exercicios resolvidos integrais múltiplas

exercicios resolvidos integrais múltiplas


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~F ·d~r+
\u222b
C2
~F ·d~r =
\u222b 1
0
(2t, 0, t) . (1, 0, 0) dt+
\u222b pi
0
(
2 cos t\u2212 t
pi
, sin t, cos t
)
.
(
\u2212 sin t, cos t, 1
pi
)
dt =
=
\u222b 1
0
2t dt+
\u222b pi
0
\u22122 cos t sin t+ t
pi
sin t+ cos t sin t+
1
pi
cos t dt =
= 1+
[
cos2 t
2
]pi
0
+
[
1
pi
sin t
]pi
0
+
[
\u2212 t
pi
cos t\u2212
\u222b
\u2212 cos t 1
pi
dt
]pi
0
= 1+0+0+
[
\u2212 t
pi
cos t+ sin t
1
pi
]pi
0
= 1+1 = 2 .
RESOLUC¸A\u2dcO DO EXERC´ICIO 44)
O trabalho realizado pelo campo de forc¸as ~F (x, y, z) = x\u131\u2c6+ 2y\uf6be\u2c6\u2212 zk\u2c6, no deslocamento ao longo da curva
C, e´ dado pelo integral
\u222b
C
~F . d~r .
A curva C e´ representada pelas equac¸o\u2dces parame´tricas\uf8f1\uf8f4\uf8f2\uf8f4\uf8f3
x = 1
y = t
z = t4
, t \u2208 [ 0, 1 ] .
Ora \uf8f1\uf8f4\uf8f2\uf8f4\uf8f3
x = 1
y = t
z = t4
=\u21d2
\uf8f1\uf8f4\uf8f2\uf8f4\uf8f3
x\u2032 = 0
y\u2032 = 1
z\u2032 = 4 t3
.
Enta\u2dco o trabalho e´\u222b
C
~F . d~r =
\u222b 1
0
(
1, 2t,\u2212t4) . (0, 1, 4 t3) dt = \u222b 1
0
2t\u2212 4 t7 dt =
[
t2 \u2212 4 t
8
8
]1
0
= 1\u2212 1
2
=
1
2
8
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
RESOLUC¸A\u2dcO DO EXERC´ICIO 52)
Figura 8: Regia\u2dco R
A regia\u2dco R e´ simplesmente conexa. A fronteira de R e´ uma curva C simples, fechada, orientada positiva-
mente e seccionalmente de classe C1 e regular. As func¸o\u2dces M(x, y) = y e N(x, y) = ey y2 sa\u2dco de classe C1
em R2 . Pelo Teorema de Green,
K =
\u222b
C
y dx+ eyy2 dy =
\u222b\u222b
R
(
\u2202
\u2202 x
ey y2 \u2212 \u2202
\u2202 y
y
)
dx dy =
\u222b\u222b
R
0\u2212 1 dx dy =
= \u2212
\u222b\u222b
R
1 dx dy = \u2212A(R) .
Resulta que a a´rea de R e´ dada por
A(R) = \u2212K .
RESOLUC¸A\u2dcO DO EXERC´ICIO 55)
A regia\u2dco R
Figura 9: Regia\u2dco R
e´ simplesmente conexa. A fronteira de R e´ uma curva C , orientada no sentido positivo, simples, fechada
e seccionalmente de classe C1 e regular.
As func¸o\u2dces M(x, y) = x+ 2y2 e N(x, y) = xy + y2 sa\u2dco func¸o\u2dces polinomiais, logo de classe C1 em R2 .
9
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
Pelo Teorema de Green,\u222b
C
(x+ 2y2) dx+ (xy + y2) dy =
\u222b\u222b
R
(
\u2202
\u2202 x
(
xy + y2
)\u2212 \u2202
\u2202 y
(
x+ 2y2
))
dx dy =
\u222b\u222b
R
y \u2212 4y dx dy =
=
\u222b\u222b
R
\u22123y dx dy = \u22123
\u222b\u222b
R
y dx dy = \u22123K
o que prova o pretendido.
RESOLUC¸A\u2dcO DO EXERC´ICIO 56)
Figura 10: Regia\u2dco R
A regia\u2dco R e´ simplesmente conexa. A fronteira de R e´ uma curva C , orientada no sentido positivo,
simples, fechada e seccionalmente de classe C1 e regular.
As func¸o\u2dces M(x, y) = \u2212y2 e N(x, y) = \u2212x2 sa\u2dco func¸o\u2dces polinomiais, logo de classe C1 em R2 .
Pelo Teorema de Green,\u222b
C
\u2212y2 dx\u2212 x2 dy =
\u222b\u222b
R
(
\u2202
\u2202 x
(\u2212x2)\u2212 \u2202
\u2202 y
(\u2212y2)) dx dy = \u222b\u222b
R
\u22122x+ 2y dx dy =
=
\u222b\u222b
R
2 (y \u2212 x) dx dy = 2
\u222b\u222b
R
y \u2212 x dx dy .
