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Integrales triplesIntegrales iteradas

Integrales triples sobre regiones generalesAplicaciones de las Integrales

Cambio de variables en integrales

Gilberto ARENAS DÍAZ

Escuela de MatemáticasUniversidad Industrial de Santander

Segundo Semestre de 2011

G A D � g a r e n a s d @ u i s . e d u .c o (U IS ) C á lc u lo I I I S e g u n d o S em e s t r e d e 2 0 1 1 1 / 5 1

Integrales triples

Sea B una caja rectangular en R3B = f(x; y; z) j a � x � b; c � y � d; g � z � hg

y sea f : B �! R(x; y; z) 7�! f(x; y; z)

una función dada.

[a; b] se subdivide en l subintervalos [xi�1; xi]de ancho �x.[c; d] se subdivide en m subintervalos [yj�1; yj ]de ancho �y.[g; h] se subdivide en n subintervalos [zk�1; zk]de ancho �z.Así, B queda dividida en lmn cajitas Bijk devolumen �V = �x�y�z

Bijk = [xi�1; xi]� [yj�1; yj ]� [zk�1; zk]

Sea (xijk; yijk; zijk) un punto de Bijk y formemos la suma triple de RiemannlXi=1

mXj=1

nXk=1

f (xijk; yijk; zijk)�V:


Integrales triples


y sea f : B �! R(x; y; z) 7�! f(x; y; z)

una función dada.[a; b] se subdivide en l subintervalos [xi�1; xi]de ancho �x.

[c; d] se subdivide en m subintervalos [yj�1; yj ]de ancho �y.[g; h] se subdivide en n subintervalos [zk�1; zk]de ancho �z.Así, B queda dividida en lmn cajitas Bijk devolumen �V = �x�y�z



mXj=1

nXk=1



Integrales triples


y sea f : B �! R(x; y; z) 7�! f(x; y; z)

una función dada.[a; b] se subdivide en l subintervalos [xi�1; xi]de ancho �x.[c; d] se subdivide en m subintervalos [yj�1; yj ]de ancho �y.

[g; h] se subdivide en n subintervalos [zk�1; zk]de ancho �z.Así, B queda dividida en lmn cajitas Bijk devolumen �V = �x�y�z



mXj=1

nXk=1



Integrales triples


y sea f : B �! R(x; y; z) 7�! f(x; y; z)

una función dada.[a; b] se subdivide en l subintervalos [xi�1; xi]de ancho �x.[c; d] se subdivide en m subintervalos [yj�1; yj ]de ancho �y.[g; h] se subdivide en n subintervalos [zk�1; zk]de ancho �z.

Así, B queda dividida en lmn cajitas Bijk devolumen �V = �x�y�z



mXj=1

nXk=1



Integrales triples


y sea f : B �! R(x; y; z) 7�! f(x; y; z)

una función dada.[a; b] se subdivide en l subintervalos [xi�1; xi]de ancho �x.[c; d] se subdivide en m subintervalos [yj�1; yj ]de ancho �y.[g; h] se subdivide en n subintervalos [zk�1; zk]de ancho �z.Así, B queda dividida en lmn cajitas Bijk devolumen �V = �x�y�z



mXj=1

nXk=1



La integral triple de f sobre B esZZZB

f (x; y; z) dV = l��ml;m;n!1

lXi=1

mXj=1

nXk=1

f (xijk; yijk; zijk)�V

si este límite existe.

Si f es continua en B el límite existe, y la integral se calcula de acuerdo con elTeorema de FubiniZZZ

B

f (x; y; z) dV =

Z b

a

Z d

c

Z h

g

f (x; y; z) dz dy dx

Observe que hay 5 órdenes más para realizar la integral anterior.ZZZB

f (x; y; z) dV=

Z h

g|{z}Z b

a

dZ d

c

f (x; y; z) bdydx dz|{z} = � � �=Z h

g

Z d

c

Z b

a

f(x; y; z) dx dy dz:

Esta expresión también se conoce como integral iterada.



f (x; y; z) dV = l��ml;m;n!1

lXi=1

mXj=1

nXk=1


si este límite existe.Si f es continua en B el límite existe, y la integral se calcula de acuerdo con elTeorema de FubiniZZZ

B

f (x; y; z) dV =

Z b

a

Z d

c

Z h

g



f (x; y; z) dV=

Z h

g|{z}Z b

a

dZ d

c

f (x; y; z) bdydx dz|{z} = � � �=Z h

g

Z d

c

Z b

a





f (x; y; z) dV = l��ml;m;n!1

lXi=1

mXj=1

nXk=1



B

f (x; y; z) dV =

Z b

a

Z d

c

Z h

g



f (x; y; z) dV=

Z h

g|{z}Z b

a

dZ d

c

f (x; y; z) bdydx dz|{z}

= � � �=Z h

g

Z d

c

Z b

a





f (x; y; z) dV = l��ml;m;n!1

lXi=1

mXj=1

nXk=1



B

f (x; y; z) dV =

Z b

a

Z d

c

Z h

g



f (x; y; z) dV=

Z h

g|{z}Z b

a

dZ d

c

f (x; y; z) bdydx dz|{z} = � � �=Z h

g

Z d

c

Z b

a




Ejemplo

CalcularZZZ

E

�x2 + yz

�dV , donde

E = f(x; y; z) j 0 � x � 2; � 3 � y � 0; � 1 � z � 1g :

ZZZE

�x2 + yz

�dV =

Z 2

0

Z 0

�3

Z 1

�1

�x2 + yz

�dz dy dx

=

Z 2

0

Z 0

�3

��x2z + y

z2

2

��z=1z=�1

dy dx

=

Z 2

0

Z 0

�32x2 dy dx

=

Z 2

0

�2yx2

�y=0y=�3 dx

=

Z 2

0

6x2 dx

=�2x3�x=2x=0

= 16:


Ejemplo

CalcularZZZ

E

�x2 + yz

�dV , donde

E = f(x; y; z) j 0 � x � 2; � 3 � y � 0; � 1 � z � 1g :ZZZE

�x2 + yz

�dV

=

Z 2

0

Z 0

�3

Z 1

�1

�x2 + yz

�dz dy dx

=

Z 2

0

Z 0

�3

��x2z + y

z2

2

��z=1z=�1

dy dx

=

Z 2

0

Z 0

�32x2 dy dx

=

Z 2

0

�2yx2

�y=0y=�3 dx

=

Z 2

0

6x2 dx

=�2x3�x=2x=0

= 16:


Ejemplo

CalcularZZZ

E

�x2 + yz

�dV , donde

E = f(x; y; z) j 0 � x � 2; � 3 � y � 0; � 1 � z � 1g :ZZZE

�x2 + yz

�dV =

Z 2

0

Z 0

�3

Z 1

�1

�x2 + yz

�dz dy dx

=

Z 2

0

Z 0

�3

��x2z + y

z2

2

��z=1z=�1

dy dx

=

Z 2

0

Z 0

�32x2 dy dx

=

Z 2

0

�2yx2

�y=0y=�3 dx

=

Z 2

0

6x2 dx

=�2x3�x=2x=0

= 16:


Integración triple sobre regiones más generales

Región del tipo A. E = f(x; y; z) j (x; y) 2 D; �1 (x; y) � z � �2 (x; y)gZZZE

f (x; y; z) dV =

ZZD

"Z �2(x;y)

�1(x;y)

f (x; y; z) dz

#dA:

Si D = f(x; y) j a � x � b; g1 (x) � y � g2 (x)g entoncesZZZE

f (x; y; z) dV =

Z b

a

Z g2(x)

g1(x)

Z �2(x;y)

�1(x;y)

f (x; y; z) dz dy dx:

Si D = f(x; y) j c � y � d; h1 (y) � x � h2 (y)g entoncesZZZE

f (x; y; z) dV =

Z d

c

Z h2(y)

h1(y)

Z �2(x;y)

�1(x;y)

f (x; y; z) dz dx dy:


Ejemplo

CalcularZZZ

E

y dV , donde E está bajo el plano z = x+ 2y y encima de la

región del plano xy acotada por las curvas y = x2, y = 0 y x = 1.

E =�(x; y; z) j 0 � x � 1; 0 � y � x2; 0 � z � x+ 2y

:ZZZ

E

y dV =

Z 1

0

Z x2

0

Z x+2y

0

y dz dy dx

=

Z 1

0

Z x2

0

[yz]x+2y0 dy dx

=

Z 1

0

Z x2

0

y (x+ 2y) dy dx

=

Z 1

0

�xy2

2+2

3y3�x20

dx

=

Z 1

0

�1

2x5 +

2

3x6�dx =

�x6

12+2

21x7�10

=1

12+2

21=5

28:


Ejemplo

CalcularZZZ

E


región del plano xy acotada por las curvas y = x2, y = 0 y x = 1.E =

�(x; y; z) j 0 � x � 1; 0 � y � x2; 0 � z � x+ 2y

:

ZZZE

y dV =

Z 1

0

Z x2

0

Z x+2y

0

y dz dy dx

=

Z 1

0

Z x2

0

[yz]x+2y0 dy dx

=

Z 1

0

Z x2

0

y (x+ 2y) dy dx

=

Z 1

0

�xy2

2+2

3y3�x20

dx

=

Z 1

0

�1

2x5 +

2

3x6�dx =

�x6

12+2

21x7�10

=1

12+2

21=5

28:


Ejemplo

CalcularZZZ

E



�(x; y; z) j 0 � x � 1; 0 � y � x2; 0 � z � x+ 2y

:ZZZ

E

y dV

=

Z 1

0

Z x2

0

Z x+2y

0

y dz dy dx

=

Z 1

0

Z x2

0

[yz]x+2y0 dy dx

=

Z 1

0

Z x2

0

y (x+ 2y) dy dx

=

Z 1

0

�xy2

2+2

3y3�x20

dx

=

Z 1

0

�1

2x5 +

2

3x6�dx =

�x6

12+2

21x7�10

=1

12+2

21=5

28:


Ejemplo

CalcularZZZ

E



�(x; y; z) j 0 � x � 1; 0 � y � x2; 0 � z � x+ 2y

:ZZZ

E

y dV =

Z 1

0

Z x2

0

Z x+2y

0

y dz dy dx

=

Z 1

0

Z x2

0

[yz]x+2y0 dy dx

=

Z 1

0

Z x2

0

y (x+ 2y) dy dx

=

Z 1

0

�xy2

2+2

3y3�x20

dx

=

Z 1

0

�1

2x5 +

2

3x6�dx =

�x6

12+2

21x7�10

=1

12+2

21=5

28:


Ejemplo

CalcularZZZ

E



�(x; y; z) j 0 � x � 1; 0 � y � x2; 0 � z � x+ 2y

:ZZZ

E

y dV =

Z 1

0

Z x2

0

Z x+2y

0

y dz dy dx

=

Z 1

0

Z x2

0

[yz]x+2y0 dy dx

=

Z 1

0

Z x2

0

y (x+ 2y) dy dx

=

Z 1

0

�xy2

2+2

3y3�x20

dx

=

Z 1

0

�1

2x5 +

2

3x6�dx

=

�x6

12+2

21x7�10

=1

12+2

21=5

28:


Ejemplo

CalcularZZZ

E



�(x; y; z) j 0 � x � 1; 0 � y � x2; 0 � z � x+ 2y

:ZZZ

E

y dV =

Z 1

0

Z x2

0

Z x+2y

0

y dz dy dx

=

Z 1

0

Z x2

0

[yz]x+2y0 dy dx

=

Z 1

0

Z x2

0

y (x+ 2y) dy dx

=

Z 1

0

�xy2

2+2

3y3�x20

dx

=

Z 1

0

�1

2x5 +

2

3x6�dx =

�x6

12+2

21x7�10

=1

12+2

21=5

28:


Región del tipo B. E = f(x; y; z) j (y; z) 2 D; 1 (y; z) � x � 2 (y; z)gZZZE

f (x; y; z) dV =

ZZD

"Z 2(y;z)

1(y;z)

f (x; y; z) dx

#dA:

Si D = f(y; z) j c � y � d; �1 (y) � z � �2 (y)g entoncesZZZE

f (x; y; z) dV =

Z d

c

Z �2(y)

�1(y)

Z 2(y;z)

1(y;z)

f (x; y; z) dx dz dy:

Si D = f(y; z) j � � z � �; �1 (z) � y � �2 (z)g entoncesZZZE

f (x; y; z) dV =

Z �

�

Z �2(z)

�1(z)

Z 2(y;z)

1(y;z)

f (x; y; z) dx dy dz:

¿Cómo queda la integral cuandoE = f(x; y; z) j (x; z) 2 D; �1 (x; z) � y � �2 (x; z)g?


Región del tipo B. E = f(x; y; z) j (y; z) 2 D; 1 (y; z) � x � 2 (y; z)gZZZE

f (x; y; z) dV =

ZZD

"Z 2(y;z)

1(y;z)

f (x; y; z) dx

#dA:

Si D = f(y; z) j c � y � d; �1 (y) � z � �2 (y)g entoncesZZZE

f (x; y; z) dV =

Z d

c

Z �2(y)

�1(y)

Z 2(y;z)

1(y;z)

f (x; y; z) dx dz dy:

Si D = f(y; z) j � � z � �; �1 (z) � y � �2 (z)g entoncesZZZE

f (x; y; z) dV =

Z �

�

Z �2(z)

�1(z)

Z 2(y;z)

1(y;z)

f (x; y; z) dx dy dz:

¿Cómo queda la integral cuandoE = f(x; y; z) j (x; z) 2 D; �1 (x; z) � y � �2 (x; z)g?G A D � g a r e n a s d @ u i s . e d u .c o (U IS ) C á lc u lo I I I S e g u n d o S em e s t r e d e 2 0 1 1 7 / 5 1

EjemploEscriba la integral

I =

Z 1

0

Z 1

px

Z 1�y

0


como una integral iterada equivalente en los otros 5 órdenes.

I =

Z 1

0

Z y2

0

Z 1�y

0

f (x; y; z) dz dx dy

=

Z 1

0

Z 1�y

0

Z y2

0

f (x; y; z) dx dz dy

=

Z Z Zf (x; y; z) d d d


Cálculo de volúmenes

Si f (x; y; z) = 1 para (x; y; z) 2 E, entonces

V ol (E) =

ZZZE

dV:

Observe que si E es del tipo A entonces

V ol (E) =

ZZZE

dV

=

ZZD

"Z �2(x;y)

�1(x;y)

dz

#dA

=

ZZD

[�2 (x; y)� �1 (x; y)] dA

Volumen del sólido localizado entre las super�cies z = �1 (x; y) y z = �2 (x; y).


EjemploHallar el volumen de la región acotada por las super�cies z = x2 + 3y2 yz = 16� 3x2 � y2.

Primero miramos donde coinciden las dos ecuaciones

x2 + 3y2 = 16� 3x2 � y2 =) x2 + y2 = 4:

V ol (E) =

ZZZE

dV =

ZZC

"Z 16�3x2�y2

x2+3y2dz

#dA

=

ZZC

��16� 3x2 � y2

��x2 + 3y2

��dA

=

ZZC

�16� 4

�x2 + y2

��dA =

Z 2�

0

Z 2

0

�16� 4r2

�r dr d�

=

Z 2�

0

�8r2 � r4

�20d� =

Z 2�

0

16 d� = [16�]2�0 = 32�:

EjemploHallar el volumen de la región acotada por las super�cies z = x2 + 3y2 yz = 9� x2. (Propuesto).

