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141
¥
b a
x
y
(
)
(
)
2
=
xdx
Solution: а)
ò
0
;
2
4
+x
а) Improper integral of the 1st order.
2
+=
xt
4
=
dxxdt
2
1
¥¥ b
xdx
=
2
+
x
0
4
1
=
dtxdx
2
=Û=
tx
40
11
¥=Û-¥=
tx
22
dt
2
lim
¥®
b
t
1
1
-
2
===
dtt
òòò
2
44
{ to solve this, let’s use formula (2)}
1
ö
æ
b
2
÷
ç
=
lim
ç
è
==
t
lim
÷
4
ø
ö
æ
b
÷
ç
t
÷
ç
4
ø
è
¥®¥®¥®
bbb
( )
b
the integral is diver-
¥=-=
2lim
gent.
Exercise: Calculate:
а) The area of a shape, defined by: у = х2 and у = 2 – х2;
Solution:
а) There are several formulas for calculating areas of flat shapes.
xfy1=
xfy
Fig.6
Fig.6 – graphic representation of the area of a shape, defined in Cartesian rectangular coordinates, and limited by у = f1(x) from above, by
у = f2(x) from below, by х = а from the left, and by x = b from the right.
The area of a shape, defined in Cartesian rectangular coordinates, and
limited by у = f1(x) from above, by у = f2(x) from below, by x = a from
the left, and by x = b from the right is defined by the formula
b
() ()
[ ]
-=
ò
a
21
dxxfxfS
(14);
142
In our case the lines that define our shape are given in Cartesian co-
Û
1
=
-
=
=
1 -1
x
y
p
££=
=
ordinates, that is why we will use formula (14).
ì
Let’s find crossover points:
1
-=x
1
( )( ) ( ) ( )
1
æ
ç
2
-=
x
ç
è
1
x
2
;
3
ö
1
x
÷
÷
3
-
ø
Þ
1
a
22
æ
ç
12
ç
è
ï
í
ï
î
1
;
1
1
æ
1
ç
ç
3
è
2
=
xy ,
Þ
2
-=
2 xy
1
b
.
1
2
---
3
( )
1
-
1
----=
3
2
12222 dxxdxxdxxxS
òòò
1
ö
ö
8
÷
÷
=
÷
÷
3
ø
ø
22
2 xx -=
Û
=-=-=--=
;
2
xy =
2
2
2 xy -=
1
2
1
=x
Fig.7
Fig.7 – graphic representation of the area of a shape, defined by у = х2
and у = 2 – х2.
Exercise: Calculate the volume of a object made by rotating the shape
, around the Ох axis.
xyxy 0 ;0 ;sin
If function у = f(x) is continuous at segment x ∈ [а; b], then the vol-
ume of object made by rotating a curvilinear trapezoid around the Ох
axis, limited from above by у = f(x), and from below by Ох, is calculated
by the formula:
b
x
=
2
p
ò
a
(20).
()
dxxfV
143
0
x y
(
)
x y
(
)
a
xfy =
b
Fig.8
Fig.8 – graphic representation of the volume of a object made by rotating the shape around the Ох axis.
If a curvilinear trapezoid is limited by a continuous function x = φ(y)
and straight lines x = 0, у = с, у = d (c < d), then the volume of a object
made by rotating this trapezoid around Оy, similarly to formula(20), is
equal to:
d
c
ò
c
d
2
jp
=
y
0
Fig.9
Fig.9 – graphic representation of the volume
of a object made by rotating trapezoid around О
(21).
()
dyyV
yx
j
=
.
y
Under the terms of our problem y = sin x; a = 0; b = π.
144
p
2
2
=
xy
p
x
=
ò
0
-
2cos1
p
p
æ
ç
2
è
p
p
x
1
-= xx
2
2
p
.
=×=
ì
22
sinsin
í
î
pp
æ
p
ç
ç
22
è
p
p
ö
2sin
÷
0
ø
æ
ç
ç
2
è
-
2cos1
x
ü
==-=
xxdxV
2
1
-=
2cos
òòò
000
1
æ
-=
ç
2
è
=
ý
þ
ö
÷
=
dxxdxdx
÷
ø
.
pp
--
ç
÷
ø
2
è
xy sin
1
æ
ö
02sin
ö
ö
0sin
=÷÷
÷
ø
ø
0
Fig.10
Fig.10 – graphic representation of the volume of a object made by rotating the shape y = sin x; y = 0; 0 ≤ x ≤ π, around the Оx axis
Try to do this yourself:
1) Calculate areas of shapes, limited by: y = x2, y = 2 – x2;
2) Calculate the improper integral or establish its divergence
¥
ò
0
2
-
x
×
dxex
π
145
Section 5. Integral calculation
of multi-variable functions
Topic 17. Multiple integrals
Introduction. This topic shows how to integrate functions of two or
more variables. The key idea is to replace a double integral by two ordinary "single" integrals.
