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Real Functions in Several Variables Examples of Maximum and Minimum Integration and Vector Analysis Calculus 2b

Year: 2006
Language: english
Pages: 167
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```LEIF MEJLBRO

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Leif Mejlbro

Real Functions in Several Variables
Examples of Maximum and Minimum Integration
and Vector Analysis
Calculus 2b

Calculus 2b – Real Functions in Several Variables
© 2006 Leif Mejlbro & Ventus Publishing ApS
ISBN 87-7681-207-3

Real Functions in Several Variables

Contents

Contents
1. Preface

6

2. The range of a function in several variables
2.1
Maximum and minimum
2.2
Extremum

7
7
29

3. The plane integral
3.1
Rectangular coordinates
3.2
Polar coordinates

37
37
44

4. The space integral
4.1
Rectangular coordinates
4.2
Semi-polar coordinates
4.3
Spherical coordinates

51
51
54
58

5. The line integral

66

6. The surface integral

72

7. Transformation theorems

84

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4

Real Functions in Several Variables

Contents

8. Improper integrals

95
111
111
118
142
151

A Formulæ
A.1
Squares etc.
A.2
Powers etc.
A.3
Differentiation
A.4
Special derivatives
A.5
Integration
A.6
Special antiderivatives
A.7
Trigonometric formulæ
A.8
Hyperbolic formulæ
A.9
Complex transformation formulæ
A.10 Taylor expansions
A.11 Magnitudes of functions

155
155
155
156
157
158
160
162
164
166
166
167

9. Vector analysis
9.1
9.2
Flux and divergence of a vector ﬁeld; Gauss’s theorem
9.3
Rotation of a vector ﬁeld; Stokes’s theorem
9.4
Potentials

5

Real Functions in Several Variables

1

Preface

Preface

The purpose of this volume is to present some worked out examples from the theory of Functions in
Several Variables in the following topics:
1) Maximum and minimum of a function.
2) Integration in the plane and in the space.
3) Vector analysis.
As an experiment I shall here use the following generic diagram for solving problems:
A. For Awareness. What is the problem?
Try to formulate the problem in your own words, thereby identifying it.
D. For Decision. What are we going to do with it?
Are there any reasonable solution procedure available? If so, which one should be chosen?
I. For Implementation. Here we do all the necessary calculations for solving the task after the
choice of the previous D.
At high school one usually starts here, but the problems may now be so complex that we need the
previous analysis as well.
C. For Control. Whenever it is possible, one should check the solution.
Note, however, that this is not always possible, so in many cases we have to skip this point.
Notice that A, D, I can always be eﬀectuated, no matter whether the problem is a mathematical
exercise, or construction of some building, or any other problem which should be solved. The model is
in this sense generic. It was ﬁrst presented for me in Telecommunication for over 15 years ago, where
I added C, the control of the solution. I hope that these simple guidelines will help the students as
much as it has helped me. Notice also that if one during the I, Implementation, comes across a new
and unforeseen problem, then one may iterate this simple model.
The intension is not to write a textbook, but only instead to give some hints of how to solve problems
in this ﬁeld. It therefore cannot replace any given textbook, but it may be used as a supplement to
such a book on Functions in Several Variables.
The chapters are only consisting of examples without any further mathematical theory, which one
must get from an ordinary textbook. On the other hand, it should be possible to copy the methods
given here in similar exercises.
In Appendix A the reader will ﬁnd a collection of formulæ which otherwise tacitly are assumed to be
known from high school. It is highly recommended that the student learns these by heart during the
course, because they form the backbone of the elementary part of Calculus, which should be mastered,
before one may proceed to more advanced parts of Mathematics.

The text is a continuation of Studentensupport: Calculus 1, Real Functions in One Variable and of
Studentensupport: Calculus 2a, Real Functions in Several Variables, Methods of Solution.
The text is based on my experiences in my teaching of students in this course. I realized that there was
absolutely a need for a practical description of how to solve explicit problems.
Leif Mejlbro

6

Real Functions in Several Variables

2

The range of a function in several variables

The range of a function in several variables

2.1

Maximum and minimum

Example 2.1
A Let A be a closed and bounded (i.e. compact) subset of the plane where the boundary ∂A is a closed
curve of the parametric representation

 1
2
2
1
t ∈ [0, 1].
r(t) = 4t 3 (1 − t) 3 , 4t 3 (1 − t) 3 ,
Find the maximum and minimum in A for the C ∞ -function
f (x, y) = x3 + y 3 − 3xy,

(x, t) ∈ A.

2

1.5

1

0.5

0

0.5

1

1.5

2

Figure 1: The closed and bounded domain A.

D Standard procedure:
1) Sketch the domain A and apply the second main theorem for continuous functions, from which
we conclude the existence of a maximum and a minimum.
2) Identify the exceptional points in A◦ , if any, and calculate the values f (x, y) in these points.
3) Set up the equations for the stationary points; ﬁnd these – which quite often is a fairly diﬃcult
task, because the system of equations is usually non-linear. Finally, compute the values f (x, y)
in all stationary points.
4) Examine the function on the boundary, i.e. restrict the function f (x, y) to the boundary and
repeat the investigation above to a set which is of lower dimension. Then ﬁnd the maximum
and minimum on the boundary.
5) Collect all the candidates for a maximum and a minimum found previously in 2)–4). Then the
maximum S and the minimum M are found by a simple numerical comparison.
Remark 2.1 Note that by using this method there is no need to use the complicated (r, s, t)method, which only should be applied when we shall ﬁnd local extrema in the plane. Here we are
dealing with global maxima and minima in a set A. ♦

7

Real Functions in Several Variables

The range of a function in several variables

Remark 2.2 Sometimes it is alternatively easy to identify the level curves f (x, y) = c for the
function f . In such a case, sketch a convenient number of the level curves, from which it may be
easy to ﬁnd the largest and the smallest constant c, for which the corresponding level curve has
points in common with the set A. Then these values of c are automatically the maximum S, resp.
the minimum M for f on A.
Notice, however, that this alternative method is demanding some experience before one can use
it as a standard method of solution. It was once used with success by a brilliant student at an
examination, summer 2000. ♦
I The level curves f (x, y) = x3 + y 3 − 3xy = c do not look to promising, so we stick to the standard
procedure.
1) The domain A has already been sketched. Since A is closed and bounded, and f (x, y) is
continuous on A, it follows from the second main theorem for continuous functions that the
function f has a maximum and a minimum on the set A.
2) Since f is of class C ∞ in A◦ , there are no exceptional points.

1

0.5

–1

–0.5

0

0.5

1

–0.5

–1

8

Real Functions in Several Variables

The range of a function in several variables

Figure 2: The stationary points are the intersections between the curves y = x 2 and x = y 2 .
3) The stationary points satisfy the two equations
∂f
∂x

= 3x2 − 3y = 0,

i.e.

y = x2 ,

∂f
= 3y 2 − 3x = 0, i.e. x = y 2 .
∂y
When we look at the graph we obtain the two solutions:
(0, 0) ∈ ∂A

and

(1, 1) ∈ A◦ .

Alternatively one inserts y = x2 into the second equation
0 = y 2 − x = x4 − x = x(x3 − 1) = x(x − 1)(x2 + x + 1).
Here x2 + x + 1 has only complex roots, hence the only real roots are x = 0 (with y = x 2 = 0)
and x = 1 (with y = x2 = 1), corresponding to
(0, 0) ∈ ∂A

and

(1, 1) ∈ A◦ .

Since (0, 0) is a boundary point, we see that (1, 1) ∈ A◦ is the only stationary point for f in A◦ .
We transfer the value
f (1, 1) = 1 + 1 − 3 = −1.
to the collection of all values in 5) below.
4) The Boundary. When we apply the parametric representation
(x, y) = r(t),

t ∈ [0, 1],

we get the restriction to the boundary
 1

2
2
1
g(t) = f (r(t)) = f 4t 3 (1 − t) 3 , 4t 3 (1 − t) 3
 1
  2

 


2
1
=
64t(1 − t)2 + 64t2 (1 − t) − 3 · 4t 3 (1 − t) 3 · 4t 3 (1 − t) 3
= 64 t(1 − t)2 + 64 t2 (1 − t) − 48 t(1 − t)
= 16 t(1 − t){4(1 − t) + 4t − 3} = 16 t(1 − t),

t ∈ [0, 1].

9

Real Functions in Several Variables

The range of a function in several variables

We have now reduced the problem to a problem known from high school
g  (t) = 12(2t − 1) = 0

for t =

1
,
2

corresponding to



1
2
2
1
1
g
= f 4 · 2− 3 · 2− 3 , 4 · 2− 3 · 2− 3 = f (2, 2) = 4.
2
At the end points of the interval, t = 0 and t = 1, we get
g(0) = g(1) = f (0, 0) = 0.
5) We collect all the candidates:
exceptional points:

None,

[from 2)]

Stationary point:

f (1, 1) = −1,

[from 3)]

Boundary points:

f (0, 0) = 0 and f (2, 2) = 4,

[from 4)].

By a numerical comparison we get
• The minimum is f (1, 1) = −1 (a stationary point),
• The maximum is f (2, 2) = 4 (a boundary point).
6) A typical addition: Since A is connected, and f is continuous, it also follows from the ﬁrst main
theorem for continuous functions, that the range is an interval (i.e. connected), hence
f (A) = [M, S] = [−1, 4].

♦

Example 2.2
A. Find maximum and minimum of the C ∞ -function
f (x, y) = x4 + 4x2 y 2 + y 4 − 4x3 − 4y 3
in the set A given by x2 + y 2 ≤ 4 = 22 .

2

y
1

–2

–1

0

2

1
x

–1

–2

Figure 3: The domain A.

10

Real Functions in Several Variables

The range of a function in several variables

50
40
30
20

–2

2

10

–1

1

–10
–1

1
2

–2

Figure 4: The graph of f (x, y) over A. Notice that a consideration of the graph does not give any
hint.

D. Even if the rewriting of the function
f (x, y) = (x2 + y 2 )2 + 2x2 y 2 − 4(x3 + y 3 )
looks reasonably nice it is still not tempting to apply an analysis of the level curves f (x, y) = c, so
we shall again use the standard method as described in the previous example, to which we refer
for the description.

I. 1) The domain A has been sketched already. Since A is closed and bounded, and f (x, y) is continuous on A, it follows from the second main theorem for continuous functions that f (x, y) has
a maximum and a minimum on A.

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11

Real Functions in Several Variables

The range of a function in several variables

While we are dealing with theoretical considerations we may aside mention that since A is
obviously connected, it follows from the ﬁrst main theorem for continuous functions that the
range is connected, i.e. an interval, which necessarily is given by
f (A) = [M, S].
2) Since f (x, y) is of class C ∞ , there is no exceptional point.
3) The stationary points (if any) satisﬁes the system of equations
0 =

∂f
= 4x3 + 8xy 2 − 12x2 = 4x(x2 + 2y 2 − 3x),
∂x

0 =

∂f
= 8x2 y + 4y 3 − 12y 2 = 4y(2x2 + y 2 − 3y).
∂y

Note that it is extremely important to factorize the expressions as much as possible in order to
solve the system. In fact, when this is done, we can reduce the system to
∂f
=0:
∂x

x=0

or x2 + 2y 2 − 3x = 0,

∂f
=0:
∂y

y=0

or

2x2 + y 2 − 3y = 0.

These conditions are now paired in 2 · 2 = 4 ways which are handled one by one.
a) When x = 0 and y = 0, we get (0, 0) ∈ A◦ , i.e. (0, 0) is a stationary point with the value of
the function
f (0, 0) = 0.
b) When x = 0 and 2x2 + y 2 − 3y = 0, we get
0 + y 2 − 3y = y(y − 3) = 0,

hence y = 0 or y = 3.

Thus, we have two possibilities: (0, 0) ∈ A◦ , which has already been found previously, and
(0, 3) ∈
/ A, so this point does not participate in the competition. We therefore do not get
further points in this case.
c) When y = 0 and x2 + 2y 2 − 3x = 0, we get by an interchange of letters (x, y) → (y, x) that
/ A. Hence we get no further
the candidates are (0, 0) ∈ A◦ [found previously] and (3, 0) ∈
point in this case.

3

2

1

–1

0

1

2

3

–1

12

Real Functions in Several Variables

The range of a function in several variables

Figure 5: The ellipses x2 + 2y 2 − 3x = 0 and 2x2 + y 2 − 3y = 0 and the line of symmetry y = x.
d) It still remains the last possibility
x2 + 2y 2 − 3x = 0 and

2x2 + y 2 − 3y = 0.

From the rewriting (cf. e.g. Linear Algebra)
 2


2
3
3
3
2
2
+ 2y =
and 2x + y −
x−
2
2
2



2

=

3
2

2

it is seen that the stationary points are the intersections of the two ellipses. It follows from
the symmetry that the points must lie on the line y = x. By eliminating y we get
0 = x2 + 2y 2 − 3x = 3x2 − 3x = 3x(x − 1).
Hence we get either x = 0, corresponding to (0, 0) ∈ A◦ [found previously] or x = 1
corresponding to (1, 1) ∈ A◦ , which is a new candidate with the value
f (1, 1) = 1 + 4 + 1 − 4 − 4 = −2.
Summarizing we get the stationary points (0, 0) and (1, 1) with the corresponding values of the
function
f (0, 0) = 0

og

f (1, 1) = −2.