G A D � g a r e n a s d @ u i s . e d u .c o (U IS ) C á lc u lo I I I S e g u n d o S em e s t r e d e 2 0 1 1 1 0 / 5 1

EjemploHallar el volumen de la región acotada por las super�cies z = x2 + 3y2 yz = 16� 3x2 � y2.Primero miramos donde coinciden las dos ecuaciones

x2 + 3y2 = 16� 3x2 � y2 =) x2 + y2 = 4:

V ol (E) =

ZZZE

dV =

ZZC

"Z 16�3x2�y2

x2+3y2dz

#dA

=

ZZC

��16� 3x2 � y2

��x2 + 3y2

��dA

=

ZZC

�16� 4

�x2 + y2

��dA =

Z 2�

0

Z 2

0

�16� 4r2

�r dr d�

=

Z 2�

0

�8r2 � r4

�20d� =

Z 2�

0

16 d� = [16�]2�0 = 32�:




x2 + 3y2 = 16� 3x2 � y2

=) x2 + y2 = 4:

V ol (E) =

ZZZE

dV =

ZZC

"Z 16�3x2�y2

x2+3y2dz

#dA

=

ZZC

��16� 3x2 � y2

��x2 + 3y2

��dA

=

ZZC

�16� 4

�x2 + y2

��dA =

Z 2�

0

Z 2

0

�16� 4r2

�r dr d�

=

Z 2�

0

�8r2 � r4

�20d� =

Z 2�

0

16 d� = [16�]2�0 = 32�:




x2 + 3y2 = 16� 3x2 � y2 =) x2 + y2 = 4:

V ol (E) =

ZZZE

dV =

ZZC

"Z 16�3x2�y2

x2+3y2dz

#dA

=

ZZC

��16� 3x2 � y2

��x2 + 3y2

��dA

=

ZZC

�16� 4

�x2 + y2

��dA =

Z 2�

0

Z 2

0

�16� 4r2

�r dr d�

=

Z 2�

0

�8r2 � r4

�20d� =

Z 2�

0

16 d� = [16�]2�0 = 32�:




x2 + 3y2 = 16� 3x2 � y2 =) x2 + y2 = 4:

V ol (E) =

ZZZE

dV

=

ZZC

"Z 16�3x2�y2

x2+3y2dz

#dA

=

ZZC

��16� 3x2 � y2

��x2 + 3y2

��dA

=

ZZC

�16� 4

�x2 + y2

��dA =

Z 2�

0

Z 2

0

�16� 4r2

�r dr d�

=

Z 2�

0

�8r2 � r4

�20d� =

Z 2�

0

16 d� = [16�]2�0 = 32�:




x2 + 3y2 = 16� 3x2 � y2 =) x2 + y2 = 4:

V ol (E) =

ZZZE

dV =

ZZC

"Z 16�3x2�y2

x2+3y2dz

#dA

=

ZZC

��16� 3x2 � y2

��x2 + 3y2

��dA

=

ZZC

�16� 4

�x2 + y2

��dA =

Z 2�

0

Z 2

0

�16� 4r2

�r dr d�

=

Z 2�

0

�8r2 � r4

�20d� =

Z 2�

0

16 d� = [16�]2�0 = 32�:




x2 + 3y2 = 16� 3x2 � y2 =) x2 + y2 = 4:

V ol (E) =

ZZZE

dV =

ZZC

"Z 16�3x2�y2

x2+3y2dz

#dA

=

ZZC

��16� 3x2 � y2

��x2 + 3y2

��dA

=

ZZC

�16� 4

�x2 + y2

��dA

=

Z 2�

0

Z 2

0

�16� 4r2

�r dr d�

=

Z 2�

0

�8r2 � r4

�20d� =

Z 2�

0

16 d� = [16�]2�0 = 32�:




x2 + 3y2 = 16� 3x2 � y2 =) x2 + y2 = 4:

V ol (E) =

ZZZE

dV =

ZZC

"Z 16�3x2�y2

x2+3y2dz

#dA

=

ZZC

��16� 3x2 � y2

��x2 + 3y2

��dA

=

ZZC

�16� 4

�x2 + y2

��dA =

Z 2�

0

Z 2

0

�16� 4r2

�r dr d�

=

Z 2�

0

�8r2 � r4

�20d� =

Z 2�

0

16 d� = [16�]2�0 = 32�:




x2 + 3y2 = 16� 3x2 � y2 =) x2 + y2 = 4:

V ol (E) =

ZZZE

dV =

ZZC

"Z 16�3x2�y2

x2+3y2dz

#dA

=

ZZC

��16� 3x2 � y2

��x2 + 3y2

��dA

=

ZZC

�16� 4

�x2 + y2

��dA =

Z 2�

0

Z 2

0

�16� 4r2

�r dr d�

=

Z 2�

0

�8r2 � r4

�20d�

=

Z 2�

0

16 d� = [16�]2�0 = 32�:




x2 + 3y2 = 16� 3x2 � y2 =) x2 + y2 = 4:

V ol (E) =

ZZZE

dV =

ZZC

"Z 16�3x2�y2

x2+3y2dz

#dA

=

ZZC

��16� 3x2 � y2

��x2 + 3y2

��dA

=

ZZC

�16� 4

�x2 + y2

��dA =

Z 2�

0

Z 2

0

�16� 4r2

�r dr d�

=

Z 2�

0

�8r2 � r4

�20d� =

Z 2�

0

16 d�

= [16�]2�0 = 32�:




x2 + 3y2 = 16� 3x2 � y2 =) x2 + y2 = 4:

V ol (E) =

ZZZE

dV =

ZZC

"Z 16�3x2�y2

x2+3y2dz

#dA

=

ZZC

��16� 3x2 � y2

��x2 + 3y2

��dA

=

ZZC

�16� 4

�x2 + y2

��dA =

Z 2�

0

Z 2

0

�16� 4r2

�r dr d�

=

Z 2�

0

�8r2 � r4

�20d� =

Z 2�

0

16 d� = [16�]2�0 = 32�:




x2 + 3y2 = 16� 3x2 � y2 =) x2 + y2 = 4:

V ol (E) =

ZZZE

dV =

ZZC

"Z 16�3x2�y2

x2+3y2dz

#dA

=

ZZC

��16� 3x2 � y2

��x2 + 3y2

��dA

=

ZZC

�16� 4

�x2 + y2

��dA =

Z 2�

0

Z 2

0

�16� 4r2

�r dr d�

=

Z 2�

0

�8r2 � r4

�20d� =

Z 2�

0

16 d� = [16�]2�0 = 32�:

EjemploHallar el volumen de la región acotada por las super�cies z = x2 + 3y2 yz = 9� x2. (Propuesto).G A D � g a r e n a s d @ u i s . e d u .c o (U IS ) C á lc u lo I I I S e g u n d o S em e s t r e d e 2 0 1 1 1 0 / 5 1

Aplicaciones de las integrales triples

Masa. Si � (x; y; z) es la densidad (unidad de masa / unidad de volumen) en(x; y; z) de un sólido que ocupa la región E de R3, su masa m es dada deforma análoga que en el caso de integrales dobles por

m =

ZZZE

� (x; y; z) dV;

y sus momentos respecto a los tres planos coordenados son

Myz =

ZZZE

x� (x; y; z) dV; Mxz =

ZZZE

y� (x; y; z) dV y

Mxy =

ZZZE

z� (x; y; z) dV;

y el centro de masa es (�x; �y; �z), donde

�x =Myz

m; �y =

Mxz

my �z =

Mxy

m:

Si la densidad es constante el centro de masa del sólido se llama centroide deE.




m =

ZZZE

� (x; y; z) dV;


Myz =

ZZZE

x� (x; y; z) dV;

Mxz =

ZZZE

y� (x; y; z) dV y

Mxy =

ZZZE

z� (x; y; z) dV;


�x =Myz

m; �y =

Mxz

my �z =

Mxy

m:





m =

ZZZE

� (x; y; z) dV;


Myz =

ZZZE


ZZZE

y� (x; y; z) dV

y

Mxy =

ZZZE

z� (x; y; z) dV;


�x =Myz

m; �y =

Mxz

my �z =

Mxy

m:





m =

ZZZE

� (x; y; z) dV;


Myz =

ZZZE


ZZZE

y� (x; y; z) dV y

Mxy =

ZZZE

z� (x; y; z) dV;


�x =Myz

m; �y =

Mxz

my �z =

Mxy

m:





m =

ZZZE

� (x; y; z) dV;


Myz =

ZZZE


ZZZE

y� (x; y; z) dV y

Mxy =

ZZZE

z� (x; y; z) dV;

y el centro de masa es (�x; �y; �z),

donde

�x =Myz

m; �y =

Mxz

my �z =

Mxy

m:





m =

ZZZE

� (x; y; z) dV;


Myz =

ZZZE


ZZZE

y� (x; y; z) dV y

Mxy =

ZZZE

z� (x; y; z) dV;


�x =Myz

m; �y =

Mxz

my �z =

Mxy

m:





m =

ZZZE

� (x; y; z) dV;


Myz =

ZZZE


ZZZE

y� (x; y; z) dV y

Mxy =

ZZZE

z� (x; y; z) dV;


�x =Myz

m; �y =

Mxz

my �z =

Mxy

m:

Si la densidad es constante el centro de masa del sólido se llama centroide deE.G A D � g a r e n a s d @ u i s . e d u .c o (U IS ) C á lc u lo I I I S e g u n d o S em e s t r e d e 2 0 1 1 1 1 / 5 1

Los momentos de inercia alrededor de los tres ejes coordenados son

Ix =

ZZZE

�y2 + z2

�� (x; y; z) dV;

Iy =

ZZZE

�x2 + z2

�� (x; y; z) dV;

Iz =

ZZZE

�x2 + y2

�� (x; y; z) dV:

La carga eléctrica de un sólido que ocupa una región E y tiene densidad de carga� (x; y; z) es

Q =

ZZZE

� (x; y; z) dV:

EjemploEncontrar Ix para un cubo de densidad constante k y longitud L, si uno de susvertices está en el origen y tres de sus aristas se localizan a lo largo de los ejescoordenados.

Ix =

ZZZE

�y2 + z2

�k dV = k

Z L

0

Z L

0

Z L

0

�y2 + z2

�dx dy dz

= k

Z L

0

Z L

0

��y2 + z2

�x�L0dy dz = kL

Z L

0

Z L

0

�y2 + z2

�dy dz

= kL

Z L

0

�y3

3+ z2y

�L0

dz = kL

Z L

0

�L3

3+ z2L

�dz

= kL

�L3

3z + L

z3

3

�L0

= kL

�L4

3+L4

3

�=2k

3L5:



Ix =

ZZZE

�y2 + z2

�� (x; y; z) dV; Iy =

ZZZE

�x2 + z2

�� (x; y; z) dV;

Iz =

ZZZE

�x2 + y2

�� (x; y; z) dV:


Q =

ZZZE

� (x; y; z) dV:


Ix =

ZZZE

�y2 + z2

�k dV = k

Z L

0

Z L

0

Z L

0

�y2 + z2

�dx dy dz

= k

Z L

0

Z L

0

��y2 + z2

�x�L0dy dz = kL

Z L

0

Z L

0

�y2 + z2

�dy dz

= kL

Z L

0

�y3

3+ z2y

�L0

dz = kL

Z L

0

�L3

3+ z2L

�dz

= kL

�L3

3z + L

z3

3

�L0

= kL

�L4

3+L4

3

�=2k

3L5:



Ix =

ZZZE

�y2 + z2

�� (x; y; z) dV; Iy =

ZZZE

�x2 + z2

�� (x; y; z) dV;

Iz =

ZZZE

�x2 + y2

�� (x; y; z) dV:


Q =

ZZZE

� (x; y; z) dV:


Ix =

ZZZE

�y2 + z2

�k dV

= k

Z L

0

Z L

0

Z L

0

�y2 + z2

�dx dy dz

= k

Z L

0

Z L

0

��y2 + z2

�x�L0dy dz = kL

Z L

0

Z L

0

�y2 + z2

�dy dz

= kL

Z L

0

�y3

3+ z2y

�L0

dz = kL

Z L

0

�L3

3+ z2L

�dz

= kL

�L3

3z + L

z3

3

�L0

= kL

�L4

3+L4

3

�=2k

3L5:



Ix =

ZZZE

�y2 + z2

�� (x; y; z) dV; Iy =

ZZZE

�x2 + z2

�� (x; y; z) dV;

Iz =

ZZZE

�x2 + y2

�� (x; y; z) dV:


Q =

ZZZE

� (x; y; z) dV:


Ix =

ZZZE

�y2 + z2

�k dV = k

Z L

0

Z L

0

Z L

0

�y2 + z2

�dx dy dz

= k

Z L

0

Z L

0

��y2 + z2

�x�L0dy dz = kL

Z L

0

Z L

0

�y2 + z2

�dy dz

= kL

Z L

0

�y3

3+ z2y

�L0

dz = kL

Z L

0

�L3

3+ z2L

�dz

= kL

�L3

3z + L

z3

3

�L0

= kL

�L4

3+L4

3

�=2k

3L5:



Ix =

ZZZE

�y2 + z2

�� (x; y; z) dV; Iy =

ZZZE

�x2 + z2

�� (x; y; z) dV;

Iz =

ZZZE

�x2 + y2

�� (x; y; z) dV:


Q =

ZZZE

� (x; y; z) dV:


Ix =

ZZZE

�y2 + z2

�k dV = k

Z L

0

Z L

0

Z L

0

�y2 + z2

�dx dy dz

= k

Z L

0

Z L

0

��y2 + z2

�x�L0dy dz

= kL

Z L

0

Z L

0

�y2 + z2

�dy dz

= kL

Z L

0

�y3

3+ z2y

�L0

dz = kL

Z L

0

�L3

3+ z2L

�dz

= kL

�L3

3z + L

z3

3

�L0

= kL

�L4

3+L4

3

�=2k

3L5:



Ix =

ZZZE

�y2 + z2

�� (x; y; z) dV; Iy =

ZZZE

�x2 + z2

�� (x; y; z) dV;

Iz =

ZZZE

�x2 + y2

�� (x; y; z) dV:


Q =

ZZZE

� (x; y; z) dV:


Ix =

ZZZE

�y2 + z2

�k dV = k

Z L

0

Z L

0

Z L

0

�y2 + z2

�dx dy dz

= k

Z L

0

Z L

0

��y2 + z2

�x�L0dy dz = kL

Z L

0

Z L

0

�y2 + z2

�dy dz

= kL

Z L

0

�y3

3+ z2y

�L0

dz = kL

Z L

0

�L3

3+ z2L

�dz

= kL

�L3

3z + L

z3

3

�L0

= kL

�L4

3+L4

3

�=2k

3L5:



Ix =

ZZZE

�y2 + z2

�� (x; y; z) dV; Iy =

ZZZE

�x2 + z2

�� (x; y; z) dV;

Iz =

ZZZE

�x2 + y2

�� (x; y; z) dV:


Q =

ZZZE

� (x; y; z) dV:


Ix =

ZZZE

�y2 + z2

�k dV = k

Z L

0

Z L

0

Z L

0

�y2 + z2

�dx dy dz

= k

Z L

0

Z L

0

��y2 + z2

�x�L0dy dz = kL

Z L

0

Z L

0

�y2 + z2

�dy dz

= kL

Z L

0

�y3

3+ z2y

�L0

dz

= kL

Z L

0

�L3

3+ z2L

�dz

= kL

�L3

3z + L

z3

3

�L0

= kL

�L4

3+L4

3

�=2k

3L5:



Ix =

ZZZE

�y2 + z2

�� (x; y; z) dV; Iy =

ZZZE

�x2 + z2

�� (x; y; z) dV;

Iz =

ZZZE

�x2 + y2

�� (x; y; z) dV:


Q =

ZZZE

� (x; y; z) dV:


Ix =

ZZZE

�y2 + z2

�k dV = k

Z L

0

Z L

0

Z L

0

�y2 + z2

�dx dy dz

= k

Z L

0

Z L

0

��y2 + z2

�x�L0dy dz = kL

Z L

0

Z L

0

�y2 + z2

�dy dz

= kL

Z L

0

�y3

3+ z2y

�L0

dz = kL

Z L

0

�L3

3+ z2L

�dz

= kL

�L3

3z + L

z3

3

�L0

= kL

�L4

3+L4

3

�=2k

3L5:



Ix =

ZZZE

�y2 + z2

�� (x; y; z) dV; Iy =

ZZZE

�x2 + z2

�� (x; y; z) dV;

Iz =

ZZZE

�x2 + y2

�� (x; y; z) dV:


Q =

ZZZE

� (x; y; z) dV:


Ix =

ZZZE

�y2 + z2

�k dV = k

Z L

0

Z L

0

Z L

0

�y2 + z2

�dx dy dz

= k

Z L

0

Z L

0

��y2 + z2

�x�L0dy dz = kL

Z L

0

Z L

0

�y2 + z2

�dy dz

= kL

Z L

0

�y3

3+ z2y

�L0

dz = kL

Z L

0

�L3

3+ z2L

�dz

= kL

�L3

3z + L

z3

3

�L0

= kL

�L4

3+L4

3

�

=2k

3L5:



Ix =

ZZZE

�y2 + z2

�� (x; y; z) dV; Iy =

ZZZE

�x2 + z2

�� (x; y; z) dV;

Iz =

ZZZE

�x2 + y2

�� (x; y; z) dV:


Q =

ZZZE

� (x; y; z) dV:


Ix =

ZZZE

�y2 + z2

�k dV = k

Z L

0

Z L

0

Z L

0

�y2 + z2

�dx dy dz

= k

Z L

0

Z L

0

��y2 + z2

�x�L0dy dz = kL

Z L

0

Z L

0

�y2 + z2

�dy dz

= kL

Z L

0

�y3

3+ z2y

�L0

dz = kL

Z L

0

�L3

3+ z2L

�dz

= kL

�L3

3z + L

z3

3

�L0

= kL

�L4

3+L4

3

�=2k

3L5:


EjemploHallar el centro de masa del sólido acotado por el cilindro z = 1� y2 y losplanos x+ z = 1, x = 0, z = 0 y tiene densidad constante � (x; y; z) = 4.

