clases de calculo 3´ · clases de calculo 3´ roberto carlos cabrales dpto. de ciencias basicas´...

Vectores y la geometrıa del espacio Integracion multiple Calculo vectorial

Clases de Calculo 3

Roberto Carlos Cabrales

Dpto. de Ciencias BasicasU. del Bıo-Bıo, Chile.

[email protected]://rcabrales.wordpress.com

1er semestre de 2016.Ultima actualizacion: Miercoles 11 de Agosto de 2016.


Contenido

Vectores y la geometrıa del espacio

Integracion multipleIntegrales doblesIntegrales triplesAplicaciones de las integrales multiplesCambio de variable en integrales multiplesAplicaciones de la formula del cambio de variables

Calculo vectorialIntegrales de lineaEl teorema fundamental para integrales de lıneaEl teorema de GreenEl rotacional y la divergencia


El producto interno

Sean a = (aj ),b = (bj ) ∈ Rn. Se define el producto interno de a y b como

〈a,b〉 = a1b1 + · · ·+ anbn =

n∑j=1

aj bj .

Si α es el angulo entre dos vectores a,b, entonces 〈a,b〉 = ‖a‖ · ‖b‖ cos(α).Los vectores a y b se dicen ortogonales si 〈a,b〉 = 0.Los angulos directores de un vector no nulo a = (a1,a2,a3) ∈ R3 son los angulosα, β, γ que forma a con la parte positiva de los ejes coordenadas. Los cosenos deestos angulos se llaman cosenos directores. De hecho:

cos(α) =a1

‖a‖, cos(β) =

a2

‖a‖, cos(γ) =

a3

‖a‖.

La proyeccion escalar de b sobre a se define como compab =〈a,b〉‖a‖

.

El vector proyeccion de b sobre a se define como proyab =〈a,b〉‖a‖2

a.


El producto vectorial (o cruz)

Si a,b ∈ R3, se define su producto vectorial como el vector a × b =

a2b3 − b2a3a3b1 − b3a1a1b2 − b1a2

.Si a,b,c ∈ R3 y α ∈ R, se verifican las siguientes propiedades:

1. a × b = −b × a.2. (αa) × b = α(a × b) = a × (αb).

3. a × (b + c) = a × b + a × c.

4. (a + b) × c = a × c + b × c.

5. a · (b × c) = (a × b) · c.

6. a × (b × c) = (a · c)b − (a · b)c.


Ecuaciones de rectas y planos


Formulas para regiones de tipo I del plano

Una region R del plano, se llama una region de tipo I del plano si existen funcionesy = g1(x) e y = g2(x) tales que

E = {(x , y) ∈ R2 : a ≤ x ≤ b,g1(x) ≤ y ≤ g2(x)}.

Entonces

If F is integrable over R, then we define the double integral of over D by

Definition 2 makes sense because R is a rectangle and so has been previ-ously defined in Section 15.1. The procedure that we have used is reasonable because thevalues of are 0 when lies outside and so they contribute nothing to the inte-gral. This means that it doesn’t matter what rectangle we use as long as it contains .

In the case where , we can still interpret as the volume of thesolid that lies above and under the surface (the graph of ). You can see thatthis is reasonable by comparing the graphs of and in Figures 3 and 4 and remember-ing that is the volume under the graph of .

Figure 4 also shows that is likely to have discontinuities at the boundary points of Nonetheless, if is continuous on and the boundary curve of is “well behaved”

(in a sense outside the scope of this book), then it can be shown that existsand therefore exists. In particular, this is the case for the following typesof regions.

A plane region is said to be of type I if it lies between the graphs of two continuousfunctions of , that is,

where and are continuous on . Some examples of type I regions are shown inFigure 5.

In order to evaluate when is a region of type I, we choose a rectanglethat contains , as in Figure 6, and we let be the function given by

Equation 1; that is, agrees with on and is outside . Then, by Fubini’s Theorem,

Observe that if or because then lies outside .Therefore

yd

c F!x, y" dy ! yt2!x"

t1!x" F!x, y" dy ! yt2!x"

t1!x" f !x, y" dy

D!x, y"y ! t2!x"y " t1!x"F!x, y" ! 0

yyD

f !x, y" dA ! yyR

F!x, y" dA ! yb

a yd

c F!x, y" dy dx

D0FDfFFDR ! #a, b$ # #c, d $

DxxD f !x, y" dA

FIGURE 5 Some type I regions

0

y

xba

D

y=g™(x)

y=g¡(x)0

y

xba

D

y=g™(x)

y=g¡(x)

0

y

xba

D

y=g™(x)

y=g¡(x)

#a, b$t2t1

D ! %!x, y" & a $ x $ b, t1!x" $ y $ t2!x"'

xD

xxD f !x, y" dAxxR F!x, y" dA

DDfD.F

FxxR F!x, y" dAFf

fz ! f !x, y"DxxD f !x, y" dAf !x, y" % 0

DRD!x, y"F!x, y"

xxR F!x, y" dA

where F is given by Equation 1yyD

f !x, y" dA ! yyR

F!x, y" dA2

f

966 | | | | CHAPTER 15 MULTIPLE INTEGRALS

y0

z

xD

graph of f

FIGURE 4

y0

z

xD

graph of F

FIGURE 3

FIGURE 6

d

0 x

y

bxa

cy=g¡(x)

D

y=g™(x)

"R

f (x , y)dA =

∫ b

a

∫ g2(x)

g1(x)f (x , y)dydx


Formulas para regiones de tipo II del plano

R es una region de tipo II del plano si existen funciones x = h1(y) e x = h2(y) talesque

E = {(x , y) ∈ R2 : h1(y) ≤ x ≤ h2(y), c ≤ y ≤ d}.

Entonces

because when . Thus we have the following formulathat enables us to evaluate the double integral as an iterated integral.

If is continuous on a type I region D such that

then

The integral on the right side of (3) is an iterated integral that is similar to the ones weconsidered in the preceding section, except that in the inner integral we regard as beingconstant not only in but also in the limits of integration, and

We also consider plane regions of type II, which can be expressed as

where and are continuous. Two such regions are illustrated in Figure 7.Using the same methods that were used in establishing (3), we can show that

where D is a type II region given by Equation 4.

EXAMPLE 1 Evaluate , where is the region bounded by the parabolas and .

SOLUTION The parabolas intersect when , that is, , so . Wenote that the region , sketched in Figure 8, is a type I region but not a type II regionand we can write

Since the lower boundary is and the upper boundary is , Equation 3gives

M ! !3

x 5

5!

x 4

4" 2

x 3

3"

x 2

2" x!

