Bai giang math4

H ANOI U NIVERSITY OF S CIENCE AND T ECHNOLOGY
S CHOOL OF A PPLIED M ATHEMATICS AND I NFORMATICS

N GUYEN T HI T HU H UONG AND T RAN M INH T OAN

Lecture on
M ATH 4
M ULTIPLE I NTEGRAL , I NTEGRAL THAT DEPENDS ON A PARAMETER ,
L INE I NTEGRAL , S URFACE I NTEGRAL , F IELD T HEORY AND S ERIES

Summary, Examples, Exercises and Solutions

Ha Noi - 2008


C ONTENTS
Contents.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

Chapter 1 . Multiple Integral . . . . . . . . . . . . . . . . . . . . . . . .

5

1

Double Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1
Calculation of a double integral in Cartesian coordinate system

1.2
Change of variables in double integrals, polar coordinate . . . .
1.3
Applications of double integrals . . . . . . . . . . . . . . . . . . .
1.4
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Triple Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1
Calculation of a triple integral in Cartesian coordinate system .
2.2
Change of variables in triple integrals . . . . . . . . . . . . . . .
2.3
Calculate the triple integrals in cylindrical coordinate . . . . .
2.4
Calculate the triple integrals in spherical coordinate . . . . . .
2.5
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 2 . Integrals that depend on a parameter . . . . . . . . . . .

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5
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12
14
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29
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35
. 39

Line integral of the first kind . . . . . . . . . . . . . . . . . . . . . . . . . . .

39

1

The definite integrals that depend on a parameter . . .
1.1
Definition . . . . . . . . . . . . . . . . . . . . . . .
1.2
Properties . . . . . . . . . . . . . . . . . . . . . .
2
The generalized integarls that depend on a parameter .
2.1
The uniformly convergent integrals . . . . . . . .
2.2
Properties . . . . . . . . . . . . . . . . . . . . . .
2.3
Euler’s integrals . . . . . . . . . . . . . . . . . . .
3
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 3 . Line integral . . . . . . . . . . . . . . . .

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2

CONTENTS
1.1
Definition . . . . . . . . . . . . . . . . . . .
1.2
Calculation formulae . . . . . . . . . . . .
2
Line integral of the second kind . . . . . . . . . .
2.1

Definition . . . . . . . . . . . . . . . . . . .
2.2
Calculation formulae . . . . . . . . . . . .
2.3
Theorem of four equivalent propositions .
2.4
Area of a plane domain . . . . . . . . . . .
3
Exercises . . . . . . . . . . . . . . . . . . . . . . .
4
Solution . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 4 . Surface integral . . . . . . . . . . . .
1

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39
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46

46
49
. 53

Surface integral of the first kind .
1.1
Definition . . . . . . . . . . .
1.2
Calculation formulae . . . .
2
Surface integral of the second kind
2.1
Definition . . . . . . . . . . .
2.2
Calculation formulae . . . .
2.3
Stokes’ formula . . . . . . .
3
Exercises . . . . . . . . . . . . . . .
4
Solution . . . . . . . . . . . . . . . .
Chapter 5 . Field theory . . . . . . . .

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53
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54
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55
58
59
60
. 63

1
Scalar field
2
Vector field
3
Exercises .
4
Solution . .
Chapter 6 . Series

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1

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63
64
66
66
. 69

Number series . . . . . . . .
1.1
Definition . . . . . . .
1.2
Convergent criterion
1.3
Exercises . . . . . . .
1.4
Solution . . . . . . . .
Function series . . . . . . .
2.1
Function sequence .
2.2
Function series . . .
2.3
Power series . . . . .
2.4
Exercises . . . . . . .
2.5
Solution . . . . . . . .
Fourier series . . . . . . . .

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69

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85

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CONTENTS
3.1
3.2
3.3

3
Decomposition theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

85
89
89


4

CONTENTS

CHAPTER

1

M ULTIPLE I NTEGRAL
§1. D OUBLE I NTEGRAL
1.1 Calculation of a double integral in Cartesian
coordinate system
Consider the integral
I=

f ( x, y)dxdy.