m =

Z 1

�1

Z 1�y2

0

Z 1�z

0

4 dx dz dy

=

Z 1

�1

Z 1�y2

0

[4x]x=1�zx=0 dz dy =

Z 1

�1

Z 1�y2

0

4 (1� z) dz dy

= 4

Z 1

�1

�z � z2

2

�z=1�y2z=0

dy = 4

Z 1

�1

"�1� y2

��1� y2

�22

#dy

= 2

Z 1

�1

�1� y4

�dy = 2

�y � y5

5

�1�1

= 2

��1� 1

5

��1�

��15

��=16

5:



m =

Z 1

�1

Z 1�y2

0

Z 1�z

0

4 dx dz dy

=

Z 1

�1

Z 1�y2

0

[4x]x=1�zx=0 dz dy

=

Z 1

�1

Z 1�y2

0

4 (1� z) dz dy

= 4

Z 1

�1

�z � z2

2

�z=1�y2z=0

dy = 4

Z 1

�1

"�1� y2

��1� y2

�22

#dy

= 2

Z 1

�1

�1� y4

�dy = 2

�y � y5

5

�1�1

= 2

��1� 1

5

��1�

��15

��=16

5:



m =

Z 1

�1

Z 1�y2

0

Z 1�z

0

4 dx dz dy

=

Z 1

�1

Z 1�y2

0


Z 1

�1

Z 1�y2

0

4 (1� z) dz dy

= 4

Z 1

�1

�z � z2

2

�z=1�y2z=0

dy = 4

Z 1

�1

"�1� y2

��1� y2

�22

#dy

= 2

Z 1

�1

�1� y4

�dy = 2

�y � y5

5

�1�1

= 2

��1� 1

5

��1�

��15

��=16

5:



m =

Z 1

�1

Z 1�y2

0

Z 1�z

0

4 dx dz dy

=

Z 1

�1

Z 1�y2

0


Z 1

�1

Z 1�y2

0

4 (1� z) dz dy

= 4

Z 1

�1

�z � z2

2

�z=1�y2z=0

dy

= 4

Z 1

�1

"�1� y2

��1� y2

�22

#dy

= 2

Z 1

�1

�1� y4

�dy = 2

�y � y5

5

�1�1

= 2

��1� 1

5

��1�

��15

��=16

5:



m =

Z 1

�1

Z 1�y2

0

Z 1�z

0

4 dx dz dy

=

Z 1

�1

Z 1�y2

0


Z 1

�1

Z 1�y2

0

4 (1� z) dz dy

= 4

Z 1

�1

�z � z2

2

�z=1�y2z=0

dy = 4

Z 1

�1

"�1� y2

��1� y2

�22

#dy

= 2

Z 1

�1

�1� y4

�dy = 2

�y � y5

5

�1�1

= 2

��1� 1

5

��1�

��15

��=16

5:



m =

Z 1

�1

Z 1�y2

0

Z 1�z

0

4 dx dz dy

=

Z 1

�1

Z 1�y2

0


Z 1

�1

Z 1�y2

0

4 (1� z) dz dy

= 4

Z 1

�1

�z � z2

2

�z=1�y2z=0

dy = 4

Z 1

�1

"�1� y2

��1� y2

�22

#dy

= 2

Z 1

�1

�1� y4

�dy

= 2

�y � y5

5

�1�1

= 2

��1� 1

5

��1�

��15

��=16

5:



m =

Z 1

�1

Z 1�y2

0

Z 1�z

0

4 dx dz dy

=

Z 1

�1

Z 1�y2

0


Z 1

�1

Z 1�y2

0

4 (1� z) dz dy

= 4

Z 1

�1

�z � z2

2

�z=1�y2z=0

dy = 4

Z 1

�1

"�1� y2

��1� y2

�22

#dy

= 2

Z 1

�1

�1� y4

�dy = 2

�y � y5

5

�1�1

= 2

��1� 1

5

��1�

��15

��=16

5:



m =

Z 1

�1

Z 1�y2

0

Z 1�z

0

4 dx dz dy

=

Z 1

�1

Z 1�y2

0


Z 1

�1

Z 1�y2

0

4 (1� z) dz dy

= 4

Z 1

�1

�z � z2

2

�z=1�y2z=0

dy = 4

Z 1

�1

"�1� y2

��1� y2

�22

#dy

= 2

Z 1

�1

�1� y4

�dy = 2

�y � y5

5

�1�1

= 2

��1� 1

5

��1�

��15

��

=16

5:



m =

Z 1

�1

Z 1�y2

0

Z 1�z

0

4 dx dz dy

=

Z 1

�1

Z 1�y2

0


Z 1

�1

Z 1�y2

0

4 (1� z) dz dy

= 4

Z 1

�1

�z � z2

2

�z=1�y2z=0

dy = 4

Z 1

�1

"�1� y2

��1� y2

�22

#dy

= 2

Z 1

�1

�1� y4

�dy = 2

�y � y5

5

�1�1

= 2

��1� 1

5

��1�

��15

��=16

5:


Ejemplo

�x =Myz

m

=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4x dx dz dy =5

16� 87=5

14

�y =Mxz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4y dx dz dy =5

16� 0 = 0

�z =Myx

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4z dx dz dy =5

16� 3235=2

7


Ejemplo

�x =Myz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4x dx dz dy

=5

16� 87=5

14

�y =Mxz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4y dx dz dy =5

16� 0 = 0

�z =Myx

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4z dx dz dy =5

16� 3235=2

7


Ejemplo

�x =Myz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4x dx dz dy =5

16� 87=5

14

�y =Mxz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4y dx dz dy =5

16� 0 = 0

�z =Myx

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4z dx dz dy =5

16� 3235=2

7


Ejemplo

�x =Myz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4x dx dz dy =5

16� 87=5

14

�y =Mxz

m

=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4y dx dz dy =5

16� 0 = 0

�z =Myx

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4z dx dz dy =5

16� 3235=2

7


Ejemplo

�x =Myz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4x dx dz dy =5

16� 87=5

14

�y =Mxz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4y dx dz dy

=5

16� 0 = 0

�z =Myx

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4z dx dz dy =5

16� 3235=2

7


Ejemplo

�x =Myz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4x dx dz dy =5

16� 87=5

14

�y =Mxz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4y dx dz dy =5

16� 0 = 0

�z =Myx

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4z dx dz dy =5

16� 3235=2

7


Ejemplo

�x =Myz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4x dx dz dy =5

16� 87=5

14

�y =Mxz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4y dx dz dy =5

16� 0 = 0

�z =Myx

m

=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4z dx dz dy =5

16� 3235=2

7


Ejemplo

�x =Myz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4x dx dz dy =5

16� 87=5

14

�y =Mxz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4y dx dz dy =5

16� 0 = 0

�z =Myx

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4z dx dz dy

=5

16� 3235=2

7


Ejemplo

�x =Myz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4x dx dz dy =5

16� 87=5

14

�y =Mxz

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4y dx dz dy =5

16� 0 = 0

�z =Myx

m=5

16

Z 1

�1

Z 1�y2

0

Z 1�z

0

4z dx dz dy =5

16� 3235=2

7


Cambio de variables e integrales múltiples

Sustitución �Cambio de variablesZ b

a

f (u (x)) u0 (x) dx =

Z u(b)

u(a)

f (u) du

Cambio de variables a coordenadas polares

La región en el plano r� corresponde a la región R del plano xyZZR

f (x; y) dA =

ZZP

f (r cos �; r sen �) r dr d�


Más generalmente, si un cambio de variables está dado por unatransformación T del plano uv en el plano xy, es decir,

T (u; v) = (x; y) = (g (u; v) ; h (u; v)) ; x = g (u; v) ; y = h (u; v)

entonces T transforma una región S del plano uv en una R del plano xy.

Si (x; y) = T (u; v) ; (x; y) se denomina imagen de (u; v) y R se denominaimagen de S (R = T (S)).T es 1�1 si puntos distintos de S tienen imágenes distintas en R. En este casoT tiene una transformación inversa y podemos despejar a u y v en términos dex y y.Trabajaremos con transformaciones T tales que T y T�1 son de clase C1.




entonces T transforma una región S del plano uv en una R del plano xy.Si (x; y) = T (u; v) ; (x; y) se denomina imagen de (u; v) y R se denominaimagen de S (R = T (S)).

T es 1�1 si puntos distintos de S tienen imágenes distintas en R. En este casoT tiene una transformación inversa y podemos despejar a u y v en términos dex y y.Trabajaremos con transformaciones T tales que T y T�1 son de clase C1.




entonces T transforma una región S del plano uv en una R del plano xy.Si (x; y) = T (u; v) ; (x; y) se denomina imagen de (u; v) y R se denominaimagen de S (R = T (S)).T es 1�1 si puntos distintos de S tienen imágenes distintas en R. En este casoT tiene una transformación inversa y podemos despejar a u y v en términos dex y y.Trabajaremos con transformaciones T tales que T y T�1 son de clase C1.G A D � g a r e n a s d @ u i s . e d u .c o (U IS ) C á lc u lo I I I S e g u n d o S em e s t r e d e 2 0 1 1 1 6 / 5 1

Ejemplo. Encuentre la imagen del triángulo con vertices en (0; 0), (2; 0) y(0; 2) mediante la transformación lineal u = y � x; v = y + x.

�u = y � xv = y + x

=)

8>>>><>>>>:

�y = u+ xy = v � x =) y =

u+ v

2�x = y � ux = v � y =) x =

v � u2

(>) 0 � x � 2; y = 0 =)�u = 0� xv = 0 + x

=)�v = �u ;�2 � u � 0

(�) 0 � x � 2; x+ y = 2 =)�u = (2� x)� xv = (2� x) + x

=)�u = 2� 2xv = 2

=)��2 � u � 2v = 2

(o) x = 0; 0 � y � 2 =)�u = y � 0v = y + 0

=)�v = u ; 0 � u � 2:


Ejemplo. Determine las curvas u y v de las transformaciones lineales T cuyainversa T�1 está dada por u = xy; v = x2 � y2.

�xy = u = c1

x2 � y2 = v = c2

=)(

y =c1x

x2 � y2 = c2

Ejemplo. Las ecuaciones x = u2 � v2, y = 2uv de�nen una transformación T .Encuentre T (S) si S = f(u; v) j 0 � u � 1; 0 � v � 1g :

Si 0 � u � 1; v = 0 entonces x = u2, y = 0.

Si u = 0; 0 � v � 1 entonces x = 1� v2, y = 2v,al reemplazar se obtiene x = 1� 1

4y2:

Si 0 � u � 1; v = 1 entonces x = u2 � 1, y = 2u,al reemplazar se obtiene x = 1

4y2 � 1:

Si u = 0; 0 � v � 1 entonces x = �v2, y = 0.


Ejemplo. Determine las curvas u y v de las transformaciones lineales T cuyainversa T�1 está dada por u = xy; v = x2 � y2.�

xy = u = c1x2 � y2 = v = c2

=)(

y =c1x

x2 � y2 = c2




4y2:


4y2 � 1:

Si u = 0; 0 � v � 1 entonces x = �v2, y = 0.


Ejemplo. Determine las curvas u y v de las transformaciones lineales T cuyainversa T�1 está dada por u = xy; v = x2 � y2.�

xy = u = c1x2 � y2 = v = c2

=)(

y =c1x

x2 � y2 = c2




4y2:


4y2 � 1:

Si u = 0; 0 � v � 1 entonces x = �v2, y = 0.G A D � g a r e n a s d @ u i s . e d u .c o (U IS ) C á lc u lo I I I S e g u n d o S em e s t r e d e 2 0 1 1 1 8 / 5 1

Deducción de la fórmula de cambio de variable

~ru=fu(u0;v0) {̂+gu(u0;v0) |̂;~rv=fv(u0; v0) {̂+gv(u0;v0) |̂;

~a=~r(u0+�u; v0)�~r(u0; v0) � ~ru(u0; v0)��u~b=~r(u0; v0+�v)�~r(u0; v0) � ~rv(u0; v0)��v

�A � ~a�~b = k(~ru (u0; v0) ��u)� (~rv (u0; v0) ��v)k

= k~ru (u0; v0)� ~rv (u0; v0)k�u ��v:

~ru � ~rv =

��{̂ |̂ k̂@ux @uy 0@vx @vy 0

�� =�� @ux @uy@vx @vy

�� k̂





�A � ~a�~b

= k(~ru (u0; v0) ��u)� (~rv (u0; v0) ��v)k

= k~ru (u0; v0)� ~rv (u0; v0)k�u ��v:

~ru � ~rv =

��{̂ |̂ k̂@ux @uy 0@vx @vy 0

�� =�� @ux @uy@vx @vy

�� k̂






= k~ru (u0; v0)� ~rv (u0; v0)k�u ��v:

~ru � ~rv =

��{̂ |̂ k̂@ux @uy 0@vx @vy 0

�� =�� @ux @uy@vx @vy

�� k̂






= k~ru (u0; v0)� ~rv (u0; v0)k�u ��v:

~ru � ~rv =

��{̂ |̂ k̂@ux @uy 0@vx @vy 0

�� =

�� @ux @uy@vx @vy

�� k̂






= k~ru (u0; v0)� ~rv (u0; v0)k�u ��v:

~ru � ~rv =

��{̂ |̂ k̂@ux @uy 0@vx @vy 0

�� =�� @ux @uy@vx @vy

�� k̂G A D � g a r e n a s d @ u i s . e d u .c o (U IS ) C á lc u lo I I I S e g u n d o S em e s t r e d e 2 0 1 1 1 9 / 5 1

De�niciónEl jacobiano de la transformación lineal T dada por x = f (u; v) e y = g (u; v)es

J (u; v) =@ (x; y)

@ (u; v)=


�� :

Observe que [J (x; y)]�1 = J (u; v) :

�A � jJ (u; v)j ��v ��u

ZZR

F (x; y) dA = l��mn;m!1

nXi=0

mXj=0

F (xij ; yij)�A

= l��mn;m!1

nXi=0

mXj=0

F (f (uij ; vij) ; g (uij ; vij)) jJ (u; v)j ��v ��u

=

ZZS

F (f (u; v) ; g (u; v)) jJ (u; v)j dv du


De�niciónEl jacobiano de la transformación lineal T dada por x = f (u; v) e y = g (u; v)es

J (u; v) =@ (x; y)

@ (u; v)=


�� :Observe que [J (x; y)]�1 = J (u; v) :

�A � jJ (u; v)j ��v ��u

ZZR

F (x; y) dA = l��mn;m!1

nXi=0

mXj=0

F (xij ; yij)�A

= l��mn;m!1

nXi=0

mXj=0

F (f (uij ; vij) ; g (uij ; vij)) jJ (u; v)j ��v ��u

=

ZZS

F (f (u; v) ; g (u; v)) jJ (u; v)j dv du


TeoremaSea F : R � R2 �! R una función continua de dos variables x e y de�nidasobre R2. Sea G : S � R2 �! R2, G (u; v) = (x; y) = (f (u; v) ; g (u; v)) unafunción que manda de manera inyectiva los puntos (u; v) de S del plano uv enlos puntos (x; y) de R2 del plano xy. Suponga que G es de clase C1 (S) y quela derivada G0 (u; v) = J (u; v) es una matriz invertible para todo (u; v) 2 S.Entonces se tiene la fórmula de cambio de variables en integrales doblesZZ

R

F (x; y) dx dy =

ZZS

F (f (u; v) ; g (u; v)) jJ (u; v)j dv du:

Cambio a coordenadas polares�x = r cos �y = r sen �

=)J (r; �)=@ (x; y)

@ (r; �)=

�� cos � sen ��r sen � r cos �

�� =r�cos2 �+sen2 ��=rZZ

R

F (x; y) dx dy =

ZZS

F (r cos �; r sen �) jJ (r; �)j dr d�

=

ZZS

F (r cos �; r sen �) r dr d�:



R

F (x; y) dx dy =

ZZS



=)J (r; �)=@ (x; y)

@ (r; �)

=



R

F (x; y) dx dy =

ZZS


=

ZZS




R

F (x; y) dx dy =

ZZS



=)J (r; �)=@ (x; y)

@ (r; �)=


��

=r�cos2 �+sen2 �

�=r

ZZR

F (x; y) dx dy =

ZZS


=

ZZS




R

F (x; y) dx dy =

ZZS



=)J (r; �)=@ (x; y)

@ (r; �)=


�� =r�cos2 �+sen2 ��=r

ZZR

F (x; y) dx dy =

ZZS


=

ZZS




R

F (x; y) dx dy =

ZZS



=)J (r; �)=@ (x; y)

@ (r; �)=



R

F (x; y) dx dy

=

ZZS


=

ZZS




R

F (x; y) dx dy =

ZZS



=)J (r; �)=@ (x; y)

@ (r; �)=



R

F (x; y) dx dy =

ZZS


=

ZZS



Ejemplo. CalcularZZ

R

(x+ y + 1) dA, donde R es la región acotada por las

rectas y � x = 1; y � x = �1; y + x = 1 e y + x = 2:(CuadriláteroABCD, A (0; 1) ; B (1; 0) ; C (3=3; 1=2) ; D (1=2; 3=2)).