Por outro lado, o volume de E e´ dado por
V (E) =
\u222b\u222b
R
y \u2212 x dx dy .
Enta\u2dco
K =
\u222b
C
\u2212y2 dx\u2212 x2 dy = 2V (E) .
Consequentemente
V (E) =
K
2
.
10
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
4 Integral de Superf´\u131cie
RESOLUC¸A\u2dcO DO EXERC´ICIO 64 (a)
A porc¸a\u2dco do parabolo´ide situada acima do plano XOY e´ a superf´\u131cie S definida por
S =
{
(x, y, z) \u2208 R3 : z = 9\u2212 x2 \u2212 y2 e z \u2265 0} = {(x, y, z) \u2208 R3 : z = 9\u2212 x2 \u2212 y2 e x2 + y2 \u2264 9} .
\u2022 Definic¸a\u2dco da orientac¸a\u2dco de S
Pretende-se orientar S com a normal unita´ria, n\u2c6 , dirigida para cima isto e´ com a componente segundo k\u2c6
positiva.
Figura 11: Superf´\u131cie S
Seja H(x, y, z) = x2 + y2 + z \u2212 9 . Enta\u2dco \u2207H(x, y, z) = (2x, 2y, 1) e o campo de vectores
n\u2c6(x, y, z) =
\u2207H(x, y, z)
\u2016\u2207H(x, y, z)\u2016 =
(2x, 2y, 1)\u221a
4x2 + 4y2 + 1
determina em S a orientac¸a\u2dco pretendida.
\u2022 Ca´lculo de ~F · n\u2c6
~F · n\u2c6 = (y,\u2212x, 8) · (2x, 2y, 1)\u221a
4x2 + 4y2 + 1
=
2xy \u2212 2xy + 8\u221a
4x2 + 4y2 + 1
=
8\u221a
4x2 + 4y2 + 1
.
\u2022 Ca´lculo de \u222b\u222bS ~F d~S \u222b\u222b
S
~F d~S =
\u222b\u222b
S
~F · n\u2c6 dS =
\u222b\u222b
S
8\u221a
4x2 + 4y2 + 1
dS .
A superf´\u131cie S e´ o gra´fico da func¸a\u2dco g(x, y) = 9\u2212x2\u2212y2 definida na regia\u2dco R = {(x, y) \u2208 R2 : x2 + y2 \u2264 9} .
Tem-se que gx(x, y) = 2x e gy(x, y) = 2y .
Enta\u2dco
\u222b\u222b
S
8\u221a
4x2 + 4y2 + 1
dS =
\u222b\u222b
R
8\u221a
4x2 + 4y2 + 1
\u221a
4x2 + 4y2 + 1 dx dy =
\u222b\u222b
R
8 dx dy = 8A(R) = 8pi 32 = 72pi .
Logo \u222b\u222b
S
~F d~S = 72pi .
11
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
RESOLUC¸A\u2dcO DO EXERC´ICIO 64 (b)
A porc¸a\u2dco do parabolo´ide z = x2 + y2 limitada pelas superf´\u131cies x = 1\u2212 y2 e x = y2 \u2212 1 , e´ a superf´\u131cie S
definida por
S =
{
(x, y, z) \u2208 R3 : z = x2 + y2 e y2 \u2212 1 \u2264 x \u2264 1\u2212 y2} = {(x, y, z) \u2208 R3 : z = x2 + y2 e (x, y) \u2208 R} ,
com
R =
{
(x, y) \u2208 R2 : y2 \u2212 1 \u2264 x \u2264 1\u2212 y2} .
\u2022 Definic¸a\u2dco da orientac¸a\u2dco de S
De acordo com as condic¸o\u2dces do enunciado, a componente da normal unita´ria, n\u2c6 , segundo k\u2c6 tem que ser
negativa.
Figura 12: Superf´\u131cie S
Seja H(x, y, z) = x2 + y2 \u2212 z . O vector
n\u2c6(x, y, z) =
\u2207H(x, y, z)
\u2016\u2207H(x, y, z) \u2016 =
(2x, 2y,\u22121)\u221a
4x2 + 4y2 + 1
define a orientac¸a\u2dco pretendida.
\u2022 Ca´lculo de ~F · n\u2c6
~F · n\u2c6 = (x,\u22121, 2x2) · (2x, 2y,\u22121)\u221a
4x2 + 4y2 + 1
=
2x2 \u2212 2y \u2212 2x2\u221a
4x2 + 4y2 + 1
=
\u22122y\u221a
4x2 + 4y2 + 1
.
\u2022 Ca´lculo de \u222b\u222bS ~F d~S \u222b\u222b
S
~F d~S =
\u222b\u222b
S
~F · n\u2c6 dS =
\u222b\u222b
S
\u22122y\u221a
4x2 + 4y2 + 1
dS .
A superf´\u131cie S e´ o gra´fico da func¸a\u2dco g(x, y) = x2 + y2 definida na regia\u2dco R .
Tem-se que gx(x, y) = 2x e gy(x, y) = 2y .
