/2

dx.

Chapter 2

Line Integrals

Definition 2.1 Let [a, b] ⊆ R. A continuous function γ : [a, b] → Rn is called a

path in Rn.

Remark 2.2 Sometimes, it is convenient to take γ : [0, 1] → Rn.

We shall usually consider the cases n = 2 or n = 3.

Remark 2.3 Let n = 2. We can define a path (a contour) in R2 in two ways:

a) - directly (explicitly ), by giving the direct equation of our path, i.e.

y = φ(x), α ≤ x ≤ β, φ continuous.

b) - by giving the parametric equations of our path:

γ(t) = (φ(t), ψ(t)), t ∈ [a, b], φ, ψ continuous,

i.e. x = φ(t),

y = ψ(t),t ∈ [a, b] ⊆ R.

γ(a) is called the initial point (the beginning point) of the path γ

γ(b) is called the terminal point (the end point) of the path γ

37


Definition 2.4 A path γ : [a, b] → R2 is called closed if γ(a) = γ(b).

Definition 2.5 A path γ : [a, b] → R2 is called smooth if γ is differentiable, γ′is

continuous and γ′(t) = 0. A path γ is called piecewise smooth if it is continuous

and splits into a finite number of smooth pieces with no common interior points.

Definition 2.6 The path γ−1 : [a, b] → Rn,

γ−1(t) = γ(a+ b− t),

is called the inverse or the opposite path of γ.

Usually, if we take γ : [0, 1] → Rn, then γ−1 : [0, 1] → Rn,

γ−1(t) = γ(1− t).

Definition 2.7 Let γ1, γ2 : [0, 1] → Rn be two contours such that γ1(1) = γ2(0). We

shall define their compound path (the sum or the composition of our two contours)

γ : [0, 1] → Rn as being:

γ(t) = (γ1 ∨ γ2)(t) =

γ1(2t), 0 ≤ t ≤ 1

2

γ2(2t− 1),1

2≤ t ≤ 1.

In a similar manner:γ : [a, b+ (d− c)] → Rn,

γ(t) = (γ1 ∨ γ2)(t) =

γ1(t), t ∈ [a, b]

γ2(t− b+ c), t ∈ [b, b+ (d− c)].

Remark 2.8 We note that γ ∨ γ−1 is a closed path.

Definition 2.9 Let γ : [a, b] → Rn and γ : [α, β] → Rn be two smooth contours.

We shall say that γ and γ are equivalent (γ ∼ γ) if there exists φ : [a, b] → [α, β]

bijective, of class C1, with φ′(t) = 0 such that γ = γ φ, i.e.

γ(t) = γ(φ(t)), ∀t ∈ [a, b].

LINE INTEGRALS 39

Remark 2.10 The above relation is an equivalence relation on the set of all the

contours of class C1. Each corresponding equivalence class is called a curve.

Definition 2.11 Let γ : [a, b] → R2 be a smooth path, γ(t) = (φ(t), ψ(t)). The

length of the path γ is:

l(γ) =

∫ b

a

√(φ′(t))2 + (ψ′(t))2 dt.

Example 2.12 The line segment joining the points A(a1, a2) and B(b1, b2) is de-

fined as being: γ : [0, 1] → R2,

γ(t) = (1− t)a+ tb

or

γ(t) = a+ t(b− a).

Obviously, γ is smooth.

Example 2.13 The circle of radius a, centered at the origin, is defined as being:

γ : [0, 2π] → R2,

γ(t) = aeit

or x = a cos t,

y = a sin t, t ∈ [0, 2π].


Example 2.14 The ellipse with semi-axes a and b is defined as being: γ : [0, 2π] →R2,

γ(t) = (a cos t, b sin t)

or x = a cos t,

y = b sin t, t ∈ [0, 2π].



Definition 2.15 Let γ : [a, b] → R2 be a smooth path, γ(t) = (φ(t), ψ(t)), and let

f : γ([a, b]) → R be a continuous function. The line integral of the first kind of the

function f along the path γ is defined as follows:

∫γf(x, y) dl =

∫ b

af(φ(t), ψ(t))

√(φ′(t))2 + (ψ′(t))2 dt.

Sometimes, we use the similar notation:

I =

∫γf ds.

If the path γ is given explicitly by the formula y = φ(x), x ∈ [α, β], then

∫γf(x, y) dl =

∫ β

αf(x, φ(x))

√(1 + φ′(x))2 dx.

Remark 2.16 1) The line integral I, sometimes called the line integral with respect

to the arc length, is independent of the sense in which the curve γ is travelled.

2) The above definition can be obviously extended to the case n = 3 or to the

case of a piecewise smooth curve.

3) Physical interpretation: the line integral of the first kind gives the mass of a

material curve (line) having the linear density f(x, y).

Proposition 2.17 The following properties for the line integral of the first kind

hold true:

1)

∫γ(αf + βg) dl = α

∫γf dl + β

∫γg dl.

2)

∫γ1∨γ2

f dl =

∫γ1

f dl +

∫γ2

f dl.

3) |∫γf dl |≤

∫γ| f | dl.

LINE INTEGRALS 41

4)

∫γ1

fdl =

∫γ2

f dl, γ1 ∼ γ2.

5) (Mean-value formula) If f is continuous along γ, then there exists M∗ on

the support of the path γ such that∫γf(x, y) dl = f(M∗)l(γ).

For a material thread of linear density ρ, we can define the coordinates of its

center of gravity by:

xG =

∫γxρ(x, y) dl∫γρ(x, y) dl

, yG =

∫γyρ(x, y) dl∫γρ(x, y) dl

.

Also, we can define the moments of inertia:

IO =

∫γ(x2 + y2)ρ(x, y) dl,

IOx =

∫γy2ρ(x, y) dl, IOy =

∫γx2ρ(x, y) dl.

Definition 2.18 Let γ : [a, b] → R2 be a smooth path, γ(t) = (φ(t), ψ(t)), and let

F : γ([a, b]) → R2, f(x, y) = (P (x, y), Q(x, y)) be a continuous function. The line

integral of the second kind of the function F along the path γ is defined as follows:

I =

∫γP (x, y) dx+Q(x, y) dy =

∫ b

a[P (φ(t), ψ(t))φ

′(t) +Q(φ(t), ψ(t))ψ

′(t)] dt.

Similar notation: I =

∫γ

−→F · d−→r .

Remark 2.19 1) The line integral I depends on the sense in which the curve γ is

travelled and reverses the sign: ∫γ−1

= −∫γ.


2) The above definition can be obviously extended to the case n = 3 or to the

case of a piecewise smooth curve.

3) Physical interpretation: the line integral of the second kind gives the work

done by a force with components P and Q on a particle to move it along the path γ

from A to B:

I =

∫AB

P dx+Qdy.