�u = y � xv = y + x

=)

8<: x =v � u2

y =u+ v

2

J (x; y) =

�� 1 11 1

�� = �2;J (u; v) =

�� 1=2 1=21=2 1=2

�� = �12 :ZZR

(x+ y + 1) dA =

Z 1

�1

Z 2

1

(v + 1)

�1

2

�dv du

=1

2

Z 1

�1

�v2

2+ v

�21

du =1

2

Z 1

�1

5

2du =

5

4u

�1�1

=5

2:


Ejemplo. CalcularZZ

R



�u = y � xv = y + x

=)

8<: x =v � u2

y =u+ v

2

J (x; y) =

�� 1 11 1

�� = �2;

J (u; v) =

�� 1=2 1=21=2 1=2

�� = �12 :ZZR

(x+ y + 1) dA =

Z 1

�1

Z 2

1

(v + 1)

�1

2

�dv du

=1

2

Z 1

�1

�v2

2+ v

�21

du =1

2

Z 1

�1

5

2du =

5

4u

�1�1

=5

2:


Ejemplo. CalcularZZ

R



�u = y � xv = y + x

=)

8<: x =v � u2

y =u+ v

2

J (x; y) =

�� 1 11 1

�� = �2;J (u; v) =

�� 1=2 1=21=2 1=2

�� = �12 :

ZZR

(x+ y + 1) dA =

Z 1

�1

Z 2

1

(v + 1)

�1

2

�dv du

=1

2

Z 1

�1

�v2

2+ v

�21

du =1

2

Z 1

�1

5

2du =

5

4u

�1�1

=5

2:


Ejemplo. CalcularZZ

R



�u = y � xv = y + x

=)

8<: x =v � u2

y =u+ v

2

J (x; y) =

�� 1 11 1

�� = �2;J (u; v) =

�� 1=2 1=21=2 1=2

�� = �12 :ZZR

(x+ y + 1) dA

=

Z 1

�1

Z 2

1

(v + 1)

�1

2

�dv du

=1

2

Z 1

�1

�v2

2+ v

�21

du =1

2

Z 1

�1

5

2du =

5

4u

�1�1

=5

2:


Ejemplo. CalcularZZ

R



�u = y � xv = y + x

=)

8<: x =v � u2

y =u+ v

2

J (x; y) =

�� 1 11 1

�� = �2;J (u; v) =

�� 1=2 1=21=2 1=2

�� = �12 :ZZR

(x+ y + 1) dA =

Z 1

�1

Z 2

1

(v + 1)

�1

2

�dv du

=1

2

Z 1

�1

�v2

2+ v

�21

du =1

2

Z 1

�1

5

2du =

5

4u

�1�1

=5

2:


Ejemplo. CalcularZZ

R



�u = y � xv = y + x

=)

8<: x =v � u2

y =u+ v

2

J (x; y) =

�� 1 11 1

�� = �2;J (u; v) =

�� 1=2 1=21=2 1=2

�� = �12 :ZZR

(x+ y + 1) dA =

Z 1

�1

Z 2

1

(v + 1)

�1

2

�dv du

=1

2

Z 1

�1

�v2

2+ v

�21

du

=1

2

Z 1

�1

5

2du =

5

4u

�1�1

=5

2:


Ejemplo. CalcularZZ

R



�u = y � xv = y + x

=)

8<: x =v � u2

y =u+ v

2

J (x; y) =

�� 1 11 1

�� = �2;J (u; v) =

�� 1=2 1=21=2 1=2

�� = �12 :ZZR

(x+ y + 1) dA =

Z 1

�1

Z 2

1

(v + 1)

�1

2

�dv du

=1

2

Z 1

�1

�v2

2+ v

�21

du =1

2

Z 1

�1

5

2du

=5

4u

�1�1

=5

2:


Ejemplo. CalcularZZ

R



�u = y � xv = y + x

=)

8<: x =v � u2

y =u+ v

2

J (x; y) =

�� 1 11 1

�� = �2;J (u; v) =

�� 1=2 1=21=2 1=2

�� = �12 :ZZR

(x+ y + 1) dA =

Z 1

�1

Z 2

1

(v + 1)

�1

2

�dv du

=1

2

Z 1

�1

�v2

2+ v

�21

du =1

2

Z 1

�1

5

2du =

5

4u

�1�1

=5

2:


Ejemplo. CalculeZZ

R

x2y2 dA, donde R es la región acotada por xy = 2,

xy = 1, x = 2y y 3x = y:

�u = xyv = x=y

J (x; y) =

�� y 1=yx �x=y2

��= �2x

y= �2v

J (u; v) = � 1

2v

ZZR

x2y2 dA =

Z 2

1

Z 2

1=3

u2�1

2v

�dv du

=1

2

Z 2

1

u2�ln v�21=3

du =1

2[ln 2� ln (1=3)]

�u3

3

�21

=7

6ln 6:


Ejemplo. CalculeZZ

R


xy = 1, x = 2y y 3x = y:

�u = xyv = x=y

J (x; y) =

�� y 1=yx �x=y2

��

= �2xy= �2v

J (u; v) = � 1

2v

ZZR

x2y2 dA =

Z 2

1

Z 2

1=3

u2�1

2v

�dv du

=1

2

Z 2

1

u2�ln v�21=3

du =1

2[ln 2� ln (1=3)]

�u3

3

�21

=7

6ln 6:


Ejemplo. CalculeZZ

R


xy = 1, x = 2y y 3x = y:

�u = xyv = x=y

J (x; y) =

�� y 1=yx �x=y2

��= �2x

y= �2v

J (u; v) = � 1

2v

ZZR

x2y2 dA =

Z 2

1

Z 2

1=3

u2�1

2v

�dv du

=1

2

Z 2

1

u2�ln v�21=3

du =1

2[ln 2� ln (1=3)]

�u3

3

�21

=7

6ln 6:


Ejemplo. CalculeZZ

R


xy = 1, x = 2y y 3x = y:

�u = xyv = x=y

J (x; y) =

�� y 1=yx �x=y2

��= �2x

y= �2v

J (u; v) = � 1

2v

ZZR

x2y2 dA

=

Z 2

1

Z 2

1=3

u2�1

2v

�dv du

=1

2

Z 2

1

u2�ln v�21=3

du =1

2[ln 2� ln (1=3)]

�u3

3

�21

=7

6ln 6:


Ejemplo. CalculeZZ

R


xy = 1, x = 2y y 3x = y:

�u = xyv = x=y

J (x; y) =

�� y 1=yx �x=y2

��= �2x

y= �2v

J (u; v) = � 1

2v

ZZR

x2y2 dA =

Z 2

1

Z 2

1=3

u2�1

2v

�dv du

=1

2

Z 2

1

u2�ln v�21=3

du =1

2[ln 2� ln (1=3)]

�u3

3

�21

=7

6ln 6:


Ejemplo. CalculeZZ

R


xy = 1, x = 2y y 3x = y:

�u = xyv = x=y

J (x; y) =

�� y 1=yx �x=y2

��= �2x

y= �2v

J (u; v) = � 1

2v

ZZR

x2y2 dA =

Z 2

1

Z 2

1=3

u2�1

2v

�dv du

=1

2

Z 2

1

u2�ln v�21=3

du

=1

2[ln 2� ln (1=3)]

�u3

3

�21

=7

6ln 6:


Ejemplo. CalculeZZ

R


xy = 1, x = 2y y 3x = y:

�u = xyv = x=y

J (x; y) =

�� y 1=yx �x=y2

��= �2x

y= �2v

J (u; v) = � 1

2v

ZZR

x2y2 dA =

Z 2

1

Z 2

1=3

u2�1

2v

�dv du

=1

2

Z 2

1

u2�ln v�21=3

du =1

2[ln 2� ln (1=3)]

�u3

3

�21

=7

6ln 6:


Ejemplo. CalculeZZ

R

x�3 dA, donde R es la región acotada por y = 2x2,

y = x2, x = y2 y x = 2y2:

�u = y=x2

v = y2=x

J (x; y) =

�� 2y=x3 �y2=x21=x2 2y=x

��= �3y

2

x4= �3u2

J (u; v) = � 1

3u2

ZZR

x�3 dA =

ZZR

� yx2

�2� x

y2

�dA =

Z 2

1

Z 1

1=2

u2

v

�1

3u2

�dv du

=1

3

Z 2

1

du

Z 1

1=2

dv

v=1

3(2� 1)

�ln 1� ln 12

�=1

3ln 2:


Ejemplo. CalculeZZ

R


y = x2, x = y2 y x = 2y2:

�u = y=x2

v = y2=x

J (x; y) =

�� 2y=x3 �y2=x21=x2 2y=x

��

= �3y2

x4= �3u2

J (u; v) = � 1

3u2

ZZR

x�3 dA =

ZZR

� yx2

�2� x

y2

�dA =

Z 2

1

Z 1

1=2

u2

v

�1

3u2

�dv du

=1

3

Z 2

1

du

Z 1

1=2

dv

v=1

3(2� 1)

�ln 1� ln 12

�=1

3ln 2:


Ejemplo. CalculeZZ

R


y = x2, x = y2 y x = 2y2:

�u = y=x2

v = y2=x

J (x; y) =

�� 2y=x3 �y2=x21=x2 2y=x

��= �3y

2

x4= �3u2

J (u; v) = � 1

3u2

ZZR

x�3 dA =

ZZR

� yx2

�2� x

y2

�dA =

Z 2

1

Z 1

1=2

u2

v

�1

3u2

�dv du

=1

3

Z 2

1

du

Z 1

1=2

dv

v=1

3(2� 1)

�ln 1� ln 12

�=1

3ln 2:


Ejemplo. CalculeZZ

R


y = x2, x = y2 y x = 2y2:

�u = y=x2

v = y2=x

J (x; y) =

�� 2y=x3 �y2=x21=x2 2y=x

��= �3y

2

x4= �3u2

J (u; v) = � 1

3u2

ZZR

x�3 dA

=

ZZR

� yx2

�2� x

y2

�dA =

Z 2

1

Z 1

1=2

u2

v

�1

3u2

�dv du

=1

3

Z 2

1

du

Z 1

1=2

dv

v=1

3(2� 1)

�ln 1� ln 12

�=1

3ln 2:


Ejemplo. CalculeZZ

R


y = x2, x = y2 y x = 2y2:

�u = y=x2

v = y2=x

J (x; y) =

�� 2y=x3 �y2=x21=x2 2y=x

��= �3y

2

x4= �3u2

J (u; v) = � 1

3u2

ZZR

x�3 dA =

ZZR

� yx2

�2� x

y2

�dA

=

Z 2

1

Z 1

1=2

u2

v

�1

3u2

�dv du

=1

3

Z 2

1

du

Z 1

1=2

dv

v=1

3(2� 1)

�ln 1� ln 12

�=1

3ln 2:


Ejemplo. CalculeZZ

R


y = x2, x = y2 y x = 2y2:

�u = y=x2

v = y2=x

J (x; y) =

�� 2y=x3 �y2=x21=x2 2y=x

��= �3y

2

x4= �3u2

J (u; v) = � 1

3u2

ZZR

x�3 dA =

ZZR

� yx2

�2� x

y2

�dA =

Z 2

1

Z 1

1=2

u2

v

�1

3u2

�dv du

=1

3

Z 2

1

du

Z 1

1=2

dv

v=1

3(2� 1)

�ln 1� ln 12

�=1

3ln 2:


Ejemplo. CalculeZZ

R


y = x2, x = y2 y x = 2y2:

�u = y=x2

v = y2=x

J (x; y) =

�� 2y=x3 �y2=x21=x2 2y=x

��= �3y

2

x4= �3u2

J (u; v) = � 1

3u2

ZZR

x�3 dA =

ZZR

� yx2

�2� x

y2

�dA =

Z 2

1

Z 1

1=2

u2

v

�1

3u2

�dv du

=1

3

Z 2

1

du

Z 1

1=2

dv

v

=1

3(2� 1)

�ln 1� ln 12

�=1

3ln 2:


Ejemplo. CalculeZZ

R


y = x2, x = y2 y x = 2y2:

�u = y=x2

v = y2=x

J (x; y) =

�� 2y=x3 �y2=x21=x2 2y=x

��= �3y

2

x4= �3u2

J (u; v) = � 1

3u2

ZZR

x�3 dA =

ZZR

� yx2

�2� x

y2

�dA =

Z 2

1

Z 1

1=2

u2

v

�1

3u2

�dv du

=1

3

Z 2

1

du

Z 1

1=2

dv

v=1

3(2� 1)

�ln 1� ln 12

�

=1

3ln 2:


Ejemplo. CalculeZZ

R


y = x2, x = y2 y x = 2y2:

�u = y=x2

v = y2=x

J (x; y) =

�� 2y=x3 �y2=x21=x2 2y=x

��= �3y

2

x4= �3u2

J (u; v) = � 1

3u2

ZZR

x�3 dA =

ZZR

� yx2

�2� x

y2

�dA =

Z 2

1

Z 1

1=2

u2

v

�1

3u2

�dv du

=1

3

Z 2

1

du

Z 1

1=2

dv

v=1

3(2� 1)

�ln 1� ln 12

�=1

3ln 2:


Ejemplo. Se quiere calcular el área del anillo elíptico acotado por

x2 + 4y2 = 4 y x2 + 4y2 = 16:

�x = 2r cos �y = r sen �

J (r; �) =@ (x; y)

@ (r; �)=

�� 2 cos � sen ��2r sen � r cos �

�� = 2r �cos2 � + sen2 �� = 2rA =

Z 2�

0

Z 2

1

2r dr d� = 2� (4� 1) = 6�:



x2 + 4y2 = 4 y x2 + 4y2 = 16:


J (r; �) =@ (x; y)

@ (r; �)

=


�� = 2r �cos2 � + sen2 �� = 2rA =

Z 2�

0

Z 2

1

2r dr d� = 2� (4� 1) = 6�:



x2 + 4y2 = 4 y x2 + 4y2 = 16:


J (r; �) =@ (x; y)

@ (r; �)=


��

= 2r�cos2 � + sen2 �

�= 2r

A =

Z 2�

0

Z 2

1

2r dr d� = 2� (4� 1) = 6�:



x2 + 4y2 = 4 y x2 + 4y2 = 16:


J (r; �) =@ (x; y)

@ (r; �)=


�� = 2r �cos2 � + sen2 �� = 2r

A =

Z 2�

0

Z 2

1

2r dr d� = 2� (4� 1) = 6�:



x2 + 4y2 = 4 y x2 + 4y2 = 16:


J (r; �) =@ (x; y)

@ (r; �)=


�� = 2r �cos2 � + sen2 �� = 2rA =

Z 2�

0

Z 2

1

2r dr d�

= 2� (4� 1) = 6�:



x2 + 4y2 = 4 y x2 + 4y2 = 16:


J (r; �) =@ (x; y)

@ (r; �)=


�� = 2r �cos2 � + sen2 �� = 2rA =

Z 2�

0

Z 2

1

2r dr d� = 2� (4� 1) = 6�:


Cambio a coordenadas polares generalizadas

�x = ar cos �y = br sen �

J (r; �) =@ (x; y)

@ (r; �)=

�� a cos � b sen ��ar sen � br cos �

�� = abr

ZZR

F (x; y) dx dy =

ZZS

F (ar cos �; br sen �) abr dr d�:

Ejemplo. Encuentre el área de la elipsex2

a2+y2

b2= 1.

A =

ZZR

dx dy =

Z 2�

0

Z 1

0

abr dr d� = ab�:




J (r; �) =@ (x; y)

@ (r; �)

=


�� = abr

ZZR

F (x; y) dx dy =

ZZS



a2+y2

b2= 1.

A =

ZZR

dx dy =

Z 2�

0

Z 1

0





J (r; �) =@ (x; y)

@ (r; �)=


��

= abr

ZZR

F (x; y) dx dy =

ZZS



a2+y2

b2= 1.

A =

ZZR

dx dy =

Z 2�

0

Z 1

0





J (r; �) =@ (x; y)

@ (r; �)=


�� = abr

ZZR

F (x; y) dx dy =

ZZS



a2+y2

b2= 1.

A =

ZZR

dx dy =

Z 2�

0

Z 1

0





J (r; �) =@ (x; y)

@ (r; �)=


�� = abr

ZZR

F (x; y) dx dy

=

ZZS



a2+y2

b2= 1.

A =

ZZR

dx dy =

Z 2�

0

Z 1

0





J (r; �) =@ (x; y)

@ (r; �)=


�� = abr

ZZR

F (x; y) dx dy =

ZZS



a2+y2

b2= 1.