4) The boundary. The simplest version is the following alternative to the standard procedure: A
parametric representation of the boundary curve is
(x, y) = r(ϕ) = (2 cos ϕ, 2 sin ϕ),

ϕ ∈ [0, 2π],

(evt. ϕ ∈ R),

where we note that

dx dy
,
(1)
= r (ϕ) = (−2 sin ϕ, 2 cos ϕ) = (−y, x).
dϕ dϕ
If we put g(ϕ) = f (r(ϕ)), where
f (x, y) = x4 + 4x2 y 2 + y 4 − 4x3 − 4y 3 ,
then we get by the chain rule, that the maximum and the minimum on the boundary should
be searched among the points on the boundary
x2 + y 2 = 4,
for which (apply (1)),

13

Real Functions in Several Variables

The range of a function in several variables

2

1

–3

–2

–1

1

2

0

–1

–2

–3

Figure 6: The intersections of the circle and the lines x = 0, y = 0, y = x and x + y + 3 = 0.

∂f dx
∂f dy
·
+
·
0 = g  (ϕ) =
∂x dϕ ∂y dϕ



 3
=
4x +8xy 2 −12x2 · (−y) + 8x2 y+4y 3 −12y 2 x




= 4x x2 + 2y 2 − 3x (−y) + 4y 2x2 + y 2 − 3y x


= 4xy −x2 − 2y 2 + 3x + 2x2 + y 2 − 3y


= 4xy x2 − y 2 + 3(x − y)

= 4xy(x − y){3 + x + y}.

14

Real Functions in Several Variables

The range of a function in several variables

Hence we shall ﬁnd the intersections between the circle x2 + y 2 = 4 = 22 and the lines
x = 0,

y = 0,

y = x and x + y + 3 = 0.

It follows immediately that these intersections are
√ √
√
√
(2, 0), ( 2, 2), (0, 2), (−2, 0), (− 2, − 2),

(0, −2).

We take a note of the values
f (2, 0) = f (0, 2) = 16 − 32 = −16,
f (−2, 0) = f (0, −2) = 16 + 32 = 48,
√ √
√
√
f ( 2, 2) = 6 · 4 − 2 · 4 · 2 2 = 24 − 16 2,
√
√
√
f (− 2, − 2) = 24 + 16 2.
5) Summarizing we shall compare numerically
exceptional points:

none,

stationary points:

f (0, 0) = 0,

boundary points:

f (2, 0) = f (0, 2) = −16,

f (1, 1) = −2,

f (−2, 0) = f (0, −2) = 48,
√
√ √
f ( 2, 2) = 24 − 16 2,
√
√
√
f (− 2, − 2) = 24 + 16 2.
√
3
Since 16 2 < 16 · = 24, it follows that
2
the minimum is

M = f (2, 0) = f (0, 2) = −16,

the maximum is

S = f (−2, 0) = f (0, −2) = 48,

and that both the minimum and the maximum are lying on the boundary.
6) Finally, we get from 1) that due to the ﬁrst main theorem for continuous functions the range
is the interval
f (A) = [M, S] = [−16, 48].

♦

Example 2.3
A. Find maximum and minimum for the function
f (x, y) =

x2 + 16y 2 − y 4

in the set
A = {(x, y) | x2 + 36y 2 ≤ 81}.

15

Real Functions in Several Variables

The range of a function in several variables

D. In this case one might ﬁnd the level curves f (x, y) = c, which by using that
a2 − b2 = (a + b)(a − b)
can be rewritten as
x2 = y 4 + c

2

− 16y 2 = y 4 + 4y + c

y 4 − 4y + c .

This expression still looks too diﬃcult to analyze, so we shall again stick to the standard procedure
as described in the ﬁrst example.

2
y
–10

–8

–6

–4

–2

1
0

2

6

4

8

10

–1
x

–2

Figure 7: The closed and bounded domain A.

I. 1) Using some Linear Algebra, the set A is written as
 x 2  y 2
+ 3
≤ 1,
9
2
which shows that at A is a closed ellipsoidal disc, cf. the ﬁgure.
Since the set A is closed and bounded, and even connected, and f (x, y) is continuous on A, it
follows from the second main theorem for continuous functions that f has a minimum M and a
maximum S on A. It follows furthermore from the ﬁrst main theorem for continuous functions
that the range is connected, i.e. an interval, which necessarily is
f (A) = [M, S].
2) Since the square root is not diﬀerentiable at 0, it follows that (0, 0) is an exceptional point! We
make a note for 5) of the value
f (0, 0) = 0.

16

Real Functions in Several Variables

The range of a function in several variables

3) The stationary points in A◦ \ {(0, 0)}, if any, must satisfy the system of equations
∂f
=
∂x

x
x2

+

16y 2

= 0 and

∂f
=
∂y

16y
x2

+

16y 2

− 4y 3 = 0.

The ﬁrst equation is only fulﬁlled for x = 0. Thus any stationary point must lie on the y-axis.
Since (0, 0) is an exceptional point, we must have y = 0 for any stationary point. When we
put x = 0 into the second equation, we get (NB: y 2 = |y|)


1
16y
y 
− y2 = 4
− 4y 3 = 4y
0=
1 − |y|3 .
|y|
|y|
16y 2
Since y =
 0, we must have |y| = 1, i.e. y = ±1. Hence the stationary points are (0, 1) and
(0, −1). We make a note for 5) of the value
√
f (0, 1) = f (0, −1) = 16 − 1 = 3.

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17

Real Functions in Several Variables

The range of a function in several variables

4) The boundary. On the boundary we get x2 + 36y 2 = 81, i.e.
x2 = 81 − 36y 2 .
Since f (x, y) only contains x in the form x2 , we can use this equation to eliminate x2 when we
write down the restriction,
f (y)

x2 + 16y 2 − y 4 =

=

81 −

=

20y 2

−y

81 − 36y 2 + 16y 2 − y 4


3 3
y∈ − ,
.
2 2

4



9
It follows immediately that g(y) is decreasing in the new variable t = y 2 ∈ 0, , hence the
4
maximum on the boundary is
g(0) = f (−9, 0) = f (9, 0) = 9,
and the minimum on the boundary is


3
g ±
2


=f

3
0,
2


=f

3
0, −
2


=

16 ·

9 81
81
15
−
=6−
=
.
4 16
16
16

5) A numerical comparison of
exceptional point: f (0, 0) = 0,
stationary points:

f (0, 1) = f (0, −1) = 3,


boundary points:

f

3
0,
2


=f

0, −

3
2

=

15
,
16

f (−9, 0) = f (9, 0) = 9,
gives
maximum:

f (−9, 0) = f (9, 0) = 9,

(boundary points),

minimum:

f (0, 0) = 0,

(exceptional point).

6) According to 1) the range is given by
f (A) = [M, S] = [0, 9],
where we have used the ﬁrst main theorem for continuous functions. ♦

18

Real Functions in Several Variables

The range of a function in several variables

Example 2.4
A. Consider the function
f (x, y) = x + 3y − 2 ln(1 + 4xy)
deﬁned on the triangle A with its vertices (1, 0), (4, 0) and (1, 1). Find the maximum and minimum
of f (x, y) on A.

1.4
1.2
1
y

0.8
0.6
0.4
0.2
0

1

3

2

4

–0.2
x

Figure 8: The closed and bounded domain A.

D. Here it is totally out of question to ﬁnd the level curves, so we apply the standard procedure as
described in Example 2.1.
I. 1) We ﬁrst sketch A. Since f (x, y) is continuous on the closed and bounded triangle A (note in
particular that 1 + 4xy > 0), it follows from the second main theorem for continuous functions
that f (x, y) has both a maximum S and a minimum M on A. Since A is also connected, it
follows from the ﬁrst main theorem for continuous functions that the range is connected, i.e.
an interval, and we have necessarily
f (A) = [M, S].
2) Since f everywhere in A◦ is of class C ∞ , it follows that f (x, y) has no exceptional point.
3) The stationary points, if any, must satisfy the equations
∂f
8y
=1−
= 0 and
∂x
1 + 4xy

∂f
8x
=3−
= 0,
∂y
1 + 4xy

i.e.
8y = 1 + 4xy

and

8x = 3(1 + 4xy).

When 1 + 4xy > 0 is eliminated we get 8x = 3 · 8y, from which x = 3y, which is a condition
that the stationary points necessarily must satisfy.
By insertion of x = 3y we get
8y = 1 + 4xy = 1 + 12y 2 ,

19

Real Functions in Several Variables

The range of a function in several variables

which is rewritten as

1
0 = 12y 2 − 8y + 1 = 12 y −
6
From this we either get y =


y−

1
2

.

1
1
1
, corresponding to x = 3 · = , i.e.
6
6
2



1 1
,
2 6

∈
/ A, or y =

1
,
2

corresponding to

3 1
,
∈ A◦ .
2 2

We only ﬁnd one stationary point

f

3 1
,
2 2

3 1
,
. We make a note of the value for 5) below,
2 2


3
3 3
= + − 2 ln 1 + 4 · · 12
2 2
2

= 3 − 2 ln 4 = 3 − 4 ln 2.

4) The investigation of the boundary is divided into three cases:
a) On the line x = 1, y ∈ [0, 1], we get the restriction
g1 (y) = 1 + 3y − 2 ln(1 + 4y),
where
g1 (y) = 3 −





8
8
5
= 0 for 1 + 4y = , i.e. y =
∈ [0, 1],
1 + 4y
3
12







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20

Real Functions in Several Variables

corresponding to


5
5
f 1,
= g1
12
12

The range of a function in several variables

=1+


5
5
− 2 ln 1 +
4
3

=

9
− 2 ln
4



8
3

.

NB: We must not forget the endpoints of the line:
f (1, 0) = g1 (0) = 1 + 0 − 2 ln(1 + 4 · 0) = 1,
f (1, 1) = g1 (1) = 1 + 3 − 2 ln(1 + 4 · 1) = 4 − 2 ln 5.
b) On the line y = 0, x ∈ [1, 4], we get the restriction
g2 (x) = x − 2 ln(1 + 4 · x · 0) = x,
which obviously is increasing. Therefore we shall only make a note on the values at the
endpoints,
f (1, 0) = 1

and

f (4, 0) = 4.

c) On the line x + 3y = 4, i.e. x = 4 − 3y, y ∈ [0, 1], the restriction is given by
g3 (y) = 4 − 2 ln(1 + 4(4 − 3y)y) = 4 − 2 ln(1 + 16y − 12y 2 ).
Here we get
2
∈ [0, 1],
3

2
2
corresponding to x = 4 − 3 · = 2. The interesting point is 2,
3
3



2
2
4
2
= 4 − 2 ln 1 + 16 · − 12 ·
f 2,
= g3
3
3
9
3

32 16
19
= 4 − 2 ln 1 +
−
= 4 − 2 ln .
3
3
3
We have already treated the two endpoints earlier.
g3 (y) = −

2
(16−24y) = 0
1+16y−12y 2

for y =

∈ ∂A with the value

5) Finally we shall compare numerically
exceptional points:

none,


3 1
,
2 2

stationary point:

f

boundary a):


5
f 1,
12

= 3 − 4 ln 2 ≈ 0, 23,
9
= − 2 ln
4



8
3

≈ 0, 29,

f (1, 1) = 4 − 2 ln 5 ≈ 0, 79,
f (1, 0) = 1,
boundary b):

f (4, 0) = 4,

boundary c):


2
f 2,
3

= 4 − 2 ln

19
≈ 0, 31.
3

21

Real Functions in Several Variables

The range of a function in several variables

By a comparison we see that
the maximum is

S = f (0, 4) = 4,

the minimum is

M =f



3 1
,
2 2

(boundary point),

= 3 − 4 ln 2,

(stationary point).

Remark 2.3 Note that the comparison is made approximatively, while the result is given in
an exact form. ♦
6) According to 1) we ﬁnally get by the ﬁrst main theorem for continuous functions that the range
is
f (A) = [M, S] = [3 − 4 ln 2, 4].

♦

Example 2.5 A nasty example which usually is not given in any textbook, is given by the following.
It illustrates that the usual division of cases in the textbooks is not exhaustive.
Let A = K(0; 1) be the open unit disc, and consider the function

1
,
x2 + y 2 < 1.
f (x, y) = x2 + y 2 cos
1 − x2 − y 2
Then f (x, y) is bounded on A,
|f (x, y)| ≤ x2 + y 2 < 1

for (x, y) ∈ A,

and we see that f (x, y) has no continuous extension to any point on the boundary.

1

y
0.5

–1

–0.5

0

0.5

1
x

–0.5

–1

Figure 9: The set A is the open unit disc.