4) If γ ∼ γ with the same orientation, i.e. γ = γ φ, with φ′ > 0, then∫γP dx+Qdy =

∫γP dx+Q dy.

Remark 2.20 In fact, the expression P dx+Qdy is a differential form.

Definition 2.21 Let U ∈ Rn, U open. A map ω : U → L(Rn,R) of class Ck is

called a differential form of degree one and class Ck.

Dk1(U) = Ck(U,L(Rn,R))

ω =n∑i=1

ωi pi, ωi ∈ Ck(U,R).

For n = 2, we get

ω = P dx+Qdy.

Definition 2.22 A form ω is called closed if∂ωi∂xj

=∂ωj∂xi

, ∀ i = j.

Definition 2.23 A form ω is called exact or a total differential form if ∃f ∈Ck+1(U) such that ω = df .

Remark 2.24 If k ≥ 1, then any exact form is closed. The converse implication is

false.

Definition 2.25 A curve without self-intersection points is called simple.

LINE INTEGRALS 43

Definition 2.26 Let D ⊆ R2 be a domain and γ1, γ2 : [0, 1] → D be two contours

in D. γ1 can be continuously deformed into γ2 without leaving D if there exists

φ : [0, 1]×[0, 1] → D continuous such that φ(0, t) = γ1(t) and φ(1, t) = γ2(t), for any

t ∈ [0, 1]. In fact, for any s ∈ [0, 1], we get a curve γs : [0, 1] → D, γs(t) = φ(s, t).

Remark 2.27 A domain D is said to be singly connected if any piecewise smooth,

non self-intersecting, closed curve in D bounds a domain all of whose points are

in D. Equivalently, D is simply (singly) connected if any simple, piecewise smooth,

closed path can be continuously deformed into a point (into the constant path γ0(t) =

γ0(a), ∀t ∈ [a, b]), without leaving D.

Theorem 2.28 (Poincare) If D ⊆ R2 is a simply connected domain, then any

closed form in D is a total differential form.

Theorem 2.29 Let D ⊆ R2 be a simply connected domain and let P,Q ∈ C1(D).

The following statements are equivalent:

a) The form ω = P dx+Qdy is an exact form.

b) The form ω = P dx+Qdy is closed.

c) For any piecewise smooth closed curve γ in D,

∫γω = 0.

d)

∫γω is independent of the path γ joining two fixed points A and B (the

integral is path-independent, i.e. it depends only on γ(a) and γ(b)).

From the point of view of mechanics, the path-independence of the integral

∫γω

in the case in which P and Q are the projections of a force field F means that the

work of this field does not depend on the shape of the trajectory of the motion and

depends solely on its initial and terminal points.

If ω = P dx+Q dy is an exact form in a simply connected domain D, it follows

that there exists F such that ω = dF . Obviously, P =∂F

∂xand Q =

∂F

∂y. The

question arising now is the following one: how can we determine F? On one hand,


choosing an arbitrary fixed point (x0, y0) and using Leibniz-Newton’s formula, we

get:

I =

∫ (x,y)

(x0,y0)P dx+Qdy =

∫ (x,y)

(x0,y0)dF = F (x, y)− F (x0, y0). (2.1)

On the other hand, using the property of path-independence for such integrals, we

can choose as a contour of integration joining the points (x0, y0) and (x, y) a path

consisting of two line segments parallel to the coordinate axes. We get:

I =

∫ (x,y)

(x0,y0)P dx+Qdy =

∫ (x,y0)

(x0,y0)P (x, y0) dx+

∫ (x,y)

(x,y0)Q(x, y) dy. (2.2)

Therefore, from (2.1)-(2.2) we obtain:

F (x, y) =

∫ x

x0

P (x, y0) dx+

∫ y

y0

Q(x, y) dy + F (x0, y0). (2.3)

In a similar way, we can obtain:

F (x, y) =

∫ x

x0

P (x, y) dx+

∫ y

y0

Q(x0, y) dy + F (x0, y0). (2.4)

So, given P and Q we can determine the potential function F associated to the

differential form ω = P dx+Qdy (up to an additive constant). In practical calcula-

tions, the arbitrary initial point (x0, y0) should be chosen such that the element of

integration becomes as simple as possible.

I =

∫ (x,y)

(x0,y0)ω =

∫ (x,y)

(x0,y0)dF = F (x, y)− F (x0, y0).

The work in a potential field of force is equal to the potential difference between the

initial and terminal points.

Example 2.30 For the gravitational field, we have P = 0 and Q = mg. Then,

F (x, y) = mgy + C.

We shall give now Green’s theorem which provides a formula connecting a line

integral over a closed contour with a double integral over the domain bounded by

the contour.

LINE INTEGRALS 45

Theorem 2.31 Let D ⊆ R2 be a compact domain bounded by a finite number of

smooth simple paths. If D is simple with respect to both axes of coordinates and if

F : D → R2, F (x, y) = (P (x, y), Q(x, y)) is a C1 function, then∫∂D

P dx+Q dy =

∫∫D

(∂Q

∂x− ∂P

∂y

)dx dy. (2.5)

Green’s formula holds true for any counter-clockwise oriented parametrization of the

path ∂D.

Green’s formula provides a convenient way for evaluating some difficult to com-

pute line integrals by transforming them into double integrals which may be easier

to compute.

Remark 2.32 A similar formula can be obtained for domains in R2 which are

unions of finite numbers of simple domains (without common interior points) or

even for general domains, for which their bounding contour is a piecewise smooth

simple closed curve. But in this last case, the proof involves more sophisticated

considerations.

Also, Green’s formula is valid even for the case in which P,Q are continuous and

∃ ∂P∂y

,∂Q

∂xcontinuous on D.

Green’s theorem has many profound implications. One of them is related to the

notion of conservative vector fields. A vector field is conservative (i.e. is given by a

gradient) if and only if its curl is identically zero. Thus, from Green’s formula, we

see that the line integral along a closed curve of a conservative vector field is equal

to zero.

Remark 2.33 We can compute the area of a compact domain by evaluating a line

integral of the second kind along its boundary:

µ(D) =1

2

∫∂D

−y dx+ x dy.

Exercise 2.34 Compute the integral

I =

∫γ−x2y dx+ xy2 dy,


where γ is the circle x2 + y2 = 9.


I =

∫γy2 dx+ x2 dy,

where γ is the positively oriented boundary of the domain

D = (x, y) ∈ R2 | x2 + y2 ≤ 4, x, y ≥ 0.


I =

∫γ(x+ y) dx+ xy dy,

where γ is the positively oriented boundary of the domain

D = (x, y) ∈ R2 | x2

4+ y2 ≤ 1, y ≥ 0.


I =

∫γ(2x+ 3 y) dl,

where γ is the segment connecting the points (1, 1) and (3, 4).