A =

ZZR

dx dy =

Z 2�

0

Z 1

0



Cambio de variables en integrales triples

8<: x = f (u; v; w)y = g (u; v; w)z = h (u; v; w)

=) J (u; v; w) =

��@ux @uy @uz@vx @vy @vz@wx @wy @wz

��ZZZ

R

F (x; y; z) dx dy dz =

ZZZS

G (u; v; w) jJ (u; v; w)j du dv dw

Ejemplo. Coordenadas cilíndricas8<: x = r cos �y = r sen �z = z

J (r; �; z) =

��cos � sen � 0�r sen � r cos � 00 0 1

�� = r

ZZZR

F (x; y; z) dV =

ZZZS

G (r; �; z) r dz dr d�:




=) J (u; v; w) =


��

ZZZR


ZZZS



J (r; �; z) =

��cos � sen � 0�r sen � r cos � 00 0 1

�� = r

ZZZR

F (x; y; z) dV =

ZZZS

G (r; �; z) r dz dr d�:




=) J (u; v; w) =


��ZZZ

R


ZZZS



J (r; �; z) =

��cos � sen � 0�r sen � r cos � 00 0 1

�� = r

ZZZR

F (x; y; z) dV =

ZZZS

G (r; �; z) r dz dr d�:


Ejemplo.Coordenadas esféricas8<: x = � sen� cos �y = � sen� sen �z = � cos�

J (�; �; �) =

��sen� cos � sen� sen � cos�

�� sen� sen � � sen� cos � 0� cos� cos � � cos� sen � �� sen�

�� = ��2 sen�ZZZ

R

F (x; y; z) dV =

ZZZS

G (�; �; �) �2 sen�d� d� d�:



J (�; �; �) =



�� = ��2 sen�

ZZZR

F (x; y; z) dV =

ZZZS

G (�; �; �) �2 sen�d� d� d�:



J (�; �; �) =



�� = ��2 sen�ZZZ

R

F (x; y; z) dV

=

ZZZS

G (�; �; �) �2 sen�d� d� d�:



J (�; �; �) =



�� = ��2 sen�ZZZ

R

F (x; y; z) dV =

ZZZS

G (�; �; �) �2 sen�d� d� d�:


Ejemplo. Determine el volumen del toro T obtenido al girar alrededor del ejez el disco circular (y � b)2 + z2 � a2, 0 < a < b:

0 � � � a; 0 � � 2�; 0 � � � 2�x = (b+ � cos ) cos �; y = (b+ � cos ) sen �; z = � sen :


J (�; ; �) =

��cos cos � cos sen � sen

�� sen cos � �� sen sen � � cos � (b+ � cos ) sen � (b+ � cos ) cos � 0

��

= �� (b+ � cos ) :

V =

ZZZR

dV =

Z 2�

0

Z 2�

0

Z a

0

� (b+ � cos ) d� d d�

=

Z 2�

0

Z 2�

0

�b�2

2+�3

3cos

�a0

d d�

=

�ba2

2 +

a3

3sen

�2�0

[�]2�0

= a2b� � [2�] = 2�2a2b = (2�b)��a2�:


J (�; ; �) =

��cos cos � cos sen � sen

�� sen cos � �� sen sen � � cos � (b+ � cos ) sen � (b+ � cos ) cos � 0

��= �� (b+ � cos ) :

V =

ZZZR

dV =

Z 2�

0

Z 2�

0

Z a

0

� (b+ � cos ) d� d d�

=

Z 2�

0

Z 2�

0

�b�2

2+�3

3cos

�a0

d d�

=

�ba2

2 +

a3

3sen

�2�0

[�]2�0

= a2b� � [2�] = 2�2a2b = (2�b)��a2�:


Coordenadas cilíndricas 8<: x = r cos �y = r sen �z = z

E = f(x; y; z) : (x; y) 2 D; u1 (x; y) � z � u2 (x; y)g ;D = f(r; �) : � � � � �; h1 (�) � r � h2 (�)g

J (r; �; z) = r

ZZZR

F (x; y; z) dV =

Z �

�

Z h2(�)

h1(�)

Z u2(r cos �;r sen �)

u1(r cos �;r sen �)

F (r cos �; r sen �; z) r dz dr d�:


Ejemplo. Encuentre la masa del sólido acotado por �=6 � � � �=3, r = cos �,z = 0 y z = r, si la densidad es dada por � (r; �; z) = 3r.

m =

Z �=3

�=6

Z cos �

0

Z r

0

(3r) � r dz dr d� =Z �=3

�=6

Z cos �

0

3r2 � [z]r0 dr d�

=

Z �=3

�=6

Z cos �

0

3r3 dr d� =

Z �=3

�=6

�3

4r4�cos �0

d� =3

4

Z �=3

�=6

cos4 �d�

=3

4

�3

8� +

1

4sen 2� +

1

32sen 4�

��=3�=6

=3

128

�2� �

p3�:



m =

Z �=3

�=6

Z cos �

0

Z r

0

(3r) � r dz dr d�

=

Z �=3

�=6

Z cos �

0

3r2 � [z]r0 dr d�

=

Z �=3

�=6

Z cos �

0

3r3 dr d� =

Z �=3

�=6

�3

4r4�cos �0

d� =3

4

Z �=3

�=6

cos4 �d�

=3

4

�3

8� +

1

4sen 2� +

1

32sen 4�

��=3�=6

=3

128

�2� �

p3�:



m =

Z �=3

�=6

Z cos �

0

Z r

0

(3r) � r dz dr d� =Z �=3

�=6

Z cos �

0

3r2 � [z]r0 dr d�

=

Z �=3

�=6

Z cos �

0

3r3 dr d� =

Z �=3

�=6

�3

4r4�cos �0

d� =3

4

Z �=3

�=6

cos4 �d�

=3

4

�3

8� +

1

4sen 2� +

1

32sen 4�

��=3�=6

=3

128

�2� �

p3�:



m =

Z �=3

�=6

Z cos �

0

Z r

0

(3r) � r dz dr d� =Z �=3

�=6

Z cos �

0

3r2 � [z]r0 dr d�

=

Z �=3

�=6

Z cos �

0

3r3 dr d�

=

Z �=3

�=6

�3

4r4�cos �0

d� =3

4

Z �=3

�=6

cos4 �d�

=3

4

�3

8� +

1

4sen 2� +

1

32sen 4�

��=3�=6

=3

128

�2� �

p3�:



m =

Z �=3

�=6

Z cos �

0

Z r

0

(3r) � r dz dr d� =Z �=3

�=6

Z cos �

0

3r2 � [z]r0 dr d�

=

Z �=3

�=6

Z cos �

0

3r3 dr d� =

Z �=3

�=6

�3

4r4�cos �0

d�

=3

4

Z �=3

�=6

cos4 �d�

=3

4

�3

8� +

1

4sen 2� +

1

32sen 4�

��=3�=6

=3

128

�2� �

p3�:



m =

Z �=3

�=6

Z cos �

0

Z r

0

(3r) � r dz dr d� =Z �=3

�=6

Z cos �

0

3r2 � [z]r0 dr d�

=

Z �=3

�=6

Z cos �

0

3r3 dr d� =

Z �=3

�=6

�3

4r4�cos �0

d� =3

4

Z �=3

�=6

cos4 �d�

=3

4

�3

8� +

1

4sen 2� +

1

32sen 4�

��=3�=6

=3

128

�2� �

p3�:


Ejemplo. Encuentre la región acotada por el cilindro x2 + y2 = 4x y por laesfera x2 + y2 + z2 = 16:

V =

Z �

0

Z 4 cos �

0

Z p16�r2

�p16�r2

r dz dr d�


V =

Z �

0

Z 4 cos �

0

Z p16�r2

�p16�r2

r dz dr d�

= 2

Z �

0

Z 4 cos �

0

Z p16�r2

0

r dz dr d�

= 2

Z �

0

Z 4 cos �

0

rp16� r2 dr d� = 2

Z �

0

��13

�16� r2

�3=2�4 cos �0

d�

=2

3

Z �

0

h64�

�16� 16 cos2 �

�3=2id�

=128

3

Z �

0

h1�

�1� cos2 �

�3=2id� =

128

3

Z �

0

�1� sen3 �

�d�

=128

3

�� +

3

4cos � � 1

12cos 3�

��0

=128

9(3� � 4) :


V =

Z �

0

Z 4 cos �

0

Z p16�r2

�p16�r2

r dz dr d�

= 2

Z �

0

Z 4 cos �

0

Z p16�r2

0

r dz dr d�

= 2

Z �

0

Z 4 cos �

0

rp16� r2 dr d�

= 2

Z �

0

��13

�16� r2

�3=2�4 cos �0

d�

=2

3

Z �

0

h64�

�16� 16 cos2 �

�3=2id�

=128

3

Z �

0

h1�

�1� cos2 �

�3=2id� =

128

3

Z �

0

�1� sen3 �

�d�

=128

3

�� +

3

4cos � � 1

12cos 3�

��0

=128

9(3� � 4) :


V =

Z �

0

Z 4 cos �

0

Z p16�r2

�p16�r2

r dz dr d�

= 2

Z �

0

Z 4 cos �

0

Z p16�r2

0

r dz dr d�

= 2

Z �

0

Z 4 cos �

0

rp16� r2 dr d� = 2

Z �

0

��13

�16� r2

�3=2�4 cos �0

d�

=2

3

Z �

0

h64�

�16� 16 cos2 �

�3=2id�

=128

3

Z �

0

h1�

�1� cos2 �

�3=2id� =

128

3

Z �

0

�1� sen3 �

�d�

=128

3

�� +

3

4cos � � 1

12cos 3�

��0

=128

9(3� � 4) :


V =

Z �

0

Z 4 cos �

0

Z p16�r2

�p16�r2

r dz dr d�

= 2

Z �

0

Z 4 cos �

0

Z p16�r2

0

r dz dr d�

= 2

Z �

0

Z 4 cos �

0

rp16� r2 dr d� = 2

Z �

0

��13

�16� r2

�3=2�4 cos �0

d�

=2

3

Z �

0

h64�

�16� 16 cos2 �

�3=2id�

=128

3

Z �

0

h1�

�1� cos2 �

�3=2id�

=128

3

Z �

0

�1� sen3 �

�d�

=128

3

�� +

3

4cos � � 1

12cos 3�

��0

=128

9(3� � 4) :


V =

Z �

0

Z 4 cos �

0

Z p16�r2

�p16�r2

r dz dr d�

= 2

Z �

0

Z 4 cos �

0

Z p16�r2

0

r dz dr d�

= 2

Z �

0

Z 4 cos �

0

rp16� r2 dr d� = 2

Z �

0

��13

�16� r2

�3=2�4 cos �0

d�

=2

3

Z �

0

h64�

�16� 16 cos2 �

�3=2id�

=128

3

Z �

0

h1�

�1� cos2 �

�3=2id� =

128

3

Z �

0

�1� sen3 �

�d�

=128

3

�� +

3

4cos � � 1

12cos 3�

��0

=128

9(3� � 4) :


Ejemplo. Encuentre el momento de inercia Iz del sólido acotadoinferiormente por el plano xy, superiormente por la esfera de radio a ylateralmente por el cilindro r = a cos �.