!1

1

!3215

! y1

!1 "!3x 4 ! x 3 " 2x 2 " x " 1# dx

! y1

!1 $x"1 " x 2 # " "1 " x 2 #2 ! x"2x 2 # ! "2x 2 #2 % dx

! y1

!1 [xy " y 2]y!2x2

y!1"x2

dx

yyD

"x " 2y# dA ! y1

!1 y1"x2

2x2 "x " 2y# dy dx

y ! 1 " x 2y ! 2x 2

D ! &"x, y# ' !1 # x # 1, 2x 2 # y # 1 " x 2(

Dx ! $1x 2 ! 12x 2 ! 1 " x 2

y ! 1 " x 2y ! 2x 2DxxD "x " 2y# dAV

yyD

f "x, y# dA ! yd

c yh2" y#

h1" y# f "x, y# dx dy5

h2h1

D ! &"x, y# ' c # y # d, h1"y# # x # h2"y#(4

t2"x#.t1"x#f "x, y#x

yyD

f "x, y# dA ! yb

a yt2"x#

t1"x# f "x, y# dy dx

D ! &"x, y# ' a # x # b, t1"x# # y # t2"x#(f3

t1"x# # y # t2"x#F"x, y# ! f "x, y#

SECTION 15.3 DOUBLE INTEGRALS OVER GENERAL REGIONS | | | | 967

FIGURE 7Some type II regions

d

0 x

y

c

x=h¡(y)

x=h¡(y)

D x=h™(y)

x=h™(y)

d

0 x

y

c

D

x1_1

y

(_1,!2) (1,!2)

Dy=2"

y=1+"

FIGURE 8

"R

f (x , y)dA =

∫ d

c

∫ h2(x)

h1(x)f (x , y)dxdy


Formulas para regiones de tipo 1 del espacio

Sea R ⊂ R3 y D la proyeccion de D en el plano xy . R es una region de tipo 1 delespacio si existen funciones z = u1(x , y) e z = u2(x , y) tales que

E = {(x , y , z) ∈ R3 : (x , y) ∈ D,u1(x , y) ≤ z ≤ u2(x , y)}.

FUBINI’S THEOREM FOR TRIPLE INTEGRALS If is continuous on the rectan-gular box , then

The iterated integral on the right side of Fubini’s Theorem means that we integrate firstwith respect to (keeping and fixed), then we integrate with respect to (keeping fixed), and finally we integrate with respect to . There are five other possible orders inwhich we can integrate, all of which give the same value. For instance, if we integrate withrespect to , then , and then , we have

EXAMPLE 1 Evaluate the triple integral , where is the rectangular boxgiven by

SOLUTION We could use any of the six possible orders of integration. If we choose to integrate with respect to , then , and then , we obtain

M

Now we define the triple integral over a general bounded region E in three-dimensional space (a solid) by much the same procedure that we used for double integrals(15.3.2). We enclose in a box of the type given by Equation 1. Then we define a function so that it agrees with on but is 0 for points in that are outside . By definition,

This integral exists if is continuous and the boundary of is “reasonably smooth.” Thetriple integral has essentially the same properties as the double integral (Properties 6–9 inSection 15.3).

We restrict our attention to continuous functions and to certain simple types of regions.A solid region is said to be of type 1 if it lies between the graphs of two continuous func-tions of and , that is,

where is the projection of onto the -plane as shown in Figure 2. Notice that theupper boundary of the solid is the surface with equation , while the lowerboundary is the surface .z ! u1!x, y"

z ! u2!x, y"ExyED

E ! #!x, y, z" $ !x, y" ! D, u1!x, y" ! z ! u2!x, y"%5

yxE

f

Ef

yyyE

f !x, y, z" dV ! yyyB

F!x, y, z" dV

EBEfFBE

! y3

0 3z2

4 dz !

z3

4 &0

3

!274

! y3

0 y2

"1 yz2

2 dy dz ! y3

0

' y 2z2

4 &y!"1

y!2

dz

yyyB

xyz2 dV ! y3

0 y2

"1 y1

0 xyz2 dx dy dz ! y3

0

y2

"1

' x 2yz2

2 &x!0

x!1

dy dz

zyx

B ! #!x, y, z" $ 0 ! x ! 1, "1 ! y ! 2, 0 ! z ! 3%

BxxxB xyz2 dVV

yyyB

f !x, y, z" dV ! yb

a ys

r yd

c f !x, y, z" dy dz dx

xzy

zzyzyx

yyyB

f !x, y, z" dV ! ys

r yd

c yb

a f !x, y, z" dx dy dz

B ! (a, b) # (c, d ) # (r, s)f4

SECTION 15.6 TRIPLE INTEGRALS | | | | 991

FIGURE 2A type 1 solid region

z

0

x yD

Ez=u™!(x,!y)

z=u¡!(x,!y)

Entonces$R

f (x , y , z)dV =

"D

∫ u2(x ,y)

u1(x ,y)f (x , y , z)dz

dA

Interpretando D como una region de tipo 1 o una de tipo 2 de R2, tenemos:$R

f (x , y , z)dV =

∫ b

a

∫ g2(x)

g1(x)

∫ u2(x ,y)

u1(x ,y)f (x , y , z)dzdydx ,

$R

f (x , y , z)dV =

∫ d

c

∫ h2(x)

h1(x)

∫ u2(x ,y)

u1(x ,y)f (x , y , z)dzdxdy .



Sea R ⊂ R3 y D la proyeccion de D en el plano xy . R es una region de tipo 2 delespacio si existen funciones z = u1(y , z) e z = u2(y , z) tales que

E = {(x , y , z) ∈ R3 : (y , z) ∈ D,u1(y , z) ≤ x ≤ u2(y , z)}.

boundary is the plane (or ), so we use andin Formula 7. Notice that the planes and

intersect in the line (or ) in the -plane. So the projection of isthe triangular region shown in Figure 6, and we have

This description of as a type 1 region enables us to evaluate the integral as follows:

M

A solid region is of type 2 if it is of the form

where, this time, is the projection of onto the -plane (see Figure 7). The back sur-face is , the front surface is , and we have

Finally, a type 3 region is of the form

where is the projection of onto the -plane, is the left surface, andis the right surface (see Figure 8). For this type of region we have

yyyE

f !x, y, z" dV ! yyD

#yu2!x, z"

u1!x, z" f !x, y, z" dy$ dA11

y ! u2!x, z"y ! u1!x, z"xzED

E ! %!x, y, z" & !x, z" ! D, u1!x, z" ! y ! u2!x, z"'

FIGURE 8 A type 3 region

z

y=u™(x,!z)

y=u¡(x,!z)x

0

y

DE

0

z

yx E

D

x=u¡(y,!z)

x=u™(y,!z)


yyyE


#yu2! y, z"

u1! y, z" f !x, y, z" dx$ dA10

x ! u2!y, z"x ! u1!y, z"yzED

E ! %!x, y, z" & !y, z" ! D, u1!y, z" ! x ! u2!y, z"'E

! 16 y1

0 !1 " x"3 dx !

16

#"!1 " x"4

4 $0

1

!1

24

! 12 y1

0

#"!1 " x " y"3

3 $y!0

y!1"x

dx! 12 y1

0 y1"x

0 !1 " x " y"2 dy dx

! y1

0

y1"x

0

# z2

2 $z!0

z!1"x"y

dy dx yyyE

z dV ! y1

0 y1"x

0 y1"x"y

0 z dz dy dx

E

E ! %!x, y, z" & 0 ! x ! 1, 0 ! y ! 1 " x, 0 ! z ! 1 " x " y'9

Exyy ! 1 " xx # y ! 1z ! 0x # y # z ! 1u2!x, y" ! 1 " x " y

u1!x, y" ! 0z ! 1 " x " yx # y # z ! 1


Entonces$R

f (x , y , z)dV =

"D

∫ u2(y ,z)

u1(y ,z)f (x , y , z)dx

dA


f (x , y , z)dV =

∫ b

a

∫ g2(y)

g1(y)

∫ u2(y ,z)

u1(y ,z)f (x , y , z)dxdzdy ,

$R

f (x , y , z)dV =

∫ d

c

∫ h2(z)

h1(z)

∫ u2(y ,z)

u1(y ,z)f (x , y , z)dxdydz.