(1.1)

D

1. (The Corollary of Fubini’s theorem)
Suppose that D = [ a, b] × [c, d] and f : D → R is a continuous function on D. Then
I=

f ( x, y)dy =

dx
a

b

d

d

b

c

c

f ( x, y)dx

dy
a


a ≤ x ≤ b
2. If D is described as follows: D =
,
 ϕ( x ) ≤ y ≤ ψ( x )
where y = ϕ( x ), y = ψ( x) are continuous and have continuous derivatives on [ a, b]
ψ( x )

b

then I =

a




ϕ( x )


f ( x, y)dy dx or

ψ( x )

b

I=

f ( x, y)dy.

dx
a

ϕ( x )

5

(1.2)


6

Chapter 1. Multiple Integral

c ≤ y ≤ d
3. If D is described as follows: D =
,

 ϕ(y) ≤ x ≤ ψ(y)
where x = ϕ(y), x = ψ(y) are continuous and have continuous derivatives on [c, d]
then
ψ(y)

d

I=

f ( x, y)dx.

dy
c

(1.3)

ϕ(y)

Example 1.1. Calculate the double integral
x2 ydxdy,

I=
D

where D = [0, 1] × [0, 2].
Solution. We have
1
2

I=

x ydxdy =
D
1

=
0

2

x2

x ydy =

dx
0

1
2

0

0

y2
2

2
0

dx

x3 1 2
4
= .
x2 dx = 2.
2
3 0 3

x3 + xy dxdy where D is bounbed

Example 1.2. Calculate the double integral I =

by the curves y = x2 and y =

√

D

x.

Solution. We have the region D = 0 ≤ x ≤ 1, x2 ≤ y ≤

√

x (Figure 1.1).

y
y = x2
y=

√

x

1

O

1
Figure 1.1

x


1. Double Integral

7

Therefore
√

1

I=

x

0

x2
1

=
0

3

x + xy dy =

dx

√

y2
x y+x
2

3

x

x2

dx

√
1
5
1

x3 x − x5 + x2 − x5 dx = .
2
2
36

Example 1.3. Interchange the order of the following integrals:

i) I =

Solution.

2x

2
0

dx

x

f ( x, y)dy;

i) We have D =

e

ii) I =





 x = 0, x = 2
y=x



y = 2x

1

ln y

dy

0

f ( x, y)dx.

(Figure 1.2)

y

4

2

O

2

4

x

Figure 1.2
From above figure, we have
y

2

I=

ii) We have D =











f ( x, y)dx +

dy
0

4

y/2

y = 1, y = e
x=0
x = ln y

(Figure 1.3).

2

dy
2

y/2

f ( x, y)dx.


8

Chapter 1. Multiple Integral
y
e

1

O

x

1
Figure 1.3

Hence
e

1

I=

f ( x, y)dy.

dx
ex

0

1.2 Change of variables in double integrals, polar
coordinate
1. In general case
Put I =

f ( x, y)dxdy.
D

To calculate I, we can perform the tranformation

 x = x (u, v)
 y = y(u, v).

(1.4)

The two equations in (1.4) define a mapping which carries a point ( x, y) ∈ D ⊂ Oxy
′ (or inversion).
to (u, v) ∈ D ⊂ Ouv
We shall consider mapping for which the functions x = x (u, v), y = y(u, v) are continuous and have continuous partial derivatives on D. Then
I=
D

∂x
where J = ∂u
∂y
∂u

f ( x (u, v), y(u, v)) | J | dudv,

f ( x, y)dxdy =

∂x
∂v = D ( x, y) = 0.
∂y
D (u, v)
∂v

D


1. Double Integral

9

2. Polar coordinate
In this case we write r and ϕ instead of u and v and discrible the mapping by the
two equations

 x = r cos ϕ
; | J | = r ≥ 0.
(1.5)
y = r sin ϕ
Then

I=

f ( x, y)dxdy =

DOxy

f (r cos ϕ, r sin ϕ) rdrdϕ.

D Orϕ

Example 1.4. Calculate I =
D

y
dxdy, where the region D is bounded by
x

y = x, y = 2x, xy = 1, xy = 3 ( x > 0).

y
Solution. Because x > 0, put = u, xy = v (u > 0, v > 0). Therefore we perform the
x
tranformation
√

√
v 1
1
v

√ √
−
. √
x = √
1
D ( x, y)
u√ 2 u 2 √v u
u
=− .
=⇒ J =
=
√
√
v
u

D (u, v)
u
y = v. u

√
√
2 u
2 v

1 ≤ u ≤ 2
The region D → D =
.
1 ≤ v ≤ 3
Hence

2

3

I=

dv
1

1

u −

1
du =
u

v

3
1

u

2

= 2.