22

Real Functions in Several Variables

The range of a function in several variables

Then note that
1) f (x, y) = 1 −

1
1
for x2 + y 2 = 1 −
, p ∈ N,
2pπ
2pπ

2) f (x, y) = −1 +

1
1
for x2 + y 2 = 1 −
,
(2p + 1)π
(2p + 1)π

p ∈ N,

from which we conclude that f has neither a maximum nor a minimum in the open set A.
However, since f (x, y) is continuous on the connected set A, it follows from the ﬁrst main theorem
for continuous functions that f (A) also is connected, i.e. an interval.
According to 1) the function f (x, y) attains values smaller than 1, though we can get as close to 1 as
we wish.
According to 2) the function f (x, y) attains values bigger than −1, though we can get as close to −1
as we wish.
Hence we conclude that the range is given by
f (A) = ] − 1, 1[.

♦

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23

Real Functions in Several Variables

The range of a function in several variables

Example 2.6
Let A be the open triangle
A = {(x, y) | 0 < x < 1, −x < y < 4x},
and let the function f (x, y) on A be given by
f (x, y) = 2xy + 3 ln(1 − x),

(x, y) ∈ A.

Find the range f (A).
A.
4

3

y

2

1

0 0.2 0.6

11.2

x
–1

Figure 10: The open and bounded domain A.

D. Here it is possible to ﬁnd the level curves. In fact, since x > 0 in A, we get that
f (x, y) = 2xy + 3 ln(1 − x) = c,

(x, y) ∈ A,

is equivalent to
y = ϕc (x) =

c
3 ln(1 − x)
− ·
.
2x 2
x

Although the expression looks very complicated, it is actually possible to analyze these level
curves. The reader is referred to section I 2 which, however, may be considered a bit advanced
for a common use.
We therefore start with the standard procedure in section I 1 with some necessary modiﬁcations.
First we exploit the theoretical main theorems as much as possible. Then we extend f to the parts
of the boundary where it is possible, and we discuss what happens at the boundary points where
such a continuous extension of f is not possible.
We see that both methods have a common theoretical start, which we here call section I.

24

Real Functions in Several Variables

The range of a function in several variables

I. Since f (x, y) is continuous on the connected set A, it follows from the ﬁrst main theorem for
continuous functions that the range f (A) is connected, i.e. an interval.
Since A is bounded, though not closed, we cannot apply the second main theorem for continuous
functions. We shall ﬁrst ﬁnd out whether f (x, y) has a continuous extension to (parts of) the
boundary of A.
It follows immediately that f (x, y) can be continuously extended to the lines
y = 4x and y = −x,

x ∈ [0, 1[,

with the same formal expression of the function, i.e. the extension is given by
f (x, y) = 2xy + 3 ln(1 − x)

for 0 ≤ x < 1,

−x ≤ y ≤ 4x.

On the other hand, we cannot extend to the vertical line x = 1, because
lim f (x, y) = 2y + 3 lim ln(1 − x) = −∞.

x→1−

x→1−

However, we see that the lower bound is −∞, so f (A) must be a semi-inﬁnite, i.e. either ] − ∞, a[
or ] − ∞, a], because the theorems do not assure that the upper bound a actually belongs to f (A).
This question can only be decided by an explicit analysis.
It follows that we shall only search the maximum in
B = {(x, y) | 0 ≤ x < 1, −x ≤ y ≤ 4x}.
Since we also have f (x, y) → −∞ for x → 1−, in B, there exists an ε ∈ ]0, 1[, such that
f (x, y) < S

for (x, y) ∈ B and 1 − ε ≤ x < 1.

The maximum S is therefore attained in the closed and bounded and truncated domain
Bε = {(x, y) | 0 ≤ x ≤ 1 − ε, −x ≤ y ≤ 4x},
where we of course assume that S exists and S < +∞.
This follows, however, from the second main theorem for continuous functions, applied on B ε .
Since we only want to ﬁnd the maximum, the standard procedure is hereafter the same as for closed
and bounded domains. The only modiﬁcation is that we shall not go through an investigation of
the boundary on the line x = 1 − ε.
I 1. Standard procedure.
1) We have already sketched a ﬁgure and quoted and applied the second main theorem.
2) Since f (x, y) belongs to the class C ∞ in A, there is no exceptional point.
3) The stationary points in A, if any, must satisfy the equations
∂f
3
= 2y −
= 0 amd
∂x
1−x

∂f
= 2x = 0.
∂y

It follows from the latter equation that x = 0; but since x > 0 in A, we see that we have no
stationary point in A for the function f .

25

Real Functions in Several Variables

The range of a function in several variables

4) Modiﬁed investigation of the boundary.
a) For y = 4x we get the restriction
g1 (x) = 8x2 + 3 ln(1 − x),

for x ∈ [0, 1[,

where
g1 (x) = 16x −

3
.
1−x

Hence, g1 (x) = 0 for
0 = 16x2 − 16x + 3 = (4x − 3)(4x − 1),
i.e. for x =

1
3
or x = .
4
4

By applying high school calculus it is seen that the maximum is either attained for x = 0,
3
corresponding to g1 (0) = f (0, 0) =, or for x = , corresponding to
4



3
3
8·9
3
9
,3 =
+ 3 ln 1 −
g1
= f
= − 6 ln 2
4
4
16
4
2
≥ 4, 5 − 6 · 0, 7 = 0, 3 > 0.

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26

Real Functions in Several Variables

The range of a function in several variables

b) For y = −x we get the restriction
g2 (x) = −2x2 + 3 ln(1 − x),

for x ∈ [0, 1[,

where
g2 (x) = −4x −

3
< 0.
1−x

Hence, g2 (x) is decreasing. The maximum on this line is therefore g2 (0, 0) = f (0, 0) = 0.
c) Numerical comparison. When we compare the values of the candidates above it follows that
the maximum in B is

3
9
, 3 = − 6 ln 2 > 0.
f
4
2

3
, 3 , so
This value is only attained at the boundary point
4


9
• f (B) = −∞, − 6 ln 2 ,
2
and


9
• f (A) = −∞, − 6 ln 2 ,
2
because A is obtained by removing all boundary points from B.

4

3

y

2

1

0 0.2 0.6

11.2

x
–1

Figure 11: The level curves for c =

1
(below) and c = 1 (above).
2

I 2. The method of level curves. The level curve
y = ϕc (x) =

c
3 ln(1 − x)
− ·
2x 2
x

is deﬁned in the strip 0 < x < 1 as the graph of a function. If c = 0, then both x = 0 and x = 1
are asymptotes. It follows that
lim ϕc (x) = +∞

x→1−

for all c ∈ R,

27

Real Functions in Several Variables

The range of a function in several variables

and that
lim ϕc (x) = +∞

for c > 0,

lim ϕc (x) = −∞

for c < 0.

x→0+

and
x→0+

The curves are characterized by f (x, y) being constant c along y = ϕc (x). We have sketched two
level curves on the ﬁgure (where c > 0), from which it is seen that the curved “move upwards”,
when c increases.
Hence we are looking for the biggest c, for which y = ϕc (x) just is contacting the boundary of B
without intersecting B. This is not possible for c < 0, and at the same time we get the line y = −x
excluded. Thus the maximum can only lie on the line y = 4x. Since y = ϕc (x) only touches this
line, the following two conditions must be fulﬁlled:
1) The curves must go through the same point, i.e. y = 4x = ϕc (x), or
4x = −

1
{−c + 3 ln(1 − x)},
2x

from which
−c + 3 ln(1 − x) = −8x2 .
2) The curves must have the same slope at this point, i.e.

1
3x

4 = ϕc (x) = 2 −c + 3 ln(1 − x) +
.
2x
1−x
The ugly terms −c + 3 ln(1 − x) in 2) can be eliminated by applying 1), hence
8x2 = −8x2 +

3x
,
1−x

which is rewritten as
0 = 16x2 (1 − x) − 3x = x{16x − 16x2 − 3} = −x(4x − 1)(4x − 3).
1
3
From this we get the solutions x = 0, x =
and x = , and since y = 4x, we ﬁnally get the
4
4
candidates


3
1
,3 ,
,1 ,
(0, 0),
4
4
with the corresponding function values for the extended function,


3
9
1
1
4
, 3 = − 6 ln 2.
, 1 = − 3 ln , f
f (0, 0) = 0, f
2
4
2
3
4

28

Real Functions in Several Variables

The range of a function in several variables


By a numerical comparison we get that the maximum is attained at the point

3
, 3 . Hence we
4

conclude that



9
f (B) = − −∞, − 6 ln 2 .
2
Finally, when we remove the boundary points from B, we obtains as previously that


9
f (A) = −∞, − 6 ln 2 .
♦
2

2.2

Extremum

Example 2.7 In this example we produce some functions in R2 , which all have (0, 0) as a stationary
point and value zero.
We supply the investigation with sketches of the graphs and discussions of the sign of the function in
the neighbourhood whenever this is necessary. Concerning the graphs the reader is also referred to
Linear Algebra.

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29

Real Functions in Several Variables

The range of a function in several variables

1
–1

–1

0.5
–0.5

–0.5
0

0.5

0.5

1

1

Figure 12: The graph of z = x2 + y 2 .

1) z = f1 (x, y) = x2 + y 2 .
The graph of f1 is a paraboloid of revolution.
Since
f1 (x, y) > 0 = f (0, 0)

for (x, y) = (0, 0),

the function f1 has a true minimum at (0, 0). It is easily seen that this is also the global minimum
of the function.
2) z = f2 (x, y) = x2 − 4xy + 4y 2 = (x − 2y)2 .
The graph of f2 is a parabolic cylinder. It follows immediately that
f2 (x, y) ≥ 0 = f (0, 0);

0.25
0.2
0.15
0.1
–0.4

–0.4

0.05
–0.2

–0.2
t
0.2
0.4

0.2 s
0.4

Figure 13: The graph of z = (x − 2y)2 .

30

Real Functions in Several Variables

but since
 x
f x,
= 0 = f (0, 0)
2

The range of a function in several variables

for alle x,

we see that (0, 0) is a weak local minimum. However, we also have in this case that 0 is a global
minimum.

1
–1

–1

0.5
–0.5

–0.5

–0.5 0.5

0.5

1

1
–1

Figure 14: The graph of z = x3 + y 3 sketched in MAPLE does not give the best picture.

3) z = f3 (x, y) = x3 + y 3 = (x + y)(x2 − xy + y 2 ).
Since x3 + y 3 is of odd degree 3, we take e.g. the restriction of f3 to the x-axis,
f3 (x, 0) = x3

(is both > and < 0),

in a neighbourhood of x = 0, so f3 has no extremum at (0, 0).
This can also be seen by analyzing the sign of the function. In fact, x2 − xy + y 2 ≥ 0 for all (x, y),
thus x3 + y 3 is everywhere of the same sign as x + y.

1

0.5

–1

–0.5

0.5

1

–0.5

–1

Figure 15: The restriction to the x-axis gives a better picture.

31

Real Functions in Several Variables

The range of a function in several variables

4

–2
–1

2

2
1

–1
–2

1
–2
2
–4

Figure 16: The graph of z = x2 − y 2 .

4) z = f4 (x, y) = x2 − y 2 = (x + y)(x − y).
The graph is a hyperbolic paraboloid. There is no extremum at (0, 0).
An analysis of the sign shows that f4 (x, y) is 0 on the lines x + y = 0 and x − y = 0, and that
f4 (x, y) attains both positive and negative values in any neighbourhood of (0, 0).
It is ﬁnally also possible to consider the restrictions
x − axis:
y − axis:

f4 (x, 0) = x2 > 0
f5 (0, y) = −y 2 < 0

for x = 0,
for y =
 0,

from which we arrive to the same conclusion. ♦

32

Real Functions in Several Variables

The range of a function in several variables

Example 2.8
A. Examine whether the function
f (x, y, z) = exp xy + z 2
has local extrema.
D. Here we shall not use the standard procedure but instead we suggest an alternative method. In
fact, since exp is strictly increasing, the functions
ϕ(x, y, z) = xy + z 2

and f (x, y, z) = exp(ϕ(x, y, z))

must have the same stationary points and extrema.
We therefore examine the simpler function ϕ(x, y, z).
I. The equations for the stationary points for ϕ(x, y, z) are
∂ϕ
= y = 0,
∂x

∂ϕ
= x = 0,
∂y

∂ϕ
= 2z.
∂z

Hence it follows immediately that (0, 0, 0) is the only stationary point.
2) For the other candidates the (r, s, t)-method is easier, because it is the essence of the determination of the approximative polynomial of at most degree two. One often forgets in the
applications that this is the general idea behind the (r, s, t)-method. Note that we in 1) had to
expand to the third degree, which is the reason why the (r, s, t)-method fails for (0, 0).
There is, however, also an alternative method for the other points. This will here be illustrated
on the point (1, 1).
a) First we reset, the problem, i.e. put (x, y) = (1 + h, 1 + k), so (h, k) = (0, 0) corresponds to
the point (x, y) = (1, 1) under examination.
b) Insert this in the expression for f (x, y) and write dots for terms of degree > 2:
f (x, y) = (1 + h)4 + 4(1 + h)2 (1 + k)2 + (1 + k)4 − 4(1 + h)3 − 4(1 + k)3
= 1 + 4h + 6h2 + · · · + 4(1 + 2h + h2 )(1 + 2k + k 2 )
+1 + 4k + 6k 2 + · · · − 4(1 + 3h + 3h2 + · · · )
−4(1 + 3k + 3k 2 + · · · )


= (1 + 4h + 6h2 ) + 4 1 + 2h + h2 + 2k + 4hk + k 2 + · · ·
+(1 + 4k + 6k 2 ) − 4(1 + 3h + 3h2 ) − 4(1 + 3k + 3k 2 ) + · · ·
= −2 − 2h2 + 16hk − 2k 2 + · · · ,
i.e.