Enta\u2dco \u222b\u222b
S
\u22122y\u221a
4x2 + 4y2 + 1
dS =
\u222b\u222b
R
\u22122y\u221a
4x2 + 4y2 + 1
\u221a
4x2 + 4y2 + 1 dx dy =
\u222b\u222b
R
\u22122y dx dy .
12
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
Figura 13: Regia\u2dco R
A regia\u2dco R
e´ horizontalmente simples e pode ser definida por
R =
{
(x, y) \u2208 R2 : \u22121 \u2264 y \u2264 1 \u2227 y2 \u2212 1 \u2264 x \u2264 1\u2212 y2} .
Enta\u2dco\u222b\u222b
R
\u22122y dx dy =
\u222b 1
\u22121
\u222b 1\u2212y2
y2\u22121
\u22122y dx dy =
\u222b 1
\u22121
\u22122y (1\u2212 y2 \u2212 y2 + 1) dy = \u222b 1
\u22121
\u22122y (2\u2212 2y2) dy = 0 ,
pois os limites de integrac¸a\u2dco sa\u2dco sime´tricos e a func¸a\u2dco integranda, \u22122y (2\u2212 2y2) , e´ \u131´mpar.
Logo \u222b\u222b
S
~F d~S = 0 .
RESOLUC¸A\u2dcO DO EXERC´ICIO 64 (c)
\u2022 Definic¸a\u2dco da orientac¸a\u2dco de S
De acordo com as condic¸o\u2dces do enunciado, a componente da normal unita´ria, n\u2c6 , segundo k\u2c6 tem que ser
positiva.
Figura 14: Superf´\u131cie S
Seja H(x, y, z) = \u2212y + z \u2212 3 . O vector
n\u2c6(x, y, z) =
\u2207H(x, y, z)
\u2016\u2207H(x, y, z) \u2016 =
(0,\u22121, 1)\u221a
2
,
13
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
define, em S , a orientac¸a\u2dco pretendida.
Enta\u2dco \u222b\u222b
S
~F d~S =
\u222b\u222b
S
~F · n\u2c6 dS =
\u222b\u222b
S
(x,\u2212y z,\u22123 y) · (0,\u22121, 1)\u221a
2
dS =
\u222b\u222b
S
y z \u2212 3 y\u221a
2
dS =
=
\u222b\u222b
R
y (y + 3)\u2212 3 y\u221a
2
\u221a
1 + 0 + 1 dx dy =
\u222b\u222b
R
y2 dx dy ,
onde R e´ a regia\u2dco plana definida por
R =
{
(x, y) \u2208 R2 : x2 + y2 \u2264 1} .
Usando coordenadas polares, obte´m-se\u222b\u222b
R
y2 dx dy =
\u222b\u222b
R\u2032
r2 sin2 \u3b8 r dr d\u3b8 =
\u222b 2pi
0
\u222b 1
0
r3 sin2 \u3b8 r dr d\u3b8 =
\u222b 2pi
0
sin2 \u3b8 d\u3b8 .
\u222b 1
0
r3 dr =
=
\u222b 2pi
0
1
2
\u2212 1
2
cos (2\u3b8) d\u3b8 .
[
r4
4
]1
0
=
[
1
2
\u3b8 \u2212 1
4
sin (2\u3b8)
]2pi
0
.
1
4
= pi.
1
4
=
pi
4
.
Logo \u222b\u222b
S
~F d~S =
pi
4
.
RESOLUC¸A\u2dcO DO EXERC´ICIO 73 (a)
A superf´\u131cie S esta´ orientada com a normal exterior n\u2c6 e limita uma regia\u2dco so´lida simples E definida por
E =
{
(x, y, z) \u2208 R3 : x \u2265 0 , y \u2265 0 , z \u2265 0x+ y + z \u2264 3} .
Figura 15: Superf´\u131cie S
Pelo teorema da Diverge\u2c6ncia, tem-se\u222b\u222b
S
~F d~S =
\u222b\u222b
S
~F · n\u2c6 dS =
\u222b\u222b\u222b
E
div ~F dV .
Sendo ~F (x, y, z) = (xy, yz, zx) a
div ~F (x, y, z) =
\u2202
\u2202 x
(xy) +
\u2202
\u2202 y
(yz) +
\u2202
\u2202 z
(zx) = x+ y + z .
14
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
Enta\u2dco \u222b\u222b
S
~F · n\u2c6 dS =
\u222b\u222b\u222b
E
x+ y + z dx dy dz .
A regia\u2dco E e´ definida por
E =
{
(x, y, z) \u2208 R3 : (x, y) \u2208 R \u2227 0 \u2264 z \u2264 3\u2212 x\u2212 y} ,
com
R =
{
(x, y) \u2208 R2 : 0 \u2264 x \u2264 3 \u2227 0 \u2264 y \u2264 3\u2212 x} .
Figura 16: Regia\u2dco R
Tem-se enta\u2dco\u222b\u222b
S
~F · n\u2c6 dS =
\u222b\u222b
R
\u222b 3\u2212x\u2212y
0
x+ y + z dz dx dy =
\u222b\u222b
R