1

y
y = 2x
y=x

1

xy = 3
xy = 1

O

1

Figure 1.4

x


10

Chapter 1. Multiple Integral

Example 1.5. Tranform each of the given integrals to one or more interated integrals in
polar coordinate
1

1. I =

√

1

f ( x, y)dy;

dx
0

Solution.

1

2. I =

0

dx

1− x 2

f ( x, y)dy.

1− x

0

1. We have D = [0, 1] × [0, 1] (Figure 1.5).
y

1

O

x

1
Figure 1.5

To transform to polar coordinate, we put x = r cos ϕ; y = r sin ϕ.
We devide the region D into two subregions by the line y = x : D = D1 ∪ D2 , where
1
π
; 0≤r≤
;
4
cos ϕ
π
π
1
D2 :

≤ϕ≤ ; 0≤r≤
4
2
sin ϕ

D1 : 0 ≤ ϕ ≤

Therefore
π/4

I=

1/cos ϕ

f (r cos ϕ, r sin ϕ) rdr +

dϕ
0

π/2

0

2. Rewrite the region D =






f (r cos ϕ, r sin ϕ) rdr

dϕ
π/4







1/sin ϕ

0

x = 1, x = 2
(Figure 1.5)
y = 1−x
√
y = 1 − x2


1. Double Integral

11
y

1

O

x

1
y = 1−x

Figure 1.6


 x = r cos ϕ;
Put
y = r sin ϕ
Hence


π

0 ≤ ϕ ≤
2
, we have
1

≤r≤1

sin ϕ + cos ϕ
π/2

I=

1

f (r cos ϕ, r sin ϕ) rdr.

dϕ
0

.

1/(sin ϕ+cos ϕ)

Note: Some regions used frequently in polar coordinate
i) If DOxy = x2 + y2 ≤ R2 then
D Orϕ = {0 ≤ ϕ ≤ 2π; 0 ≤ r ≤ R} .
ii) If D = x2 + y2 ≤ 2ax; a > 0 = ( x − a)2 + y2 ≤ a2

− π ≤ ϕ ≤ π
2
2
D=
0 ≤ r ≤ 2a cos ϕ.

iii) If D = x2 + y2 ≤ 2ay; a > 0 =

0 ≤ ϕ ≤ π
D=
0 ≤ r ≤ 2a sin ϕ.

x 2 + ( y − a )2 ≤ a2

then

then

iv) If

D=

x2 + y2 ≤ 2ax + 2ay, a > 0, b > 0

= ( x − a )2 + ( y − b )2 ≤ a2 + b2


12

Chapter 1. Multiple Integral

− arctg a ≤ ϕ ≤ − arctg a + π
b
b
then D =
0 ≤ r ≤ 2 ( a cos ϕ + b sin ϕ) .

v) If D = x2 + y2 = −2ax, a > 0 = ( x + a)2 + y2 = a2 then

 π ≤ ϕ ≤ 3π
2
D= 2
0 ≤ r ≤ −2a cos ϕ

vi) If D is ellipse

x 2 y2
+ 2 = 1 then we perform the transformation
a2
b

 x = ar cos ϕ
, | J | = abr.
 y = br sin ϕ


0 ≤ ϕ ≤ 2π
Then D → D =
0 ≤ r ≤ 1

and

2π

I=

1

f ( ar cos ϕ, br sin ϕ) abrdr.

dϕ
0

(1.6)

0

1.3 Applications of double integrals
1. To compute the area of a plane domain D
The area of the domain D in the plane Oxy is computed by the formula:
S=

(1.7)

dxdy
D

Example 1.6. Compute the area of the domain D bounded by the curves
xy = a2 , x + y =

5
a.
2

5a
Solution. The curve xy = a2 cuts the line x + y =
at two points that have abscis2
a
sas x = and x = 2a, respectively (Figure 1.7).
2
Therefore the area of D is
2a

S=

dx
a/2

5a/2− x

dy =

a2/x

15
− 2 ln 2 a2
8


1. Double Integral

13

y

O

1

x

4

Figure 1.7

2. To compute the area of a curved surface
Suppose that S is a curved surface whose equation is z = f ( x, y) (Figure 1.6) and D
is the projection of S on the plane Oxy. Then the area of S is
′

′

1 + z x2 + zy2 dxdy

S=

(1.8)

D

z = f ( x, y)

z

y

O
D

x
Figure 1.8
3. To compute the volume of a object
Suppose that the object Ω is bounded by the smooth curves
z = f ( x, y), z = g( x, y), f ≥ g ∀( x, y) ∈ D

14

Chapter 1. Multiple Integral
and the surrounding cylinder has the directrix that is the border of the region D :
ϕ( x, y) = 0 and has the element that is parallel with Oz.
Then the volume of Ω is
V=
D

[ f ( x, y) − g( x, y)] dxdy.