I =

∫γ(x2 + y2) dl,

where γ is the segment joining the points (1, 0) and (2, 3).

Exercise 2.39 For the vector field

−→F =

1

2(−y, x2),

LINE INTEGRALS 47

compute the line integral

I =

∫γ

−→F · d−→r

along the square path, traveled in the counter clockwise direction, composed of the

four line segments joining the points (0, 0) and (2, 0), (2, 0) and (2, 2), (2, 2) and

(0, 2), and, respectively, (2, 2) and (0, 0).

Exercise 2.40 Compute the line integral

∫γ(x+ y − z) dl, where γ is the line seg-

ment from (0, 0, 0) to (1, 1, 1).

Exercise 2.41 Determine whether the following fields are conservative and, if so,

find the corresponding scalar potentials:

1)−→F = (y, x+ z,−y);

2)−→F = (z + y, z, x+ y).


I =

∫γy dx+ (x+ z) dy − y dz

where γ is the line segment joining the points (0, 1, 1) and (1, 1, 0).

Exercise 2.43 Prove that the following integral is path-independent and calculate

its value:

I =

∫γ2 cos y dx+

(1

y− 2x sin y

)dy +

1

zdz.

where γ is a smooth path joining the points (0, 2, 1) and (1, π/2, 2).

Exercise 2.44 Compute

I =

∫ (2,2,2)

(−2,−2,−2)

2x dx+ 2y dy + 2z dz

x2 + y2 + z2.

Chapter 3

Multiple Integrals

3.1 Double Integrals

We introduce now the definition of a double integral. Let A = [a, b] × [c, d] be a

rectangle. We shall divide the closed interval [a, b] into n subdivisions using the

points a = x0 < x1 < · · · < xn = b. Also, let us divide the closed interval [c, d]

into m subdivisions using the points c = y0 < y1 < · · · < ym = d. Let Ajk =

[xj , xj+1]× [yk, yk+1], j = 0, n− 1, k = 0,m− 1. All these n×m rectangles form a

subdivision P of the rectangle A. The diagonal djk =√

(xj+1 − xj)2 + (yk+1 − yk)2

is called the diameter of the rectangle Ajk and the greatest diameter of all the

rectangles Ajk, denoted by ∆ = ∥P∥, is called the norm of the division P .

Definition 3.1 Let A = [a, b]× [c, d] be a rectangle, f : A→ R and P a subdivision

of the rectangle A. For each subrectangle Ajk of P we shall choose an intermediate

point (ξjk, ηjk) ∈ Ajk. The sum

S(f, P, (ξjk, ηjk)) =n−1∑j=0

m−1∑k=0

f(ξjk, ηjk)(xj+1 − xj)(yk+1 − yk)

is called the integral sum or the Riemann sum associated with f, P and the given

choice of the intermediate points (ξjk, ηjk).

49


Definition 3.2 Let A = [a, b] × [c, d] be a rectangle and f : A → R. The function

f is called integrable (in the sense of Riemann) if there exists I ∈ R such that

∀ε > 0, ∃ δε > 0 such that for any division P of the rectangle A, with ∥P∥ < δε and

for any system of intermediate points (ξjk, ηjk) we have:

| S(f, P, (ξjk, ηjk))− I |< ε.

I is called the integral of f on A.


I =

∫Af =

∫∫Af(x, y) dxdy.

If f is integrable,then I is unique and is called the double integral of f over A.

Exactly like for a single definite integral, it can be established that any integrable

function f on a rectangle A is bounded on that rectangle and, therefore, we shall

consider here only bounded functions (for the case of proper integrals).

Proposition 3.3 Let A = [a, b]× [c, d] be a rectangle and f : A→ R. The function

f is integrable (in the sense of Riemann) on A if and only if ∀ε > 0, ∃ δε > 0 such

that for any divisions P, P′of the rectangle A, with ∥P∥ < δε,

∣∣∣P ′∣∣∣ < δε and for

any intermediate points (ξjk, ηjk) and (ξ′jk, η

′jk) we have:

| S(f, P, (ξjk, ηjk))− S(f, P′, (ξ

′jk, η

′jk)) |< ε.

Theorem 3.4 Any continuous function f : A→ R is integrable on A.

Proposition 3.5 The following property holds true:∫∫A(αf(x, y) + βg(x, y)) dxdy = α

∫∫Af(x, y) dxdy + β

∫∫Ag(x, y) dx dy.

Proposition 3.6 If f is integrable and f(x, y) ≥ 0, then∫∫Af(x, y) dxdy ≥ 0.

MULTIPLE INTEGRALS 51

Proposition 3.7 If f is integrable and m ≤ f(x, y) ≤M on A, then

m(b− a)(d− c) ≤∫∫

Af(x, y) dxdy ≤M(b− a)(d− c).

Proposition 3.8 If f and | f | are integrable, then

∣∣∣ ∫∫Af(x, y) dxdy

∣∣∣ ≤ ∫∫A| f(x, y) | dxdy.

Proposition 3.9 If the function f : A → R is continuous on A, then there exists

at least one point (ξ, η) ∈ A such that∫∫Af(x, y) dxdy = f((ξ, η))(b− a)(d− c).

Proposition 3.10 If f(x, y) ≤ g(x, y) on A, then∫∫Af(x, y) dxdy ≤

∫∫Ag(x, y) dx dy.

Remark 3.11 If fn are integrable functions fn : A → R and fn → f uniformly on

A, then f is integrable on A and∫∫Af = lim

n→∞

∫∫Afn.

Remark 3.12 If the domain of integration is split into two domains D1 and D2

having no interior points in common, then∫∫Df(x, y) dx dy =

∫∫D1

f(x, y) dx dy +

∫∫D2

f(x, y) dxdy.

Remark 3.13 The area of the rectangle A is µ(A) = (b− a)(d− c).

Definition 3.14 A set M ⊆ R2 is said to be a set of zero area if it can be covered

with a number of rectangles having the sum of areas less than any ε > 0.


Remark 3.15 1) Any finite set of points has zero area.

2) Any sequence of points from R2 has zero area.

3) The image of a path of class C1 is of zero area.

4) If f is continuous on A \∪mj=1 γj, where γj are the images of m contours of

class C1, then f is integrable on the rectangle A.

5) Let f : A → R and γ ⊂ A be the image of a path of class C1. If f(x, y) =

0, ∀(x, y) not belonging to γ, then f is integrable on A and∫∫Af = 0.

6) If f, g : A→ R, f(x, y) = g(x, y), for (x, y) ∈ A \ γ and f is integrable, then

g is integrable and ∫∫Af =

∫∫Ag.

Evaluating Double Integrals

Proposition 3.16 Let f : A → R, f(x, y) = g(x)h(y). If g : [a, b] → R and

h : [c, d] → R are integrable functions, then f is integrable on A = [a, b]× [c, d] and∫Af =

∫ b

a

(∫ d

cf(x, y) dy

)dx =

∫ d

c

(∫ b

af(x, y) dx

)dy.