Sea R ⊂ R3 y D la proyeccion de D en el plano xy . R es una region de tipo 2 delespacio si existen funciones z = u1(x , z) e z = u2(x , z) tales que

E = {(x , y , z) ∈ R3 : (x , z) ∈ D,u1(x , z) ≤ y ≤ u2(x , z)}.

boundary is the plane (or ), so we use andin Formula 7. Notice that the planes and

intersect in the line (or ) in the -plane. So the projection of isthe triangular region shown in Figure 6, and we have

This description of as a type 1 region enables us to evaluate the integral as follows:

M

A solid region is of type 2 if it is of the form

where, this time, is the projection of onto the -plane (see Figure 7). The back sur-face is , the front surface is , and we have

Finally, a type 3 region is of the form

where is the projection of onto the -plane, is the left surface, andis the right surface (see Figure 8). For this type of region we have

yyyE


#yu2!x, z"

u1!x, z" f !x, y, z" dy$ dA11

y ! u2!x, z"y ! u1!x, z"xzED

E ! %!x, y, z" & !x, z" ! D, u1!x, z" ! y ! u2!x, z"'


z

y=u™(x,!z)

y=u¡(x,!z)x

0

y

DE

0

z

yx E

D

x=u¡(y,!z)

x=u™(y,!z)


yyyE


#yu2! y, z"

u1! y, z" f !x, y, z" dx$ dA10

x ! u2!y, z"x ! u1!y, z"yzED

E ! %!x, y, z" & !y, z" ! D, u1!y, z" ! x ! u2!y, z"'E

! 16 y1

0 !1 " x"3 dx !

16

#"!1 " x"4

4 $0

1

!124

! 12 y1

0

#"!1 " x " y"3

3 $y!0

y!1"x

dx! 12 y1

0 y1"x

0 !1 " x " y"2 dy dx

! y1

0

y1"x

0

# z2

2 $z!0

z!1"x"y

dy dx yyyE

z dV ! y1

0 y1"x

0 y1"x"y

0 z dz dy dx

E

E ! %!x, y, z" & 0 ! x ! 1, 0 ! y ! 1 " x, 0 ! z ! 1 " x " y'9

Exyy ! 1 " xx # y ! 1z ! 0x # y # z ! 1u2!x, y" ! 1 " x " y

u1!x, y" ! 0z ! 1 " x " yx # y # z ! 1


Entonces$R

f (x , y , z)dV =

"D

∫ u2(x ,z)

u1(x ,z)f (x , y , z)dy

dA


f (x , y , z)dV =

∫ b

a

∫ g2(x)

g1(x)

∫ u2(x ,z)

u1(x ,z)f (x , y , z)dydzdx ,

$R

f (x , y , z)dV =

∫ d

c

∫ h2(z)

h1(z)

∫ u2(x ,z)

u1(x ,z)f (x , y , z)dydxdz.


Integrales doble

Considere una lamina del plano con densidadρ(x , y). La integral doble sirve paracalcular el area de la lamina y tambienlas coordenadas de su centro de masa.

SOLUTION From Equation 2 and Figure 3 we have

Thus the total charge is C. M

MOMENTS AND CENTERS OF MASS

In Section 8.3 we found the center of mass of a lamina with constant density; here we con-sider a lamina with variable density. Suppose the lamina occupies a region and has den-sity function . Recall from Chapter 8 that we defined the moment of a particle aboutan axis as the product of its mass and its directed distance from the axis. We divide intosmall rectangles as in Figure 2. Then the mass of is approximately , so wecan approximate the moment of with respect to the -axis by

If we now add these quantities and take the limit as the number of subrectangles becomeslarge, we obtain the moment of the entire lamina about the x-axis:

Similarly, the moment about the y-axis is

As before, we define the center of mass so that and . The physi-cal significance is that the lamina behaves as if its entire mass is concentrated at its centerof mass. Thus the lamina balances horizontally when supported at its center of mass (seeFigure 4).

The coordinates of the center of mass of a lamina occupying the regionD and having density function are

where the mass is given by

m ! yyD

!!x, y" dA

m

y !Mx

m!

1m

yyD

y !!x, y" dAx !My

m!

1m

yyD

x !!x, y" dA

!!x, y"!x, y"5

my ! Mxmx ! My!x, y "

My ! lim m, nl"

#m

i!1 #

n

j!1 xij* !!xij*, yij*" #A ! yy

D

x !!x, y" dA4

Mx ! lim m, nl"

#m

i!1 #

n

j!1 yij* !!xij*, yij*" #A ! yy

D

y !!x, y" dA3

$!!xij*, yij*" #A% yij*

xRij

!!xij*, yij*" #ARij

D!!x, y"

D

524

! 12 y1

0 !2x 2 $ x 3 " dx !

12

&2x 3

3$

x 4

4 '0

1

!5

24

! y1

0

&x y 2

2 'y!1$x

y!1

dx ! y1

0 x2

$12 $ !1 $ x"2 % dx

Q ! yyD

%!x, y" dA ! y1

0 y1

1$x xy dy dx

SECTION 15.5 APPLICATIONS OF DOUBLE INTEGRALS | | | | 981

FIGURE 3

1

y

0 x

(1,!1)y=1

y=1-x

D

FIGURE 4

D(x,!y)

De hecho, el area es dada por

A =

"D

dA,

y las coordenadas (x , y) del centro de masa son dadas por

x =My

m=

"D

xρ(x , y)dA

"D

ρ(x , y)dA, y =

Mx

m=

"D

yρ(x , y)dA

"D

ρ(x , y)dA,

donde m es la masa del objeto y My ,Mx son los momentos enrelacion al eje y y al eje x , respectivamente.


Integrales triples

Considere un solido E del espacio con densidad ρ(x , y , z). La integral triple sirve paracalcular el volumen del solido y tambien las coordenadas de su centro de masa. Dehecho, su volumen V y las coordenadas (x , y , z) del centro de masa son dadas por

V =

$E

dV , x =Myz

m=

$E

xρ(x , y , z)dV

$E

ρ(x , y , z)dV,

y =Mxzm =

$E

yρ(x , y , z)dV$E

ρ(x , y , z)dV, z =

Mxym =

$E

zρ(x , y , z)dV$E

ρ(x , y , z)dV,

donde m es la masa del objeto y Myz ,Mxz ,Mxy son los momentos enrelacion al plano yz, al plano xz y al plano xy , respectivamente.


Formula para integrales dobles

Sea T : S ⊂ R2→ R ⊂ R2 tal que T (u, v) = (x(u, v), y(u, v)) una transformacion de

clase C1 con una inversa T−1.

A change of variables can also be useful in double integrals. We have already seen oneexample of this: conversion to polar coordinates. The new variables and are related tothe old variables and by the equations

and the change of variables formula (15.4.2) can be written as

where is the region in the -plane that corresponds to the region in the -plane.More generally, we consider a change of variables that is given by a transformation

from the -plane to the -plane:

where and are related to and by the equations

or, as we sometimes write,

We usually assume that is a C transformation, which means that and have contin-uous first-order partial derivatives.

A transformation is really just a function whose domain and range are both subsetsof . If , then the point is called the image of the point .If no two points have the same image, is called one-to-one. Figure 1 shows the effect ofa transformation on a region in the -plane. transforms into a region in the

-plane called the image of S, consisting of the images of all points in .