(1.9)

z = f ( x, y)

z

Ω
z = g( x, y)
y

O
D

x

Figure 1.9

Example 1.7. Use the double integral to compute the volume of the object bounded by

z = 1 + x + y, z = 0, x + y = 1, x = 0, y = 0.
Solution. We have
1

I=

dx
0

1− x

1

(1 + x + y) dy =

0

0

1
1
5
1 1
4 − (1 + x )2 dx = 2 − . (1 + x )3 = .
2
2 3
6
0

1.4 Exercises

Exercise 1.1. Interchange the order of the following integrals
1

1.

dx

− 1− x 2
1+

1

√

1− y2

f ( x, y)dx;

dy
0

f ( x, y)dy;

√

−1

2.

1− x 2

2− y


1. Double Integral
√

2

3.

2x

f ( x, y)dy;

dx
√

0

√

2x − x2

5

4.

y

dy
0

15

0

√

2

f ( x, y)dx +

4− y2

f ( x, y)dx.

dy
√

2

0

Exercise 1.2. Calculate the following integrals
1.
D

2.
D

3.
D

x sin ( x + y) dxdy, D = ( x, y) ∈ R2 : 0 ≤ x ≤

π
π
;0 ≤ y ≤
;
2
2

x2 (y − x ) dxdy, where D is bounded by curves y = x2 and x = y2 ;

| x + y| dxdy, D = ( x, y) ∈ R2 : | x | ≤ 1, |y| ≤ 1 ;
|y − x2 |dxdx, D = {| x | ≤ 1, 0 ≤ y ≤ 1};

4.
D

(| x | + |y|) dxdy.

5.
| x |+|y|≤1

Exercise 1.3. Transform the integral to polar coordinate and compute its value
√

R

1.

dx
0

dx
0

3.
D

4.
D

ln 1 + x2 + y2 dy, ( R > 0);

0

√

R

2.

R2 − x 2

Rx − x2

√

− Rx − x2

Rx − x2 − y2 dy, ( R > 0);

xydxdy, where D = ( x, y) ∈ R2 : ( x − 2)2 + y2 ≤ 1, y ≥ 0 .
xy2 dxdy, where D is bounded by the curves x2 + (y − 1)2 = 1 and x2 + y2 − 4y = 0.

Exercise 1.4. Calculate these following integrals:

4y ≤ x2 + y2 ≤ 8y
dxdy
1.
,
where
D
:
√
2
 x ≤ y ≤ 3x
( x 2 + y2 )
D

;


16

Chapter 1. Multiple Integral

2.

D

3.
D

4.
D

5.
D



x2 + y2 ≤ 12




 x2 + y2 ≥ 2x
xy
dxdy, where D :
√
2 + y2 ≥ 2 3y

x 2 + y2

x





x ≥ 0, y ≥ 0

;

1 − x 2 − y2
dxdy, where D : x2 + y2 ≤ 1;
1 + x 2 + y2

9x2 − 4y2 dxdy, where D :

4x2 − 2y2

x 2 y2
+
≤ 1;
4
9


1 ≤ xy ≤ 4
dxdy, where D :
 x ≤ y ≤ 4x



x


y = 2

Exercise 1.5. Compute the area of the domain D bounded by y = 2− x



y = 4.


y = 0; y2 = 4ax
Exercise 1.6. Compute the area of the domain D bounded by
 x + y = 3a; y ≤ 0, ( a > 0).


 x2 + y2 = 2x; x2 + y2 = 4x
Exercise 1.7. Compute the area of the domain D bounded by
 x = y, y = 0

Exercise 1.8. Compute the volume of the object bounded by the surfaces




3x + y ≥ 1
3x + 2y ≤ 2



y ≥ 0, 0 ≤ z ≤ 1 − x − y.

Exercise 1.9. Compute the volume of the object bounded by the surfaces


0 ≤ z ≤ 1 − x 2 − y2
√
y ≥ x, y ≤ x 3.