Remark 3.17 1) If P (x, y) is a polynomial, then∫AP =

∫ b

a

(∫ d

cP (x, y) dy

)dx.

2) If f is continuous, then there exists a sequence of polynomials (Pn)n such

that Pn → f uniformly on A and∫Af =

∫ b

a

(∫ d

cf(x, y) dy

)dx =

∫ d

c

(∫ b

af(x, y) dx

)dy.


Double integrals on arbitrary domains

Definition 3.18 Let D ⊆ R2 be a bounded domain. The function χD : R2 → R

defined by

χD(x, y) =

1, if (x, y) ∈ D

0, if (x, y) /∈ D

is called the characteristic function of the domain D.

Definition 3.19 Let D ⊆ R2 be a bounded domain and let f : D → R. The

function f is called integrable on D if for a rectangle A with D ⊆ A, the function

fχD : A→ R is integrable on A. Here,

f =

f, on D

0, elsewhere .

Then, by definition ∫Df =

∫AfχD.

The above definition is independent on the rectangle A.

Theorem 3.20 Let D ⊆ R2 be a compact set. If its boundary consists of the union

of the images of a finite number of paths of class C1 and f : D → R is a continuous

function, then f is integrable on D.

Remark 3.21 One has ∫Df =

∫Int(D)

f.

Remark 3.22 Let D ⊆ R2 be a compact set. If f is bounded on D and has in D

points of discontinuity contained only on the union of the images of a finite number

of paths of class C1, then f is integrable on D and∫Df =

∫Int(D)

f.


Theorem 3.23 Let D ⊆ R2 be a compact set,

D = D1

∪D2 · · ·

∪Dp,

with

Int(Di)∩Int(Dj) = ∅, i = j.

Let us suppose that ∂Dj ⊆ R2 consists of the union of the images of a finite number

of paths of class C1. If f : D → R is bounded and continuous on each Int(Dj), then

f is integrable on D and ∫Df =

p∑j=1

∫Dj

f.

Definition 3.24 a) A domain D ⊆ R2 is called simple with respect to the Ox axis

if ∃φ,ψ : [a, b] → R, continuous on [a, b] and of class C1 on (a, b), such that

D = (x, y) | a ≤ x ≤ b, φ(x) ≤ y ≤ ψ(x).

b) A domain D ⊆ R2 is called simple with respect to the Oy axis if ∃ g, h : [c, d] → R,

continuous on [c, d] and of class C1 on (c, d), such that

D = (x, y) | c ≤ y ≤ d, g(y) ≤ x ≤ h(y).

Theorem 3.25 a) Let D ⊆ R2 be a simple domain with respect to the Ox axis,

i.e.

D = (x, y) | a ≤ x ≤ b, φ(x) ≤ y ≤ ψ(x).

If f : D → R is continuous, then f is integrable on D and

∫Df =

∫ b

a(

∫ ψ(x)

φ(x)f(x, y) dy) dx.

b) Let D ⊆ R2 be a simple domain with respect to the Oy axis, i.e.

D = (x, y) | c ≤ y ≤ d, g(y) ≤ x ≤ h(y).


If f : D → R is continuous, then f is integrable on D and∫Df =

∫ d

c

(∫ h(y)

g(y)f(x, y) dx

)dy.

Remark 3.26 If D = D1∪D2 · · ·

∪Dp, with Dj simple, then∫

Df =

p∑j=1

∫Dj

f.

Definition 3.27 Let D ⊆ R2 be a bounded domain such that its boundary consists

of the union of the images of a finite number of paths of class C1. We shall define

the area or the measure of D as being:

area(D) = |D| =∫ ∫

D1 dxdy.

If D is simple, we get

area(D) =

∫ ∫D1 dxdy =

∫ b

a(ψ(x)− φ(x)) dx

or

area(D) =

∫ ∫D1 dx dy =

∫ d

c(h(y)− g(y)) dy.

We shall also denote the area of a domain D by µ(D).

3.2 Applications of double integrals to problems of me-chanics

1) Let us consider a nonhomogeneous plate (lamina) D ⊆ R2 having the density

ρ(x, y) (the plate is supposed to be sufficiently thin so that the variation of the mass

density in the direction perpendicular to the plane Oxy can be neglected). The mass

of this plate is:

M =

∫∫Dρ(x, y) dxdy.


2) The coordinates of the center of gravity of the plate D are defined by:

xG =

∫∫Dxρ(x, y) dxdy∫ ∫Dρ(x, y) dxdy

and

yG =

∫ ∫Dyρ(x, y) dx dy∫∫

Dρ(x, y) dxdy

.

3) The moments of inertia of the plate D with respect to the coordinate axes and

to the origin are defined by:

MOx =

∫∫Dy2ρ(x, y) dx dy,

MOy =

∫∫Dx2ρ(x, y) dxdy

and

MO =

∫∫D(x2 + y2)ρ(x, y) dx dy.

Change of variables in the double integral

An appropriate change of variables may essentially simplify the computation of

the given integrals. For instance, sometimes a good change of variables may lead to

constant limits of integration in the transformed integral.

Theorem 3.28 Let D ⊆ R2 be a bounded domain whose boundary is the union of

the images of a finite number of paths of class C1. Let T : D → R2 be an injective

function such that T ∈ C1 and detT′ = 0 on D. If f : T (D) → R is integrable, then∫

T (D)f =

∫D(f T )|detT ′|. (3.1)


Remark 3.29 1) If T is linear, then T′= T .

2) The above theorem holds true for f continuous on T (D).

Example 3.30 (Polar coordinates) Let us apply the general formula (3.1) to the

change of variables from Cartesian coordinates x and y to polar coordinates ρ and

θ, respectively: x = ρ cos θ,

y = ρ sin θ,(3.2)

where ρ ≥ 0 and θ ∈ [0, 2π].

It is not difficult to see that the map T : (0,∞)× (0, 2π) → R2 \ [0,∞)× 0 is a

C1-diffeomorphism. In this case,

detT ′ =

∣∣∣∣∣∣cos θ − ρ sin θ

sin θ ρ cos θ

∣∣∣∣∣∣ = ρ.

|J | = ρ = 0

and ∫∫Ωf(x, y) dx dy =

∫∫Ω∗f(ρ cos θ, ρ sin θ)ρdρ dθ.

Notice that Ω∗ is a rectangle.

Example 3.31 (Elliptic coordinates) We apply the general formula (3.1) to the

change of variables from Cartesian coordinates x and y to elliptic coordinates ρ and

θ, respectively: x = a ρ cos θ,

y = b ρ sin θ,(3.3)

where a, b > 0 are the semi-axes of our ellipse.