If is a one-to-one transformation, then it has an inverse transformation from the-plane to the -plane and it may be possible to solve Equations 3 for and in terms

of and :

EXAMPLE 1 A transformation is defined by the equations

Find the image of the square , .

SOLUTION The transformation maps the boundary of into the boundary of the image. Sowe begin by finding the images of the sides of . The first side, , is given by v ! 0S1S

S

0 ! v ! 1!S ! "#u, v$ % 0 ! u ! 1

y ! 2uvx ! u 2 " v2

V

v ! H#x, y$u ! G#x, y$yx

vuuvxyT"1T

FIGURE 10

!

0

y

u x

(u¡,"!¡)(x¡,"y¡)

S RT –!

T

SxyRSTuvST

T#u1, v1$#x1, y1$T#u1, v1$ ! #x1, y1$! 2

T

ht1T

y ! y#u, v$x ! x#u, v$

y ! h#u, v$x ! t#u, v$3

vuyx

T#u, v$ ! #x, y$

xyuvT

xyRr#S

yyR

f #x, y$ dA ! yyS

f #r cos #, r sin #$ r dr d#

y ! r sin #x ! r cos #

yx#r

SECTION 15.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS | | | | 1013

Entonces, la formula del cambio de variable para una integral doble es"D

f (x , y)dA =

"S

f (x(u, v), y(u, v))

∣∣∣∣∣∂(x , y)

∂(u, v)

∣∣∣∣∣dudv ,

donde∂(x , y)

∂(u, v)= det

∂x∂u

∂x∂v

∂y∂u

∂y∂v

es el jacobiano de la transformacion.


Formula para integrales triples

Analogamente, para una integral triple la formula del cambio de variable es$D

f (x , y)dA =

$S

f (x(u, v ,w), y(u, v ,w), z(u, v ,w))

∣∣∣∣∣ ∂(x , y , z)

∂(u, v ,w)

∣∣∣∣∣dudvdw ,

donde

∂(x , y , z)

∂(u, v ,w)= det

∂x∂u

∂x∂v

∂x∂w

∂y∂u

∂y∂v

∂y∂w

∂y∂u

∂y∂v

∂y∂w

es el jacobiano de la transformacion.


Integrales dobles en coordenadas polares

Si el par (r , θ) representa las coordenadas polares de un punto de R2, entonces, parauna region R del plano"

R

f (x , y)dA =

∫ β

α

∫ b

af (r cos(θ), r sin(θ))rdrdθ.

DOUBLE INTEGRALS IN POLAR COORDINATES

Suppose that we want to evaluate a double integral , where is one of theregions shown in Figure 1. In either case the description of in terms of rectangular coor-dinates is rather complicated but is easily described using polar coordinates.

Recall from Figure 2 that the polar coordinates of a point are related to the rect-angular coordinates by the equations

(See Section 10.3.)The regions in Figure 1 are special cases of a polar rectangle

which is shown in Figure 3. In order to compute the double integral , whereis a polar rectangle, we divide the interval into subintervals of equal

width and we divide the interval into subintervals ofequal width . Then the circles and the rays divide the polarrectangle R into the small polar rectangles shown in Figure 4.

r=ri-1

FIGURE 3 Polar rectangle FIGURE 4 Dividing R into polar subrectangles

O

!å

r=a ¨=å

¨=!r=b

RÎ¨

¨=¨j

(ri*,"¨j*)

r=ri

Rij

O

¨=¨j-1

! ! ! jr ! ri"! ! !# $ %"#n$!j$1, !j%n$%, #%"r ! !b $ a"#m

$ri$1, ri%m$a, b%RxxR f !x, y" dA

R ! &!r, !" ' a & r & b, % & ! & #(

y ! r sin !x ! r cos !r 2 ! x 2 ' y 2

!x, y"!r, !"

FIGURE 1

x0

y

R#+¥=1

(a) R=s(r,"¨) | 0¯r¯1, 0¯¨¯2$d

x0

y

R

#+¥=4

#+¥=1

(b) R=s(r,"¨) | 1¯r¯2, 0¯¨¯$d

RR

RxxR f !x, y" dA

15.4


O

y

x¨

x

yr

P(r,"¨)=P(x,"y)

FIGURE 2


Integrales triples en coordenadas cilındricas

Para una region E de tipo 1 del espacio cuya proyeccion D en el plano xy se describeen coordenadas polares, entonces$

R

f (x , y)dA =

"D

∫ u2(x ,y)

u1(x ,y)f (x , y , z)dz

dA

=

∫ β

α

∫ h2(θ)

h1(θ)

∫ u2(r cos(θ),r sen(θ))

u1(r cos(θ),r sen(θ))f (r cos(θ), r sen(θ), z)rdzdrdθ.

CYLINDRICAL COORDINATES

In the cylindrical coordinate system, a point in three-dimensional space is representedby the ordered triple , where and are polar coordinates of the projection of onto the -plane and is the directed distance from the -plane to . (See Figure 2.)

To convert from cylindrical to rectangular coordinates, we use the equations

whereas to convert from rectangular to cylindrical coordinates, we use

EXAMPLE 1(a) Plot the point with cylindrical coordinates and find its rectangular coordinates.(b) Find cylindrical coordinates of the point with rectangular coordinates .

SOLUTION(a) The point with cylindrical coordinates is plotted in Figure 3. FromEquations 1, its rectangular coordinates are

Thus the point is in rectangular coordinates.

(b) From Equations 2 we have

so

Therefore one set of cylindrical coordinates is . Another is. As with polar coordinates, there are infinitely many choices. M

Cylindrical coordinates are useful in problems that involve symmetry about an axis, andthe -axis is chosen to coincide with this axis of symmetry. For instance, the axis of thecircular cylinder with Cartesian equation is the -axis. In cylindrical coordi-nates this cylinder has the very simple equation . (See Figure 4.) This is the reason for the name “cylindrical” coordinates.

r ! czx 2 ! y 2 ! c 2

z

(3s2 , "#!4, "7)(3s2 , 7#!4, "7)

z ! "7

$ !7#

4! 2n#tan $ !

"33

! "1

r ! s32 ! ""3#2 ! 3s2

("1, s3 , 1) z ! 1

y ! 2 sin 2#

3! 2$s3

2 % ! s3

x ! 2 cos 2#

3! 2$"

12% ! "1

"2, 2#!3, 1#

"3, "3, "7#

"2, 2#!3, 1#

z ! ztan $ !yx

r 2 ! x 2 ! y 22

z ! zy ! r sin $x ! r cos $1

PxyzxyP$r"r, $, z#

P

SECTION 15.7 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES | | | | 1001

Or

z

¨

(r,! ,!0)

P(r,! ,!z)

FIGURE 2The cylindrical coordinates of a point

x

z

y

FIGURE 3

”2,!!!!!!!,!1’2"3

0

2"3

2

1

xy

z

FIGURE 4r=c, a cylinder

0

z

y

x

(0,!c,!0)

(c,!0,!0)

EXAMPLE 2 Describe the surface whose equation in cylindrical coordinates is .