P2 (h, k) = −2 − 2(h2 − 8hk + k 2 ) = −2 − 2 (h − 4k)2 − 15k 2 .
Since P2 (h, k) + 2 attains both negative values (for k = 0 and h = 0) and positive values
(for h = 4k) in any neighbourhood of (h, k) = (0, 0), we conclude that (x, y) = (1, 1) is not
an extremum. ♦

33

Real Functions in Several Variables

The range of a function in several variables

Example 2.10
A. Examine whether the function
f (x, y) = 1 − 4x2 − 4y 2 + x2 y 2 ,

(x, y) ∈ R2 .

has any extremum. Find the range f (R2 ).
D. When we apply the standard procedure we are guided through the usual examination of the
exceptional points (there are none) and of the stationary points. It is, however, here possible to
make a shortcut by noticing that f (x, y) only is a function in u = x2 and v = y 2 , i.e.
f (x, y)1 − 4x2 − 4y 2 + x2 y 2 = g(u, v) = 1 − 4u − 4v + uv,

u, v ≥ 0.

I. The stationary points for f (x, y), if any, must fulﬁl the equations
∂f
= −2x(4 − y 2 ) = 0 and
∂x

∂f
= −2y(4 − x2 ) = 0.
∂y

This system is split into
x = 0 or y = 2 or y = −2,
and
y = 0 or x = 2

or x = −2.

Formally we get 3 · 3 possibilities, but four of them are not possible (e.g. x cannot at the same
time be 0 and 2 or −2). We therefore get ﬁve stationary points,
(0, 0),

(2, 2),

(−2, −2),

(2, −2).

(−2, 2),

1
0.8
0.6
0.4
–0.4

–0.4

0.2
–0.2

0
0.2

0.4

–0.2

0.2
0.4

Figure 17: The graph of z = 1 − 4x2 − 4y 2 .

34

Real Functions in Several Variables

The range of a function in several variables

1) In (0, 0) we get
f (x, y) ≈ P2 (x, y) = 1 − 4x2 − 4y 2 .
The graph of z = P2 (x, y) is an elliptic paraboloid of revolution. Obviously we have a proper
local maximum at (0, 0).
2) We take a shortcut by considering
u = x2 ,

g(u, v) = 1 − 4u − 4v + uv,

v = y2 ,

instead. First, all four stationary points for f are seen to correspond to the only point (u, v) =
(4, 4). It is therefore suﬃcient to examine g(u, v) in the neighbourhood of (4, 4).
The approximating polynomial for g(u, v) expanded from (4, 4) of at most degree 2 is found by
using:
g(u, v) = 1 − 4u − 4v + uv, g(4, 4) = −15,
∂g
= −4 + v,
∂u

gu (4, 4) = 0,

∂g
= −4 + u,
∂v

gv (4, 4) = 0,

∂2g
∂v = 1,
∂u
,

∂2g
∂2g
=
= 0,
∂u2
∂v 2

35

Real Functions in Several Variables

The range of a function in several variables

–3

–3
–2

–2
–1

10 –1

1
2
3

1
–10

2
3

–20
–30

Figure 18: The graph of f (x, y) = 1 − 4x2 − 4y 2 + x2 + y 2 .

hence
P3 (u, v) = −15 +

1
· (u − 4)(v − 4) = −15 + (u − 4)(v − 4).
2

It follows that P2 (u, v) in the neighbourhood of (4, 4) attains values which are both > −15 and
< −15. Thus g(u, v) does not have an extremum at (4, 4). This implies that f (x, y) does not
have an extremum at (±2, ±2) (all four possible combinations of the sign).
3) The function f (x, y) has only one local maximum,
f (0, 0) = 1.
However, this value is not the global maximum. In fact, by rewriting
f (x, y) = 1 − 4x2 − 4y 2 + x2 y 2 = (x2 − 4)(y 2 − 4) − 15,
we see that the restriction to the line y = x gives
f (x, x) = x2 − 4

2

− 15 → +∞

for x → ±∞.

4) Note also that
f (x, 0) = 1 − 4x2 → −∞

for x → ±∞.

Thus, since f is continuous on the connected set R2 , it follows from the ﬁrst main theorem for
continuous functions that the range is
f (R2 ) = R.

♦

36

Real Functions in Several Variables

3

The plane integral

The plane integral

3.1

Rectangular coordinates

Example 3.1

A. Calculate B xy dS, where B is given on the ﬁgure.

1.2

1

0.8
y

0.6

0.4

0.2

0

0.5

1

–0.2

1.5

2

x

Figure 19: The domain B is the upper triangle.

D. We have two possibilities for the reduction:
D1. We ﬁrst integrate horizontally.
D2. We ﬁrst integrate vertically.
We shall treat both possibilities so we can compare the calculations in the two cases.

1.2

1

0.8
y

0.6

0.4

0.2

0
–0.2

0.5

1

1.5

2

x

Figure 20: The domain B with the vertical integration line from y = 1 −

1
2

x to y = 1.

D 1. We ﬁrst integrate vertically.

37

Real Functions in Several Variables

The plane integral

I 1. In this case we write the domain in the form
B = {(x, y) ∈ R2 | 0 ≤ x ≤ 2, 1 −

1
x ≤ y ≤ 1}.
2

Notice that the outer variable x must always lie between two constants, 0 ≤ x ≤ 2. Then we use
the ﬁgure for any ﬁxed x to ﬁnd the integration interval for the inner variable of integration y,
1
i.e. in this particular case 1 − x ≤ y ≤ 1.
2
Then write down the double integral:


 
2

(2)

xy dS =
B

0

1

1− 12 x

xy dy



dx =



2

x
0

Calculate the inner integral,


1

 1
1 2
1
1
y
y dy =
=
1− 1− x
1
2
2
2
1
1− 2 x
1− x



1

1− 12 x

2

y dy


=

1−

1 2
x
4


=

1
1
x − x2 .
2
8

2

1
2

dx.

38

Real Functions in Several Variables

The plane integral

By insertion in (2), we get


 2
1
1
1 2 1 3
x − x2 dx =
x − x dx
2
8
2
8
0
0

2
1 3
1 4
8 1
5
x −
=
x
= − = .
6
32
6
2
6
0



xy dS
B

=

2

x

1.2

1

0.8
y

0.6

0.4

0.2

0

0.5

1

–0.2

1.5

2

x

Figure 21: The domain B with the horizontal line of integration from x = 2 − 2y to x = 2.
D 2. Here we ﬁrst integrate horizontally.
I 2. The domain is written
B = {(x, y) ∈ R2 | 0 ≤ y ≤ 1, 2 − 2y ≤ x ≤ 2},
because y ∈ [0, 1] is now the outer variable of integration (lying between two constants), and where
2 − 2y ≤ x ≤ 2 for the inner variable of integration x for any ﬁxed y.
The double integral becomes here

 1  2


xy dS =
xy dx dy =
(3)
0

B

2−2y

1



2


x dx dy.

0

2−2y

Let us ﬁrst calculate the inner integral,




2

x dx =
2(1−y)

1 2
x
2

2
=
2(1−y)
2


1 2 2
· 2 1 − (1 − y)2 = 2(2y − y 2 )
2

= 4y − 2y .

When this result is put into (3), we get by another calculation



xy dS

1

=

B


=

0



2

y(4y − 2y ) dy =

4 3 1 4
y − y
3
2

1
0

5
= .
6

0

1

(4y 2 − 2y 3 ) dy
♦

39

Real Functions in Several Variables

The plane integral

Example 3.2

A. Calculate B x exp y 3 dS, where B is given on the ﬁgure for a = 1.
1.2

1

0.8

y

0.6

0.4

0.2

–0.2

0

0.2

0.4

0.6

0.8

1

1.2

x
–0.2

Figure 22: The domain B for a = 1.

D. We shall again examine the two possibilities of the order of integrations.
1.2

1

0.8

y

0.6

0.4

0.2

–0.2

0

0.2

0.4

0.6

0.8

1

1.2

x
–0.2

Figure 23: The domain B for a = 1 with a vertical line of integration from y = x to y = 1.

D 1. Let us ﬁrst try to integrate vertically for ﬁxed x, just to see what happens.
I 1. Here the domain is written (note the order of x and y):
B = {(x, y) ∈ R2 | 0 ≤ x ≤ a, x ≤ y ≤ a}.
Then we can write down the double integral:

 a  a


x exp y 3 dS =
x exp y 3 dy dx =
B

0

x

0



a

a

x


exp y 3 dy dx.

x

The inner integral,
 a
exp y 3 dy,
x

40

Real Functions in Several Variables

The plane integral

does not just look impossible to calculate; it is impossible to calculate with our arsenal of functions! Therefore we give up this variant. Instead we examine, if we shall be more successful by
interchanging the order of integration.
1.2

1

0.8

y

0.6

0.4

0.2

–0.2

0

0.2

0.4

0.6

0.8

1

1.2

x
–0.2

Figure 24: The domain B for a = 1 with a horizontal line of integration from x = 0 to x = y where y
is kept ﬁxed.

D 2. The domain is here written (note again the order of x and y):
B = {(x, y) ∈ R2 | 0 ≤ y ≤ a, 0 ≤ x ≤ y}.

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41

Real Functions in Several Variables

The plane integral

Then we turn to the double integral,

 a  y


3
3
x exp y dS =
x exp y dx dy =
(4)
0

B

0

0

a

exp y

3



y


x dx dy.

0

The inner integral is calculated in the following way:

y
 y
1 2
1
x
x dx =
= y2 .
2
2
0
0
By insertion in (4) followed by the substitution t = y 3 and dt = 3y 2 dy, where y 2 dy already can
be found in the integrand, we get

x exp y

3


dS

=

B

=

 a3
1 2
1 1
exp y · y dy =
exp(t) · · dt
2
2 3
0
0

1  t a3
1 
e 0 =
exp a3 − 1 .
6
6
a

3

Remark 3.1 In this case one of the two variants cannot be calculated, while the second one is
easy to perform. ♦
Example 3.3
A. Find the value of

y
E = Lz
dS,
2 + y 2 + z 2 )2
(x
B

where B = [0, a] × [0, b] and z > 0.

D. Here we can expect a quite a few diﬃculties, no matter which version we are choosing. In fact,
the integrand is suited for the polar coordinates, while the domain B in the (x, y)-plane is best
described in rectangular coordinates. By experience, such a mixture of polar and rectangular
coordinates will always give computational problems.
Then we notice that if we start by ﬁrst integrating after x, then we shall immediately run into
troubles with this ﬁrst integral

dx
.
2
(x + y 2 + z 2 )2
It is possible to go through with the calculations, but they are far from elementary. On the other
1
hand, if we ﬁrst integrate with respect to y, we shall beneﬁt from the fact that y dy = d y 2 ,
2
where y 2 already can be found in the integrand. For that reason we choose ﬁrst to integrate with
respect to y.

42

Real Functions in Several Variables

The plane integral

I. The double integral is here

(5) E = Lz
B

y
dS = Lz
(x2 + y 2 + z 2 )2


0

a


0

b

y
dy
(x2 + y 2 + z 2 )2


dx.

By the calculations of the inner integral we put for convenience c = x2 + z 2 = equal a constant,
and we apply the substitution t = y 2 where dt = 2y dy, and where y dy already is a factor in the
1
integrand, i.e. y dy = dt. Thus,
2

0

b

y
dy
(x2 + y 2 + z 2 )2


=
=

b

1
y dy =
2 + c)2
y
0

1 1
1
−
=
2 c b2 + c


b2
 2
dt
1 b
1
1
=
−
2 0 (t + c)2
2
t + c t=0

1
1
1
− 2
.
x + b2 + z 2
2 x2 + z 2

The result of this calculation is then inserted into (5), by which
 
1
1
Lz a
− 2
dx.
(6) E =
2
2
x +z
x + b2 + z 2
2 0
Notice here that it is not a good idea to put everything in the same fraction with a common
denominator, so we keep the form above.
A new calculation shows that if k 2 > 0, then we get by the change of variable t =


dx
2
x + k2

=

=

x
that
k


1
1
dt
1
 x 2 dx =
k
k
1 + t2
1+
k
x
1
1
Arctan t = Arctan
.
k
k
k
1
k



The trick in this type of calculation is by division to get the constant 1 in the denominator plus
some square.
When this calculation is applied with
k1 = z

and

k2 =

b2 + z 2 ,

we get by insertion into (6) that
E

=
=
=



a

x
x
Lz 1
1
Arctan
Arctan
−
k1
k2 0
2 k1
k2




a
Lz 1
a
1
Arctan
−√
Arctan √
2
z
z
b2 + z 2
b2 + z 2




a
L
a
z
Arctan √
.
Arctan
−√
2
z
b2 + z 2
b2 + z 2

♦

43

Real Functions in Several Variables

3.2

The plane integral

Polar coordinates

Example 3.4
A. Calculate

I = (x2 + y 2 ) dS,
b

where B is described in polar coordinates by



ϕ

A = (, ϕ)  a ≤  ≤ 2a,
≤ϕ≤
.
2a
a

Note that B has a “weird” form in (x, y)-plane, while the parameter domain A in the (, ϕ)-plane
is “straightened out”, so one can apply the rectangular version in the (.ϕ)-plane. The price for
this is that one must add the weight function  to the integrand.