In this case, it is easy to see that

J = a b ρ.


Example 3.32 (Generalized polar coordinates) Let us apply the general formula

(3.1) to the change of variables from Cartesian coordinates x and y to generalized

polar coordinates ρ and θ, respectively:x = a ρ cosα θ,

y = b ρ sinα θ.(3.4)

In this case, it is easy to see that

J = αa b ρ cosα−1 θ sinα−1 θ.

3.3 Triple and n-fold multiple integrals

All the above considerations can be easily extended for triple and n-fold multiple

integrals. We can start by considering a n-dimensional rectangular parallelepiped,

a simple domain, and, then, a general one.

I =

∫∫∫Df(x, y, z) dxdy dz.

V ol(D) = µ(D) =

∫∫∫D1 dxdy dz.

Definition 3.33 A domain D ⊆ R3 is called simple with respect to the Oxy plane

(or to the Oz axis) if ∃ g, h : Ω ⊆ R2 → R, continuous on Ω and of class C1 on

Int(Ω), such that

D = (x, y, z) | g(x, y) ≤ z ≤ h(x, y), (x, y) ∈ Ω,

where Ω is a compact domain in R2 with the boundary made up by a finite number

of images of smooth curves.


If the domain D ⊆ R3 is simple with respect to the Oxy plane, f is continuous on

D and ∃∫ h(x,y)

g(x,y)f(x, y, z) dz, then

I =

∫∫∫Df(x, y, z) dxdy dz =

∫∫Ω

(∫ h(x,y)

g(x,y)f(x, y, z) dz

)dxdy.

∫T (D)

f =

∫D(f T )

∣∣∣ detT ′∣∣∣. (3.5)

Example 3.34 (Spherical coordinates) Let us apply the general formula (3.5) to

the change of variables from Cartesian coordinates x, y, z to spherical coordinates

ρ, θ, φ: x = ρ sin θ cosφ,

y = ρ sin θ sinφ,

z = ρ cos θ,

(3.6)

where ρ ≥ 0, θ ∈ [0, π] and φ ∈ [0, 2π].

We shall work on the open set (0,∞)× (0, π)× (0, 2π). In this case,

| J |= ρ2 sin θ = 0.

Example 3.35 (Ellipsoidal coordinates) We apply now the general formula (3.5)

to the change of variables from Cartesian coordinates x, y, z to elliptic coordinates

ρ, θ, φ: x = a ρ sin θ cosφ,

y = b ρ sin θ sinφ,

z = c ρ cos θ,

(3.7)

where a, b, c > 0 are the semi-axes of our ellipsoid. In this case, it is easy to see that

J = a b c ρ2 sin θ.


Example 3.36 (Cylindrical coordinates) Letx = ρ cos θ,

y = ρ sin θ,

z = z.

(3.8)

We shall work on (0,∞)× (0, 2π)× R. It is easy to see that

J = ρ.

Applications of triple integrals

1) Let us consider a nonhomogeneous body D ⊆ R3 having the volume density

ρ(x, y, z) The mass of this body is:

M =

∫∫∫Dρ(x, y, z) dxdy dz.

2) The coordinates of the center of gravity of the body D are defined by:

xG =

∫∫∫Dxρ(x, y, z) dxdy dz∫∫∫

Dρ(x, y, z) dxdy dz

,

yG =

∫∫∫Dyρ(x, y, z) dxdy dz∫∫∫


,

and

zG =

∫∫∫Dzρ(x, y, z) dxdy dz∫∫∫


,


3) The moments of inertia of the body D with respect to the coordinate axes,

planes and to the origin are defined by:

IOx =

∫∫∫D(y2 + z2)ρ(x, y, z) dx dy dz,

IOy =

∫∫∫D(x2 + z2)ρ(x, y, z) dx dy dz,

IOz =

∫∫∫D(x2 + y2)ρ(x, y, z) dxdy dz,

IOxy =

∫∫∫Dz2ρ(x, y, z) dx dy dz,

IOxz =

∫∫∫Dy2ρ(x, y, z) dxdy dz,

IOyz =

∫∫∫Dx2ρ(x, y, z) dxdy dz,

IO =

∫∫∫D(x2 + y2 + z2)ρ(x, y, z) dxdy dz.

Remark 3.37 Similar results hold true for n-fold multiple integrals.

Exercise 3.38 Evaluate each of the following iterated integrals:

a)

∫ 2

0

∫ 1

−1(6x2 + 2y) dy dx;

b)

∫ π/2

0

∫ π

0(x sin y + y cosx) dy dx.

Exercise 3.39 Draw the domain of integration and, then, change the order of in-

tegration in the integral

I =

∫ 1

0

∫ x

x2(x+ y) dy dx.

Exercise 3.40 Sketch the region of integration and change the order of integration

in the integral

I =

∫ 2

−2

∫ 8−x2

x2(2x+ y) dy dx.


Exercise 3.41 Let

D = (x, y) ∈ R2 | 1 ≤ x2 + y2 ≤ 4, y ≤ x, x ≥ 0.

Draw the domain D and compute the integral

I =

∫∫Dy dx dy.

Exercise 3.42 Compute the volume of the solid D lying inside the sphere x2+y2+

z2 = 64 and outside the sphere x2 + y2 + z2 = 1.

Exercise 3.43 Compute the volume of the solidD bounded by x2+y2 = 1, x2+y2 =

9, z = 0 and z = 4. Draw the domain of integration.

Exercise 3.44 Evaluate the following iterated integral, by changing the order of

integration (also, sketch the region D):

I =

∫ 8

0

(∫ 3√y

0ex

4dx

)dy.

Hint. We have

I =

∫ 2

0

(∫ x3

0ex

4dy

)dx =

∫ 2

0x3ex

4dx =

1

4

(e16 − 1

).

Exercise 3.45 Using polar coordinates, evaluate the area of the finite domain

bounded by the Oy-axis, the line y = 3, and a quarter of the circle of radius 3

with center at the point (3, 0). Sketch the domain of integration.

Exercise 3.46 Find the volume of the solid ball x2 + y2 + z2 = 4.

Exercise 3.47 Evaluate the following triple integral:

I =

∫∫∫Dz dxdy dz,

where D is the volume inside the sphere x2+y2+z2 = 1 and the cone z =√x2 + y2.


Hint. Use spherical coordinates.

Exercise 3.48 Find the volume of a solid hemisphere of radius 3 and center at

(0, 0, 0) lying in the upper half space z ≥ 0.

