Calculo 2 De dos variables_9na Edición - Ron Larson

1110 Chapter 15 Vector Analysis
1110 1110 1110 Chapter CAPÍTULO 15 15 Vector Vector 15 Analysis
Análisis vectorial
1110 Chapter 15 Vector Analysis
1110 Chapter 15 Vector Analysis
Surface of Revolution In Exercises 31–34, write a set of Area In Exercises 39– 4
1110 Chapter 15 Vector Analysis
Surface Superficie 1110 of of Revolution
de Chapter revolución 15 In In Exercises Vector En los Analysis ejercicios 31–34, 31–34, write 31 write a 34, a a set dar set parametric of un of con-Arejunto
Surface de equations ecuaciones of Revolution for for the paramétricas the surface In Exercises of of para revolution 31–34, la superficie obtained write de revolving byrevolu-
set of given the given graph Area sobre region. region. of la In the Use región Exercises Use function a computer a dada. 39– about Utilizar 46, algebra the find given un the sistema system area axis. system of algebraico to the to verify surface verify por your results. over your compu-
the
equations AreaÁrea
In In Exercises for En the los 39– surface ejercicios 39– 46, 46, find of find revolution 39 the a the area 46, area hallar of obtained of the the el surface área by de over over la given the superficie the region. Use a com
parametric revolving Surface
ción the obtenida the graph
of
graph
Revolution
por of of the revolución the function In Exercises
de about la about gráfica the the
31–34,
given de given la axis.
write
función axis.
a set of
en tornoresults.
Area In Exercises 39– 46, find the area of the surface over the
parametric equations for the surface of revolution obtained by given tadora region. y verificar Use los computer resultados. algebra system to verify your
parametric Surface of equations Revolutionfor In the Exercises surface of 31–34, revolution write obtained set Función by of given Area region. In Exercises Use a 39– Eje computer 46, de revolución find algebra the area system of the surface to verify 39. over The your the part of the
al Surface revolving eje dado. of the Revolution graph of the In function Exercises about 31–34, the given write axis. a set of Area results. In Exercises 39– 46, find the area of the surface over the
Función revolving parametric the equations graph of for the the function surface Eje Eje de about de revolución of revolution the given obtained axis. by
x39. 39. The results. given 39. The part La region. part parte of of the Use del the plano plane plane computer r(u, r u, r v) vu, algebra v= 4ui 4ui – system vj vj + vk,
to vk, verify where donde where 0your
u 2 y 0 v
parametric equations for the surface of revolution obtained
31. y
by
x
0 u 2 y 0 v 1
31. y
eje x
2 , 0 x 6 0 u 2 y 0 v eje 1 x
x
31. y
eje x
2 , 0 x 6 2 , 0 given x 6 region. Use a computer algebra system to verify your
revolving Función the graph of the function about Eje de the revolución given axis.
results.
revolving 39. The part of the plane u, 4ui vj vk, 40. where
Función the graph of the function Eje about de revolución the given axis.
39. results. The part of the plane r u, v 4ui vj vk, where The part of the parab
40. 40. The La part parte of the del paraboloide rr(u, vv) = 2u cos vi + 2u sen vj
+ u 2 k,
32.
40. The part of the eje
r u, v 2u cos vi 2u sen vj u 2 k, where 0 u
31.
Función
x
eje
y x, 0 x 4
x
32. eje
u 2 k, donde where 0 u 2 y 0 v 2
32. y y x, x, 0 , 0 x x 4 4
Eje de revolución
39. 0The upart 2 yof 0 the v plane 1
u, 4ui vj vk, where
31. y
Función
eje x eje x x
u 2 40. k, where
part 0 of u the 2paraboloid
y 0 v 2u, 2u cos vi 2u
33. eje
41. sen The vj
2 , 0 x 6 Eje de revolución
39. The part of the plane r v 4ui vj vk, where
x sen z, 40. 0 The z part of the paraboloid
z r u, v 2u cos vi 2u sen vj part of the cyli
31. x
eje
33. x sen z, z
eje z
41. 41. The La part parte of del the cilindro ru, r u, vv a cos ui sin a sen uj vk,
33. x 32. sen z, 0 z
eje z
41. The part 2 k, of where the cylinder
r v a cos x, 0 u 2 y 0 v 1
ui a uj vk, where donde0 u 2 a
32.