Exercise 1. Transform a double integral into an iterated integral
and change its order of integration, if integration domain is D: y = x2;
y = 2 – x; x ≥ 0.
Solution. Integration domain D is proper (plain) in the direction of
ОУ, because any line, parallel to ОУ, crosses the limits of D in no more
than 2 points. Let’s name the first point the line crosses the limit y = x2 an
entry point (the line y = x2 is called an entry line). The second crossover
point y = 2 – x we will call an exit point (the line y = 2 – x is called an
exit line). The iterated integral at the right part is comprised of two defined ones: the first is taken by у, the axes of which ОУ are parallel to
the secant lines, is called an inner integral. The limits of integration of it
depend on х and coincide with the ordinates of crossover points of secant
lines with the entrance line (at the lower limit) and the exit line (at the
upper line). When integrating internally, the variable х is considered a
constant, that is why its result is a function, which (after substituting the
limits of integration) depends of х. The second integral of х is taken from
this function by variable х, and the limits of its integration are the lowest
(for the lower) and highest (for the upper) value of the domain D‘s points
projected onto ОХ axis:
146
-
x
21
òò òò
D
=
2
0
x
dyyxfdxdxdyyxf
),(),(
Fig.11
Fig.11 – graphic representation of the transforming of double integral into an iterated integral with integration domain D: y = x2; y = 2 – x; x ≥ 0.
When changing the order of integration, the entry line to domain D
has the equation x = 0, and the exit line splits into two parts, the first
looks like õ = ó, and the second: õ = 2 = ó. According to the additive
property of a double integral it splits into two, where there was an substi-
tution in each one to an iterative integral with internal integration by variable х, and external integration by variable у:
y
1
D
0
y
-+=2
2
dxyxfdydxyxfdydxdyyxf
òòòò òò
0
10
),(),(),(
Fig.12
Fig.12 – graphic representation of the idea to replace a double integral by two ordinary "single" integrals.
147
(
)
2
2
2
2
Exercise 2. Solve a double integral at domain D, limited by these
functions
2
òò
Fig.13 – graphic representation of conditions of exercise 2
Solution. Integration domain D is proper (plain) in the direction of
ОУ, so we can change the double integral with an iterative one, with an
internal integral by y and an external integral by x . Entry line to D is
1
õó
=
, exit line is a parabola õ = ó. Let’s calculate the internal inte-
gral with a constant х by using the Newton-Leibniz formula with lower
1
limit
õ
and upper limit
D
Fig.13
We now can find the crossover points of
.õ
1
;;
.
xyxydxdyxyx
==-
2
the parabola and the line by solving the system:
õó
=
1
õó
=
The abscissas of the points we just got are the limits of integration in
the lower integral. The process of transforming a double into a two-time
iterated integral goes like this:
1
2
4,0,04,
===-= õõõõõõ
21
148
x
(
)
2
( ) ( )
D
4
æ
ç
ç
è
0
æ
2
ç
ç
7
è
xxx
2
1
7
6
4
0
x
æ
ç
2
è
ö
3
43
32
÷
÷
ø
-- dxxxx
2
1
x
2
1
---
2
4
0
1
2
7
412
128
4
æ
2
ç
-=-=-
òòòòò
ç
è
0
4
ö
ö
222
÷
÷
÷
ø
òò
ø
0
1
64
32
6
xyxdyxyxdxdxdyxyx
5
æ
2
ç
ç
è
3
256
x
2
ö
y
÷
÷
2
ø
1
2
40
=--=
=
dx
1
x
2
ö
3
32
÷
--=
8
=
dxxxxdxxxxx
÷
ø
.
21
Try to do this yourself:
1. Change the order of integration
0
1
-
+
2
-
2
+-
2. Solve a double integral
xy
ì
these conditions
ï
í
ï
î
00
dxyxfdydxyxfdy
òòòò
1
-
--
=
1
.
