Hassani S. Mathematical Methods: For Students of Physics and Related Fields
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92 Integration: Formalism
3.3.1 Reduction to Single Integrals
Most integrals encountered in physics are multidimensional. Thus, it is impor-
tant to know how to evaluate multiple integrals. Let us concentrate on triple
integration, and for definiteness, let us assume that the integration variables
are actual coordinates in a Cartesian coordinate system.
10
The most general
integral, namely Equation (3.4), will then be rewritten as
##
Ω
f(r, r
) dQ(r
) ≡
##
Ω
f(r,x
,y
,z
) dQ(x
,y
,z
)
=
##
V
#
f(r,x
,y
,z
)ρ
Q
(x
,y
,z
) dx
dy
dz
,
where we have reexpressed dQ in terms of some density. The region of integra-
tion V may have to be divided into a number of other more easily integrable
regions. However, in most applications, by a good choice of the order of inte-
gration, one can avoid such division. Let us assume that by integrating the
z
variable first, we will not need to divide the region. The z
integration is a
single integral and is done by keeping x
and y
constant. To find the upper
limit of this integral, we pick an arbitrary point
11
in the region, fix its first
two coordinates, move “up” until we hit the boundary of V at a point. The
third coordinate of this boundary point,whenexpressedintermsofx
and
y
, will be the upper limit of the z
integration. The lower limit is obtained
similarly. In most cases, V is bounded by a given upper surface of the form
z = g(x, y), and a lower surface of the form z = h(x, y) as shown in Figure 3.3.
Figure 3.3: The limits of the first integration of a triple integral are defined by two
surfaces.
10
Recall that the integration variables, although considered as “coordinates of a point,”
need not be an actual geometric point in space. They could, for instance, be a set of
thermodynamical variables describing a thermodynamical system.
11
A common mistake at this stage is to pick a special point. To make sure that you have
picked an arbitrary point, go through the following process using the point chosen, then
pick a different point, go through the process and see if you obtain the same result for the
upper and lower limits of the integral.
3.3 Guidelines for Calculating Integrals 93
Thus, since the first two coordinates of the boundary points are x
and y
,
the upper limit will be g(x
,y
) and the lower limit will be h(x
,y
). We thus
write the integral as
##
Ω
f(r, r
) dQ(r
)=
##
S
dx
dy
#
g(x
,y
)
h(x
,y
)
f(r,x
,y
,z
)ρ
Q
(x
,y
,z
) dz
,
where S is the projection of V on the xy-plane. For S to be useful, it must
have the following property: Every point of the upper and lower boundaries of
V has one and only one image in S, and no two points of the upper (or lower)
boundary project onto the same point in S. If this property is not fulfilled,
then we must choose another coordinate as our first integration variable, or,
if this does not work, divide the region of integration into pieces for each one
of which this property holds.
Let us assume that the property holds for S, and that we can do the
integral in z
. The result of this integration is a complete elimination of the
z
-coordinate and the reduction of the triple integral down to a double integral.
To be more specific, assume that the primitive of the integrand, as a function
of z
,isF (r,x
,y
,z
), i.e., that
∂F
∂z
= f (r,x
,y
,z
)ρ
Q
(x
,y
,z
).
Then, the z
integration yields
#
g(x
,y
)
h(x
,y
)
f(r,x
,y
,z
)ρ
Q
(x
,y
,z
) dz
= F (r,x
,y
,g(x
,y
)) −F (r,x
,y
,h(x
,y
)) ≡ G(r,x
,y
),
where the last line defines G. The triple integration has now been reduced to
adoubleintegral,andwehave
##
Ω
f(r, r
) dQ(r
)=
##
S
dx
dy
G(r,x
,y
).