31. y
x, 0 x 4 eje
eje
x
34. z y 2 40.
1, where where 0
uThe 2 k,
0 y
where of
≤0 u 2
0the paraboloid u
≤u 2 22y
and 0and
eje
≤0 y
2 y 0 vu, 2 2u cos vi 2u sen vj
2 , 0 x 6 sen
40.
v0 ≤vb
v b b
34. 34. z 33. z y 2 y 2 1, 1, 0 y y 2 2 eje eje y
41. The sen z, y
part of the cylinder
u, cos ui sen 42. uj The vk, sphere r u, v
33.
32.
x sen z, z
eje
41. The 2 part of the paraboloid r u, v 2u cos vi 2u sen vj
x, z
k, where
part of the cylinder
r u, a cos ui a sen uj vk,
32. y x, 0 x 4 eje x
42. 42. The 42.
u
Tangent Plane The In sphere where La sphere
2 k,
esfera
where
Exercises r 0
u, r vu, 35–38, vru, u
a sen v a sen
find
and 2
a
y
u cos u sin cos an viu vi equation cos a vi sen a sen u sen a
of u sin vj the
u vj sin cos vj a cos a uk, uk, cos where uk, 0 u an
34. 2 1, Tangent 34.
33. where 0 u 2 and 0 v b
Plane z Plane y 2 In 1, In Exercises 0 y 35–38, 2
eje
41. The part of the cylinder
sen
0 v 2
u, cos
sen
ui
sen
sen z, 35–38, find find an y
sen uj vk,
33. x sen z, 0 z
eje z an equation of of the the where 41.
tangent plane to 42. where The
the donde 0surface 0sphere
part u
u≤of represented and u, the ≤and
0cylinder
y 00 vsen ≤
by vv 2r the
≤cos u, 22
v
vector-valued
vi a cos sen ui sen a vj sen
43. ujcos vk,
The uk,
42. The
where
sphere
r u, v and
a sen u cos vi part of the co
tangent Plano
34. plane plane tangente
Tangent to Plane
2 to the 1, the surface En los represented ejercicios eje by a by 38, the the hallar vector-valued
una ecuación43. 43. The 43.
where
In Exercises 35–38, find an equation function of theat the given The part La part parte
0
point. of of the del
u
the cone cono
2
cone rand
ru,
and
r v u, v
0
v au v
au cos b a sen u sen vj a cos uk,
cos vi vi au au sin sen vj vj uk,
uk, where donde0 u b and
function Tangent
34. z y
para at el at the plano the
Plane
2 1,
given given tangente point.
In
0
point.
Exercises
y 2
35–38,
eje
find
y
an equation of the 42. where The sphere 0 u u, and
0sen vcos vi 2
sen sen
sen vj cos uk,
tangent plane to the surface a la superficie represented dada by por the la función vector-valued vectorial,
Tangent function el Plane at punto the given indicado. In Exercises point. 35–38, find an equation of the
where 42. 43. where The 0 ≤0 sphere
part u ≤u bof ry
and bu, 0the and v≤ 0 vcone
0≤a v sen 2 v 2u u, cos 2
viau cos a sen viu sen au vj sen a
35.