5
=+
yx
),(),(
yy
òò
D
.
at domain D, defined by
dydxyx
2
Topic 18. Contour and surface integrals
Introduction. This workshop is devoted to the discussion of the following concepts: a line contour integral, calculation of work along a
curve and flow across a curve, the surface integral.
Example.
1) Solve a contour integral of the 1st kind:
где .10;:
3
££= xxyL
3
ò
L
Solution. Solving a contour integral of the 1st kind can come down to
solving a definite integral, the way of transforming it into a definite inte-
,
+
dlyx
149
gral depends on the way the integration curve L is presented. If L is pre-
()(
)
(
)
bjajr
r
££=
(
)
(
)
[
]
(
)
sented by an equation y = φ(x), a ≤ x ≤ b where function y = φ(x) has a
continuous derivative φ'(x) for õ ∈ [à, b], then
b
( ) ()( ) ()
ò ò
L
a
If L is presented by parameters:
+=
2
( )
jj
dxxxxfdlyxf
'1,,
where
,,,
ba
££== ttyytõõ
functions x(t) è y(t) have a continuous derivatives õ'(t), y'(t), for x ∈[a, β]
then
( ) () ()( ) ()( ) ()( )
ò ò
L
b
a
22
dttytxtytxfdlyxf
'',, +=
If L is presented by polar coordinates in an equation
,
jr
'
for
ò ò
L
and function
baj
,Î
, then
b
( ) ( ) () ()( )
a
jr
has a continuous derivative
2
+=
2
dfdlyxf
jjrjrjrjr
'sin,cos,
In our case we have an obvious presentation of cure L by an equation
y = x3. That is why we will use the first method of transforming the inte-
gral into a defined one. At the end we get:
1
( ) ( ) ( ) ( ) ( )
L
0
10
1
2
ò ò
18
1
333
++=+
1
ò
0
912
1
9
2
¢
ö
æ
3
÷
ç
ø
è
43
=+=
dxxx
10
22
dttdttt
1
1
++=
òòò
0
42
+=
91
xt
3
=
362
dxxtdt
=Þ=
1
2
1
tx
=Þ=
tx
10
3
t
39
1
27
1
121 dxxdlyxdxxxxdlyx
=
10
101
( )
-===
2
¢
ö
æ
33
ç
è
=
÷
ø
.11000
2) Calculate the work (A) of a force (F)
2
when
jyxixF -+=
moving a material point on a curve y = x2 from point А(0;0) to point
В(1;1).
150
Solution. Work of a variable force
()()(
)
r
(
)
()(
)
r
r
r
¶
¶
¶
(
)
r
(
)
r
material point on a flat curve L with the equation
on moving a
yxQyxPF ,,,
is calculated
xy
j
=
via a contour integral of the 2nd kind by coordinates
ò
L
+=
dyyxQdxyxPA ,,,
Which comes down to a definite integral (keeping in mind the way
that L is presented). In our case, L is clearly presented by an equa-
tion
we simply need to change
2
. That is why, similarly to the previous example,
10,
££= xxy
1
( )
ò
0
23
2
==
LL
æ
32
-=-=
ç
è
and we will get:
xdxdyxy 2,
1
1
ö
43
xxdxxx
2
1
÷
ø
0
1
=-=
1
.
2
( )
òòò
0
1
2
222
=-+=-+=+=
2)(),(),( xdxxxdxxdyyxdxxdyyxQdxyxPA
Example.
Find the surface integral of the 2nd order
2
òò
s
ydxdyxzdydzdydzzI ++=
outer side of a part of a paraboloid’s surface ,4:
and also of a part of a plane
where closed surface σ consists of an
,
s
1
.0:
s
=z
2
22
-=+ zyx
z ≥ 0,
Solution. Let’s apply the Ostrogradsky-Gauss Divergence formula to
surface integral of the 2nd kind интегралу I:
æ
¶
¶
Q
P
ç
=++=
òò òòò ÷
s
RdxdyQdxdzPdydzI
V
+
ç
è
¶
¶
y
x
ö
¶
R
÷
+
dxdydz
¶
z
ø
.
In vector form the Ostrogradsky-Gauss Divergence formula looks
like this:
òò òòò
s
Where, in the left part, there is a flux П of vector field
==
s
n
dVadivdaÏ ,
V
a
through the
closed surface σ, and
R
Q
P
r
But then
adiv
=
x
¶
where vector field
=
òòò
V
,dVMadivI
+
+
¶
z
y
¶
Ma
looks like:
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