We follow the same procedure as above to do the double integral. Once
again, the region of integration S mayhavetobedividedintoanumberof
other more easily integrable regions. However, let us assume that by inte-
grating the x
variable first, we will not need to divide the region. The x
integration is again a single integral and is done by keeping y
constant. To
find the upper limit of this integral, we pick an arbitrary point in S,fixits
second coordinate, and move “to the right” until we hit the boundary of S at
a point. The first coordinate of this boundary point, when expressed in terms
of y
, will be the upper limit of the x
integration. The lower limit is obtained
by “moving to the left.” Once again, in most cases, S is bounded by a given
upper curve of the form x = v(y), and a lower curve of the form x = u(y)(see
Figure 3.4). Thus, since the second coordinate of both boundary points is y
,
94 Integration: Formalism
x
y
v (
y
)
u ( y )
a
b
I
'
'
(x , y
)
'
'
Figure 3.4: The limits of the second integration of a triple integral are defined by two
curves.
the upper limit will be v(y
) and the lower limit will be u(y
). We thus write
the integral as
##
Ω
f(r, r
) dQ(r
)=
#
I
dy
#
v(y
)
u(y
)
dx
G(r,x
,y
),
where I is the projection of S on the y-axis. For I to be useful, it must have
thesamepropertyasS, namely: Every point of the right and left boundaries
of S has one and only one image in I, and no two points of the right (or left)
boundary project onto the same point in I. If this property is not fulfilled,
then we must choose y
as our first integration variable, or, if this does not
work, divide the region of integration into pieces for each one of which this
property holds. Assuming that I satisfies this property, and that the primitive
of the integrand, as a function of x
,isW (r,x
,y
), i.e, that
∂W
∂x
= G(r,x
,y
),
we get
#
v(y
)
u(y
)
G(r,x
,y
) dx
= W (r,v(y
),y
) −W (r,u(y
),y
) ≡ H(r,y
),
where the last line defines H. The triple integration has now been reduced to
a single integral, and we have
##
Ω
f(r, r
) dQ(r
)=
#
I
H(r,y
) dy
=
#
b
a
H(r,y
) dy
,
where a and b are the end points of the interval I.
Sometimes the inverse of the foregoing operation is useful whereby a single
integral is turned into a multiple integral. This happens when the integrand
is given in terms of an integral. To be specific, suppose in the integral
3.3 Guidelines for Calculating Integrals 95
I ≡
#
b
a
g(x) dx,
g(x)isgivenby
g(x)=
#
v
u
h(x, t) dt,
where u and v could be functions of x. Then, the original integral can be
written as
I =
#
b
a
(
#
v
u
h(x, t) dt
)
dx =
#
b
a
#
v
u
h(x, t) dt dx.
Example 3.3.1.
A historical example of this inverse operation is the evaluation
of the integral integral of e
−x
2
over the positive
real line
I ≡
#
∞
0
e
−x
2
dx.
As the reader attempting to solve this integral will soon find out, it is impossible to
find a primitive of the integrand. However, with
I
2
=
#
∞
0
e
−x
2
dx
=I
#
∞
0
e
−y
2
dy
=I
=
#
∞
0
#
∞
0
e
−x
2
e
−y
2
=e
−(x
2
+y
2
)
dx dy
weendupwithanintegrationoverthefirstquadrantofthexy-plane which opens up
the possibility of using other coordinate systems. In polar coordinates, the integrand
becomes e
−r
2
and the Cartesian element of area dx dy becomes the element of area
in polar coordinates, namely rdrdθ. The limits of integration correspond to the
first quadrant, with the range of θ being from 0 to π/2 and that of r being from 0
to infinity. This leads to
I
2
=
#
π/2
0
#
∞
0
e
−r
2
rdrdθ =
#
π/2
0
dθ
=π/2
#
∞
0
e
−r
2
rdr.
=−
1
2
e
−r
2
∞
0
This shows that I
2
= π/4 and, therefore, I =
√
π/2. The reader may verify that
#
∞
−∞
e
−x
2
dx =
√
π (3.26)
by either invoking the evenness of the integrand or starting from scratch as done
above.
3.3.2 Components of Integrals of Vector Functions
Many calculations involve an integrand which is a vector and whose integra-
tion also leads to a vector. Let us write this as
finding the
components of the
vector resulting
from integration
of another vector
F(r)=
##
Ω
A(r, r
) dQ(r
)
=
##
Ω
[A
1
(r, r
)
ˆ
e
1
(r
)+A
2
(r, r
)
ˆ
e
2
(r
)+A
3
(r, r
)
ˆ
e
3
(r
)] dQ(r
),
96 Integration: Formalism
where A
1
, A
2
,andA
3
are the components of the vector A along the mutually
perpendicular unit vectors
ˆ
e
1
,
ˆ
e
2
,and
ˆ
e
3
, respectively.