44. vjcos uk,
The tangent Tangent plane Planeto In the Exercises surface represented 35–38, find by an the equation vector-valued of r u, thev u43. vThe where
i part u of v jthe vk, cone
and
1, r u, 1, 1
au cos vi au sen vj uk, torus r u, v a
35. r u, v u v i u v j vk, 1, 1, 1
44. 44.
where
The El torus toro r u, vru, va a b cos b cos vcos ui ui a a b cos b vcos sen vsin uj uj
35. r u, v u v i u v j vk, 1, 1, 1
44. The torus
0
r u, u
v a and and
b cos 0
v cos v
ui 2
function tangent plane at the to given the point. surface represented by the vector-valued 43. a b cos v sen uj b sen vk, where a > b
z where The part 0 of u the b and cone
b sen b sen vk, b vk, sin where vk, where donde a > a b, > a 0b,
> 0b,
z 0 u, v 2 au cos vi au sen vj
sen
uk,
tangent plane to the surface represented by the vector-valued
u 0u≤2 u2 , ≤and
, 2, and 0 y 00v ≤vv2≤22
function 35. ru, v at the z u given
z vi ipoint.
43.
u vj z
z 1, 1, 1
44. The vk, torus
u, cos cos ui cos (1, −1, 1)
45. sen The uj surface of revolu
35.
44. The
where
senpart of the cone r u, v au cos vi au sen vj uk,
torus
r u, v and
a b cos v function r u, vat the ugiven v i point. u v j vk, 1, 1, 1
45. 45.
where
The La surface superficie of de revolución r u, ru, v u cos u vi cos vi u sen u vj sin vj
(1, (1, −1, −1, 1) 1)
45. The 2 surface sen vk, 0
of where u
revolution
b
and
rb,
0
u, v cos
2
v 2 ui a b cos v sen uj
u cos , and vi u sen vj uk, where 0 u 4
35.
z
z
44. b The sen torus vk, where
u, a > b,
0 cos u cos 2 , ui and 0 v cos 2
sen
sen
u, vk, 1, 1, uj
35. r 2 2
uk, uk, where donde 0 0u≤ u4≤and
4 y
0 ≤vv ≤ 2
2 2
uk, 45. where
0surface u of 4revolution
0 v u, 2
(1,
u,
−1,
v
1)
u z v i u v j vk, z 1, 1, 1
44. The torus r u, v a b cos v cos ui cos a vib cos v sen
−2
46. uj
The vj
45. The sen surface of r
46. 46. The La −1
surface
vk, where
of revolution
b,
r u, v and
u cos vi (1, −1, 1) z
z
b superficie of de revolución ru, r u, vv sin sen u u cos cos vi vi uj uj sen u
−2
46. The uk, sen
surface where vk, where
of a > b, and 0 u 2 ,
2
and 0 v 2u sen vj
z
sen u cos vi uj
−1
−2
z 2
(1, 1, 1)
sen u sen vk, where 0
−12
45. uk, The where surface 0 of urevolution
4 and 1 0 u, sen
(1, −1, 1)
2
cos vi sen vj
2
(1, −1, 1)
(1, (1, 1, 1)
2 sen sin 45. sen u sen u sen vk, vk, where donde where 000≤u
uu≤
and y and 00≤ 0v v≤ 22
2
−2
1 1
1)
46. u The surface of revolution
u, y sen cos vi uj
2 2
−1
x 46. The
uk,
sen sin
where
vk, surface of revolution r u, v u cos vi u sen vj
2
2
surface of revolution
and
r u, 2 −2
v sen u cos vi
y y
sen sen vk, where
and
WRITING uj
−1
2
uk, where 0 u 4 and 0 v 2
ABOUT C
(1, 1, 1)
x x
1
2 2
WRITING 2
2
−2
(1, 1, 1)
46. sen The u
Desarrollo ABOUT sen surface vk, where of
CONCEPTS de 0revolution
u and
u, 0 v sen 2
cos vi uj
−1
1
conceptos
2 2 2 −2
y 46. The surface of revolution r u, v sen u cos vi
−2
47. uj Define a parametric
x
−1
y
sen sen y vk, where
and
1 (1, 1, 1)
x −2 −2
47. Define a y y
2 47. Define WRITING sen a u parametric sen vk, ABOUT where surface. 0CONCEPTS
u1
and 0 v 2
2
2
1 (1, 1, 1)
47. Definir una superficie paramétrica.
48. Give the double in
1 1
2 y WRITING ABOUT CONCEPTS
x
2
2
y
2
−2 y
48. 48. Give 47. Give the Define the double double parametric integral that surface.