SOLUTION The equation says that the -value, or height, of each point on the surface is thesame as r, the distance from the point to the -axis. Because doesn’t appear, it canvary. So any horizontal trace in the plane is a circle of radius k. Thesetraces suggest that the surface is a cone. This prediction can be confirmed by convertingthe equation into rectangular coordinates. From the first equation in (2) we have

We recognize the equation (by comparison with Table 1 in Section 12.6) asbeing a circular cone whose axis is the -axis. (See Figure 5.) M

EVALUATING TRIPLE INTEGRALS WITH CYLINDRICAL COORDINATES

Suppose that is a type 1 region whose projection on the -plane is convenientlydescribed in polar coordinates (see Figure 6). In particular, suppose that is continuousand

where is given in polar coordinates by

We know from Equation 15.6.6 that

But we also know how to evaluate double integrals in polar coordinates. In fact, combin-ing Equation 3 with Equation 15.4.3, we obtain

Formula 4 is the formula for triple integration in cylindrical coordinates. It says thatwe convert a triple integral from rectangular to cylindrical coordinates by writing

, , leaving as it is, using the appropriate limits of integration for ,, and , and replacing by . (Figure 7 shows how to remember this.) It is

worthwhile to use this formula when is a solid region easily described in cylindricalcoordinates, and especially when the function involves the expression .

EXAMPLE 3 A solid lies within the cylinder , below the plane ,and above the paraboloid . (See Figure 8.) The density at any point isproportional to its distance from the axis of the cylinder. Find the mass of .E

z ! 1 ! x 2 ! y 2z ! 4x 2 " y 2 ! 1EV

x 2 " y2f !x, y, z"E

r dz dr d#dV#rzzy ! r sin #x ! r cos #

yyyE

f !x, y, z" dV ! y$

% yh2!#"

h1!#" yu2!r cos #, r sin #"

u1!r cos #, r sin #" f !r cos #, r sin #, z" r dz dr d#4

yyyE


#yu2!x, y"

u1!x, y" f !x, y, z" dz$ dA3

D ! %!r, #" & % & # & $, h1!#" & r & h2!#"'

D

E ! %!x, y, z" & !x, y" ! D, u1!x, y" & z & u2!x, y"'

fxyDE

zz2 ! x 2 " y 2

z2 ! r 2 ! x 2 " y 2

z ! k !k ' 0"#z

z

z ! rV


z

dz

drr!d¨

d¨r

FIGURE 7Volume element in cylindricalcoordinates: dV=r!dz!dr!d¨

FIGURE 5z=r, a cone

0

z

x

y

FIGURE 6

¨=b¨=a

z

xy

0

D

r=h¡(¨)

r=h™(¨)

z=u™(x,!y)

z=u¡(x,!y)


Integrales triples en coordenadas esfericas

En el sistema de coordenadas esfericas la contraparte de una caja rectangular es unacuna esferica E definida como

E = {(ρ, θ, φ) : a ≤ ρ ≤ b, α ≤ θ ≤ β, c ≤ φ ≤ d}

Entonces$R

f (x , y , z)dV =

∫ d

c

∫ β

α

∫ b

af (ρ sen(φ) cos(θ), ρ sen(φ) sen(θ), ρ cos(θ))ρ2 sen(φ)dρdθdφ.

The spherical coordinate system is especially useful in problems where there is symmetryabout a point, and the origin is placed at this point. For example, the sphere with center theorigin and radius has the simple equation (see Figure 2); this is the reason for thename “spherical” coordinates. The graph of the equation is a vertical half-plane (seeFigure 3), and the equation represents a half-cone with the -axis as its axis (seeFigure 4).

The relationship between rectangular and spherical coordinates can be seen from Fig-ure 5. From triangles and we have

But and , so to convert from spherical to rectangular coordinates,we use the equations

Also, the distance formula shows that

We use this equation in converting from rectangular to spherical coordinates.

EXAMPLE 1 The point is given in spherical coordinates. Plot the pointand find its rectangular coordinates.

SOLUTION We plot the point in Figure 6. From Equations 1 we have

Thus the point is in rectangular coordinates. M(s3!2 , s3!2 , 1)"2, !!4, !!3#

z ! " cos # ! 2 cos !

3! 2(1

2 ) ! 1

y ! " sin # sin $ ! 2 sin !

3 sin

!

4! 2$s3

2 %$ 1s2 % ! &3

2

x ! " sin # cos $ ! 2 sin !

3 cos

!

4! 2$s3

2 %$ 1s2 % ! &3

2

"2, !!4, !!3#V

"2 ! x 2 % y 2 % z22

z ! " cos #y ! " sin # sin $x ! " sin # cos $1

y ! r sin $x ! r cos $

r ! " sin #z ! " cos #

OPP&OPQ

FIGURE 2 !=c, a sphere FIGURE 3 ¨=c, a half-plane

0

z

c

xy

0

z

xy

FIGURE 4 ˙=c, a half-cone

0

z

c

0<c<"/2

y

x

0

z

c

"/2<c<"

y

x

z# ! c$ ! c

" ! cc


FIGURE 6

O2

"3

"4

(2,#"/4,#"/3)

z

xy

FIGURE 5

P(x,#y,#z)P(!,#¨,#˙)

P ª(x,#y,#0)

O

¨

y

x

z˙

r

!

xy

z

˙

Q


Longitud de arco

Deseamos calcular la longitud de la curva y = f (x) de la figura. Entonces, dividimos elintervalo [a,b] en n intervalos y obtenemos

ARC LENGTH

What do we mean by the length of a curve? We might think of fitting a piece of string tothe curve in Figure 1 and then measuring the string against a ruler. But that might be difficult to do with much accuracy if we have a complicated curve. We need a precise definition for the length of an arc of a curve, in the same spirit as the definitions we devel-oped for the concepts of area and volume.

If the curve is a polygon, we can easily find its length; we just add the lengths of theline segments that form the polygon. (We can use the distance formula to find the distancebetween the endpoints of each segment.) We are going to define the length of a generalcurve by first approximating it by a polygon and then taking a limit as the number of seg-ments of the polygon is increased. This process is familiar for the case of a circle, wherethe circumference is the limit of lengths of inscribed polygons (see Figure 2).

Now suppose that a curve is defined by the equation , where f is continuousand . We obtain a polygonal approximation to by dividing the interval into n subintervals with endpoints and equal width . If , then the point lies on and the polygon with vertices , , . . . , , illustrated inFigure 3, is an approximation to .

The length L of is approximately the length of this polygon and the approximationgets better as we let n increase. (See Figure 4, where the arc of the curve between and

has been magnified and approximations with successively smaller values of areshown.) Therefore we define the length of the curve with equation ,

, as the limit of the lengths of these inscribed polygons (if the limit exists):

Notice that the procedure for defining arc length is very similar to the procedure weused for defining area and volume: We divided the curve into a large number of small parts.We then found the approximate lengths of the small parts and added them. Finally, we tookthe limit as .

The definition of arc length given by Equation 1 is not very convenient for compu-tational purposes, but we can derive an integral formula for in the case where has a continuous derivative. [Such a function is called smooth because a small change in produces a small change in .]