12
Note that these unit
vectors are, in general, functions of the variables of integration, and that
Box 3.3.1. The geometry of the distribution of the source determines the
most convenient variables of integration (coordinate variables).
To find the component of F(r) along any unit vector
ˆ
e
a
,onesimplytakes
the dot product of F(r)with
ˆ
e
a
.Thus,
F
a
(r) ≡
ˆ
e
a
· F(r)=
ˆ
e
a
·
##
Ω
A(r, r
) dQ(r
)=
##
Ω
[
ˆ
e
a
·A(r, r
)] dQ(r
)
≡
##
Ω
[A
1
(r, r
)f
1
(r
)+A
2
(r, r
)f
2
(r
)+A
3
(r, r
)f
3
(r
)] dQ(r
), (3.27)
where f
1
(r
) ≡
ˆ
e
a
·
ˆ
e
1
, f
2
(r
) ≡
ˆ
e
a
·
ˆ
e
2
,andf
3
(r
) ≡
ˆ
e
a
·
ˆ
e
3
. Once these dot
products are expressed in terms of the variables of integration, the integral
becomes an ordinary integral which, in principle, can be performed using the
guidelines above.
Box 3.3.2. In practice,
ˆ
e
a
is one of the unit vectors of some convenient
coordinate system which need not be the same as the coordinate system
used for integration.
For example, one may be interested in the Cartesian components of the grav-
itational field of a spherical distribution of mass. In that case, one uses spher-
ical coordinates for integration and the unit vectors inside the integral, and
ˆ
e
x
,
ˆ
e
y
,or
ˆ
e
z
for
ˆ
e
a
. We shall illustrate this point extensively with numerous
examples scattered throughout this chapter.
Historical Notes
By the time Newton entered the scene, an immense amount of knowledge of calculus
had accumulated. In his book Lectiones Geometricae, Barrow, for example, shows
a method of finding tangents, theorems on the differentiation of products and quo-
tients of functions, change of variables in a definite integral, and even differentiation
of implicit functions. So, why, one may wonder, is the word “calculus” so much
attached to Newton and Leibniz? The answer is in these two men’s recognition of
the generality of the methods of calculus, and, more importantly, their emphasis on
the newly discovered analytic geometry.
Isaac Newton was born in the hamlet of Woolsthorpe, England, two months
after his father’s death. His mother, in need of help for the management of the fam-
ily farm, wanted Isaac to pursue a farming career. However, Isaac’s uncle persuaded
him to enter Trinity College, Cambridge University. Newton took the entrance exam
12
These unit vectors are usually those of a convenient coordinate system.
3.3 Guidelines for Calculating Integrals 97
and was accepted to the College in 1661 with a deficiency in Euclidean geometry.
Apparently receiving very little stimulation from his teachers, except possibly Bar-
row, he studied Descartes’s G´eom´etrie,aswellastheworksofCopernicus, Kepler,
Galileo, Wallis,andBarrow, by himself.
Isaac Newton
1642–1727
Upon his graduation, Newton had to leave Cambridge due to the widespread
plague in the London area to spend the next eighteen months, during 1665 and
1666, in the quiet of his family farm at Woolsthorpe. These eighteen months were
the most productive of his (as well as any other scientist’s) life. In his own words:
In the beginning of 1665 I found the . . . rule for reducing any dignity
[power] of binomial to a series.
13
The same year, in May, I found the method
of tangents . . . and in November the direct method of Fluxions [the elements
of what is now called differential calculus], and the next year in January had
the theory of Colours, and in May following I had entrance into the inverse
method of Fluxions [integral calculus], and in the same year I began to think
of gravity extending to the orb of the Moon . . . and . . . compared the force
requisite to keep the Moon in her orb with the force of gravity at the surface
of the Earth.
Newton spent the rest of his scientific life developing and refining the ideas
conceived at his family farm. At the age of 26 he became the second Lucasian
professor of mathematics at Cambridge replacing Isaac Barrow who stepped aside
in favor of Newton. At 30 he was elected a Fellow of the Royal Society, the highest
scientific honor in England.