2
WRITING that yields yields the the surface surface area area of of a a parametric surface o
−2
2 2
47. 48. Define Dar la a integral
ABOUT parametric doble
CONCEPTS
x
2
surface.
y
con las que x se obtiene el área de la
1
2
WRITING ABOUT CONCEPTS
2
surface over an open region D.
x x
parametric 48. Give surface over an open region D.
Figure for 35 superficie the double de una integral
Figure superficie that
for 36 paramétrica yields the surface sobre una area región of −2
1
2
48.
47.
Give
Define
the parametric
double integral
surface.
y
that yields the surface area of a
−2
47. Define parametric a parametric
Figure Figure for for 35 35 Figure Figure for for 36 36 x
abierta D. surface over surface.
y
1 2
an open region D.
1
48. parametric Give the double surface integral over an that open yields region the D. surface area 49. of Show that the cone in
2 x
36. r u, v ui 48. vj Give uv the k, double 1, 1, integral 1 that yields the surface area of a
49. 49. Show Show that that the the cone cone in Example in 3 can 3 can be be represented parametrically
cos by Show
cally by r u, v u c
36. 36. r u, r vu, Figure
Figura v ui ui
for 35
para vj vj35 uv uv k, k, 1, 1, 1, 1
Figure for 2 36
parametric surface over an open region D.
x
Figure for Figura Figure para for 36 36 37. r u, v 2u cally 49.
49. Mostrar vi
parametric
by r u, r3u that vu, sen que vthe vj
surface
u se
cone cos u puede cos uvi 2 in
k,
over
vi Example
representar u 0,
an u 6,
open vj 4
vj can
el
region
uk, cono
D.
x
be uk, where represented where del 0ejemplo 0 u parametrically
by u, cos vi sen vj uk, where and
and u 3 and de 0 manera
v paramétrica 2 . mediante r(u, v) = u cos vi + u sen vj + uk,
v 2 .
37. 37. r 36. u, r vu, Figure vu, 2u 2u cos for cos ui vi 35vi vj 3u 3u sen sen vj uv vj k, u 2 k, u1, 2 k, 1, Figure 0, 6, 0, 46, for 4 36
0
36. 36.
0
49.
v
Show
z2 that
.
the cone in Example 3 can be represented parametrically
Show by that r the u, vcone u in cos Example vi u sen can vjbe represented uk, where 0parametri-
.
CAPSTONE u and
ru,
rFigure u, v v for
ui
ui35 vj
vj
uv
uv
k,
k,
1,
1,
1,
1, Figure
1 1 for 36
37.
2u z
donde 0 u y 0 v 2p.
zcos vi 3u sen vj 2 k, 0, 6, 49.
37.
36.
u, 2u
ui vj
vi 3u
uv
sen
k,
vj
1, 1,
u 2 k, 0, 6, 4
49. Show that the cone in Example 3 can be represented parametrically
by r u, v u cos vi u sen vj uk, where 0 u and
36. r u, v ui vj uv k, 1, 1, 1
CAPSTONE
0cally 6 vby 2 u, .
cos vi sen vj uk, where and
37.
u, 2u cos z
50. The four figures
6 6
vi 3u sen vj 2 k, 0, 6, 37. r u, v 2u cos viz
3u sen vj u 2 k, 0, 6, 4
5
50. 50. The CAPSTONE Para
The 0four four vdiscusión
figures 2 figures . below below are are graphs graphs of of the the surface
r u, v ui sen u
5 5 6
CAPSTONE
6
z
z
rCAPSTONE
u, 50. r vu, The Las v ui cuatro ui sen sen u cos u cos vj vj sen sen u sen u sen vk, vk,
(0, four 6, 4) figuras figures son below gráficas are de graphs la superficie of the surface
5
50. The four figures below are graphs of the surface 0 u 2, 0
5
6
CAPSTONE
(0, (0, 6, 4) 6, 4)
ru, v ui sen sin u cos vj sen sin u sen sin vk,
6
050. 0 ur The u, u vfour 2, ui 2, figures 0 sen 0 vu cos vbelow 2 vj 2 . . are sen graphs sen vk, of the surface
5
50. The four figures below are graphs of the surface
5 (0, 6, 4)
Match each of the f
(0, 6, 4)
−6 Match Match 0u, ≤ u
each each ≤ ui sen
of of the the
2, 0 ≤ four four vcos ≤ vj
graphs graphs 2. sen sen vk,
with with .