Iz

=

Z �

0

Z a cos �

0

Z pa2�r2

0

�x2 + y2

�r dz dr d�

=

Z �

0

Z a cos �

0

Z pa2�r2

0

r3 dz dr d�

=

Z �

0

Z a cos �

0

r3pa2 � r2dr d�

=

Z �

0

Z a cos �

0

r�r2 � a2 + a2

� �a2 � r2

�1=2dr d�

=

Z �

0

Z a cos �

0

rha2�a2 � r2

�1=2 � �a2 � r2�3=2i dr d�=

Z �

0

��13a2�a2 � r2

�3=2+1

5

�a2 � r2

�5=2�a cos �0

d�

=

Z �

0

��13a2�a2�a2cos2�

�3=2+1

5

�a2�a2cos2�

�5=2+1

3a2�a2�3=2� 1

5

�a2�5=2�

d�



Iz =

Z �

0

Z a cos �

0

Z pa2�r2

0

�x2 + y2

�r dz dr d�

=

Z �

0

Z a cos �

0

Z pa2�r2

0

r3 dz dr d�

=

Z �

0

Z a cos �

0

r3pa2 � r2dr d�

=

Z �

0

Z a cos �

0

r�r2 � a2 + a2

� �a2 � r2

�1=2dr d�

=

Z �

0

Z a cos �

0

rha2�a2 � r2

�1=2 � �a2 � r2�3=2i dr d�=

Z �

0

��13a2�a2 � r2

�3=2+1

5

�a2 � r2

�5=2�a cos �0

d�

=

Z �

0

��13a2�a2�a2cos2�

�3=2+1

5

�a2�a2cos2�

�5=2+1

3a2�a2�3=2� 1

5

�a2�5=2�

d�



Iz =

Z �

0

Z a cos �

0

Z pa2�r2

0

�x2 + y2

�r dz dr d�

=

Z �

0

Z a cos �

0

Z pa2�r2

0

r3 dz dr d�

=

Z �

0

Z a cos �

0

r3pa2 � r2dr d�

=

Z �

0

Z a cos �

0

r�r2 � a2 + a2

� �a2 � r2

�1=2dr d�

=

Z �

0

Z a cos �

0

rha2�a2 � r2

�1=2 � �a2 � r2�3=2i dr d�

=

Z �

0

��13a2�a2 � r2

�3=2+1

5

�a2 � r2

�5=2�a cos �0

d�

=

Z �

0

��13a2�a2�a2cos2�

�3=2+1

5

�a2�a2cos2�

�5=2+1

3a2�a2�3=2� 1

5

�a2�5=2�

d�



Iz =

Z �

0

Z a cos �

0

Z pa2�r2

0

�x2 + y2

�r dz dr d�

=

Z �

0

Z a cos �

0

Z pa2�r2

0

r3 dz dr d�

=

Z �

0

Z a cos �

0

r3pa2 � r2dr d�

=

Z �

0

Z a cos �

0

r�r2 � a2 + a2

� �a2 � r2

�1=2dr d�

=

Z �

0

Z a cos �

0

rha2�a2 � r2

�1=2 � �a2 � r2�3=2i dr d�=

Z �

0

��13a2�a2 � r2

�3=2+1

5

�a2 � r2

�5=2�a cos �0

d�

=

Z �

0

��13a2�a2�a2cos2�

�3=2+1

5

�a2�a2cos2�

�5=2+1

3a2�a2�3=2� 1

5

�a2�5=2�

d�


=

Z �

0

��a

5

3

�1�cos2 �

�3=2+a5

5

�1�cos2 �

�5=2+1

3a2�a2�3=2� 1

5

�a2�5=2�

d�

=

Z �

0

��a

5

3sen3 � +

a5

5sen5 � +

1

3a5 � 1

5a5�d�

=a5

15

Z �

0

�3 sen5 � � 5 sen3 � + 2

�d�

=a5

15

�2� +

Z �

0

�3 sen4 � � 5 sen2 �

�sen �d�

�=

a5

15

�2� +

Z �

0

h3�1� cos2 �

�2 � 5 (1� cos �)i sen �d��=

a5

15

�2� +

Z �

0

�3�1� 2 cos2 � + cos4 �

�� 5

�1� cos2 �

��sen �d�

�=

a5

15

�2� +

Z �

0

��2� cos2 � + 3 cos4 �

�sen �d�

�=

a5

15

�2� +

�2 cos � +

1

3cos3 � � 3

5cos5 �

��0

�=2a5

15

�� 26

15

�


=

Z �

0

��a

5

3

�1�cos2 �

�3=2+a5

5

�1�cos2 �

�5=2+1

3a2�a2�3=2� 1

5

�a2�5=2�

d�

=

Z �

0

��a

5

3sen3 � +

a5

5sen5 � +

1

3a5 � 1

5a5�d�

=a5

15

Z �

0

�3 sen5 � � 5 sen3 � + 2

�d�

=a5

15

�2� +

Z �

0

�3 sen4 � � 5 sen2 �

�sen �d�

�

=a5

15

�2� +

Z �

0

h3�1� cos2 �

�2 � 5 (1� cos �)i sen �d��=

a5

15

�2� +

Z �

0

�3�1� 2 cos2 � + cos4 �

�� 5

�1� cos2 �

��sen �d�

�=

a5

15

�2� +

Z �

0

��2� cos2 � + 3 cos4 �

�sen �d�

�=

a5

15

�2� +

�2 cos � +

1

3cos3 � � 3

5cos5 �

��0

�=2a5

15

�� 26

15

�


=

Z �

0

��a

5

3

�1�cos2 �

�3=2+a5

5

�1�cos2 �

�5=2+1

3a2�a2�3=2� 1

5

�a2�5=2�

d�

=

Z �

0

��a

5

3sen3 � +

a5

5sen5 � +

1

3a5 � 1

5a5�d�

=a5

15

Z �

0

�3 sen5 � � 5 sen3 � + 2

�d�

=a5

15

�2� +

Z �

0

�3 sen4 � � 5 sen2 �

�sen �d�

�=

a5

15

�2� +

Z �

0

h3�1� cos2 �

�2 � 5 (1� cos �)i sen �d��

=a5

15

�2� +

Z �

0

�3�1� 2 cos2 � + cos4 �

�� 5

�1� cos2 �

��sen �d�

�=

a5

15

�2� +

Z �

0

��2� cos2 � + 3 cos4 �

�sen �d�

�=

a5

15

�2� +

�2 cos � +

1

3cos3 � � 3

5cos5 �

��0

�=2a5

15

�� 26

15

�


=

Z �

0

��a

5

3

�1�cos2 �

�3=2+a5

5

�1�cos2 �

�5=2+1

3a2�a2�3=2� 1

5

�a2�5=2�

d�

=

Z �

0

��a

5

3sen3 � +

a5

5sen5 � +

1

3a5 � 1

5a5�d�

=a5

15

Z �

0

�3 sen5 � � 5 sen3 � + 2

�d�

=a5

15

�2� +

Z �

0

�3 sen4 � � 5 sen2 �

�sen �d�

�=

a5

15

�2� +

Z �

0

h3�1� cos2 �

�2 � 5 (1� cos �)i sen �d��=

a5

15

�2� +

Z �

0

�3�1� 2 cos2 � + cos4 �

�� 5

�1� cos2 �

��sen �d�

�

=a5

15

�2� +

Z �

0

��2� cos2 � + 3 cos4 �

�sen �d�

�=

a5

15

�2� +

�2 cos � +

1

3cos3 � � 3

5cos5 �

��0

�=2a5

15

�� 26

15

�


=

Z �

0

��a

5

3

�1�cos2 �

�3=2+a5

5

�1�cos2 �

�5=2+1

3a2�a2�3=2� 1

5

�a2�5=2�

d�

=

Z �

0

��a

5

3sen3 � +

a5

5sen5 � +

1

3a5 � 1

5a5�d�

=a5

15

Z �

0

�3 sen5 � � 5 sen3 � + 2

�d�

=a5

15

�2� +

Z �

0

�3 sen4 � � 5 sen2 �

�sen �d�

�=

a5

15

�2� +

Z �

0

h3�1� cos2 �

�2 � 5 (1� cos �)i sen �d��=

a5

15

�2� +

Z �

0

�3�1� 2 cos2 � + cos4 �

�� 5

�1� cos2 �

��sen �d�

�=

a5

15

�2� +

Z �

0

��2� cos2 � + 3 cos4 �

�sen �d�

�

=a5

15

�2� +

�2 cos � +

1

3cos3 � � 3

5cos5 �

��0

�=2a5

15

�� 26

15

�


=

Z �

0

��a

5

3

�1�cos2 �

�3=2+a5

5

�1�cos2 �

�5=2+1

3a2�a2�3=2� 1

5

�a2�5=2�

d�

=

Z �

0

��a

5

3sen3 � +

a5

5sen5 � +

1

3a5 � 1

5a5�d�

=a5

15

Z �

0

�3 sen5 � � 5 sen3 � + 2

�d�

=a5

15

�2� +

Z �

0

�3 sen4 � � 5 sen2 �

�sen �d�

�=

a5

15

�2� +

Z �

0

h3�1� cos2 �

�2 � 5 (1� cos �)i sen �d��=

a5

15

�2� +

Z �

0

�3�1� 2 cos2 � + cos4 �

�� 5

�1� cos2 �

��sen �d�

�=

a5

15

�2� +

Z �

0

��2� cos2 � + 3 cos4 �

�sen �d�

�=

a5

15

�2� +

�2 cos � +

1

3cos3 � � 3

5cos5 �

��0

�

=2a5

15

�� 26

15

�


=

Z �

0

��a

5

3

�1�cos2 �

�3=2+a5

5

�1�cos2 �

�5=2+1

3a2�a2�3=2� 1

5

�a2�5=2�

d�

=

Z �

0

��a

5

3sen3 � +

a5

5sen5 � +

1

3a5 � 1

5a5�d�

=a5

15

Z �

0

�3 sen5 � � 5 sen3 � + 2

�d�

=a5

15

�2� +

Z �

0

�3 sen4 � � 5 sen2 �

�sen �d�

�=

a5

15

�2� +

Z �

0

h3�1� cos2 �

�2 � 5 (1� cos �)i sen �d��=

a5

15

�2� +

Z �

0

�3�1� 2 cos2 � + cos4 �

�� 5

�1� cos2 �

��sen �d�

�=

a5

15

�2� +

Z �

0

��2� cos2 � + 3 cos4 �

�sen �d�

�=

a5

15

�2� +

�2 cos � +

1

3cos3 � � 3

5cos5 �

��0

�=2a5

15

�� 26

15

�G A D � g a r e n a s d @ u i s . e d u .c o (U IS ) C á lc u lo I I I S e g u n d o S em e s t r e d e 2 0 1 1 3 6 / 5 1

Coordenadas esféricas 8<: x = � sen� cos �y = � sen� sen �z = � cos�

� 2 I� � [0;1) ; � 2 I� � [0; 2�] ; � 2 I� � [0; �]

�Vijk � (��) (�i��) (�i sen�k��)

J (�; �; �) = ��2 sen�ZZZR

F (x; y; z) dV =

ZZZE

G (�; �; �) �2 sen�d� d� d�:



� 2 I� � [0;1) ; � 2 I� � [0; 2�] ; � 2 I� � [0; �]

�Vijk � (��) (�i��) (�i sen�k��)

J (�; �; �) = ��2 sen�

ZZZR

F (x; y; z) dV =

ZZZE

G (�; �; �) �2 sen�d� d� d�:



� 2 I� � [0;1) ; � 2 I� � [0; 2�] ; � 2 I� � [0; �]

�Vijk � (��) (�i��) (�i sen�k��)

J (�; �; �) = ��2 sen�ZZZR

F (x; y; z) dV =

ZZZE

G (�; �; �) �2 sen�d� d� d�:


Ejemplos

Ejemplo. Determine el volumen y elcentroide del cono de helado C queestá acotado por el cono � = �=6 ypor la esfera � = 2a cos�.

V =

Z 2�

0

Z �=6

0

Z 2a cos�

0

�2 sen� d� d� d�

=8

3a3Z 2�

0

Z �=6

0

cos3 � sen� d� d�

=8

3a3Z 2�

0

��14cos4 �

��=60

d� =8

12a3

0@1� p32

!41AZ 2�

0

d�

=8

12a3�1� 9

16

�[2�] =

7

12�a3


Ejemplos


V =

Z 2�

0

Z �=6

0

Z 2a cos�

0

�2 sen� d� d� d�

=8

3a3Z 2�

0

Z �=6

0


=8

3a3Z 2�

0

��14cos4 �

��=60

d�

=8

12a3

0@1� p32

!41AZ 2�

0

d�

=8

12a3�1� 9

16

�[2�] =

7

12�a3


Ejemplos


V =

Z 2�

0

Z �=6

0

Z 2a cos�

0

�2 sen� d� d� d�

=8

3a3Z 2�

0

Z �=6

0


=8

3a3Z 2�

0

��14cos4 �

��=60

d� =8

12a3

0@1� p32

!41AZ 2�

0

d�

=8

12a3�1� 9

16

�[2�] =

7

12�a3


Ejemplos


V =

Z 2�

0

Z �=6

0

Z 2a cos�

0

�2 sen� d� d� d�

=8

3a3Z 2�

0

Z �=6

0


=8

3a3Z 2�

0

��14cos4 �

��=60

d� =8

12a3

0@1� p32

!41AZ 2�

0

d�

=8

12a3�1� 9

16

�[2�]

=7

12�a3


Ejemplos


V =

Z 2�

0

Z �=6

0

Z 2a cos�

0

�2 sen� d� d� d�

=8

3a3Z 2�

0

Z �=6

0


=8

3a3Z 2�

0

��14cos4 �

��=60

d� =8

12a3

0@1� p32

!41AZ 2�

0

d�

=8

12a3�1� 9

16

�[2�] =

7

12�a3


�x = �y = 0,

�z =1

V

Z 2�

0

Z �=6

0

Z 2a cos�

0

(� cos�) �2 sen� d� d� d�

=12

7�a3

Z 2�

0

Z �=6

0

Z 2a cos�

0

�3 cos� sen� d� d� d�

=12

7�a3

Z 2�

0

Z �=6

0

�1

4�4�2a cos�0

cos� sen� d� d�

=12 � 4a47�a3

Z 2�

0

Z �=6

0


=48a4

7�a3

Z 2�

0

��16cos6 �

��=60

d� =8a4

7�a3

241� p32

!635Z 2�

0

d�

=8a4

7�a3

�1� 27

64

�2� =

37

28a


�x = �y = 0,

�z =1

V

Z 2�

0

Z �=6

0

Z 2a cos�

0

(� cos�) �2 sen� d� d� d�

=12

7�a3

Z 2�

0

Z �=6

0

Z 2a cos�

0


=12

7�a3

Z 2�

0

Z �=6

0

�1

4�4�2a cos�0


=12 � 4a47�a3

Z 2�

0

Z �=6

0


=48a4

7�a3

Z 2�

0

��16cos6 �

��=60

d�

=8a4

7�a3

241� p32

!635Z 2�

0

d�

=8a4

7�a3

�1� 27

64

�2� =

37

28a


�x = �y = 0,

�z =1

V

Z 2�

0

Z �=6

0

Z 2a cos�

0

(� cos�) �2 sen� d� d� d�

=12

7�a3

Z 2�

0

Z �=6

0

Z 2a cos�

0


=12

7�a3

Z 2�

0

Z �=6

0

�1

4�4�2a cos�0


=12 � 4a47�a3

Z 2�

0

Z �=6

0


=48a4

7�a3

Z 2�

0

��16cos6 �

��=60

d� =8a4

7�a3

241� p32

!635Z 2�

0

d�

=8a4

7�a3

�1� 27

64

�2� =

37

28a


�x = �y = 0,

�z =1

V

Z 2�

0

Z �=6

0

Z 2a cos�

0

(� cos�) �2 sen� d� d� d�

=12

7�a3

Z 2�

0

Z �=6

0

Z 2a cos�

0


=12

7�a3

Z 2�

0

Z �=6

0

�1

4�4�2a cos�0


=12 � 4a47�a3

Z 2�

0

Z �=6

0


=48a4

7�a3

Z 2�

0

��16cos6 �

��=60

d� =8a4

7�a3

241� p32

!635Z 2�

0

d�

=8a4

7�a3

�1� 27

64

�2�

=37

28a


�x = �y = 0,

�z =1

V

Z 2�

0

Z �=6

0

Z 2a cos�

0

(� cos�) �2 sen� d� d� d�

=12

7�a3

Z 2�

0

Z �=6

0

Z 2a cos�

0


=12

7�a3

Z 2�

0

Z �=6

0

�1

4�4�2a cos�0


=12 � 4a47�a3

Z 2�

0

Z �=6

0


=48a4

7�a3

Z 2�

0

��16cos6 �

��=60

d� =8a4

7�a3

241� p32

!635Z 2�

0

d�

=8a4

7�a3

�1� 27

64

�2� =

37

28a


Ejemplo. Halle la masa de una esfera de radio R si la densidad en cada puntoes inversamente proporcional a su distancia desde el centro de la esfera.(Esto es � (�; �; �) = k=�).

m =

Z 2�

0

Z �

0

Z R

0

k

��2 sen� d� d� d�

= 2�k

Z �

0

Z R

0

� sen� d� d�

= �kR2Z �

0

sen� d� = �kR2 [� cos�]�0 = 2�kR2:

�densidad �nita aunque en el centro es in�nita�

¿Cuál es el centro de masa?



m =

Z 2�

0

Z �

0

Z R

0

k

��2 sen� d� d� d�

= 2�k

Z �

0

Z R

0


= �kR2Z �

0

sen� d�

= �kR2 [� cos�]�0 = 2�kR2:





m =

Z 2�

0

Z �

0

Z R

0

k

��2 sen� d� d� d�

= 2�k

Z �

0

Z R

0


= �kR2Z �

0

sen� d� = �kR2 [� cos�]�0

= 2�kR2:





m =

Z 2�

0

Z �

0

Z R

0

k

��2 sen� d� d� d�

= 2�k

Z �

0

Z R

0


= �kR2Z �

0

sen� d� = �kR2 [� cos�]�0 = 2�kR2:




Ejemplo. Hállese el volumen del cono seccionado en la esfera � = a por elcono � = �.

V =

Z 2�

0

Z �

0

Z a

0

�2 sen� d� d� d�

= 2� [� cos�]�0��3

3

�a0

=2

3�a3 (1� cos�) :


Ejemplo. Describa la super�cie � = 2a sen� y calcule el volumen de la regiónque acota.

V =

Z 2�

0

Z �

0

Z 2a sen�

0

�2 sen� d� d� d�

= 2�

Z �

0

��3

3

�2a sen�0

sen� d� =16a3�

3

Z �

0

sen4 � d�

=32a3�

3

Z �=2

0

sen4 � d� =32a3�

3

�1

2� 34� �2

�= 2a3�2:



V =

Z 2�

0

Z �

0

Z 2a sen�

0

�2 sen� d� d� d�

= 2�

Z �

0

��3

3

�2a sen�0

sen� d�

=16a3�

3

Z �

0

sen4 � d�

=32a3�

3

Z �=2

0

sen4 � d� =32a3�

3

�1

2� 34� �2

�= 2a3�2:



V =

Z 2�

0

Z �

0

Z 2a sen�

0

�2 sen� d� d� d�

= 2�

Z �

0

��3

3

�2a sen�0

sen� d� =16a3�

3

Z �

0

sen4 � d�

=32a3�

3

Z �=2

0

sen4 � d� =32a3�

3

�1

2� 34� �2

�= 2a3�2:



V =

Z 2�

0

Z �

0

Z 2a sen�

0

�2 sen� d� d� d�

= 2�

Z �

0

��3

3

�2a sen�0

sen� d� =16a3�

3

Z �

0

sen4 � d�

=32a3�

3

Z �=2

0

sen4 � d�

=32a3�

3

�1

2� 34� �2

�= 2a3�2:



V =

Z 2�

0

Z �

0

Z 2a sen�

0

�2 sen� d� d� d�

= 2�

Z �

0

��3

3

�2a sen�0

sen� d� =16a3�

3

Z �

0

sen4 � d�

=32a3�

3

Z �=2

0

sen4 � d� =32a3�

3

�1

2� 34� �2

�

= 2a3�2:



V =

Z 2�

0

Z �

0

Z 2a sen�

0

�2 sen� d� d� d�

= 2�

Z �

0

��3

3

�2a sen�0

sen� d� =16a3�

3

Z �

0

sen4 � d�

=32a3�

3

Z �=2

0

sen4 � d� =32a3�

3

�1

2� 34� �2

�= 2a3�2:


Área de super�cie

~u = �x {̂+ fx (xi; yj)�x k̂

~v = �y |̂+ fy (xi; yj)�y k̂

~u� ~v =

��{̂ |̂ k̂�x 0 fx (xi; yj)�x0 �y fy (xi; yj)�y

��=��fx(xi; yj) {̂� fy(xi; yj) |̂+ k̂

��x�y

�Tij = k~u� ~vk =q[fx (xi; yj)]

2+ [fy (xi; yj)]

2+ 1 �A

A (S) = l��mm;n!1

nXj=1

mXi=1

�Tij

=

ZZR

q[fx (x; y)]

2+ [fy (x; y)]

2+ 1 dA



~u = �x {̂+ fx (xi; yj)�x k̂

~v = �y |̂+ fy (xi; yj)�y k̂

~u� ~v =


��

=��fx(xi; yj) {̂� fy(xi; yj) |̂+ k̂

��x�y


2+ [fy (xi; yj)]

2+ 1 �A

A (S) = l��mm;n!1

nXj=1

mXi=1

�Tij

=

ZZR

q[fx (x; y)]

2+ [fy (x; y)]

2+ 1 dA



~u = �x {̂+ fx (xi; yj)�x k̂

~v = �y |̂+ fy (xi; yj)�y k̂

~u� ~v =


��=��fx(xi; yj) {̂� fy(xi; yj) |̂+ k̂

��x�y


2+ [fy (xi; yj)]

2+ 1 �A

A (S) = l��mm;n!1

nXj=1

mXi=1

�Tij

=

ZZR

q[fx (x; y)]

2+ [fy (x; y)]

2+ 1 dA



~u = �x {̂+ fx (xi; yj)�x k̂

~v = �y |̂+ fy (xi; yj)�y k̂

~u� ~v =


��=��fx(xi; yj) {̂� fy(xi; yj) |̂+ k̂

��x�y

�Tij = k~u� ~vk

=

q[fx (xi; yj)]

2+ [fy (xi; yj)]

2+ 1 �A

A (S) = l��mm;n!1

nXj=1

mXi=1

�Tij

=

ZZR

q[fx (x; y)]

2+ [fy (x; y)]

2+ 1 dA



~u = �x {̂+ fx (xi; yj)�x k̂

~v = �y |̂+ fy (xi; yj)�y k̂

~u� ~v =


��=��fx(xi; yj) {̂� fy(xi; yj) |̂+ k̂

��x�y


2+ [fy (xi; yj)]

2+ 1 �A

A (S) = l��mm;n!1

nXj=1

mXi=1

�Tij

=

ZZR

q[fx (x; y)]

2+ [fy (x; y)]

2+ 1 dA


Ejemplos

Ejemplo. Determine el área de la elipseobtenida de la intersección del planoz = 2x+ 2y + 1 con el cilindro x2 + y2 = 1.