Exercise 3.49 Compute the following integrals:

1) I =

∫∫D

x2

1 + y2dxdy, (D) :

x = 0, x = 1

y = 0, y = 1

2) I =

∫∫D

x2

y2dxdy, (D) :

y = x,

y =1

x,

x ∈ [1, 2]

3) I =

∫∫D(x2 + y) dxdy, (D) :

y = x2,

y2 = x

4) I =

∫∫D(x− y) dx dy, (D) :

y = 2− x2,

y = 2x− 1

Exercise 3.50 Evaluate the following double integrals:

1) I =

∫∫D

dxdy, D = (x, y) ∈ R2 | x2 + y2 ≤ 4, x2 +y2

4≥ 1, x ≥ 0.

2) I =

∫∫D

dxdy,

D = (x, y) ∈ R2 | x2 + y2 ≤ 2, x ≤ y2, x ≥ −y2, y ≤ 0.

3) Change the order of integration in

I =

∫ 2

−6dx

∫ 2−x

x2

4−1f(x, y) dy.


4) I =

∫∫Dxy2 dxdy, (D) :

y2 = 2px,

x =p

2, p > 0

Exercise 3.51 Compute the following double integrals:

1) I =

∫∫D(x2 + y2)2 dxdy, D = (x, y) ∈ R2 | x2 + y2 ≤ 1, x, y ≥ 0;

2) µ(D) =

∫∫D

dxdy,

D = (x, y) ∈ R2 | a2 ≤ x2 + y2 ≤ b2, y ≥ x, b > a > 0;

3) µ(D) =

∫∫D

dxdy,

D = (x, y) ∈ R2 | xy ≥ a2, xy ≤ 2a2, y ≥ x, y ≤ 2x, x, y > 0, a > 1.

Exercise 3.52 Compute the area of the following domain:

D = (x, y) ∈ R2 | x+ y ≥ a, x+ y ≤ b, y ≥ αx, y ≤ βx,

0 < α < β, 0 < a < b.

Exercise 3.53 Evaluate the following integrals:

1) I =

∫∫De−(x+y)2 dxdy, (D) :

x+ y = 1

x = 0, y = 0;

2) I =

∫∫D

√1− x2

a2− y2

b2dxdy,

D = (x, y) ∈ R2 | x2

a2+y2

b2≤ 1, x, y ≥ 0, a, b > 0;


3) I =

∫∫D(x2 + y2) dxdy, (D) :

x2 + y2 = a2

y = x√3, y

√3 = x

;

4) I =

∫∫Dsin√x2 + y2 dx dy,

D = (x, y) ∈ R2 | π2 ≤ x2 + y2 ≤ 4π2.

Exercise 3.54 Compute the area of the following domain:

D = (x, y) ∈ R2 | (x+ y)4 ≤ ax2y, x, y ≥ 0, a > 0.

Exercise 3.55 Compute the area of the following domain (Bernoulli’s lemniscate):

D = (x, y) ∈ R2 | (x2 + y2)2 ≤ x2 − y2.

Exercise 3.56 Compute the area of the following domains:

D = (x, y) ∈ R2 | (x2 + y2)2 ≤ a2x2 − b2y2, a, b > 0;

D = (x, y) ∈ R2 | (x2

a2+y2

b2)2 ≤ x2

a2− y2

b2, x ≥ 0, a, b > 0.

Exercise 3.57 Find the volume of the following domain:

D = (x, y, z) ∈ R3 | x2 + y2 + z2 ≤ a2, a > 0.


I =

∫∫D(x2 + y2)2 dxdy,

where

D = (x, y) ∈ R2 | x2 + y2 ≤ 4, x ≥ 0.



I =

∫∫D(x+ y) dxdy,

where

D = (x, y) ∈ R2 | 1 ≤ x2 + y2 ≤ 4, y ≥ 0.

Chapter 4

Surface Integrals

Definition 4.1 Let D ⊆ R2 be a bounded domain and f, g, h : D → R, f, g, h ∈ C1.

The set of all the triplets p = (f, g, h) is called a smooth sheet in R3.

f = f(u, v),

g = g(u, v),

h = h(u, v),

(u, v) ∈ D ⊆ R2.

Remark 4.2 A curve was a class of equivalence of smooth paths. A surface will be

a class of equivalence of smooth sheets.

Definition 4.3 Two smooth sheets are equivalent (p ∼ p) if there exists a bijection

of class C1, Φ : D → D, Φ = (φ,ψ) such that

∂(φ,ψ)

∂(u, v)=

∣∣∣∣∣∣∣∣∣∂φ

∂u

∂φ

∂v

∂ψ

∂u

∂ψ

∂v

∣∣∣∣∣∣∣∣∣ = 0

and p = p Φ, i.e.(f(u, v), g(u, v), h(u, v)) =

= (f(φ(u, v), ψ(u, v)), g(φ(u, v), ψ(u, v)), h(φ(u, v), ψ(u, v))).

67


Remark 4.4 The above relationship is an equivalence relation on the set of all the

smooth sheets. Each corresponding equivalence class will be called a surface.

Definition 4.5 The image of a (smooth) sheet in R3 is called a surface in R3, i.e.

S = (x, y, z) ∈ R3 | x = f(u, v), y = g(u, v), z = h(u, v), (u, v) ∈ D ⊆ R2. (4.1)

Example 4.6 An ellipsoid is defined by the parametric equations:f(θ, φ) = a sin θ cosφ,

g(θ, φ) = b sin θ sinφ,

h(θ, φ) = c cos θ,

0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π.

Definition 4.7 A path on the surface S is defined by the functions

φ : [a, b] → R, ψ : [a, b] → R such that, for all t ∈ [a, b], we have (φ(t), ψ(t)) ∈ D.

The compound functions

t 7→ f(φ(t), ψ(t)),

t 7→ g(φ(t), ψ(t)),

t 7→ h(φ(t), ψ(t))

will define a path in R3. If we choose φ(t) = u0, ψ(t) = t and φ(t) = t, ψ(t) = v0,

for any t ∈ [a, b], we get particular paths on S, corresponding to parallels to the axes

of coordinates in the plane (u, v). These paths (curves) are called coordinate curves

on S. The tangent vectors to these curves at t = u0 and, respectively, at t = v0 are

denoted by −→τ u and −→τ v:

−→τ u(u0, v0) = (∂f

∂u(u0, v0),

∂g

∂u(u0, v0),

∂h

∂u(u0, v0))

−→τ v(u0, v0) = (∂f

∂v(u0, v0),

∂g

∂v(u0, v0),

∂h

∂v(u0, v0))

SURFACE INTEGRALS 69

The plane determined by these two vectors is called the tangent plane of S at the

point P0 = (f(u0, v0), g(u0, v0), h(u0, v0)). The tangent plane is defined only if −→τ u×−→τ v = 0, i.e. | −→τ u ×−→τ v |> 0.