  







44

Real Functions in Several Variables

The plane integral

2

1.5

y
1

0.5

0

–0.5

–1

0.5

1

1.5

x

Figure 25: The domain B in the (x, y)-plane.

2

1.5

y
1

0.5

0

0.5

1

1.5

2

x

Figure 26: The parameter domain A in the (, ϕ)-plane.
D. Apply the reduction formula in the second version, i.e. where the ϕ-integral is the inner integral.
This means that we ﬁrst integrate vertically in the parameter domain.
I. By the reduction formula in its second version we get with the weight function 



 2a  a
 2a  a
2
2
2
3
(x + y ) dS =
 dϕ  d =

dϕ d.

2a

a

B


2a

a

First calculate the inner integral,
 a



dϕ = −
=
,

a
2a
2a
2a
which is seen to be the length of the ϕ-interval.
Then by insertion,

(x2 + y 2 ) dS

2a

 2a

1
1 1 5
d =
4 d =

2a
2a a
2a 5
a
1


 31 4

1
1
5
5
5
5
32 a − a =
a .
(2a) − a =
10a
10
10a


=

B

=

2a

3 ·

♦

45

Real Functions in Several Variables

The plane integral

Example 3.5
A. Calculate

I=
(x + y) dS,
B

where B is described in polar coordinate (for a > 0) by
 π


π

A = (, ϕ)  − ≤ ϕ ≤ , 0 ≤  ≤ a ,
2
4
i.e. A is a rectangle in the (, ϕ)-plane.
D. Here we can apply both reduction formulæ, so we give two solutions.
D 1. Apply the ﬁrst reduction formula; do not forget the weight function .
I 1. From x =  cos ϕ and y =  sin ϕ, we get in the ﬁrst version, where we start by integrating
horizontally after , that


 π4  a
I=
(x + y) dS =
( cos ϕ +  sin ϕ)  d dϕ.
B

−π
2

0

Then calculate the inner integral,
 a

( cos ϕ +  sin ϕ)  d = (cos ϕ + sin ϕ)
0

a

2 d =

0

a3
(cos ϕ + sin ϕ).
3

By insertion of this result we ﬁnally get
 π4 3
π
a3
a
a3
I=
(cos ϕ + sin ϕ) dϕ =
[sin ϕ − cos ϕ]−4 π = .
2
3
3
3
−π
2
1

y

0.5

0.2

–0.2

0.4

x
0.6

0.8

1

1.2

–0.5

–1

Figure 27: The domain B for a = 1 in the (x, y)-plane.

46

Real Functions in Several Variables

The plane integral

0.5

x
0.2 0.4 0.6 0.8

–0.2

1

1.2

–0.5

y
–1

–1.5

Figure 28: The parametric domain A for a = 1 in the (, ϕ)-plane.

D 2. Apply the second reduction formula. Again, do not forget the weight function .
I 2. In the second version we just interchange the order of integration. Since the bounds are constants,
and the variables can be separated in the integrand, we can split the integral into a product of two
integrals. Then

I

a

=


0

π
4

−π
2

0

=



a

2 d ·


( cos ϕ +  sin ϕ) dϕ  d


−π
2

Student
Discounts

π
4

(cos ϕ + sin ϕ) dϕ =

+

Student
Events

π
a3
a3
· [sin ϕ − cos ϕ]−4 π = .
2
3
3

+

Money

=

♦

Happy
Days!

2009

47

Real Functions in Several Variables

The plane integral

Example 3.6
A. Calculate I =


B

x dS, where B = K

 a
,0 ;
, a > 0.
2
2

 a

0.4

0.2

0

0.2

0.6

0.4

0.8

1

–0.2

–0.4

√
√
Figure 29: The domain B for a = 1, i.e. − x − x2 ≤ y ≤ x − x2 for 0 ≤ x ≤ 1.

D. In this case it is possible to calculate the integral by using either rectangular or polar coordinates.
D 1. In rectangular coordinates the domain B is described by
B = {(x, y) | 0 ≤ 0 ≤ a, −

ax − x2 ≤ y ≤

ax − x2 }.

I 1. The rectangular double integral is given by
 a  √ax−x2 


x dS =
x
dy
dx
=
2a
I=
√
0

B

− ax−x2

a

ax − x2 dx.

The trick in problems of this type is to call the “ugly” part something diﬀerent. We put
t = ax − x2 ,

dt = (a − 2x) dx.

Then by adding the right term and subtract it again we get
 a
 a
2x ax − x2 dx = −
(a − 2x − a) ax − x2 dx
I =
0
0
 a
 a
= −
ax − x2 · (a − 2x) dx + a
ax − x2 dx
0
0
 a
 a √
t dt + a
= −
ax − x2 dx
x=0
0

a
 a
 a
3
2
2 2
2
ax − x
= −
+a
ax − x dx = 0 + a
ax − x2 dx.
3
0
0
0
a√
The integral 0 ax − x2 dx does not look nice; but the geometrical interpretation helps a lot:
The integral is the area of the domain between the x-axis and the curve
y=+

ax − x2 ,

48

Real Functions in Several Variables

The plane integral

i.e. (cf. the ﬁgure) the area of a half-disc of radius
 a 2
1
·π
2
2

I =a·


=

a
. Therefore,
2

πa3
.
8

D 2. The polar version; do not forget the weight function .
I 2. When we put x =  cos ϕ and y =  sin ϕ, the equation of the boundary curve becomes
0 = x2 + y 2 − ax = 2 − a  cos ϕ = ( − a cos ϕ).
Since  = 0 corresponds to the point (0, 0), it follows that the boundary curve is described by
med −

 = a cos ϕ

π
π
≤ϕ≤ .
2
2

The parametric

A = (, ϕ)

domain A corresponding to B is therefore
 π

π

 − ≤ ϕ ≤ , 0 ≤  ≤ a cos ϕ .
2
2
When we use the ﬁrst version of the reduction formula we get

1.5

y

1

0.5

0

0.2 0.4 0.6 0.8 1 1.2
x

–0.5

–1

–1.5

Figure 30: The parametric domain A in the (, ϕ)-plane.



x dS =
B

π
2

−π
2


0

a cos ϕ


 cos ϕ ·  d dϕ =



π
2

−π
2


cos ϕ

a cos ϕ

2



 d dϕ.
0

When we calculate the inner integral we get

a cos ϕ
 a cos ϕ
1 3
a3
2

cos3 ϕ.
 d =
=
3
3
0
0
Then by insertion

 π2
 π2
a3
a3
x dS =
cos4  dϕ = 2 ·
cos4 ϕ dϕ,
π
3
3
B
−2
0
where we use that the even function cos4 ϕ is integrated over a symmetric interval.

49

Real Functions in Several Variables

The plane integral

When we shall calculate a trigonometric integral, where the integrand is of even order, we change
variables to the double angle:
2
 2 2

1
1 
4
cos x =
(1 + cos 2x) =
1 + 2 cos 2x + cos2 2x
cos x =
2
4

1
1
=
1 + 2 cos 2x + (1 + cos 4x)
4
2
3 1
1
=
+ cos 2x + cos 4x.
8 2
8
Finally, by insertion,

 π2
2
x dS = a3
3
B
0


3 1
1
2
3 π
πa3
+ cos 2x + cos 4x dx = a3 ·
.
+0+0=
8 2
8
3
8
8 2

♦

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50

Real Functions in Several Variables

4

The space integral

The space integral

4.1

Rectangular coordinates

Example 4.1
A. Calculate the space integral

I = (3 + y − z) x dΩ,
A

where
A = {(x, y, z) ∈ R3 | (x, y) ∈ B, 0 ≤ z ≤ 2y},
and B is the upper triangle shown on the ﬁgure.

1.2

1

0.8
y

0.6

0.4

0.2

0

0.5

–0.2

1

1.5

2

x

Figure 31: The domain B, i.e. the projection of the domain A onto the (x, y)-plane.

D Apply the ﬁrst rectangular reduction theorem in 3 dimensions.
We have according to the ﬁrst rectangular reduction theorem that



 2y
(3 + y − z) dz dS.
(7) I = (3 + y − z)x dΩ =
A

B

0

Considering x and y as constants, we calculate the inner and concrete integral,

0

2y


2y
1 2
= x
(3 + y − z = dz = x (3 + y)z − z
2
0
z=0


2
= x (3 + y) · 2y − 2y = x · 2y{3 + y − y} = 6xy.


(3 + y − z)x dz

2y

When this result is inserted into (7), it is reduced to an abstract plane integral over B, i.e. of a
lower dimension,


I=
6xy dS = 6
xy dS.
B

B

51

Real Functions in Several Variables

The space integral

2.5

2

1.5

1
1
0.5

0
0.5

1

1.5

2

Figure 32: The domain A.

We have already calculated the abstract plane integral in Example 3.1,

5
xy dS = ,
6
B
hence





I=
A

(3 + y − z)x dΩ = 6

B

xy dS = 6 ·

5
= 5.
6

♦

Example 4.2
A. Calculate the space integral

(x + 2y + z) exp z 4 dΩ,
A

where
A = {(x, y, z) | z ∈ [2, 0], (x, y) ∈ B(z)}
with a cut at the height z,
 z
B(z) = [0, z] × 0,
,
2

z ∈ ]0, 2].

D. Apply the second reduction theorem in 3 dimensions.
I. When we insert into the second reduction theorem, we get

(8) I =
(x + 2y + z) exp z 4 dΩ
A



2

=
0

exp z 4

(x + 2y + z) dS

dz.

B(z)

52

Real Functions in Several Variables

The space integral

For every ﬁxed z we reduce the inner abstract plane integral,




(x + 2y + z) dS =
x dS +
2y dy + z
dS
B(z)



z

=
0
2


x dx ·

B(z)
z
2



z

dy +

0

0


dx ·

z2
1
z z
· +z·
+z· z· z
2 2
4
2

=

0

B(z)
z
2



B(z)

2y dy + z · areal B(z)
= z3.

By insertion of this result into (8), using the substitution
t = z4,

dt = 4z 3 dz,

dvs. z 3 dz =

1
dt,
4

we ﬁnally get

I=

exp z 4 · z 3 dz =


0

24

et ·

1 16
1
dt =
e −1 .
4
4

♦

0

2

53

Real Functions in Several Variables

4.2

The space integral

Semi-polar coordinates

Example 4.3
A. Calculate the space integral

I=
x2 yz dΩ,
A

where
A = {(x, y, z) ∈ R3 | x2 + y 2 ≤ a2 , y ≥ 0, 0 ≤ z ≤ h}.

1.5

1
–1
0.5

–0.5

0
0.5
–0.5

1

0.5
1

Figure 33: The domain A for a = h = 1 with a cut B(ϕ).

1.2

1

0.8
y

0.6

0.4

0.2

–1

–0.5

0

0.5

–0.2

1
x

Figure 34: The projection of A onto the (x, y)-plane (a = 1).

D. It is here possible to go through with the rectangular calculations, but we end up with the same
problems as in Example 3.6. We therefore here choose the semi-polar representation, where we
must not forget to add the weight function  as a factor.

54

Real Functions in Several Variables

The space integral

When we use semi-polar coordinates. the domain A is represented by the parametric domain
Ã = {(, ϕ, z) | 0 ≤  ≤ a, 0 ≤ ϕ ≤ π, 0 ≤ z ≤ h}.
Then we get at least two possibilities for the reduction.
I 1. For ﬁxed ϕ the domain A is cut into B(ϕ) = [0, a] × [0, h]. In this case we get the following
reduction where ϕ is kept in the outer integral,

I

π





z · 2 cos2 ϕ ·  sin ϕ ·  d dϕ

=
0



B(ϕ)
π

=


0

−

=



2

cos ϕ sin ϕ dϕ ·
π 
1 2
z
·
2
0

1
cos3 ϕ
3

h

0
h
0


z dz ·

·

1 5

5

a

4 d

0

a
=
0

2 1 2 1 5
1 2 5
· h · a =
h a .
3 2
5
15

I 2. If we instead integrate ﬁrst after z, then we get where we use that B is a half disc,

I

=
B

x2 y





h

z dz

dS

0


h  π  a
1 2
z
·
2 cos2 ϕ ·  sin ϕ ·  d dϕ
2
0
0
 a
0
1 2 π
h
cos2 ϕ sin ϕ dϕ ·
4 d
2
0
0
π 5

1 2 5
1 2
a
1
=
h a .
h − cos3 ϕ ·
5
15
2
3
0


=
=
=

C. Weak control (considerations of the dimension). Since


x ∼ a, y ∼ a, z ∼ h,
· · · dΩ = · · · dx dy dz ∼ a · a · h = a2 h
we get

I=
A

x2 y 2 z dΩ ∼ a2 · a2 · h · (a2 h) = h2 a5 ,

hence the result must be of the form constant·h2 a5 . If this is not the case, we have made an error.
On the other hand, even if we get a result like c · h2 a2 , the constant c may still be calculated
wrongly, explaining why the method is only giving a weak control. ♦

55

Real Functions in Several Variables

The space integral

Example 4.4
A. Calculate the space integral

I=
xy 2 z dΩ,
A

where
A = {(x, y, z) ∈ R3 | x2 + y 2 ≤ a2 , x ≥ 0,

x2 + y 2 ≤ z ≤ a}.