If we let , then

! Pi!1Pi ! ! s"xi ! xi!1 #2 " "yi ! yi!1 #2 ! s"#x#2 " "#yi#2

#yi ! yi ! yi!1

f $"x#xf

fL

n l %

L ! limn l %

$n

i!1 ! Pi!1Pi !1

a & x & by ! f "x#CL

#xPi

Pi!1

C

FIGURE 3

y

P¸

P¡P™

Pi-1 Pi Pn

y=ƒ

0 xi¤ i-1 bx¡a x x

CPnP1P0CPi "xi, yi#yi ! f "xi##xx0, x1, . . . , xn

%a, b&Ca & x & by ! f "x#C

8.1

525

FIGURE 1

Pi-1

Pi

Pi-1

Pi

Pi-1

Pi

Pi-1

Pi

FIGURE 4

FIGURE 2

Visual 8.1 shows an animation of Figure 2.TEC

L = lımn→∞

n∑i=1

|Pi−1Pi | = lımn→∞

n∑i=1

√1 + (f ′(x ∗i ))2∆xi

=

∫ b

a

√1 + (f ′(x))2dx

Si los puntos de C se describen parametricamente, es decir, x = x(t) e y = y(t) cont ∈ [α, β], entonces

L =

∫ β

α

√(dxdt

)2+

(dydt

)2

dt


Definicion

Sea C ⊂ R2 una curva suave dada por ecuaciones parametricas x = x(t) e y = y(t),a ≤ t ≤ b y f : C → R. La integral de linea de f a lo largo de C (tambien llamada laintegral de f con respecto a la longitud de arco) se define como

LINE INTEGRALS

In this section we define an integral that is similar to a single integral except that insteadof integrating over an interval , we integrate over a curve . Such integrals are calledline integrals, although “curve integrals” would be better terminology. They were inventedin the early 19th century to solve problems involving fluid flow, forces, electricity, andmagnetism.

We start with a plane curve given by the parametric equations

or, equivalently, by the vector equation , and we assume that is asmooth curve. [This means that is continuous and . See Section 13.3.] If wedivide the parameter interval into n subintervals of equal width and we let

and , then the corresponding points divide into subarcswith lengths (See Figure 1.) We choose any point in the thsubarc. (This corresponds to a point in .) Now if is any function of two vari-ables whose domain includes the curve , we evaluate at the point , multiply bythe length of the subarc, and form the sum

which is similar to a Riemann sum. Then we take the limit of these sums and make the fol-lowing definition by analogy with a single integral.

DEFINITION If is defined on a smooth curve given by Equations 1, thenthe line integral of f along C is

if this limit exists.

In Section 10.2 we found that the length of is

A similar type of argument can be used to show that if is a continuous function, then thelimit in Definition 2 always exists and the following formula can be used to evaluate theline integral:

The value of the line integral does not depend on the parametrization of the curve, pro-vided that the curve is traversed exactly once as t increases from a to b.

yC

f !x, y" ds ! yb

a f (x!t", y!t")#$dx

dt %2

! $dydt %2

dt

3

f

L ! yb

a

#$dxdt %2

! $dydt %2

dt

C

yC f !x, y" ds ! lim

n l " &

n

i!1 f !xi*, yi*" #si

Cf2

&n

i!1 f !xi*, yi*" #si

#si

!xi*, yi*"fCf'ti$1, ti(ti*

iPi*!xi*, yi*"#s1, #s2, . . . , #sn.nCPi !xi, yi "yi ! y!ti"xi ! x!ti"

'ti$1, ti ('a, b(r%!t" " 0r%

Cr!t" ! x!t" i ! y!t" j

a & t & by ! y!t"x ! x!t"1

C

C'a, b(

16.2

1034 | | | | CHAPTER 16 VECTOR CALCULUS

FIGURE 1

t i-1

P¸P¡P™

C

a b

x0

y

tt i

t*i

Pi-1 Pi

Pn

P*i (x*i ,!y*i )

∫C

f (x , y)ds = lımn→∞

n∑i=1

f (x ∗i , y∗

i )∆si

=

∫ b

af (x(t), y(t))

√(dxdt

)2+

(dydt

)2

dt ,

si el limite existe.

Ejemplo Calcular∫C(2 + x2y)ds, donde C es la mitad superior de la circunferencia de

ecuacion x2 + y2 = 1.


Extension a curvas suaves por tramos

If is the length of C between and , then

So the way to remember Formula 3 is to express everything in terms of the parameter Use the parametric equations to express and in terms of t and write ds as

In the special case where is the line segment that joins to , using as theparameter, we can write the parametric equations of as follows: , ,

. Formula 3 then becomes

and so the line integral reduces to an ordinary single integral in this case.Just as for an ordinary single integral, we can interpret the line integral of a positive

function as an area. In fact, if , represents the area of one side ofthe “fence” or “curtain” in Figure 2, whose base is and whose height above the point

is .

EXAMPLE 1 Evaluate , where is the upper half of the unit circle.

SOLUTION In order to use Formula 3, we first need parametric equations to represent C.Recall that the unit circle can be parametrized by means of the equations

and the upper half of the circle is described by the parameter interval (See Figure 3.) Therefore Formula 3 gives

M

Suppose now that is a piecewise-smooth curve; that is, is a union of a finite num-ber of smooth curves where, as illustrated in Figure 4, the initial point of

is the terminal point of Then we define the integral of along as the sum of theintegrals of along each of the smooth pieces of :

yC f !x, y" ds ! y

C1

f !x, y" ds ! yC2

f !x, y" ds ! " " " ! yCn

f !x, y" ds

CfCfCi .Ci!1

C1, C2, . . . , Cn,CC

! 2# ! 23

! y#

0 !2 ! cos2t sin t" dt ! #2t $

cos3t3 $

0

#

! y#

0 !2 ! cos2t sin t"ssin2 t ! cos2 t dt

yC !2 ! x 2y" ds ! y#

0 !2 ! cos2t sin t"%&dx

dt '2

! &dydt '2

dt

0 % t % #.

y ! sin tx ! cos t

x 2 ! y 2 ! 1CxC !2 ! x 2y" ds

f !x, y"!x, y"C

xC f !x, y" dsf !x, y" & 0

yC f !x, y" ds ! yb

a f !x, 0" dx

a % x % by ! 0x ! xC

x!b, 0"!a, 0"C

ds ! %&dxdt '2

! &dydt '2

dt

yxt:

dsdt

! %&dxdt '2

! &dydt '2

r!t"r!a"s!t"

SECTION 16.2 LINE INTEGRALS | | | | 1035

N The arc length function is discussed in Section 13.3.

s

FIGURE 2

f(x,!y)

(x,!y)

C y

z

x

0

FIGURE 3

0

"+¥=1(y˘0)

x

y

1_1

FIGURE 4A piecewise-smooth curve

0

C£C™

C¡

C¢C#

x

ySi la curva esta formada por la union de n tramos defunciones suaves, entonces∫

Cf (x , y)ds =

∫C1

f (x , y)ds + · · ·+

∫Cn

f (x , y)ds

Ejemplo Calcular∫C 2x ds, donde C es la union del arco de la parabola y = x2 de

(0,0) a (1,1) seguido del segmento de recta desde (1,1) a (1,2).Las integrales∫

Cf (x , y)dx =

∫ b

af (x(t), y(t))x ′(t)dt ,

∫C

f (x , y)dy =

∫ b

af (x(t), y(t))y ′(t)dt ,

se llaman integrales de f a lo largo de C en relacion a x y a lo largo de C en relacion ay , respectivamente. Es usual la notacion∫

CP(x , y)dx +

∫C

Q(x , y)dy =

∫C

P(x , y)dx + Q(x , y)dy .


Ejemplos

Evaluar∫

Cy2dx + xdy , donde C es:

1. El segmento recto que une los puntos (−5,−3) y (0,2),

2. El arco de la parabola y = 4 − y2 desde los puntos (−5,−3) y (0,2),

En generalEl valor de la integral de linea depende de:

1. los puntos inicial y final,

2. de la orientacion (la direccion en la que vamos de un punto hacia otro),

3. de la curva que los une (mas adelante veremos las hipotesis que se necesitanpara que el valor de la integral no dependa del camino).