A (S) =

ZZR

q[@xz]

2+ [@yz]

2+ 1dA

=

ZZR

q[2]

2+ [2]

2+ 1dA

=

ZZR

3dA =

ZZD

3r dr d�

=

Z 2�

0

Z 1

0

3r dr d�

= 3 [�]2�0

�r2

2

�10

= 3 � 2� � 12= 3�:


Ejemplo. Determine el área de lasuper�cie en el hemisferio superior dela esfera x2 + y2 + z2 = 2 cortado porel cilindro x2 + y2 = 1.

z =p2� x2 � y2;

@xz =�xp

2� x2 � y2;

@yz =�yp

2� x2 � y2:

A (S) =

ZZR

s��xp

2�x2�y2

�2+

��yp

2�x2�y2

�2+ 1dA

=

ZZR

rx2+y2+(2�x2�y2)

2�x2�y2 dA =

ZZR

s2

2� (x2 + y2)dA

=

Z 2�

0

Z 1

0

p2p

2� r2r dr d� =

p2 [�]

2�0

h��2� r2

�1=2i10

= 2p2��p2� 1

�= 2�

�2�

p2�:



z =p2� x2 � y2;

@xz =�xp

2� x2 � y2;

@yz =�yp

2� x2 � y2:

A (S) =

ZZR

s��xp

2�x2�y2

�2+

��yp

2�x2�y2

�2+ 1dA

=

ZZR

rx2+y2+(2�x2�y2)

2�x2�y2 dA

=

ZZR

s2

2� (x2 + y2)dA

=

Z 2�

0

Z 1

0

p2p

2� r2r dr d� =

p2 [�]

2�0

h��2� r2

�1=2i10

= 2p2��p2� 1

�= 2�

�2�

p2�:



z =p2� x2 � y2;

@xz =�xp

2� x2 � y2;

@yz =�yp

2� x2 � y2:

A (S) =

ZZR

s��xp

2�x2�y2

�2+

��yp

2�x2�y2

�2+ 1dA

=

ZZR

rx2+y2+(2�x2�y2)

2�x2�y2 dA =

ZZR

s2

2� (x2 + y2)dA

=

Z 2�

0

Z 1

0

p2p

2� r2r dr d� =

p2 [�]

2�0

h��2� r2

�1=2i10

= 2p2��p2� 1

�= 2�

�2�

p2�:



z =p2� x2 � y2;

@xz =�xp

2� x2 � y2;

@yz =�yp

2� x2 � y2:

A (S) =

ZZR

s��xp

2�x2�y2

�2+

��yp

2�x2�y2

�2+ 1dA

=

ZZR

rx2+y2+(2�x2�y2)

2�x2�y2 dA =

ZZR

s2

2� (x2 + y2)dA

=

Z 2�

0

Z 1

0

p2p

2� r2r dr d�

=p2 [�]

2�0

h��2� r2

�1=2i10

= 2p2��p2� 1

�= 2�

�2�

p2�:



z =p2� x2 � y2;

@xz =�xp

2� x2 � y2;

@yz =�yp

2� x2 � y2:

A (S) =

ZZR

s��xp

2�x2�y2

�2+

��yp

2�x2�y2

�2+ 1dA

=

ZZR

rx2+y2+(2�x2�y2)

2�x2�y2 dA =

ZZR

s2

2� (x2 + y2)dA

=

Z 2�

0

Z 1

0

p2p

2� r2r dr d� =

p2 [�]

2�0

h��2� r2

�1=2i10

= 2p2��p2� 1

�= 2�

�2�

p2�:



z =p2� x2 � y2;

@xz =�xp

2� x2 � y2;

@yz =�yp

2� x2 � y2:

A (S) =

ZZR

s��xp

2�x2�y2

�2+

��yp

2�x2�y2

�2+ 1dA

=

ZZR

rx2+y2+(2�x2�y2)

2�x2�y2 dA =

ZZR

s2

2� (x2 + y2)dA

=

Z 2�

0

Z 1

0

p2p

2� r2r dr d� =

p2 [�]

2�0

h��2� r2

�1=2i10

= 2p2��p2� 1

�

= 2��2�

p2�:



z =p2� x2 � y2;

@xz =�xp

2� x2 � y2;

@yz =�yp

2� x2 � y2:

A (S) =

ZZR

s��xp

2�x2�y2

�2+

��yp

2�x2�y2

�2+ 1dA

=

ZZR

rx2+y2+(2�x2�y2)

2�x2�y2 dA =

ZZR

s2

2� (x2 + y2)dA

=

Z 2�

0

Z 1

0

p2p

2� r2r dr d� =

p2 [�]

2�0

h��2� r2

�1=2i10

= 2p2��p2� 1

�= 2�

�2�

p2�:


Ejemplo. Hallese el área del paraboloide z = x2 + y2 seccionado por el planoz = 1.

A (S) =

ZZR

q1 + (2x)

2+ (2y)

2dA =

ZZR

p1 + 4 (x2 + y2)dA

=

Z 2�

0

Z 1

0

p1 + 4r2r dr d� =

�

6

�5p5� 1

�:

Ejemplo. Halle el área de la parte de la super�cie z = 23

�x3=2 + y3=2

�que

está sobre el rectángulo [0; 3]� [0; 3].

A (S) =

Z 3

0

Z 3

0

p1 + x+ ydy dx =

Z 3

0

�2

3(1 + x+ y)

3=2

�30

dx

=2

3

Z 3

0

h(4 + x)

3=2 � (1 + x)3=2idx

=2

3� 25

h(4 + x)

5=2 � (1 + x)5=2i30

=4

15

�75=2 � 45=2 � 45=2 + 1

�=4

15

�75=2 � 63

�:



A (S) =

ZZR

q1 + (2x)

2+ (2y)

2dA

=

ZZR

p1 + 4 (x2 + y2)dA

=

Z 2�

0

Z 1

0

p1 + 4r2r dr d� =

�

6

�5p5� 1

�:


�x3=2 + y3=2

�que


A (S) =

Z 3

0

Z 3

0

p1 + x+ ydy dx =

Z 3

0

�2

3(1 + x+ y)

3=2

�30

dx

=2

3

Z 3

0

h(4 + x)

3=2 � (1 + x)3=2idx

=2

3� 25

h(4 + x)

5=2 � (1 + x)5=2i30

=4

15

�75=2 � 45=2 � 45=2 + 1

�=4

15

�75=2 � 63

�:



A (S) =

ZZR

q1 + (2x)

2+ (2y)

2dA =

ZZR

p1 + 4 (x2 + y2)dA

=

Z 2�

0

Z 1

0

p1 + 4r2r dr d� =

�

6

�5p5� 1

�:


�x3=2 + y3=2

�que


A (S) =

Z 3

0

Z 3

0

p1 + x+ ydy dx =

Z 3

0

�2

3(1 + x+ y)

3=2

�30

dx

=2

3

Z 3

0

h(4 + x)

3=2 � (1 + x)3=2idx

=2

3� 25

h(4 + x)

5=2 � (1 + x)5=2i30

=4

15

�75=2 � 45=2 � 45=2 + 1

�=4

15

�75=2 � 63

�:



A (S) =

ZZR

q1 + (2x)

2+ (2y)

2dA =

ZZR

p1 + 4 (x2 + y2)dA

=

Z 2�

0

Z 1

0

p1 + 4r2r dr d�

=�

6

�5p5� 1

�:


�x3=2 + y3=2

�que


A (S) =

Z 3

0

Z 3

0

p1 + x+ ydy dx =

Z 3

0

�2

3(1 + x+ y)

3=2

�30

dx

=2

3

Z 3

0

h(4 + x)

3=2 � (1 + x)3=2idx

=2

3� 25

h(4 + x)

5=2 � (1 + x)5=2i30

=4

15

�75=2 � 45=2 � 45=2 + 1

�=4

15

�75=2 � 63

�:



A (S) =

ZZR

q1 + (2x)

2+ (2y)

2dA =

ZZR

p1 + 4 (x2 + y2)dA

=

Z 2�

0

Z 1

0

p1 + 4r2r dr d� =

�

6

�5p5� 1

�:


�x3=2 + y3=2

�que


A (S) =

Z 3

0

Z 3

0

p1 + x+ ydy dx =

Z 3

0

�2

3(1 + x+ y)

3=2

�30

dx

=2

3

Z 3

0

h(4 + x)

3=2 � (1 + x)3=2idx

=2

3� 25

h(4 + x)

5=2 � (1 + x)5=2i30

=4

15

�75=2 � 45=2 � 45=2 + 1

�=4

15

�75=2 � 63

�:



A (S) =

ZZR

q1 + (2x)

2+ (2y)

2dA =

ZZR

p1 + 4 (x2 + y2)dA

=

Z 2�

0

Z 1

0

p1 + 4r2r dr d� =

�

6

�5p5� 1

�:


�x3=2 + y3=2

�que


A (S) =

Z 3

0

Z 3

0

p1 + x+ ydy dx

=

Z 3

0

�2

3(1 + x+ y)

3=2

�30

dx

=2

3

Z 3

0

h(4 + x)

3=2 � (1 + x)3=2idx

=2

3� 25

h(4 + x)

5=2 � (1 + x)5=2i30

=4

15

�75=2 � 45=2 � 45=2 + 1

�=4

15

�75=2 � 63

�:



A (S) =

ZZR

q1 + (2x)

2+ (2y)

2dA =

ZZR

p1 + 4 (x2 + y2)dA

=

Z 2�

0

Z 1

0

p1 + 4r2r dr d� =

�

6

�5p5� 1

�:


�x3=2 + y3=2

�que


A (S) =

Z 3

0

Z 3

0

p1 + x+ ydy dx =

Z 3

0

�2

3(1 + x+ y)

3=2

�30

dx

=2

3

Z 3

0

h(4 + x)

3=2 � (1 + x)3=2idx

=2

3� 25

h(4 + x)

5=2 � (1 + x)5=2i30

=4

15

�75=2 � 45=2 � 45=2 + 1

�=4

15

�75=2 � 63

�:



A (S) =

ZZR

q1 + (2x)

2+ (2y)

2dA =

ZZR

p1 + 4 (x2 + y2)dA

=

Z 2�

0

Z 1

0

p1 + 4r2r dr d� =

�

6

�5p5� 1

�:


�x3=2 + y3=2

�que


A (S) =

Z 3

0

Z 3

0

p1 + x+ ydy dx =

Z 3

0

�2

3(1 + x+ y)

3=2

�30

dx

=2

3

Z 3

0

h(4 + x)

3=2 � (1 + x)3=2idx

=2

3� 25

h(4 + x)

5=2 � (1 + x)5=2i30

=4

15

�75=2 � 45=2 � 45=2 + 1

�

=4

15

�75=2 � 63

�:



A (S) =

ZZR

q1 + (2x)

2+ (2y)

2dA =

ZZR

p1 + 4 (x2 + y2)dA

=

Z 2�

0

Z 1

0

p1 + 4r2r dr d� =

�

6

�5p5� 1

�:


�x3=2 + y3=2

�que


A (S) =

Z 3

0

Z 3

0

p1 + x+ ydy dx =

Z 3

0

�2

3(1 + x+ y)

3=2

�30

dx

=2

3

Z 3

0

h(4 + x)

3=2 � (1 + x)3=2idx

=2

3� 25

h(4 + x)

5=2 � (1 + x)5=2i30

=4

15

�75=2 � 45=2 � 45=2 + 1

�=4

15

�75=2 � 63

�:


Ejemplo. Encuentre el área de la super�cie de una esfera de radio a que estáentre los planos z = t1 y z = t2.


Como x2 + y2 + z2 = a2, entonces

z =pa2 � x2 � y2;

@xz =�xp

a2 � x2 � y2; @yz =

�ypa2 � x2 � y2

�T =

vuut �xpa2 � x2 � y2

!2+

�yp

a2 � x2 � y2

!2+ 1

=

sx2 + y2 + (a2 � x2 � y2)

a2 � x2 � y2

=

sa2

a2 � x2 � y2 =ap

a2 � x2 � y2=

apa2 � r2

:

A (S) =

Z 2�

0

Z r2

r1

apa2 � r2

r dr d� =h�2�a

pa2 � r2

ir2r1

= 2�a

�qa2 � r21 �

qa2 � r22

�= 2�a (t1 � t2) = 2�ah:

Observe que esta área corresponde a la de un cilindro circular recto de radio ay altura h = t1 � t2.



z =pa2 � x2 � y2; @xz =

�xpa2 � x2 � y2

;

@yz =�yp

a2 � x2 � y2

�T =


!2+

�yp

a2 � x2 � y2

!2+ 1

=

sx2 + y2 + (a2 � x2 � y2)

a2 � x2 � y2

=

sa2

a2 � x2 � y2 =ap

a2 � x2 � y2=

apa2 � r2

:

A (S) =

Z 2�

0

Z r2

r1

apa2 � r2


pa2 � r2

ir2r1

= 2�a

�qa2 � r21 �

qa2 � r22

�= 2�a (t1 � t2) = 2�ah:




z =pa2 � x2 � y2; @xz =

�xpa2 � x2 � y2

; @yz =�yp

a2 � x2 � y2

�T =


!2+

�yp

a2 � x2 � y2

!2+ 1

=

sx2 + y2 + (a2 � x2 � y2)

a2 � x2 � y2

=

sa2

a2 � x2 � y2 =ap

a2 � x2 � y2=

apa2 � r2

:

A (S) =

Z 2�

0

Z r2

r1

apa2 � r2


pa2 � r2

ir2r1

= 2�a

�qa2 � r21 �

qa2 � r22

�= 2�a (t1 � t2) = 2�ah:




z =pa2 � x2 � y2; @xz =

�xpa2 � x2 � y2

; @yz =�yp

a2 � x2 � y2

�T =


!2+

�yp

a2 � x2 � y2

!2+ 1

=

sx2 + y2 + (a2 � x2 � y2)

a2 � x2 � y2

=

sa2

a2 � x2 � y2

=ap

a2 � x2 � y2=

apa2 � r2

:

A (S) =

Z 2�

0

Z r2

r1

apa2 � r2


pa2 � r2

ir2r1

= 2�a

�qa2 � r21 �

qa2 � r22

�= 2�a (t1 � t2) = 2�ah:




z =pa2 � x2 � y2; @xz =

�xpa2 � x2 � y2

; @yz =�yp

a2 � x2 � y2

�T =


!2+

�yp

a2 � x2 � y2

!2+ 1

=

sx2 + y2 + (a2 � x2 � y2)

a2 � x2 � y2

=

sa2

a2 � x2 � y2 =ap

a2 � x2 � y2

=ap

a2 � r2:

A (S) =

Z 2�

0

Z r2

r1

apa2 � r2


pa2 � r2

ir2r1

= 2�a

�qa2 � r21 �

qa2 � r22

�= 2�a (t1 � t2) = 2�ah:




z =pa2 � x2 � y2; @xz =

�xpa2 � x2 � y2

; @yz =�yp

a2 � x2 � y2

�T =


!2+

�yp

a2 � x2 � y2

!2+ 1

=

sx2 + y2 + (a2 � x2 � y2)

a2 � x2 � y2

=

sa2

a2 � x2 � y2 =ap

a2 � x2 � y2=

apa2 � r2

:

A (S) =

Z 2�

0

Z r2

r1

apa2 � r2


pa2 � r2

ir2r1

= 2�a

�qa2 � r21 �

qa2 � r22

�= 2�a (t1 � t2) = 2�ah:




z =pa2 � x2 � y2; @xz =

�xpa2 � x2 � y2

; @yz =�yp

a2 � x2 � y2

�T =


!2+

�yp

a2 � x2 � y2

!2+ 1

=

sx2 + y2 + (a2 � x2 � y2)

a2 � x2 � y2

=

sa2

a2 � x2 � y2 =ap

a2 � x2 � y2=

apa2 � r2

:

A (S) =

Z 2�

0

Z r2

r1

apa2 � r2

r dr d�

=h�2�a

pa2 � r2

ir2r1

= 2�a

�qa2 � r21 �

qa2 � r22

�= 2�a (t1 � t2) = 2�ah:




z =pa2 � x2 � y2; @xz =

�xpa2 � x2 � y2

; @yz =�yp

a2 � x2 � y2

�T =


!2+

�yp

a2 � x2 � y2

!2+ 1

=

sx2 + y2 + (a2 � x2 � y2)

a2 � x2 � y2

=

sa2

a2 � x2 � y2 =ap

a2 � x2 � y2=

apa2 � r2

:

A (S) =

Z 2�

0

Z r2

r1

apa2 � r2


pa2 � r2

ir2r1

= 2�a

�qa2 � r21 �

qa2 � r22

�= 2�a (t1 � t2) = 2�ah:




z =pa2 � x2 � y2; @xz =

�xpa2 � x2 � y2

; @yz =�yp

a2 � x2 � y2

�T =


!2+

�yp

a2 � x2 � y2

!2+ 1

=

sx2 + y2 + (a2 � x2 � y2)

a2 � x2 � y2

=

sa2

a2 � x2 � y2 =ap

a2 � x2 � y2=

apa2 � r2

:

A (S) =

Z 2�

0

Z r2

r1

apa2 � r2


pa2 � r2

ir2r1

= 2�a

�qa2 � r21 �

qa2 � r22

�

= 2�a (t1 � t2) = 2�ah:




z =pa2 � x2 � y2; @xz =

�xpa2 � x2 � y2

; @yz =�yp

a2 � x2 � y2

�T =


!2+

�yp

a2 � x2 � y2

!2+ 1

=

sx2 + y2 + (a2 � x2 � y2)

a2 � x2 � y2

=

sa2

a2 � x2 � y2 =ap

a2 � x2 � y2=

apa2 � r2

:

A (S) =

Z 2�

0

Z r2

r1

apa2 � r2


pa2 � r2

ir2r1

= 2�a

�qa2 � r21 �

qa2 � r22

�= 2�a (t1 � t2)

= 2�ah:




z =pa2 � x2 � y2; @xz =

�xpa2 � x2 � y2

; @yz =�yp

a2 � x2 � y2

�T =


!2+

�yp

a2 � x2 � y2

!2+ 1

=

sx2 + y2 + (a2 � x2 � y2)

a2 � x2 � y2

=

sa2

a2 � x2 � y2 =ap

a2 � x2 � y2=

apa2 � r2

:

A (S) =

Z 2�

0

Z r2

r1

apa2 � r2


pa2 � r2

ir2r1

= 2�a

�qa2 � r21 �

qa2 � r22

�= 2�a (t1 � t2) = 2�ah:




z =pa2 � x2 � y2; @xz =

�xpa2 � x2 � y2

; @yz =�yp

a2 � x2 � y2

�T =


!2+

�yp

a2 � x2 � y2

!2+ 1

=

sx2 + y2 + (a2 � x2 � y2)

a2 � x2 � y2

=

sa2

a2 � x2 � y2 =ap

a2 � x2 � y2=

apa2 � r2

:

A (S) =

Z 2�

0

Z r2

r1

apa2 � r2


pa2 � r2

ir2r1

= 2�a

�qa2 � r21 �

qa2 � r22

�= 2�a (t1 � t2) = 2�ah:

Observe que esta área corresponde a la de un cilindro circular recto de radio ay altura h = t1 � t2.G A D � g a r e n a s d @ u i s . e d u .c o (U IS ) C á lc u lo I I I S e g u n d o S em e s t r e d e 2 0 1 1 4 8 / 5 1

Ejemplo. Encuentre el área de la parte de la esfera x2 + y2 + z2 = 4z quequeda arriba del paraboloide z = x2 + y2.