1. Double Integral

17

1.5 Solutions
√

0

1. I =

Solution 1.1.

dy
−1

√

2

2. I =

dx

1−

3.

√

4. I =

f ( x, y)dx;
1− y

4− x 2

1+

√

2

2

f ( x, y)dx +

dy
1

1− y2

f ( x, y)dx;

y2/2

f ( x, y)dy.

x

0

π/2

1. I =

Solution 1.2.

π/2

√

1

dx
x2

x sin( x + y)dy =

dx
0

0

−

√

2

0

√

dx

2. I =

0

dy

y2/2

2

dy

1− y2

1

f ( x, y)dx +

√

f ( x, y)dx +

1− y

f ( x, y)dy;

1− y2

dy
0

√

1

2− x

1
1

2x − x2

−

√

1− y2

0

x

x2 y − x3 dy = −

π
;
2

1
;
504

3. Divide D into two regions D = D1 ∪ D2 , where
D1 = {−1 ≤ x ≤ 1, − x ≤ y ≤ 1}

D2 = {−1 ≤ x ≤ 1, −1 ≤ y ≤ − x } .

Then

1

1

I=

dx
−x

−1

1

( x + y)dy −

dx

−x
−1

−1

8
( x + y)dy = .
3

4. Divide D into two regions D = D1 ∪ D2 , where
D1 = −1 ≤ x ≤ 1, x2 ≤ y ≤ 1
D2 = −1 ≤ x ≤ 1, 0 ≤ y ≤ x2 .
Hence
I=

dx
−1

x2

x2

1

1

1

y − x2 dy +

dx
−1

0

x2 − ydy =

3π + 4
.
12


18

Chapter 1. Multiple Integral
5. Note that D is axisymetric and the function f ( x, y) = | x | + |y| is even with respect
to x and y. Therefore I = 4

(| x | + |y|) dxdy, where

D1

D1 = {( x, y) : 0 ≤ x ≤ 1; 0 ≤ y ≤ 1 − x } .

Hence

1

I=4

dx
0

1− x

4
( x + y)dy = .
3

0

Solution 1.3.
1. We have D = 0 ≤ x ≤ R; 0 ≤ y ≤

 x = r cos ϕ
π
Put
=⇒ 0 ≤ ϕ ≤ , 0 ≤ r ≤ R.
y = r sin ϕ
2
Hence

π/2

I=

R

ln 1 + r2 rdr =

dϕ
0

0

π
4

√

R2 − x 2 .

R2 + 1 ln( R2 + 1) − R2 .

√
√
2. We have D = 0 ≤ x ≤ R, − Rx − x2 ≤ y ≤ RX − x2 .


 x = R + r cos ϕ
0 ≤ ϕ ≤ 2π
2
.
Put

=⇒ | J | = r,
y = r sin ϕ
0 ≤ r ≤ R
2
Hence
2π

I=

R/2

dϕ
0


 x = 2 + r cos ϕ
3. Put
y = r sin ϕ

0

R2
πR3
− r2 rdr =
.
4
12


0 ≤ r ≤ 1

=⇒
0 ≤ ϕ ≤ 2π
2π

I=

. Then

1

(2 + r cos ϕ) r sin ϕrdr = 0

dϕ
0

0

Note: The domain D is Ox-axisymetric and the function f ( x, y) = xy is odd with
respect to y. Therefore we have I = 0.


 x = r cos ϕ
0 ≤ ϕ ≤ π
4. Put
=⇒
. Hence
y = r sin ϕ
2 sin ϕ ≤ r ≤ 4 sin ϕ
4 sin ϕ

π

I=

dϕ
0

2 sin ϕ

r cos ϕ (r sin ϕ)2 rdr = 0.


1. Double Integral

19

Note: The domain D is symetric to the axis Oy and the function f ( x, y) = xy2 is odd
with respect to x. Therefore we have I = 0.


 x = r cos ϕ
π ≤ ϕ ≤ π
3
. Hence
Solution 1.4.
1. Put
=⇒ 4
y = r sin ϕ
4 sin ϕ ≤ r ≤ 8 sin ϕ
8 sin ϕ

π/3

I=

dϕ
4 sin ϕ

π/4

1
3
rdr =
4
128
r

1
1− √
3

.