Extension a curvas en el espacio

Todas las formulas anteriores se extienden a curvas en el espacio definidasparametricamente como x = x(t), y = y(t) y z = z(t):

∫C

f (x , y , z)ds =

∫ b

af (x(t), y(t), z(t))

√(dxdt

)2+

(dydt

)2

+(dz

dt

)2dt ,∫

Cf (x , y , z)dx =

∫ b

af (x(t), y(t), z(t))x ′(t)dt ,∫

Cf (x , y , z)dy =

∫ b

af (x(t), y(t), z(t))y ′(t)dt ,∫

Cf (x , y , z)dz =

∫ b

af (x(t), y(t), z(t))z′(t)dt ,

Ejemplo Evaluar∫

Cy sen(z)ds, donde C es la helice circular con ecuaciones

parametricas x = cos(t), y = sen(t), z = t , con 0 ≤ t ≤ 2π.


Integrales de linea de campos vectoriales

Un campo vectorial en RM es una funcion F : D ⊂ RM→ RM ,M = 2,3. Si F es un

campo vectorial continuo definido en alguna curva C dada por una funcion vectorialr(t),a ≤ t ≤ b, entonces la integral de linea de F a lo largo de C es∫

CF · dr =

∫ b

aF (r(t)) · r ′(t)dt =

∫C

F · T ds, donde T =r ′(t)|r ′(t)|

.

Esta definicion se puede interpretar como el trabajo realizado por una fuerza F paramover una partıcula atraves de la curva C.

Ejemplo Evaluar∫

CF · dr , donde F (x , y , z) = xy i + yzj + zxk y C es la curva con

ecuaciones parametricas

x = t , y = t2, z = t3, con 0 ≤ t ≤ 1.


Campos conservativos y el teorema fundamental

Un campo vectorial F se dice conservativo si es el gradiente de alguna funcion escalarde alguna funcion escalar f , es decir, si existe f tal que

∇f = F .

En este caso f se llama una funcion potencial para F .

Teorema 1. Teorema fundamentalSea C una curva suave dada por un vector r(t),a ≤ t ≤ b. Sea f una funciondiferenciable de dos o tres variables cuyo gradiente es continuo sobre C. Entonces∫

C∇f · dr = f (r(b)) − f (r(a)).

Este teorema nos dice que podemos evaluar la integral de un campo conservativo soloconociendo el valor de f en los puntos inicial y final de C.


Independencia del camino

Sea F un campo vectorial con dominio D. Decimos que la integral de linea∫

CF · dr es

independiente del camino si para dos caminos C1,C2 con punto inicial y final igualesse tiene ∫

C1

F · dr =

∫C2

F · dr .

Ademas, se dice que una curva o camino C es cerrada si el punto inicial y el finalcoinciden, es decir r(a) = r(b). Tenemos entonces el siguiente resultado

Teorema 2. Teorema de caracterizacionLa integral

∫C

F · dr es independiente del camino en D si y solo si∫

CF · dr = 0 para

todo camino cerrado C en D.


Independencia del camino

Supongamos que D es un conjunto abierto y que es conexo, es decir cualquier par depuntos en D se pueden unir con un camino incluido en D. Entonces

Teorema 3.Supongamos que F es un campo vectorial continuo sobre una region D abierta y

conexa. Si∫

CF · dr es independiente del camino en D, entonces F es un campo

vectorial conservativo en D, es decir, existe una funcion f tal que ∇f = F .

Es decir, los unicos campos vectoriales independientes del camino son los camposvectoriales conservativos.


Caracterizacion de los campos conservativos

Teorema 4.Sean P y Q funciones continuas con primeras derivadas parciales continuas sobre D.Si F (x , y) = P(x , y)i + Q(x , y)j es un campo conservativo entonces

∂P∂y

=∂Q∂x

Para enunciar el reciproco del teorema 4, necesitamos un par de definiciones. Unacurva simple es una curva que no se intercepta a sı misma, salvo en sus puntos inicialy final. Una region D del plano se llama simplemente conexa si cualquier curvacerrada C ⊂ D encierra unicamente puntos de D.

Notice that the first of these integrals does not depend on , so

If we write , then

On , is constant, so . Using as the parameter, where , we have

by Part 1 of the Fundamental Theorem of Calculus (see Section 5.3). A similar argu-ment, using a vertical line segment (see Figure 5), shows that

Thus

which says that is conservative. M

The question remains: How is it possible to determine whether or not a vector field is conservative? Suppose it is known that is conservative, where and have continuous first-order partial derivatives. Then there is a function such that

, that is,

Therefore, by Clairaut’s Theorem,

THEOREM If is a conservative vector field,where and have continuous first-order partial derivatives on a domain , thenthroughout we have

The converse of Theorem 5 is true only for a special type of region. To explain this, wefirst need the concept of a simple curve, which is a curve that doesn’t intersect itself any-where between its endpoints. [See Figure 6; for a simple closed curve, but

when .]In Theorem 4 we needed an open connected region. For the next theorem we need a

stronger condition. A simply-connected region in the plane is a connected region such D

a ! t1 ! t2 ! br!t1 " ! r!t2 "r!a" " r!b"

"P"y

""Q"x

DDQP

F!x, y" " P!x, y" i # Q!x, y" j5

"P"y

""2 f

"y "x"

"2 f"x "y

""Q"x

Q ""f"y

andP ""f"x

F " ! ffQ

PF " P i # Q jF

F

F " P i # Q j ""f"x

i #"f"y

j " ! f

"

"y f !x, y" "

"

"y y

C2

P dx # Q dy ""

"y yy

y1

Q!x, t" dt " Q!x, y"

""

"x y x

x1

P!t, y" dt " P!x, y" "

"x f !x, y" "

"

"x y

C2

P dx # Q dy

x1 $ t $ xtdy " 0yC2

y

C2

F ! dr " y

C2

P dx # Q dy

F " P i # Q j

"

"x f !x, y" " 0 #

"

"x y

C2

F ! dr

x

SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS | | | | 1049

FIGURE 5

(a,!b)

x0

y

D

(x,!y)

C¡

C™(x,!y¡)

FIGURE 6Types of curves

simple,not closed

not simple,closed

not simple,not closed

not simple,closed

simple,closed

that every simple closed curve in encloses only points that are in . Notice from Figure7 that, intuitively speaking, a simply-connected region contains no hole and can’t consistof two separate pieces.

In terms of simply-connected regions we can now state a partial converse to Theorem 5that gives a convenient method for verifying that a vector field on is conservative. Theproof will be sketched in the next section as a consequence of Green’s Theorem.

THEOREM Let be a vector field on an open simply-connectedregion . Suppose that and have continuous first-order derivatives and

Then is conservative.

EXAMPLE 2 Determine whether or not the vector field

is conservative.

SOLUTION Let and . Then

Since , is not conservative by Theorem 5. M

EXAMPLE 3 Determine whether or not the vector field

is conservative.

SOLUTION Let and . Then

Also, the domain of is the entire plane , which is open and simply-connected. Therefore we can apply Theorem 6 and conclude that is conservative. M

In Example 3, Theorem 6 told us that is conservative, but it did not tell us how to findthe (potential) function such that . The proof of Theorem 4 gives us a clue as tohow to find . We use “partial integration” as in the following example.