Newton often worked until early morning, kept forgetting to eat his meals, and
when he appeared, once in a while, in the dining hall of the college, his shoes
were down at the heels, stockings untied, and his hair scarcely combed. Being
always absorbed in his thoughts, he was very naive and impractical concerning
daily routines. It is said that once he made a hole in the door of his house for his
cat to come in and out. When the cat had kittens, he added some smaller holes in
the door!
Newton did not have a pleasant personality, and was often involved in contro-
versy with his colleagues. He quarreled bitterly with Robert Hooke (founder of the
theory of elasticity and the discoverer of Hooke’s law) concerning his theory of color
as well as priority in the discovery of the universal law of gravitation. He was also
involved in another priority squabble with the German mathematician Gottfried Leib-
niz over the development of calculus. With Christian Huygens, the Dutch physicist,
he got into an argument over the theory of light. Astronomer John Flamsteed, who
was hardly on speaking terms with Newton, described him as “insidious, ambitious,
excessivelycovetous of praise, and impatient ofcontradictions...agood manatthe
bottom but, through his nature, suspicious.”
De Morgan says that “a morbid fear of opposition from others ruled his whole
life.” Because of this fear of criticism, Newton hesitated to publish his works.
When in 1672 he did publish his theory of light and his philosophy of science, he
was criticized by his contemporaries. Newton decided not to publish in the future,
a decision that had to be abandoned frequently.
His theory of gravity, although germinated in 1665 under the influence of works
by Hooke and Huygens, was not published until much later, partly because of his
fear of criticism. Another reason for this hesitance in publishing this result was his
13
Newton is talking about the binomial theorem here.
98 Integration: Formalism
lack of proof that the gravitational attraction of a solid sphere acts as if the sphere’s
map were concentrated at the center. So, when his friend Edmund Halley urged
him in 1684 to publish his results, he refused. However, in 1685 he showed that
a sphere whose density varies only with distance to the center does in fact attract
particles as though its mass were concentrated at the center, and agreed to write up
his work. Halley then assisted Newton editorially and paid for the publication. The
first edition of Philosophiae Naturalis Principia Mathematica appeared in 1687, and
the Newtonian age began.
3.4 Problems
3.1. Use Equation (3.7) to show that
a
a
f(t) dt =0.
3.2. In Equation (3.8), it was assumed that p<q<r. Show that the
equation holds even if q is not between p and r.
3.3. For each of the following integrals make the given change of variables:
(a)
8
0
tdt, t= y
3
. (b)
1
0
dt
1+t
2
,t=tany, 0 ≤ y ≤ π/2.
(c)
1
0
dt
1+t
,t=lny. (d)
∞
1
tdt
1+t
3
,t=
1
y
.
3.4. By a suitable change of variables, show the following integral identities:
(a)
∞
−∞
dt
(a
2
+t
2
)
3/2
=
2
a
2
π/2
0
cos tdt. (b)
∞
0
dt
(1+t)
2
=
1
0
dt.
3.5. If
g(x)=
#
sin(πx)
x
2
−1
{cos[π(t + x)]}e
−t
4
sin
2
[(π/2) ln(tx
2
+1)]
dt,
find g
(1).
3.6. Suppose that F (x)=
cos x
0
e
xt
2
dt, G(x)=
cos x
0
t
2
e
xt
2
dt,andH(x)=
G(x) − F
(x). Find H(x) in terms of elementary functions. Show that
H(π/4) = e
π/8
/
√
2.
3.7. Suppose that F (x)=
sin x
0
ln(cos
2
x + t
2
+1)dt, G(x)=
sin x
0
(cos
2
x +
t
2
+1)
−1
dt,andH(x)=F
(x)+2sinx cos xG(x). Find H(x)intermsof
elementary functions. Show that H(π/3) = ln 2/2.
3.8. Evaluate the derivative of the following integrals with respect to x at
the given values of x:
(a)
x
0
e
−t
2
dt at x =1. (b)
x
−3
cos tdt at x = π.
(c)
√
cos(x/3)
−∞
e
−t
2
dt at x = π. (d)
x
2
0
cos (
√
s) ds at x = π.