The normal vector at S is collinear with the vector product:

−→τ u ×−→τ v =

∣∣∣∣∣∣∣∣∣∣∣∣∣

−→i

−→j

−→k

∂f

∂u

∂g

∂u

∂h

∂u

∂f

∂v

∂g

∂v

∂h

∂v

∣∣∣∣∣∣∣∣∣∣∣∣∣=

=∂(g, h)

∂(u, v)(u0, v0)

−→i +

∂(h, f)

∂(u, v)(u0, v0)

−→j +

∂(f, g)

∂(u, v)(u0, v0)

−→k .


A(u, v) =∂(g, h)

∂(u, v)(u, v),

B(u, v) =∂(h, f)

∂(u, v)(u, v),

C(u, v) =∂(f, g)

∂(u, v)(u, v).

A2 +B2 + C2 = 0

We shall also use the notation:

E(u, v) =

(∂f

∂u

)2

+

(∂g

∂u

)2

+

(∂h

∂u

)2

,

F (u, v) =∂f

∂u

∂f

∂v+∂g

∂u

∂g

∂v+∂h

∂u

∂h

∂v,

G(u, v) =

(∂f

∂v

)2

+

(∂g

∂v

)2

+

(∂h

∂v

)2

.


Exercise 4.8 Prove that

A2 +B2 + C2 = EG− F 2 (Lagrange).

Hint. Use | a× b |2 + | a · b |2=| a |2| b |2.

The normal unit vector at the surface S has the coordinates:

−→ν ± = ±(

A√A2 +B2 + C2

,B√

A2 +B2 + C2,

C√A2 +B2 + C2

).

Remark 4.9 A surface can be also given by the equation z = h(x, y), , (x, y) ∈ Ω.

Then, the natural parametric representation of S will be:x = u

y = v

z = h(u, v).

(u, v) ∈ Ω. (4.2)

Definition 4.10 The area of the surface S given by the functions f, g, h is defined

as being:

A(S) =

∫ ∫D

√A2(u, v) +B2(u, v) + C2(u, v) du dv, (4.3)

or, equivalently, by:

A(S) =

∫ ∫D

√EG− F 2 du dv. (4.4)

Example 4.11 The area of the sphere of equation x2 + y2 + z2 = R2 is

Area(S) = 4πR2.

Remark 4.12 If the surface S is defined by z = h(x, y), (x, y) ∈ Ω and if we denote

p =∂h

∂xand q =

∂h

∂y, then

A(S) =

∫ ∫Ω

√1 + p2 + q2 dxdy.


Remark 4.13 Two equivalent smooth sheets have equal areas.

Definition 4.14 Let S be a smooth surface in R3 defined by 4.1. and F : S → R be

a continuous function. The surface integral of the first kind is defined as being:∫∫SF (x, y, z) dσ =

∫∫DF (f(u, v), g(u, v), h(u, v))

√A2 +B2 + C2 du dv. (4.5)

Also, we can write∫∫SF dσ =

∫∫ΩF (x, y, h(x, y))

√1 + p2 + q2 dx dy

Remark 4.15 If S ∼ S, then

∫∫SF dσ =

∫∫SF dσ.

Remark 4.16 If S is a material surface having the density F (x, y, z), then the

surface integral of the first kind gives the mass of this material surface.

Remark 4.17 Two sheets are equivalent (p ∼ p) with the same orientation if

∂(φ,ψ)

∂(u, v)> 0

and with changing the orientation if

∂(φ,ψ)

∂(u, v)< 0.

Definition 4.18 Let S be a smooth surface and let−→F : S → R3,

−→F = (P,Q,R) be

a continuous function. The surface integral of the second kind is defined as being:∫∫S

−→F · −→ν dσ =

∫∫SP dy dz +Q dz dx+R dx dy =

=

∫∫D[P (f(u, v), g(u, v), h(u, v))A(u, v) +Q(f(u, v), g(u, v), h(u, v))B(u, v)+

+R(f(u, v), g(u, v), h(u, v))C(u, v)] du dv.


Remark 4.19

∫∫S

−→F · −→ν dσ represents the flux of the vector field

−→F through the

surface S.

In fact, the aim of a surface integral of the second kind is to determine the flux of

a given vector field through a surface.

Remark 4.20 If S ∼ S with the same orientation, then∫∫S

−→F · −→ν dσ =

∫∫S

−→F · −→ν dσ.

If we consider different orientation for our surface, then the above integral differs

as sign.

Theorem 4.21 (Gauss-Ostrogradsky) Let D3 be a simple bounded domain in R3,−→F = (P,Q,R),

−→F ∈ C1(D3),

−→F ∈ C0(D3). If S = ∂D3, then:∫∫

SP dy dz +Q dz dx+R dx dy =

∫∫∫D3

(∂P

∂x+∂Q

∂y+∂R

∂z

)dx dy dz.

∫∫S

−→F · −→ν dσ =

∫∫∫D3

div−→F dx dy dz (flux− divergenceformula)

Remark 4.22 If we take P =1

3x, Q =

1

3y, R =

1

3z, then

V ol(D3) =1

3

∫∫Sxdy dz + y dz dx+ z dxdy.

Remark 4.23 If we take−→F = gradU , then∫∫

SgradU · −→ν dσ =

∫∫∫D3

∆U dx dy dz.

Remark 4.24 ∫∫S

−→ν ×−→F dσ =

∫∫∫D3

rot−→F dxdy dz.


Theorem 4.25 (Stokes) Let S be a (piecewise) smooth oriented surface whose bound-

ary γ is a (piecewise) smooth path and let−→F : Ω → R3,

−→F = (P,Q,R) be a function

of class C1 defined on a domain Ω containing S. Then,∫γP dx+Q dy +R dz =

∫∫S

(∂R

∂y− ∂Q

∂z

)dy dz +

(∂P

∂z− ∂R

∂x

)dz dx+

(∂Q

∂x− ∂P

∂y

)dxdy.

In fact, we heve ∫∂S

−→F · d−→r =

∫∫Scurl

−→F · −→ν dσ

Remark 4.26 This formula is independent of the surface S having the boundary γ.

Exercise 4.27 Think about the particular consequences of this theorem for the

case in which−→F is defined on a simply connected domain.

Exercise 4.28 Find the area of the surface obtained by cutting the hemisphere

x2 + y2 + z2 = 4, for z ≥ 0, with the cylinder x2 + y2 = 1.

Exercise 4.29 Find the exterior flux of the vector field−→F = (xz, y, 1) across the

surface of the upper cap cut from the sphere x2+y2+ z2 ≤ 36 with the plane z = 3.

Exercise 4.30 Find a parametrization of the following surface:

S = (x, y, z) ∈ R3 | z =√

1− (x2 + y2).

Exercise 4.31 Find a parametrization of the cone

S = (x, y, z) ∈ R3 | z = 3√x2 + y2, x2 + y2 ≤ 9.

Exercise 4.32 Find a parametrization of the surface

S = (x, y, z) ∈ R3 | x2 + y2 = 9, 0 ≤ z ≤ 4.