0r u, vu ui 2 ,
2, sen
0
u cos
v
vj
2 the sen
. the point u sen point vk,
space in space from from which the surface is
−6 −6
(0, (0, 6, 6, 4) 4)
(0, 6, 4)
4
2which Match
Relacionar the the surface surface each
cada is of 2, viewed. is the
una four de The graphs las The cuatro four four with points gráficas points the are point are con 10, in 10, el space
punto 0, , 0from
, en el10, 10, 0 , 0, 10
4 2 2
x
2
4 −6
10, Match 0
which
espacio u
10, 10, 0 ,
each
4the 0 desde 0, , surface 10, of 2,
10,
the
0 el , 0 cual and
four 0
, and
graphs
10, 10, 10,
with 10 .
the
.
point in space from
x x −6 2 2
6
is viewed.
se vcontempla 2 .
The four
la superficie.
points are
Los
10, 0,
cuatro
(a) ,
z
2 4 4
which Match the each surface of the
6
(a)
z
(b)
z
6
(a)
z
yis four viewed. graphs The with four the points in are space 10, 0, from 0 ,
4
10, 10, , 0, 10, ,(b)
and 10, 10, 10 . z
x −6
−6
4
2
Match puntos each son (10, of the 0, 0), four (10,10, graphs with 0), (0, the 10, point 0) y in (10, space 10, from 10).
2 y y
which 10, 10, the 0 surface , 0, 10, is 0 viewed. , and The 10, 10, four 10 points . are 10, 0, −6
x
2 4
2
which
1
6
(a)
(b)
z
4
4
the surface z is viewed. The
6 y
38. r u, v 2u cosh (a)
10, 10, 0,
vi 2u senh vjz
10, and 10, four points are z 10, 0, 0 ,
2
1 1
u2 k, 4, (b)
10, 10 z
0, 2
x 4
2 2
4
y
10, 10, 0 , 0, 10, 0 , and 10, 10, 10 .
38. 38. r u, r vu, xv 2u 2u cosh cosh vi vi 2u 2u
2
senh senh vj vj 6 2 u2 2 k, u2 k, 4, 0, 4, 20, 2
(a)
z
(b)
z
4
z
y
6
y
(a)
z
(b)
z
1
38.
u, 2u cosh viz
z2u senh vj y
12 u2 k, 4, 0, y y
38. r u, v 2u cosh vi 2u senh vj 2 u2 k, 4, 0, 2
38.
u, 2u cosh vi 2u senh z
1
vj
x x
y y
z
21
u2 y
k, 4, 0, y
38. r u, v 2u cosh vi 2u senh vj 2 u2 k, 4, 0, 2
4
x
y
4 4 z
y
x
y
x
y
z
y
(−4, 0, 2)
(c)
z
4
x
y
4 (−4, (−4, 0, 2) 0, 2)
2
(c) (c)
z z
x
(d) (d) z z
y
2 2 4
(−4, 0, 2)
4
(c)
z
(d)
z z
2 (−4, 0, 2)
(c)
z
(d)
z
2 (−4, 0, 2)
x
6 4 2
−2 −4 −6
(c)
z
(d)
z
x x
2
6 4 2
−2 −4 −6
6 4 2
−2 (−4,
(−4,
−40, 0,
2)
2) −6
(c) 4 z
(d)
z
4 4
2
y
y y
x
6 4 2 y y
−2 −4 −6
x
6 4 2 4 −2 −4 −6
x
y y
x
4
y
6 4 2
y −2 −4 −6
x
4 y
4
2
−2 −4 −6
y
6
−2 −4 −6
4
y
y
y