EXAMPLE 4(a) If , find a function such that .(b) Evaluate the line integral , where is the curve given by

0 ! t ! "r!t" ! e t sin t i # e t cos t j

CxC F ! drF ! ! ffF!x, y" ! !3 # 2xy" i # !x 2 $ 3y 2 " j

fF ! ! ff

F

F!D ! ! 2 "F

%P%y

! 2x !%Q%x

Q!x, y" ! x 2 $ 3y 2P!x, y" ! 3 # 2xy

F!x, y" ! !3 # 2xy" i # !x 2 $ 3y 2 " j

V

F%P#%y " %Q#%x

%Q%x

! 1%P%y

! $1

Q!x, y" ! x $ 2P!x, y" ! x $ y

F!x, y" ! !x $ y" i # !x $ 2" j

V

F

throughout D%P%y

!%Q%x

QPDF ! P i # Q j6

! 2

DD

1050 | | | | CHAPTER 16 VECTOR CALCULUS

FIGURE 7

simply-connected region

regions that are not simply-connected

C

10

_10

_10 10

FIGURE 8

N Figures 8 and 9 show the vector fields inExamples 2 and 3, respectively. The vectors inFigure 8 that start on the closed curve allappear to point in roughly the same direction as . So it looks as if and there-fore is not conservative. The calculation inExample 2 confirms this impression. Some of thevectors near the curves and in Figure 9point in approximately the same direction as thecurves, whereas others point in the oppositedirection. So it appears plausible that line inte-grals around all closed paths are . Example 3shows that is indeed conservative.F

0

C2C1

FxC F ! dr & 0C

C

FIGURE 9

C™C¡

2

_2

_2 2


Caracterizacion de los campos conservativos

Teorema 5. Reciproco del teorema 4Sea F (x , y) = P(x , y)i + Q(x , y)j un campo vectorial definido en una regionsimplemente conexa D. Si P y Q tienen primeras derivadas parciales continuas sobreD y

∂P∂y

=∂Q∂x

,

entonces F es conservativo.

Ejemplo Si F (x , y) = (3 + 2xy)i + (x2− 3y2)j , determinar una funcion f tal que

∇f = F . Evaluar ademas,∫C F · dr , donde C es la curva dada por

r(t) = et sen(t)i + et cos(t)j , 0 ≤ t ≤ π.


Conservacion de la energıa

Considere una fuerza F que mueve un objeto a lo largo de una curva C dada porr(t),a ≤ t ≤ b. Sean A = r(a) y B = r(b). De acuerdo a la segunda ley de NewtonF (r(t)) = mr ′′(t). Entonces, si W es el trabajo ejercido por la fuerza

W =

∫C

F · dr =

∫ b

aF (r(t)) · r ′(t)dt = m

∫ b

ar ′′(t) · r ′(t)dt =

m2

∫ b

a

ddt|r ′(t)|2dt

=m2

∫ b

a

ddt|r ′(t)|2dt =

m2|r ′(t)|2

]b

a=

m|v(b)|2

2−

m|v(a)|2

2,

donde v(t) = r ′(t) es la velocidad y 12 m|v(t)|2 es la energia cinetica. Si F es

conservativo, existe una funcion f tal que F = ∇f . La cantidad P = −f se llama energıapotencial. Entonces

W = −

∫C∇P · dr = −[P(r(b)) − P(r(a))] = P(A) − P(B),

de donde se obtiene la ley de conservacion de la energıa:

m|v(b)|2

2+ P(B) =

m|v(a)|2

2+ P(A).


GREEN’S THEOREM

Green’s Theorem gives the relationship between a line integral around a simple closedcurve and a double integral over the plane region bounded by . (See Figure 1. Weassume that consists of all points inside as well as all points on .) In stating Green’sTheorem we use the convention that the positive orientation of a simple closed curve refers to a single counterclockwise traversal of . Thus if is given by the vector func-tion , , then the region is always on the left as the point traverses .(See Figure 2.)

GREEN’S THEOREM Let be a positively oriented, piecewise-smooth, simpleclosed curve in the plane and let be the region bounded by . If and havecontinuous partial derivatives on an open region that contains , then

The notation

gC

is sometimes used to indicate that the line integral is calculated using the positive orienta-tion of the closed curve . Another notation for the positively oriented boundary curve of

is , so the equation in Green’s Theorem can be written as

Green’s Theorem should be regarded as the counterpart of the Fundamental Theorem ofCalculus for double integrals. Compare Equation 1 with the statement of the FundamentalTheorem of Calculus, Part 2, in the following equation:

In both cases there is an integral involving derivatives ( , , and ) on the leftside of the equation. And in both cases the right side involves the values of the originalfunctions ( , , and ) only on the boundary of the domain. (In the one-dimensional case,the domain is an interval whose boundary consists of just two points, and .)ba!a, b"

PQF

!P#!y!Q#!xF"

yb

a F"$x% dx ! F$b% # F$a%

yyD

&!Q!x

#!P!y ' dA ! y

!D P dx $ Q dy1

!DDC

P dx $ Q dyor!yC P dx $ Q dy

NOTE

yC P dx $ Q dy ! yy

D

&!Q!x

#!P!y ' dA

DQPCD

C

FIGURE 2 (a) Positive orientation

y

x0

D

C

(b) Negative orientation

y

x0

D

C

Cr$t%Da % t % br$t%CC

CCCDCDC

16.4

SECTION 16.4 GREEN’S THEOREM | | | | 1055

FIGURE 1

y

x0

D

C

N Recall that the left side of this equation is another way of writing , where

.F ! P i $ Q jx

C F ! dr

Una curva cerrada simple C tiene orienta-cion positiva si al caminar sobre ella en di-reccion contraria a las manecillas del reloj,tenemos la region D encerrada por C siem-pre a nuestra izquierda.

El teorema de GreenSea C una curva con orientacion positiva, suave a trozos, simple y cerrada en el planoy sea D la region encerrada por C. Si P y Q tienen derivadas parciales continuas enuna region abierta que contiene a D entonces∫

CPdx + Qdy =

"D

(∂Q∂x−∂P∂y

)dA.

Tambien se usan las notaciones∮

Cy

�C

para∫

Cy ∂D para C.


El rotacional

Sea F = Pi + Qj + Rk un campo vectorial. Se define el rotational de F como

curl(F ) = rot(F ) = ∇ × F =

(∂R∂y−∂Q∂z

)i +

(∂P∂z−∂R∂x

)j +

(∂Q∂x−∂P∂y

)k .

Notemos que si f (x , y , z) es un funcion con segundas derivadas parciales continuasentonces

curl(∇f ) = 0.

es decir, si F es un campo conservativo, entonces curl(F ) = 0. Mas aun, se tiene elsiguiente reciproco:

Si F es un campo vectorial en R3 cuyas componentes tienen derivadas parcialescontinuas y curl(F ) = 0, entonces F es un campo conservativo.


La divergencia

Sea F = Pi + Qj + Rk un campo vectorial. Se define la divergencia de F como

div(F ) = ∇ · F =∂P∂x

+∂Q∂y

+∂R∂z.

Si F = Pi + Qj + Rk es un campo vectorial en R3 cuyas componentes tienenderivadas parciales continuas entonces

div(curl(F )) = 0.

Usando estos operadores, podemos escribir el teorema de Green como∮C

F · dr =

"D

(curl(F )) · kdA.

Adicionalmente, se tiene ∮C

F · nds =

"D

div(F (x , y))dA.

clases de calculo 3´ · clases de calculo 3´ roberto carlos cabrales dpto. de ciencias basicas´...

Documents