Exercise 4.33 Find a parametrization of the part of the hemisphere x2+y2+z2 =

36, z ≥ 0, that lies inside the cylinder x2 + y2 = 6y.

Exercise 4.34 Compute the area of the surface S obtained by cutting the cylinder

x2 + y2 = 9 with z = 0 and z = 5.

Exercise 4.35 Compute

I =

∫∫Sz dσ,

where

S = (x, y, z) ∈ R3 | z = 4(x2 + y2), x2 + y2 ≤ 1.


I =

∫∫S

−→F · −→ν dσ,

where S is the unit sphere x2 + y2 + z2 = 1, −→ν is the unit outward normal to S and

−→F (x, y, z) = (1, 0, 0).

Exercise 4.37 Using Stokes formula, compute

I =

∫∫S(∇×−→

F ) · −→ν dσ,

where S is the surface of the sphere x2 + y2 + z2 = 1, with z ≤ 0, and

−→F = (y, x, xyz).

Exercise 4.38 Let−→F (x, y, z) = −→r /r3, where −→r = (x, y, z) and r2 = x2 + y2 + z2.

Prove that the flux of this vector through any closed surface S is 0 if the surface S

does not enclose the origin (0, 0, 0).

Hint. Use Gauss theorem. Notice that−→F is defined (and smooth) on the punctured

space R3 \ (0, 0, 0), which is simply connected. Therefore, since curl−→F = 0, it

follows that−→F is conservative. Moreover, a scalar potential is

U(x, y, z) =1

2ln(x2 + y2 + z2).


Exercise 4.39 Calculate the integral:

I =

∫γ

−yx2 + y2

dx+x

x2 + y2dy,

where γ is the circle x2 + y2 = 1 counterclockwise oriented.


I =

∫x2+y2=r2

−yx2 + y2

dx+x

x2 + y2dy,

where γ is the counterclockwise oriented circle x2 + y2 = r2, with r > 0, has the

same value for any r.

Hint. Use Green’s formula.

Exercise 4.41 Check if the vector field

−→F (x, y, z) = (3x2z, z2, x3 + 2yz), (x, y, z) ∈ R3,

is conservative or not.

Hint. The vector field−→F is conservative, since it is the gradient of the scalar function

U(x, y, z) = x3z + yz2.

Bibliography

[1] M. Abramowitz, I.E. Stegun, Handbook of Mathematical Functions with

Formulas, Graphs, and Mathematical Tables, Dover Publications, Inc., New

York, 1977.

[2] G. Arfken, H. Weber, Mathematical Methods for Physicists, Elsevier Aca-

demic Press, 2005.

[3] P. Bamberg, S. Sternberg, A Course in Mathematics for Students of

Physics, Cambridge University Press, 1990.

[4] G. N. Berman, A Problem Book in Mathematical Analysis, MIR Publishers,

Moscow, 1981.

[5] N. Boboc, Mathematical Analysis, Tipografia Universitatii Bucuresti, 1988

(in Romanian).

[6] A. M. Bruckner, J. B. Bruckner, B. S. Thomson, Real Analysis,

Prentice- Hall, 1997.

[7] Ya.S. Bugrov, S.M. Nikolsky, Differential Equations, Multiple Integrals,

Series, Theory of Functions of Complex Variable, Mir Publishers, Moscow,

1983.

[8] H. Cartan Calcul differentiel, Hermann, Paris, 1967.

[9] K. Ciesielski Set Theory for the Working Mathematician, Cambridge Uni-

versity Press, 1997.

77


[10] I. Colojoara, Mathematical Analysis, Editura Didactica si Pedagogica,

Bucharest, 1983 (in Romanian).

[11] N. Cotfas, L. A. Cotfas, Elements of Mathematical Analysis, Editura Uni-

versitatii din Bucuresti, 2010 (in Romanian).

[12] R. Courant, Differential and Integral Calculus, Wiley, New York, 1992.

[13] R. Courant, D. Hilbert, Methods of Mathematical Physics, Wiley, New

York, 1989.

[14] L. Elsgolts, Differential Equations and the Calculus of Variations, Mir Pub-

lishers, Moscow, 1980.

[15] D. Flondor, N. Donciu, Algebra and Mathematical Analysis. Problem Book,

Editura Didactica si Pedagogica, Bucharest, 1979 (in Romanian).

[16] D. Flondor, O. Stanasila, Lessons of Mathematical Analysis, Editura

ALL, Bucharest, 1993 (in Romanian).

[17] A. Halanay, V. Olariu, S. Turbatu, Mathematical Analysis, Editura

Didactica si Pedagogica, Bucharest, 1983 (in Romanian).

[18] P. Halmos, Naive set theory, Princeton, NJ: D. Van Nostrand Company,

1960; reprinted by Springer-Verlag, New York, 1974.

[19] O. Knill, Multivariable Calculus, Math 21a, Harvard University, 2004.

[20] S. Lang, Analysis I, Addison-Wesley Publ. Co., Reading, Massachusetts,

1968.

[21] L.H. Loomis, S. Sternberg, Advanced Calculus, Addison-Wesley Reading,

1968.

[22] C.P. Niculescu, Fundamentals of Mathematical Analysis, Vol. I, Editura

Academiei Romane, Bucharest, 1996 (in Romanian).

BIBLIOGRAPHY 79

[23] S.M. Nikolsky, A Course of Mathematical Analysis, Mir Publishers,

Moscow, 1981.

[24] V. Olariu, Mathematical Analysis, Editura Didactica si Pedagogica,


[25] M. Rosculet, Mathematical Analysis, Editura Didactica si Pedagogica,


[26] W. Rudin, Principles of Mathematical Analysis, McGraw-Hill, New York,

1964.

[27] L. Schwarz, Cours d’analyse, 2 tomes, Hermann Paris, 1967.

[28] Gh. Siretchi, Series. Set Theory. Elementary Functions, Tipografia Univer-

sitatii Bucuresti, 1984 (in Romanian).

[29] Gh. Siretchi, General Topology, Tipografia Universitatii Bucuresti, 1983 (in

Romanian).

[30] Gh. Siretchi, Differential Calculus, Tipografia Universitatii Bucuresti, 1983

(in Romanian).

[31] O. Stanasila, Mathematical Analysis, Editura Didactica si Pedagogica,


[32] P. Szekeres, A Course in Modern Mathematical Physics, Cambridge Uni-

versity Press, 2006.

[33] D. Stefanescu, Real Analysis, Tipografia Universitatii Bucuresti, 1990 (in

Romanian).

[34] C. Timofte, Differential Calculus, Editura Universitatii din Bucuresti, 2009.

[35] S. Turbatu, D. Stefanescu, Mathematical Analysis. Problem Book, Ti-

pografia Universitatii Bucuresti, 1985 (in Romanian).

integral calculus problems -...

Documents