2. Divide D into two regions D = D1 ∪ D2 , where

√
π
, 2 cos ϕ ≤ r ≤ 2 3
6
√

√
π
π
≤ ϕ ≤ , 2 3 sin ϕ ≤ r ≤ 2 3 .
6
2

D1 = 0 ≤ ϕ ≤
D2 =
Then
√
2 3

π/6

I=

dϕ

r2 cos

ϕ sin ϕ
rdr +
r2

2 cos ϕ

0



 x = r cos ϕ
3. Put
y = r sin ϕ


0 ≤ ϕ ≤ 2π
=⇒
0 ≤ r ≤ 1
2π

I=

dϕ
0


 x = 2r cos ϕ
4. Put
y = 3r sin ϕ

0

dϕ
√
2 3 sin ϕ

π/6

1

1 − r2
π2
.
rdr
=
2
1 + r2

2π


1 ≤ u ≤ 4
=⇒
1 ≤ v ≤ 4

4

du
1

|cos 2ϕ| dϕ

and x =

4

I=

0

1

r2 cos ϕ sin ϕ
11
rdr = .
2
8
r

. Then


0 ≤ ϕ ≤ 2π
=⇒ | J | = 6r,
0 ≤ r ≤ 1
I = 6 × 36


u = xy
5. Put
v = y
x

√
2 3

π/2

. Then
1

r3 dr = 216.
0

√
u
, y = uv. Therefore
v

u
1
45
4 − 2uv . dv = − .
v
2v
4


20

Chapter 1. Multiple Integral

Solution 1.5. Divide the domain D into two subdomain D = D1 ∪ D2 , where
D1 = −2 ≤ x ≤ 0, 2− x ≤ y ≤ 4 ; D2 = {0 ≤ x ≤ 2, 2x ≤ y ≤ 4} .
Then

0

S=

2

4

dy +

dx
2− x

−2

4

dx
2x

0

dy = 2 8 −

3
ln 2

.

Solution 1.6. We have
3a−y

0

S=

dx =

dy
−6a

0

3a − y −

−6a

y2/4a

y2
4a


 x = r cos ϕ
Solution 1.7. Change to polar coordinate
y = r sin ϕ

Hence

4 cos ϕ

π/4

S=

rdr =

dϕ
2 cos ϕ

0

dy = 18a2 .

0 ≤ ϕ ≤ π ,
4
=⇒
2 cos ϕ ≤ r ≤ 4 cos ϕ

.

3π 3
+ .
4
2

Solution 1.8. We have
(2−2y)/3

1

V=

dy

0

(1−y)/3

1
(1 − x − y) dx =
6

1

0

1 − 2y + y2 dy =

1
.
18

Solution 1.9. We have V = 2V1 , where
1

π/3

V1 =

dϕ
0

π/4

Hence V =

1 − r2 rdr =

π
.
48

π
.
24

§2. T RIPLE I NTEGRAL
2.1 Calculation of a triple integral in Cartesian
coordinate system
Consider the integral
I=

f ( x, y, z)dxdydz,
V

(1.10)


2. Triple Integral

21

where f ( x, y, z) is three-variables function that is continuous on V.
If

V = ( x, y, z) ∈ R3 |( x, y) ∈ D; z1 ( x, y) ≤ z ≤ z2 ( x, y) ,
where D is the projection of V on the plane Oxy and z1 , z2 are continuous on D then


I=
D



z2 ( x,y)


dxdy 

z1 ( x,y)

dxdydz

Example 2.1. Calculate the integral I =

planes x = 0, y = 0, z = 0 and x + y + z = 1.


f ( x, y, z)dz .

V

( x + y + z )3

(1.11)

, where V is bounded by the

Solution. V is tetrahedron bounded by two planes z = 0 and z = 1 − x − y, ( x, y) ∈ D,
where D is the triangle OAB in the plane Oxy (Figure 2.1). Hence we have
1− x − y

I=

dxdy
D

1
=−
2

=−

1
2

1

dx
0

0
1

0

0
1− x

dz

( x + y + z )3

=
D

1
1
−
4 (1 + x + y )2

1
3 x
− −
4 4 1+x

dx =

(1 + x + y + z ) −2
−2
1
dy = −
2

1

0

1
5
ln 2 − .
2
16

C

B

A
x

Figure 2.1

z =0

dxdy

1
1−x 1
+ −
4
2 1+x

z

O

z =1− x − y

y

dx


22

Chapter 1. Multiple Integral

2.2 Change of variables in triple integrals
Consider the tranformation:




 x = x (u, v, w)
y = y(u, v, w)



z = z(u, v, w).