By considering the dimensions we get x, y, z ∼ a and

I=
xy 2 z dΩ ∼ ·a2 · a · a3 = a7 .



· · · dΩ ∼ a3 , so

A

Therefore, the result must be of the form constant·a7 .

D. The shape of A (as a part of a body of revolution) is an invitation to use semi-polar coordinates
(it cannot be said too often: Do not forget the weight function ! ), where A is represented by



π
π

Ã = (, ϕ, z)  0 ≤  ≤ a, − ≤ ϕ ≤ ,  ≤ z ≤ a .
2
2

www.job.oticon.dk

56

Real Functions in Several Variables

The space integral

1
1
0.5
0.5
0
0.2
0.4
–0.5

0.6
0.8
1

–1

Figure 35: The domain A for a = 1.

1.2

1

0.8

y

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

1.2

x
–0.2

Figure 36: The cut in the meridian half-plane for a = 1.

I. The cut B(ϕ), which is revolved around the z-axis, must be independent of ϕ, so have in the
meridian half plane
B(ϕ) = {(, z) | 0 ≤ z ≤ a, 0 ≤  ≤ z}.
Then we get by the reduction theorem that the ϕ-integral can be factored out,

 π 
I

2

=

−π
2


=

=
=

π
2

−π
2

B(ϕ)

 cos ϕ · 2 sin2 ϕ · z ·  d dz

sin2 ϕ · cos ϕ dϕ ·



dϕ

4 z d dz

B(ϕ)

 π2

 a
 z
 a
1
2
1
sin3 ϕ
·
z
4 d dz = ·
z · z 5 dz
3
3 0
5
0
0
−π
2
 a
2
2 1 7
2 7
z 6 dz =
· a =
a .
15 0
15 7
105

C. It is seen as a weak control that the result is of the form constant·a7 as mentioned in A. ♦

57

Real Functions in Several Variables

4.3

The space integral

Spherical coordinates

Example 4.5
A. Let A be an upper half sphere of radius 2a, from which we have removed a cylinder of radius a
and then halved the resulting domain by the plane x + y = 0. We shall only consider that part for
which x + y ≥ 0. Calculate the space integral

xz dΩ.
A

2
–2

–2

1
–1

–1
0
1

1

2

2

Figure 37: The domain A for a = 1 in the (x, y, z)-space.

2
1.8
1.6
1.4
1.2
y

1
0.8
0.6
0.4
0.2
0

0.5

1

1.5

2

x

Figure 38: The cut in the meridian half-plane for a = 1, i.e. in the (, z)-half-plane.

When we consider the dimensions (i.e. a rough overview) we get

· · · dΩ ∼ a3 ,
x ∼ a, y ∼ a, z ∼ a,
A

58

Real Functions in Several Variables

The space integral

2

y

–2

1

–1

2

1
x

–1

–2

Figure 39: The projection of A onto the (x, y)-plane for a = 1.

from which A xz dΩ ∼ a5 , and thus

xz dΩ = constant · a5 .
A

D. The geometrical structure of revolution and the sphere indicate that one either should apply I 1.
semi-polar coordinates or I 2. spherical coordinates. We shall in the following go through both
possibilities for comparison.

59

Real Functions in Several Variables

The space integral

I 1. In semi-polar coordinates the domain A is represented by



3π
π
, 0 ≤ z ≤ 4a2 − 2 .
Ã = (, ϕ, z)  a ≤  ≤ 2a, − ≤ ϕ ≤
4
4
Hence by the reduction theorem (where the weight function is ),
 3π 
4
xz dΩ =


I

=

−π
4

A


=

3π
4

−π
4



cos ϕ dϕ ·

=
=
=



2a

2

 √ 2 2
4a −
0


 cos ϕ · z dz

 √ 2 2
4a −


a


 d dϕ


z dz

d

0



√4a2 −2

√ 1  2a 2
2·
 4a2 − 2 d
2
a
a
0
√  2a
√ 
2a
2
2 4 2 3 1 5
2 2
4
4a  −  d =
a  − 
2 a
2
3
5
a
√



4 2 3 1 5
2
4 2
32 5
a ·a − a
3 a · 8a3 −
a −
3
2
4
5
5
√

2 5 32 32 4 1
−
− +
a
5
3 5
3
2
√
√

28 31
2 5
47 2 a5
−
a ·
.
=
3
5
2
30

= [sin ϕ]
=

a

2a

3π
4
−π
4

·

2a

2

1 2
z
2

d =

I 2. If we instead choose spherical coordinates then
x = r sin θ cos ϕ,

y = r sin θ sin ϕ,

z = r cos θ,

where θ is measured from the z-axis (and not from the (x, y)-plane, which one might expect), and
the weight function is r 2 sin θ, and the domain A is represented by the parametric space


 π
3π π
π
a

, ≤θ≤ ,
≤ r ≤ 2a ,
Â = (r, ϕ, θ)  − ≤ ϕ ≤
4
4 6
2 sin θ
where the vertical bounding line for B0 is described by r sin θ = a, so the lower bound for r is
a
≤ r.
sin θ

60

Real Functions in Several Variables

The space integral

Then we get by the reduction theorem
 π 
 
 3π

2a
2
4
xz dΩ =
r sin θ cos ϕ · r cos θ r 2 sin θ dr dθ dϕ
π
6

−π
4

A


=

3π
4

−π
4



3π

4

=
=
=
=



cos ϕ dϕ ·

= [sin ϕ]−4 π ·
=

a
sin θ


√
2·5

π
2
π
6

π
2
π
6

π
2
π
6



2

sin θ cos θ

sin2 θ · cos θ



a
sin θ

1 5
r
5

sin2 θ cos θ · 32 a5 −



2a

4

r dr

dθ

2a

dθ
a
sin θ

a5
sin5 θ


dθ

√

 π
1
2 5 2
32 sin2 θ −
cos θ dθ
a
π
5
sin3 θ
6
√

 π
π2
1 1
2 5 32
sin3 θ +
a
2
3
2 sin θ 6
5
√



32 1
32 1 1
2 5
+
· + ·4
a
−
3
2
3 8 2
5
√
√

2 5 32 1 4
47 2 5
a
+ − −2 =
a .
5
3
2 3
60

C. We see in both variants that the result is ∼ a5 , so we get a weak control, cf. the examination of
the dimensions in A. ♦
Example 4.6
A. Let A be the spindle formed domain on the ﬁgure, which is obtained by revolving the ﬁgure in the
meridian half-plane around the z-axis. Calculate the space integral I = A z dΩ.

1

0.8

0.6

0.4

0.2
–0.3
–0.2
–0.1
0.3

0.2

–0.3
–0.2
0 –0.1
0.10.1
0.2

0.3

Figure 40: The domain A for a = 1 in the (x, y, z)-space.
An examination of the dimensions gives z ∼ a and

I=
z dΩ = c · a · a3 = c · a4 ,


A

· · · dΩ ∼ a3 , hence

A

61

Real Functions in Several Variables

The space integral

1

0.8

0.6
y

0.4

0.2

0

0.1

0.2

0.3

0.4

0.5

x

Figure 41: The meridian half-plane with the curve r =
the distance from (0, 0) to a point on the curve.

√
cos 2θ for a = 1 given spherically, i.e. r is

where the task now is to ﬁnd c.
D. Since A is a domain of revolution it is natural either to choose spherical or semi-polar coordinates.
In this particular case is the variant I 1. Spherical coordinates the easiest one to apply. We have
for comparison added I 2, so one can see what may happen if one only chooses to apply one
method on all problems.
I 1. We have in spherical coordinates that
x = r sin θ cos ϕ,

y = r sin θ sin ϕ,

z = r cos θ,

represent A by the parametric domain (cf. the ﬁgure over the meridian half-plane)

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62

Real Functions in Several Variables

The space integral

√
π
, 0 ≤ r ≤ a cos 2θ}.
4

Ã = {(r, ϕ, θ) | 0 ≤ ϕ ≤ 2π, 0 ≤ θ ≤

Since neither the bounds for θ or r, the weight function r 2 sin θ or z = r cos θ contain the variable
ϕ, we can factorize the integrations
 2π
1 dϕ = 2π,
0

hence we get by the reduction theorem,

I



π
4

= 2π

0

0



π
4

= 2π

√
a cos 2θ


2

r cos θ · r sin θ dr



√
a cos 2θ

cos θ sin θ

dθ


3

r dr

dθ.

0

0

We ﬁrst calculate the inner integral, where θ is considered as a constant,


√
a cos 2θ

0



1 4
r
r3 dr =
4

a√cos 2θ
=
0

a4
cos2 2θ.
4

Then by insertion,

I

= 2π
0

π
4

cos θ sin θ ·

a4
cos2 2θ dθ
4


 π4
 π4
πa4
1
πa4
2
2
cos 2θ sin 2θ dθ =
cos 2θ −
=
4 0
4 θ=0
2


4
4
4
πa
1
π
πa
πa
cos3 2θ
{1 − 0} =
.
=
= −
8
3
24
24
0 4

d cos 2θ

I 2. Let us now turn to the semi-polar variant.
√ The problem with this is to ﬁnd  = P (z) as a
function of z for the boundary curve r = a cos 2θ for the domain in the meridian half-plane.
When we express  by means of r and θ we get (cf. e.g. the meridian-half-plane)
hence r 2 = 2 + z 2 .
√
Then the task is to eliminate r and θ from r = a cos 2θ. By squaring this equation we get
 = r sin θ

and z = r cos θ.

r2 = a2 cos 2θ = a2 cos2 θ − sin2 θ .

63

Real Functions in Several Variables

The space integral

It follows from the expressions of  and z that we can get rid of cos2 θ and sin2 θ by multiplying
by r2 , i.e.
r2

2

= 2 + z 2

2

= a2 r2 cos2 θ − r2 sin2 θ = a2 z 2 − a2 2 .

Since r ≥ 0 and  ≥ 0, these two operations are “equivalent”, i.e. we have not obtained some
further “false solutions”.
When the equation above is rearranged we get an equation of second order in 2 ,
2

2

+ 2z 2 + a2 2 + z 4 − a2 z 2 = 0.

This is solved in high school manner, where we just put + in front of the square root, because
2 > 0,
2

=
=
=


1
−(2z 2 + a2 ) + (2z 2 + a2 )2 − 4z 4 + 4z 2 a2
2

1
4z 4 + 4z 2 a2 + a4 − 4z 4 + 4z 2 a2 − 2z 2 + a2
2

1 
a 8z 2 + a2 − (2z 2 + a2 ) .
2

A test shows that 2 = 0 for z = 0 and z = a, which is in harmony with the situation in the
meridian half-plane.
Formally A is therefore in semi-polar coordinates represented by the parametric space



 


1

a 8z 2 + a2 − (2z 2 + a2 )
Â = (, ϕ, z)  ≤ ϕ ≤ 2π, 0 ≤ z ≤ a, 0 ≤  ≤
,

2
which does not look too nice. However, it is not as bad as it seems to be, because the integrand
only depends on z. By using the “slice method”, we get by a reduction that





a

z dΩ =

I=
A

a

z
0

dS
B(z)

dz =
0

z · areal B(z) dz,

where B(z) is the disc of radius
 

1
a 8z 2 + a2 − (2z 2 + a2 ) .
 = P (z) =
2
Then
area B(z) = π {P (z)}2 =


π 
a 8z 2 + a2 − (2z 2 + a2 ) ,
2

64

Real Functions in Several Variables

The space integral

from which

 a
I =
z dΩ =
z · area B(z) dz
0
A


π a
=
a 8z 2 + a2 − 2z 2 − a2 z dz (substitute: t = z 2 , dt = 2z dz)
2 0
 2

π a 
=
a 8t + a2 − 2t − a2 dt
4 0
 a2
a2
π 2
π
t + a2 t 0
8t + a2 dt −
=
a
4
4
0
a2

3
π 1 2
π
=
·
8t + a2 2
− a4
a 8 3
2
0
π
1
π
π 4
π 4
=
a·
{27a3 − a3 } − a4 =
a (13 − 12) =
a .
4
12
2
24
24
C. Weak control. We see in both cases that the result is of the form c · a4 , as already deduced in A.
Remark 4.1 We see that this problem could be calculated in both spherical and semi-polar coordinates, though the variant of the semi-polar coordinates was far more diﬃcult than the spherical
version. Occasionally one may ﬁnd similar problems in examinations sets, where the composer of the
problem thought that it was obvious to use the spherical coordinates, while the students unfortunately
preferred the semi-polar coordinates instead. This has through the years caused a lot of frustration.
Therefore, try also to learn the spherical method as well as the semi-polar version, and examine the
role of the geometry in the choice of method. ♦





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



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65

Real Functions in Several Variables

5

The line integral

The line integral

Example 5.1
A. Find the curve length from (0, 0) of any ﬁnite piece (0 ≤ ϕ ≤ α) of the Archimedes’s spiral, given
in polar coordinates by
 = a ϕ,

0 ≤ ϕ < +∞,

hvor a > 0,

i.e. calculate the line integral
 α
=
ds.
ϕ=0

8

6

4

2

–8

–6

–4

–2

0

2

4

6

–2

–4

–6

Figure 42: A piece of the Archimedes’s spiral for a = 1.