3.4 Problems 99
3.9. Find the numerical value of the derivative of the following two integrals
at x =1:
(a)
ln x
0
e
−x(t
2
−2)
dt. (b)
x
2
+a−1
a
sin
*
πxe
−t
2
2e
−(x
2
+a−1)
2
+
dt.
3.10. Write the derivatives with respect to x of the following integrals in
terms of other integrals. Do not try to evaluate the integrals.
(a)
b
a
ln(1 + sx) ds. (b)
b
a
dt
t
2
+x
2
. (c)
1
0
√
x
2
+ a
2
− 2ax cos tdt.
3.11. Differentiate
∞
−∞
dt/(z + t
2
)=π/
√
z with respect to z to show that
(a)
∞
−∞
dt
(1+t
2
)
2
=
π
2
. (b)
∞
−∞
dt
(1+t
2
)
3
=
3π
8
.
3.12. Using the method of Example 3.2.2, find the following integrals:
(a)
b
a
t
2
sin tdt. (b)
b
a
t
3
sin tdt. (c)
b
a
t
4
sin tdt.
(d)
b
a
t
2
cos tdt. (e)
b
a
t
3
cos tdt. (f)
b
a
t
4
cos tdt.
In each case calculate the primitive of the integrand and verify your answer
by differentiating the primitive.
3.13. Find the integral
Γ(n +1)=
#
∞
0
t
n
e
−t
dt
by first evaluating the integral
#
∞
0
e
−xt
dt
and then differentiating the result n times, and setting x = 1 at the end. Can
you see why Γ(n + 1) is called the factorial function?
3.14. Sketch each of the following integrands to decide whether the approxi-
mation to the integral is good or not.
(a)
0.1
−0.1
dt
10+t
2
≈ 0.02. (b)
0.1
−0.1
dt
0.001+t
2
≈ 200.
(c)
0.1
−0.1
cos(5πx) dx ≈ 0.2. (d)
0.1
−0.1
cos
πx
10
dx ≈ 0.2.
(e)
0.1
−0.1
e
−100t
2
dt ≈ 0.2. (f)
0.1
−0.1
e
−t
2
/100
dt ≈ 0.2.
3.15. Show that if a function is even (odd), then its derivative is odd (even).
3.16. Use the result of Example 3.3.1 to show that
#
∞
−∞
e
−xt
2
dt =
π
x
.
100 Integration: Formalism
3.17. By differentiating the electrostatic potential
Φ(r)=
##
Ω
k
e
dq(r
)
|r − r
|
with respect to x, y,andz, and assuming that Ω is independent of x, y,and
z, show that the electric field
E(r)=
##
Ω
k
e
dq(r
)
|r − r
|
3
(r − r
)
can be written as
E = −
∂Φ
∂x
ˆ
e
x
−
∂Φ
∂y
ˆ
e
y
−
∂Φ
∂z
ˆ
e
z
.
Chapter 4
Integration: Applications
The preceding chapter introduced integration and dealt with its formal as-
pects. It also gave some general guidelines concerning the calculation and
manipulation of integrals, in particular how to reduce the process of multiple
integration to a number of single integrations. In this chapter, we apply the
formalism of the previous chapter to concrete examples.
4.1 Single Integrals
This section is devoted to the simple but important case of single integrals
with examples from mechanics, electrostatics and gravity, and magnetostatics.
Generally, we encounter problems which are defined and set up in a single
dimension leading to integrals that have a single variable to be integrated.
4.1.1 An Example from Mechanics
In our discussion of primitive, Equation (3.18) clearly shows that integration
can be interpreted as the inverse of differentiation. Thus, if we know the
functional form of the derivative of a quantity, we should be able to express
the quantity in terms of an integral.
Velocity is the derivative of displacement. So, we seek to write displace-
ment in terms of an integral of velocity. This is easily done as follows:
1
dr
dt
= v(t) ⇒
dr
ds
= v(s) ⇒ dr = v(s) ds.
Integrating both sides from 0 to t,weget
#
t
0
dr =
#
t
0
v(s) ds ⇒ r(t) −r
0
=
#
t
0
v(s) ds, (4.1)
where r
0
= r(0), and we used Equation (3.21).
1
As cautioned below, we change t to s becauseweanticipateusingt as the upper limit
of the integral.