Suppose that the following conditions are satisfied:
i) (u, v, w) ∈ V ′ in O′ uvw-plane and x (u, v, w), y(u, v, w), z(u, v, w) are continuous and
have continuous partial derivatives on V ′ .
ii) The vecto-valued mapping Φ : V ′ → V is one-to-one.

iii) The Jacobian determinant
′
xu′ xv′ xw
D ( x, y, z)
= y′u y′v y′w = 0 in V ′ .
J=
D (u, v, w)
z′u z′v z′w

Then
I=

f ( x, y, z)dxdydz =
V′

V

f ( x (u, v, w), y(u, v, w), z(u, v, w)) | J | dudvdw.

(1.12)

2.3 Calculate the triple integrals in cylindrical
coordinate
Here we write r, ϕ, z for u, v, w and define the mapping by the equations:
x = r cos ϕ, y = r sin ϕ, z = z.

(1.13)

In other words, we replace x and y by their polar coordinate in the plane Oxy and retain
z.

Again, to get a one-to-one mapping we must keep r > 0 and restrict ϕ to be in an interval
of the form: ϕo ≤ ϕ < ϕo + 2π.
The Jacobian determinant of the mapping in (1.13) is
cos ϕ −r sin ϕ 0
J = sin ϕ r cos ϕ 0 = r cos2 ϕ + sin2 ϕ = r > 0
0
0
1


2. Triple Integral

23

and therefore we have the tranformation formula
f ( x, y, z)dxdydz =

f (r cos ϕ, r sin ϕ, z) rdrdϕdz.

(1.14)

V′

V

Note: In some cases we use the generalized cyclindrical coordinate
x = ar cos ϕ, y = br sin ϕ, z = z
and J = abr.
x2 + y2 dxdydz to cylindrical coordinate

Example 2.2. Transform the integral I =
V

and compute its value, where V is the region bounded by the surfaces x2 + y2 = 2z and
z = 2.
Solution. Transform to cylindrical coordinate : x = r cos ϕ, y = r sin ϕ, z = z, 0 ≤ ϕ ≤ 2π.
z
2

−2

O

y

2

Figure 2.2
x
We note that the paraboloid x2 + y2 = 2z cuts the plane x2 + y2 = 4 by the circle x2 + y2 =
4, therefore 0 ≤ r ≤ 2.
r2
On the other hand on the paraboloid we have r2 cos2 ϕ + r2 sin2 ϕ = 2z =⇒ z = .
2
r2
So that in B we have
≤ z ≤ 2.
2
Therefore we have
2π

I=

2

dϕ
0

dr
0

2

2

r2/2

3

r dz = 2π
0

r3 2 −

r2
2

dr =

16π

.
3


24

Chapter 1. Multiple Integral

2.4 Calculate the triple integrals in spherical
coordinate
In this case the symbols r, θ, ϕ are used instead of u, v, w and the mapping is defined
by the equations




 x = r sin θ cos ϕ
(1.15)
y = r sin θ sin ϕ



z = r cos θ

To get a one-to-one mapping we keep r > 0, 0 ≤ ϕ < 2π and 0 ≤ θ < π.
The Jacobian determinant of the mapping is
sin θ cos ϕ r cos θ cos ϕ −r sin θ sin ϕ
J = sin θ sin ϕ r cos θ sin ϕ r sin θ cos ϕ = −r2 sin θ
cos θ
−r sin θ

0
Therefore we have the tranformation formula

f (r sin θ cos ϕ, r sin θ sin ϕ, r cos θ ) r2 sin θdrdθdϕ

f ( x, y, z)dxdydz =

(1.16)

V′

V

x2 + y2 + z2 dxdydz to spherical coordi-

Example 2.3. Transform the integral I =
V

nate and compute its value, where V is the sphere x2 + y2 + z2 ≤ z.
Solution. We have
1
x +y +z −z = x +y + z−
2
2

So that V =

2

2

( x, y, z) : x2 + y2 + z −

2

1
2

2

2

≤

2

1
− .
4

1
, i.e V is sphere whose center is the
4

1
1
and radius R = .
2
2
Transform to spherical coordinate.

point

0, 0,

2.5 Exercises
Exercise 1.10. Calculate the following triple integrals:




0 ≤ x ≤ 4
1.

zdxdydz, where the region V is defined by

V

x ≤ y ≤ 2x



0 ≤ z ≤ 1 − x 2 − y2

.