D. First ﬁnd the line element ds expressed by means of ϕ and dϕ.
We shall here meet a very unpleasant integral, which we shall calculate in three diﬀerent ways:
1) by a substitution,
2) by using partial integration,
3) by using a pocket calculator.
I. Since  = P (ϕ) = a ϕ, and since we have a description of the curve in polar coordinates, the line
element is
ds =

{P (ϕ)}2 + {P  (ϕ)}2 dϕ =

(a ϕ)2 + a2 dϕ = a

Then by a reduction,
 α
 α

=
ds =
a 1 + ϕ2 dϕ = a
ϕ=0

0

α

1 + ϕ2 dϕ.

1 + ϕ2 dϕ.

0

66

Real Functions in Several Variables

The line integral

1) Since 1 + sinh2 t = cosh2 t, we have 1 + sinh2 t = + cosh t, because both sides of the equation
sign must be positive. Thus we can remove the square root of the integrand by using the
monotonous substitution,


ϕ = sinh t,
dϕ = cosh t dt,
t = Arsinh ϕ = ln ϕ + 1 + ϕ2 .
Since t can be expressed uniquely by ϕ, the substitution must be monotonous.
Then



α

= a
0



α

= a
ϕ=0

=
=
=
=

a
2
a
2
a
2
a
2





α

1 + ϕ2 dϕ = a
cosh2 t dt = a ·
α

1
2

ϕ=0
 α
ϕ=0

1 + sinh2 t · cosh t dt
{cosh 2t + 1} dt

a
α
[(t + sinh t · cos t)]ϕ=0
2
ϕ=0
α

t + sinh t · 1 + sinh2 t
ϕ=sinh t=0

α
 
ln ϕ + 1 + ϕ2 + ϕ · 1 + ϕ2
 0


2
2
.
α 1 + α + ln α + 1 + α
1
sinh 2t + t
2

=

2) If we instead apply partial integration, then
 α
 α
2
= a
1 + ϕ dϕ = a
1 · 1 + ϕ2 dϕ
0
0
 α
α

ϕ
2
−a
ϕ·
dϕ
= a ϕ 1+ϕ
0
1 + ϕ2
0


 α 2
ϕ +1 −1
2
dϕ
= a α 1+α −
1 + ϕ2
0


 α
 α
dϕ
2
2
= a α 1+α −
1 + ϕ dϕ +
1 + ϕ2
0
0
 α



= −a
1 + ϕ2 dϕ + a α 1 + α2 + ln α + 1 + α2 .
0

α
The ﬁrst term is −a 0
1 + ϕ2 dϕ = − , so we get by adding


a 
=
α 1 + α2 + ln α + 1 + α2 .
2

and dividing by 2 that

3) This is an example where a pocket calculator will give an equivalent, though diﬀerent answer,
so it is easy to see for the teacher, whether a pocket calculator has been applied or not. It is
here illustrated by the use of a TI-89, where the command is given by

a
( (1 + t^2), t, 0, b),
because neither ϕ nor α are natural. Then the answer of the pocket calculator is

 √
√
ln( b2 + 1 + b) b b2 + 1
+
,
(9) a ·
2
2

67

Real Functions in Several Variables

The line integral

followed by writing α again instead of b.
However, if one does not apply a pocket calculator, but instead uses the standard methods of
integration, one would never state the result in the form (9). The reason for this discrepancy
is that the programs of the pocket calculator are created from specialists in Algebra, and
they do not always speak the same mathematical language as the specialists in Calculus or
Mathematical Analysis. In Calculus the priority of the terms would be (b = α)


a 
α 1 + α2 + ln α + 1 + α2 ,
2
because one would try to put as many factors as possible outside the parentheses and then
order the rest of the terms, such that the simplest is also the ﬁrst one. Obviously, this is not
the structure of (9).
The phenomenon was discovered at an examination where pocket calculators were only allowed
if one also wrote down the applied command and the type of the pocket calculator. Many
students did not do it, and yet it was discovered that they had used a pocket calculator.
The morale of this story is that even if a pocket calculator may give the right result, this result
does not have to be put in a practical form. It is even worse by applications of e.g. MAPLE

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68

Real Functions in Several Variables

The line integral

where the result is sometimes given in a form using functions which are not known by students
of Calculus.
√
Note also that pocket calculators in general do not like the operations | · | and ·, and cases
where we have got two parameters. The latter is not even one of the favorites of MAPLE
either, and it is in fact possible to obtain some very strange results by using MAPLE on even
problems from this part of Calculus. I shall therefore warn the students: Do not use pocket
calculators and computer programs like MAPLE or Mathematica uncritically! Since they exist,
they should of course be applied, but do it with care. ♦

Example 5.2


A. Find the value of the line integral I =
by
 = P (ϕ) = a(1 + cos ϕ),

K

|y| ds, where K is the cardioid given in polar coordinates

−π ≤ ϕ ≤ π.

1

0.5

0

0.5

1

1.5

2

–0.5

–1

Figure 43: The cardioid for a = 1; (καδια = heart).

Examination of dimensions: Since  ∼ a, We get
be of the form c · a · a = c · a2 .


K

· · · ds ∼ a, and since y ∼ a, The result must

Due to the numerical sign in the integrand we must be very careful. In particular, a pocket
calculator will be in big trouble here, if one does not give it a hand from time to time during the
calculations.
D. First ﬁnd the line element ds.
I. The line element is seen to be
{P (ϕ)}2 + {P  (ϕ)}2 dϕ = {a(1 + cos ϕ)}2 + (−a sin ϕ)2 dϕ

√
= a (1 + 2 cos ϕ + cos2 ϕ) + sin2 ϕ dϕ = a 2 · 1 + cos ϕ dϕ.

ds =

69

Real Functions in Several Variables

The line integral

2
1.5
3

1
2.5
2

0.5

1.5
1
0.5

0
0.1
0.2

0.3
0.4
0.5
0.6
0.7

Figure 44: The space curve x = r(t).

The line integral becomes

I

2

=
1



2

=
1


 
√
1 2
1
1
2
t
+ t dt
· ( 2 · t) + 2
·
2
t
t

 2
 2
1
1
+ t dt =
2t2 + t2 ·
(3 + 3t2 ) dt = 3 + t3 1 = 10.
t
t
1

♦

Example 5.4
A. Let a, h > 0. Consider the helix
t ∈ R.

r(t) = (x, y, z) = (a cos t, a sin t, h t),
2

2

2

This is lying on the cylinder x + y = a .
Find the natural parametric representaion of the curve from (a, 0, 0), corresponding to t = 0.

2
1.5
1
–1

–1
0.5
–0.5

–0.5

0.5

0.5
1

1

Figure 45: The helix for a = 1 and h =

1
.
5

D. Find the arc length s = s(t) as a function of the parameter t. Solve this equation t = t(s), and
put the result into the parametric representation above.

70

Real Functions in Several Variables

The line integral

I. Let us ﬁrst ﬁnd the line element ds = r (t) dt. Since
r (t) = (−a sin t, a cos t, h),
we have
r (t) =

a2 sin2 t + a2 cos2 t + h2 =

hence the arc length is
 t


s = s(t) =
r (τ ) dτ =
0

t

a2 + h2 ,

a2 + h2 dτ =

0

a2 + t2 · t.

By solving after t we get
s
t = t(s) = √
.
2
a + h2
When this is put into the parametric representation of the helix, we get
(x, y, z) = (a cos t, a sin t, h t)



s
s
=
a cos √
, a sin √
2
2
2
a +h
a + h2

,√

hs
+ h2

a2

,

s ∈ R,

which is the natural parametric representation of the helix. ♦

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71

Real Functions in Several Variables

6

The surface integral

The surface integral

Example 6.1


A. Find the surface integral I =

F

|z| dS, where F is given by the parametric representation

(x, y, z) = r(u, v) = (u sin v, u cos v, u v) = u (sin v, cos v, v),
where −1 ≤ u ≤ 1, 0 ≤ v ≤ 1.

–1
1
–0.5
–0.2
0.8

0.6

0.4

–0.6

–0.4

–0.8

0.2
0.5
1

–1

Figure 46: The surface F has two components.
If we keep u = 1 ﬁxed and let v vary, then we get an arc of the helix with a = h = 1, cf.
Example 5.4. When (0, 0, 0) is removed, the surface is split into its two components F 1 and F2 ,
which are symmetric with respect to the point (0, 0, 0). The surface F1 is obtained by drawing all
lines from (0, 0, 0) to a point on the helix.
D. The area element is given by dS = N(u, v) du dv. We ﬁrst calculate the normal vector N(u, v)
corresponding to the given parametric representation.
I. It follows from r(u, v) = u (sin v, cos v, v) that
∂r
= (sin v, cos v, v),
∂u

∂r
= u (cos v, − sin v, 1),
∂v

hence the normal vector is
N(u, v)

=


 e1
∂r
∂r 
sin v
×
=
∂u ∂v 
u cos v

e2
cos v
−u sin v

e3
v
u








= u (cos v + v sin v, v cos v − sin v, −1)
= u{(cos v, − sin v, −1) + v (sin v, cos v, 0)}.
Now
(cos v, − sin v, −1) · (sin v, cos v, 0) = 0,
69

72

Real Functions in Several Variables

The surface integral

so the two vectors are perpendicular. Then we get by Pythagoras’s theorem


N(u, v) 2 = u2 (cos v, − sin v, −1) 2 + v 2 (sin v, cos v, 0) 2


= u2 cos2 v + sin2 +1 + v 2 sin2 v + cos2 +02


= u2 2 + v 2 .
Notice that −1 ≤ u ≤ 1 shows that u may be negative. When we take the square root we get the
area element
dS = N(u, v) du dv = |u|

2 + v 2 du dv.

Putting D = [−1, 1] × [0, 1] we get by the reduction formula


|z| dS =
|u v| · |u| 2 + v 2 du dv
I =
F
D

 1  1
 1
 1
2
2
2
=
|u| |v| 2 + v dv du =
u du ·
0
−1
−1
0

1  3
3


√
1 3
1
2
1 2 3
u
t · dt = ·
t2
·
=
3
2
3
2 3
2
−1
2
√
2 √
=
(3 3 − 2 2).
♦
9

2 + v 2 · v dv

Example 6.2
A. Let F be the surface given by the graph representation
√
0 ≤ x ≤ 3,
0 ≤ y ≤ 1 + x2 ,
z = xy.

Find the surface integral F z dS.

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73

Real Functions in Several Variables

The surface integral

3.5
3
2.5

2

2
1.5

1.5
1

1

t

0.5

0.5

0
0.2
0.4
0.6
0.8
s

1
1.2
1.4
1.6

Figure 47: The surface F with its projection E.
D. The usual procedure is to consider F as a graph of the function
z = f (x, y) = xy,

(x, y) ∈ E.

We shall not do this here, but instead alternatively introduce a rectangular parametric representation (x, y, z) = r(u, v). Then afterwards we shall ﬁnd the weight function N(u, v) .

2

1.5

y
1

0.5

0

0.5

1

1.5

2

x

Figure 48: The projection E of F in the (x, y)-plane.

I. The parameters u and v are for obvious reasons not given above. They are introduced by duplicating
(x, y) by the trivial formula
(x, y) = (u, v),
i.e. we choose the parametric representation
r(u, v) = (x, y, z) = (u, v, uv),

0≤u≤

√

3,

0≤v≤

1 + u2 ,

74

Real Functions in Several Variables

The surface integral

so we can distinguish between (x, y) as the ﬁrst two coordinates on the surface in the 3-dimensional
space, and (u, v) ∈ E in the parametric domain. By experience it is always diﬃcult to understand
why we use this duplication, until one realizes that we in this way can describe two diﬀerent aspects
(as described above) of the same coordinates. This will be very useful in the following.
Since
∂r
= (1, 0, v)
∂u

∂r
= (0, 1, u),
∂v

and

the corresponding normal vector becomes


 e1 e2 e3 


∂r
∂r 
N(u, v) =
×
=  1 0 v  = (−v, −u, 1).
∂u ∂v 
0 1 u 
Hence
1 + u2 + v 2 .

N(u, v) =

When dS denotes the area element on F, and dS1 denotes the area element on E, then we have
the correspondence
dS =

1 + u2 + v 2 dS1 .