�x2 + y2 + z2 = 4z

z = x2 + y2

=)�x2 + y2 + (z � 2)2 = 4

z = r2

=) r2 +�r2 � 2

�2= 4

=) r2 + r4 � 4r2 + 4 = 4=) r2

�r2 � 3

�= 0 =) r =

p3

Por otra parte, como x2 + y2 + (z � 2)2 = 4 entonces

@z

@x= � x

z � 2 y@z

@y= � y

z � 2 ;

además(z � 2)2 = 4� x2 � y2 = 4� r2:


Ejemplo. Encuentre el área de la parte de la esfera x2 + y2 + z2 = 4z quequeda arriba del paraboloide z = x2 + y2. �

x2 + y2 + z2 = 4zz = x2 + y2

=)�x2 + y2 + (z � 2)2 = 4

z = r2

=) r2 +�r2 � 2

�2= 4

=) r2 + r4 � 4r2 + 4 = 4=) r2

�r2 � 3

�= 0 =) r =

p3


@z

@x= � x

z � 2 y@z

@y= � y

z � 2 ;

además(z � 2)2 = 4� x2 � y2 = 4� r2:



x2 + y2 + z2 = 4zz = x2 + y2

=)�x2 + y2 + (z � 2)2 = 4

z = r2

=) r2 +�r2 � 2

�2= 4

=) r2 + r4 � 4r2 + 4 = 4

=) r2�r2 � 3

�= 0 =) r =

p3


@z

@x= � x

z � 2 y@z

@y= � y

z � 2 ;

además(z � 2)2 = 4� x2 � y2 = 4� r2:



x2 + y2 + z2 = 4zz = x2 + y2

=)�x2 + y2 + (z � 2)2 = 4

z = r2

=) r2 +�r2 � 2

�2= 4

=) r2 + r4 � 4r2 + 4 = 4=) r2

�r2 � 3

�= 0

=) r =p3


@z

@x= � x

z � 2 y@z

@y= � y

z � 2 ;

además(z � 2)2 = 4� x2 � y2 = 4� r2:



x2 + y2 + z2 = 4zz = x2 + y2

=)�x2 + y2 + (z � 2)2 = 4

z = r2

=) r2 +�r2 � 2

�2= 4

=) r2 + r4 � 4r2 + 4 = 4=) r2

�r2 � 3

�= 0 =) r =

p3


@z

@x= � x

z � 2 y@z

@y= � y

z � 2 ;

además(z � 2)2 = 4� x2 � y2 = 4� r2:


De este manera se tiene que

ds =

s1 +

�@z

@x

�2+

�@z

@y

�2dA

=

s1 +

��xz � 2

�2+

��yz � 2

�2dA

=

s1 +

x2

(z � 2)2+

y2

(z � 2)2dA =

s(4� x2 � y2) + x2 + y2

4� x2 � y2 dA

=

r4

4� r2 dA =2p4� r2

dA:

Luego el área de super�cie es:

A(S) =

Z 2�

0

Z p3

0

2p4� r2

r dr d� = 2�h�2�4� r2

�1=2ip30

= 4�:



ds =

s1 +

�@z

@x

�2+

�@z

@y

�2dA =

s1 +

��xz � 2

�2+

��yz � 2

�2dA

=

s1 +

x2

(z � 2)2+

y2

(z � 2)2dA =

s(4� x2 � y2) + x2 + y2

4� x2 � y2 dA

=

r4

4� r2 dA =2p4� r2

dA:


A(S) =

Z 2�

0

Z p3

0

2p4� r2

r dr d� = 2�h�2�4� r2

�1=2ip30

= 4�:



ds =

s1 +

�@z

@x

�2+

�@z

@y

�2dA =

s1 +

��xz � 2

�2+

��yz � 2

�2dA

=

s1 +

x2

(z � 2)2+

y2

(z � 2)2dA

=

s(4� x2 � y2) + x2 + y2

4� x2 � y2 dA

=

r4

4� r2 dA =2p4� r2

dA:


A(S) =

Z 2�

0

Z p3

0

2p4� r2

r dr d� = 2�h�2�4� r2

�1=2ip30

= 4�:



ds =

s1 +

�@z

@x

�2+

�@z

@y

�2dA =

s1 +

��xz � 2

�2+

��yz � 2

�2dA

=

s1 +

x2

(z � 2)2+

y2

(z � 2)2dA =

s(4� x2 � y2) + x2 + y2

4� x2 � y2 dA

=

r4

4� r2 dA =2p4� r2

dA:


A(S) =

Z 2�

0

Z p3

0

2p4� r2

r dr d� = 2�h�2�4� r2

�1=2ip30

= 4�:



ds =

s1 +

�@z

@x

�2+

�@z

@y

�2dA =

s1 +

��xz � 2

�2+

��yz � 2

�2dA

=

s1 +

x2

(z � 2)2+

y2

(z � 2)2dA =

s(4� x2 � y2) + x2 + y2

4� x2 � y2 dA

=

r4

4� r2 dA

=2p4� r2

dA:


A(S) =

Z 2�

0

Z p3

0

2p4� r2

r dr d� = 2�h�2�4� r2

�1=2ip30

= 4�:



ds =

s1 +

�@z

@x

�2+

�@z

@y

�2dA =

s1 +

��xz � 2

�2+

��yz � 2

�2dA

=

s1 +

x2

(z � 2)2+

y2

(z � 2)2dA =

s(4� x2 � y2) + x2 + y2

4� x2 � y2 dA

=

r4

4� r2 dA =2p4� r2

dA:


A(S) =

Z 2�

0

Z p3

0

2p4� r2

r dr d� = 2�h�2�4� r2

�1=2ip30

= 4�:



ds =

s1 +

�@z

@x

�2+

�@z

@y

�2dA =

s1 +

��xz � 2

�2+

��yz � 2

�2dA

=

s1 +

x2

(z � 2)2+

y2

(z � 2)2dA =

s(4� x2 � y2) + x2 + y2

4� x2 � y2 dA

=

r4

4� r2 dA =2p4� r2

dA:


A(S) =

Z 2�

0

Z p3

0

2p4� r2

r dr d�

= 2�h�2�4� r2

�1=2ip30

= 4�:



ds =

s1 +

�@z

@x

�2+

�@z

@y

�2dA =

s1 +

��xz � 2

�2+

��yz � 2

�2dA

=

s1 +

x2

(z � 2)2+

y2

(z � 2)2dA =

s(4� x2 � y2) + x2 + y2

4� x2 � y2 dA

=

r4

4� r2 dA =2p4� r2

dA:


A(S) =

Z 2�

0

Z p3

0

2p4� r2

r dr d� = 2�h�2�4� r2

�1=2ip30

= 4�:


Ejemplo. Encuentre el volumen del sólido acotado por abajo por el planoz = 1� y y arriba por el paraboloide z = 1� x2 � y2.

�z = 1� x2 � y2

z = 1� y=) 1� x2 � y2 = 1� y=) x2 + y2 = y

=) r2 = r sen � =) r = sen �:

Luego, utilizando simetría se obtiene

V =

Z �

0

Z sen �

0

Z 1�r2

1�r sen �r dz dr d�

=

Z �

0

Z sen �

0

��1� r2

�� (1� r sen �)

�r dr d�

=

Z �

0

Z sen �

0

�r sen � � r2

�r dr d� =

Z �

0

�r3

3sen � � r4

4

�sen �0

d�

=1

12

Z �

0

sen4 � d� =1

12

�3

8� � 1

4sen 2� +

1

32sen 4�

��0

=1

12

�3

8�

�=

�

32:



�z = 1� x2 � y2

z = 1� y

=) 1� x2 � y2 = 1� y=) x2 + y2 = y

=) r2 = r sen � =) r = sen �:


V =

Z �

0

Z sen �

0

Z 1�r2


=

Z �

0

Z sen �

0

��1� r2

�� (1� r sen �)

�r dr d�

=

Z �

0

Z sen �

0

�r sen � � r2

�r dr d� =

Z �

0

�r3

3sen � � r4

4

�sen �0

d�

=1

12

Z �

0

sen4 � d� =1

12

�3

8� � 1

4sen 2� +

1

32sen 4�

��0

=1

12

�3

8�

�=

�

32:



�z = 1� x2 � y2

z = 1� y=) 1� x2 � y2 = 1� y

=) x2 + y2 = y

=) r2 = r sen � =) r = sen �:


V =

Z �

0

Z sen �

0

Z 1�r2


=

Z �

0

Z sen �

0

��1� r2

�� (1� r sen �)

�r dr d�

=

Z �

0

Z sen �

0

�r sen � � r2

�r dr d� =

Z �

0

�r3

3sen � � r4

4

�sen �0

d�

=1

12

Z �

0

sen4 � d� =1

12

�3

8� � 1

4sen 2� +

1

32sen 4�

��0

=1

12

�3

8�

�=

�

32:



�z = 1� x2 � y2

z = 1� y=) 1� x2 � y2 = 1� y=) x2 + y2 = y

=) r2 = r sen � =) r = sen �:


V =

Z �

0

Z sen �

0

Z 1�r2


=

Z �

0

Z sen �

0

��1� r2

�� (1� r sen �)

�r dr d�

=

Z �

0

Z sen �

0

�r sen � � r2

�r dr d� =

Z �

0

�r3

3sen � � r4

4

�sen �0

d�

=1

12

Z �

0

sen4 � d� =1

12

�3

8� � 1

4sen 2� +

1

32sen 4�

��0

=1

12

�3

8�

�=

�

32:



�z = 1� x2 � y2

z = 1� y=) 1� x2 � y2 = 1� y=) x2 + y2 = y

=) r2 = r sen �

=) r = sen �:


V =

Z �

0

Z sen �

0

Z 1�r2


=

Z �

0

Z sen �

0

��1� r2

�� (1� r sen �)

�r dr d�

=

Z �

0

Z sen �

0

�r sen � � r2

�r dr d� =

Z �

0

�r3

3sen � � r4

4

�sen �0

d�

=1

12

Z �

0

sen4 � d� =1

12

�3

8� � 1

4sen 2� +

1

32sen 4�

��0

=1

12

�3

8�

�=

�

32:



�z = 1� x2 � y2

z = 1� y=) 1� x2 � y2 = 1� y=) x2 + y2 = y

=) r2 = r sen � =) r = sen �:


V =

Z �

0

Z sen �

0

Z 1�r2


=

Z �

0

Z sen �

0

��1� r2

�� (1� r sen �)

�r dr d�

=

Z �

0

Z sen �

0

�r sen � � r2

�r dr d� =

Z �

0

�r3

3sen � � r4

4

�sen �0

d�

=1

12

Z �

0

sen4 � d� =1

12

�3

8� � 1

4sen 2� +

1

32sen 4�

��0

=1

12

�3

8�

�=

�

32:



�z = 1� x2 � y2

z = 1� y=) 1� x2 � y2 = 1� y=) x2 + y2 = y

=) r2 = r sen � =) r = sen �:


V =

Z �

0

Z sen �

0

Z 1�r2


=

Z �

0

Z sen �

0

��1� r2

�� (1� r sen �)

�r dr d�

=

Z �

0

Z sen �

0

�r sen � � r2

�r dr d�

=

Z �

0

�r3

3sen � � r4

4

�sen �0

d�

=1

12

Z �

0

sen4 � d� =1

12

�3

8� � 1

4sen 2� +

1

32sen 4�

��0

=1

12

�3

8�

�=

�

32:



�z = 1� x2 � y2

z = 1� y=) 1� x2 � y2 = 1� y=) x2 + y2 = y

=) r2 = r sen � =) r = sen �:


V =

Z �

0

Z sen �

0

Z 1�r2


=

Z �

0

Z sen �

0

��1� r2

�� (1� r sen �)

�r dr d�

=

Z �

0

Z sen �

0

�r sen � � r2

�r dr d� =

Z �

0

�r3

3sen � � r4

4

�sen �0

d�

=1

12

Z �

0

sen4 � d� =1

12

�3

8� � 1

4sen 2� +

1

32sen 4�

��0

=1

12

�3

8�

�=

�

32:



�z = 1� x2 � y2

z = 1� y=) 1� x2 � y2 = 1� y=) x2 + y2 = y

=) r2 = r sen � =) r = sen �:


V =

Z �

0

Z sen �

0

Z 1�r2


=

Z �

0

Z sen �

0

��1� r2

�� (1� r sen �)

�r dr d�

=

Z �

0

Z sen �

0

�r sen � � r2

�r dr d� =

Z �

0

�r3

3sen � � r4

4

�sen �0

d�

=1

12

Z �

0

sen4 � d�

=1

12

�3

8� � 1

4sen 2� +

1

32sen 4�

��0

=1

12

�3

8�

�=

�

32:



�z = 1� x2 � y2

z = 1� y=) 1� x2 � y2 = 1� y=) x2 + y2 = y

=) r2 = r sen � =) r = sen �:


V =

Z �

0

Z sen �

0

Z 1�r2


=

Z �

0

Z sen �

0

��1� r2

�� (1� r sen �)

�r dr d�

=

Z �

0

Z sen �

0

�r sen � � r2

�r dr d� =

Z �

0

�r3

3sen � � r4

4

�sen �0

d�

=1

12

Z �

0

sen4 � d� =1

12

�3

8� � 1

4sen 2� +

1

32sen 4�

��0

=1

12

�3

8�

�=

�

32:



�z = 1� x2 � y2

z = 1� y=) 1� x2 � y2 = 1� y=) x2 + y2 = y

=) r2 = r sen � =) r = sen �:


V =

Z �

0

Z sen �

0

Z 1�r2


=

Z �

0

Z sen �

0

��1� r2

�� (1� r sen �)

�r dr d�

=

Z �

0

Z sen �

0

�r sen � � r2

�r dr d� =

Z �

0

�r3

3sen � � r4

4

�sen �0

d�

=1

12

Z �

0

sen4 � d� =1

12

�3

8� � 1

4sen 2� +

1

32sen 4�

��0

=1

12

�3

8�

�

=�

32:



�z = 1� x2 � y2

z = 1� y=) 1� x2 � y2 = 1� y=) x2 + y2 = y

=) r2 = r sen � =) r = sen �:


V =

Z �

0

Z sen �

0

Z 1�r2


=

Z �

0

Z sen �

0

��1� r2

�� (1� r sen �)

�r dr d�

=

Z �

0

Z sen �

0

�r sen � � r2

�r dr d� =

Z �

0

�r3

3sen � � r4

4

�sen �0

d�

=1

12

Z �

0

sen4 � d� =1

12

�3

8� � 1

4sen 2� +

1

32sen 4�

��0

=1

12

�3

8�

�=

�

32:


matematicas.uis.edu.comatematicas.uis.edu.co/~garenasd/calculoiii/imps2011bm.pdf · la integral...

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