The abstract surface integral over F is therefore reduced to the abstract plane integral over E by


z dS =
u v 1 + u2 + v 2 dS.
F

E

Then we reduce the abstract plane integral over E in rectangular coordinates, where the v-integral
is the inner one,



 √  √ 2
z dS =
F

uv

3

1 + u2 + v 2 dS =
0

E

1+u

u

1 + u2 + v 2 v dv

du.

0

Calculate the inner integral by means of the substitution
t = v2 ,

dt = 2v dv.

From this we get


√
1+u2


1+

0



u2

+

v2

1+u2

v dv =
0

1+u2

1 + u2 + t ·

1 2
1
1 + u2 + t
=
2(1 + u2 )
2 3
3
t=0
3
1 √
2 2
= (2 2 − 1) · 1 + u
.
3

=

3
2

1
dt
2
3
2

3

− (1 + u2 ) 2



75

Real Functions in Several Variables

The surface integral

By insertion and b the substitution t = u2 , dt = 2u du we ﬁnally get



z dS

=

F

=
=
=
=
=

√

3
1 √
(2 2 − 1) · 1 + u2 2 du
3
0
 √3
3
1 √
1 + u2 2 u du
(2 2 − 1)
3
0
 3
3 1
1 √
(1 + t) 2 dt
(2 2 − 1)
2
3
0
3

√
5
1
1 2
2
(2 2 − 1) ·
(1 + t)
3
2 5
0
 5

1 √
(2 2 − 1) · 4 2 − 1
15 √
31(2 2 − 1)
.
♦
15

3

u·

Example 6.3

A. A surface of revolution O is obtained by revolving the meridian curve M given by
π
r = a(1 + sin θ),
0 ≤ θ ≤ , a > 0,
2

76

Real Functions in Several Variables

The surface integral

where θ is the angle measured from the z-axis and r is the distance to (0, 0) (an arc of a cardioid,
cf. Example 5.2). Find the surface integral

z
I=
dS.
2 + y2 + z2
x
O

1.2
1
0.8
0.6
0.4

–2

–2

0.2

–1

1

–1

1
2

2

Figure 49: The surface O for a = 1.
An examination of the dimensions shows that x, y, z ∼ a and

z
a
dS ∼ 2 · a2 = a.
2
2
2
a
O x +y +z


O

· · · dS ∼ a2 , thus

1.2

1

0.8

0.6

0.4

0.2

0

0.5

1

1.5

2

Figure 50: The meridian curve M for a = 1.

The ﬁnal result must therefore be of the form c · a, where c is the constant, we are going to ﬁnd.
D. When we look at surfaces (or bodies) of revolution one should always try either semi-polar or
spherical coordinates. Since the parametric representation of the meridian curve M is given in a
way which is very similar to the spherical coordinates, it is quite reasonable to expect that one
should use spherical coordinates.
Although it is here possible to solve the problem by a very nasty trick, it is far better for pedagogical
reasons to follow the way which most students would go. Let us analyze the reduction formula

 b  2π

f (x, y, z) dS =
F (t, ϕ) dϕ R(t) sin Θ(t) {R (t)}2 + {R(t)Θ (t)}2 dt,
O

a

0

77

Real Functions in Several Variables

The surface integral

where
F (t, ϕ) := f (R(t) sin Θ(t), cos ϕ, R(t) sin Θ(t) sin ϕ, R(r) cos Θ(t).
There is no t in A., so we must start by introducing the parameter t in a convenient form. Then
we shall identify the transformed function F (t, ϕ) as well as the weight function, and ﬁnally we
shall carry out all the integrations.
I. 1) The introduction of the parameter t. The most obvious thing is to by θ = t, i.e. M is described
by
r = R(t) = a (1 + sin t),

0≤t≤

θ = Θ(t) = t,

π
.
2

By doing this we split the diﬀerent aspects: θ belongs to the curve M, and t belongs to the
parametric interval
 π
0,
= [a, b].
2
2) Identiﬁcation of F (t, ϕ) and the weight function. Since
z = R(t) cos Θ(t) = a(1 + sin t) cos t

on M,

x2 + y 2 + z 2 = r2 = R(t)2 = a2 (1 + sin t)2

on M,

and

we obtain the integrand
f (x, y, z) =

z
a(1+sin t) cos t
cos t
= 2
=
= F (t, ϕ),
x2 +y 2 +z 2
a (1+sin t)2
a(1+sin t)

which is independent of ϕ. Since the weight function and the boundaries of does not depend
2π
on t either, the ϕ-integral becomes trivial, and we can put 0 dϕ = 2π outside the integral as
a factor.
Then we calculate the weight function,
R(t) sin Θ(t)

{R (t)}2 + {R(t)Θ (t)}2

{a cos t}2 + {a(1 + sin t) · 1}2

= a(1 + sin t) · sin t · a cos2 t + (1 + 2 sin t + sin2 t)
= a(1 + sin t) · sin t ·

= a2 (1 + sin t) · sin t · 2(1 + sin t)
√
3
= 2 a2 (1 + sin t) 2 · sin t.

78

Real Functions in Several Variables

The surface integral

3) Integration by reduction. First we note that the parametric domain is 2-dimensional,


π

.
D = (ϕ, t)  0 ≤ ϕ ≤ 2π, 0 ≤ t ≤
2

In fact, dimension corresponds to dimension, and since F is a C ∞ -surface, the parametric
domain D must necessarily be 2-dimensional. (If not we have made an error, so start from the
very beginning!)

79

Real Functions in Several Variables

The surface integral

We have now identiﬁed all functions, so we get by the reduction formula that


√
3
z
cos t
· 2 a2 (1 + sin t) 2 sin t dϕ dt
dS =
2 + y2 + z2
x
a(1
+
sin
t)
O
D
 π2
√
1
sin t(1 + sin t) 2 cos t dt
=
2 · a · 2π
√



√



√



1

= 2 2πa
0

1

0
1

= 2 2 πa

=
=
=
=
=

1

u(1 + u) 2 du

= 2 2πa

=

0

0

1

(1 + u − 1) (1 + u) 2 du


1
3
(1 + u) 2 − (1 + u) 2 du



1
5
3
2
2
(1 + u) 2 − (1 + u) 2
2 2πa
5
3
0

√
5
3 1
2 
2 2πa ·
3(1 + u) 2 − 5(1 + u) 2
15
0

 3

4πa √   5
2
2
· 2 3 2 −1 −5 2 −1
15
√
√
4πa √
2 {3(4 2 − 1) − 5(2 2 − 1)}
15
√
4πa √
8πa √ √
2 {2 2 + 2} =
2 ( 2 + 1)
15
15
√
8π(2 + 2)a
.
15
√

C. Weak control. The result has the form c · a, in agreement with A.
Since z ≥ 0 on O [cf. the ﬁgure], the result must be ≥ 0. We see that this is also the case here. ♦

Example 6.4
A. A surface of revolution O has an arc of a parable M as its meridian curve, where this is given by
the equation
z=

2
,
a

0 ≤  ≤ a,

a > 0.

Find the surface integral

x2
√
I=
dS.
a2 + 4az
O

80

Real Functions in Several Variables

The surface integral

1
0.8
0.6
0.4
–1

–1
0.2
–0.5

–0.5

0.5

0.5
1

1

Figure 51: The surface O and its projection onto the (x, y)-plane for a = 1.

1.2

1

0.8

y 0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

1.2

x

Figure 52: The meridian curve M for a = 1.

Examination of the dimensions. It follows from x,y, z ∼ a, that
x2
a2
√
∼ √ = a.
a2 + 4az
a2

Since O · · · dS ∼ a2 , we get all together

x2
√
dS ∼ a · a2 = a3 ,
a2 + 4az
O
i.e. the result must have the form

x2
√
dS = c · a3 ,
2 + 4az
a
O
where the constant c must be positive, because the integrand is ≥ 0.
D. The description invites to semi-polar coordinates I 1. For the matter of training we also add I 2.
Rectangular coordinates, which give a slightly diﬀerent variant, although we in the end is forced
back to (semi-)polar coordinates.

81

Real Functions in Several Variables

The surface integral

I 1. Semi-polar coordinates. We introduce t as a parameter by
 = P (t) = t.
Then
z = Z(t) =

1 2
t ,
a

0 ≤ t ≤ a.

Since
x = P (t) cos ϕ = t cos ϕ,

y = P (t) sin ϕ = t sin ϕ,

z = Z(t) =

1 2
t ,
a

we get the following interpretation of the integrand,
f (x, y, z) = √

t2 cos2 ϕ
x2
,
=√
a2 + 4t2
a2 + 4az

and the weight function is

P (t)

{P  (t)}2

+

{Z  (t)}2

=t


12

+

2
t
a

2

=

t
a

a2 + 4t2 .

The parametric domain is

D = {(t, ϕ) | 0 ≤ t ≤ a, 0 ≤ ϕ ≤ 2π} = [0, a] × [0, 2π].

82

Real Functions in Several Variables

The surface integral

Hence we get by a reduction

O

√

x2
dS
a2 + 4az


=
D

t2 cos2 ϕ t
√
·
a2 + 4t2 a

a2 + 4t2 dt dϕ

 2π

1 a 3
t cos ϕ dt dϕ =
t dt ·
cos2 ϕ dϕ
a 0
0
D
a  2π

1 + cos 2ϕ
πa3
1 1 4
.
dϕ =
·
t
4
2
a 4
0
0

1
a

=
=



3

2

C 1. Weak control. The result has the right dimension [a3 ], and it is positive, cf. A.
I 2. The rectangular version. In this case we interpret the surface as the graph of the function
z = f (x, y) =

1 2
(x + y 2 )
a

for (x, y) ∈ E,

where the parametric domain is the disc
E = {(x, y) | x2 + y 2 ≤ a2 }.
The weight function is



2
∂z
∂z
1+
+
∂x
∂y

2

=


1+

2x
a



2

+

2

2y
a

=

1
a

a2 + 4x2 + 4y 2 .

We have now found everything which is needed for an application of the reduction theorem:

O

x2
√
dS
a2 + 4az


E

=

1
a

x2
1
a2 + 4a · (x2 + y 2 )
a
x2
·
a2 + 4x2 + 4y 2



=


E

·

1
a

a2 + 4x2 + 4y 2 dx dy

a2 + 4x2 + 4y 2 dx dy =

1
a



x2 dx dy.

E

From this point it is again most natural to change to polar coordinates,

O

x2
√
dS
a2 + 4az

=
=


 a

1 2π
2
x dx dy =
( cos ϕ) ·  d dϕ
a 0
0
E

 a
πa3
1 2π
1
a4
=
.
cos2 ϕ dϕ ·
3 d = · π ·
4
4
a 0
a
0
1
a



2

♦

83

Real Functions in Several Variables

7

Transformation theorems

Transformation theorems

Example 7.1
A. Calculate the plane integral


y−x
cos
I=
dx dy
y+x
B
over the trapeze shown on the ﬁgure.

3

2.5

2

y
1.5

1

0.5

0

0.5

1

1.5

2

x

Figure 53: The trapeze B.

D. A direct calculation applying one of the usual reduction theorems is not possible, because none of
the forms






y−x
2y
2x
− 1 dx = cos 1 −
dy
cos
dx = cos
y+x
y+x
y+x
can be integrated within the realm of our known functions. The situation is even worse in polar
coordinates. Therefore, the only possibility left is to ﬁnd a convenient transform, such that the
integrand becomes more easy to handle.
y−x
. One idea would be to introduce the numerator
y+x
as a new variable, and the denominator as another new variable. If we do this, then we must show
that we obtain a unique correspondence between the domain B and a parametric domain D, which
also should be found. Finally we shall ﬁnd the Jacobian. When we have found all the terms in the
transformation formula, then calculate the integral.
The unpleasant thing is of course the fraction

Remark 7.1 This time we see that it is here quite helpful to start the discussion in D, which is
not common knowledge from high school. First we discuss the problem. Based on this discussion
we make a decision on the further procedure. ♦
I. According to D. we choose the numerator and the denominator as our new variables. Most people
would here choose the numerator as u and the denominator as v, so we shall do the same, although
it can be shown that we shall get simpler calculations if we interchange the deﬁnition of u and v.

84

Real Functions in Several Variables

Transformation theorems

We therefore put as the most natural choice
(10) numerator :

u = y − x and

denominator :

v = y + x.

Then we shall prove that this gives a one-to-one correspondence. This means that we for any
given u and v obtain unique solutions x and y:
u+v
v−y
and
y=
.
x=
2
2
Obviously the transform is continuous both ways. Since B is closed and bounded, the range D by
this transform is again closed and bounded, cf. the important second main theorem for continuous
functions.
Since the transform is one-to-one everywhere, the boundary ∂B is mapped one-to-one onto the
boundary ∂D. This is expressed in the following way:
1) The line x + y = 1 corresponds by (10) to v = 1.
2) The line y = x, i.e. y − x = 0, corresponds by (10) to u = 0.
3) The line y + x = 4 corresponds by (10) to v = 4.
u+v
v−u
, i.e. to v = 2u.
4) The line y = 3x corresponds to
=3
2
2

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