exercicios resolvidos integrais múltiplas

exercicios resolvidos integrais múltiplas


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Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
1 Integral duplo
RESOLUC¸A\u2dcO DO EXERC´ICIO 6 a)
O dom\u131´nio de integrac¸a\u2dco R e´ definido por
R =
{
(x, y) \u2208 R2 : 0 \u2264 x \u2264 1 \u2227 2x \u2264 y \u2264 2}
e esta´ representado na figura
Figura 1: Esboc¸o da regia\u2dco R
Para inverter a ordem de integrac¸a\u2dco temos que definir R como uma regia\u2dco horizontalmente simples, isto e´,
com as condic¸o\u2dces
0 \u2264 y \u2264 2 \u2227 0 \u2264 x \u2264 y
2
.
Podemos enta\u2dco escrever\u222b 1
0
\u222b 2
2x
ey
2
dy dx =
\u222b 2
0
\u222b y/2
0
ey
2
dx dy =
\u222b 2
0
ey
2
\u222b y/2
0
1 dx dy =
\u222b 2
0
ey
2
[x]y/20 dy =
\u222b 2
0
y
2
ey
2
dy =
1
4
\u222b 2
0
2y ey
2
dy =
1
4
[
ey
2
]2
0
=
1
4
(e4 \u2212 1) .
RESOLUC¸A\u2dcO DO EXERC´ICIO 6 b)
A partir dos limites de integrac¸a\u2dco do integral dado
\u222b 1
0
\u222b 1
u
\u221a
1\u2212 v2 dv du , podemos definir o dom\u131´nio de
integrac¸a\u2dco R com as seguintes condic¸o\u2dces
0 \u2264 u \u2264 1 \u2227 u \u2264 v \u2264 1 .
e definir R graficamente
Para podermos inverter a ordem de integrac¸a\u2dco temos que definir R como uma regia\u2dco horizontalmente
simples, isto e´, com as condic¸o\u2dces
0 \u2264 v \u2264 1 \u2227 0 \u2264 u \u2264 v .
Podemos enta\u2dco escrever\u222b 1
0
\u222b 1
u
\u221a
1\u2212 v2 dv du =
\u222b 1
0
\u222b v
0
\u221a
1\u2212 v2 du dv =
\u222b 1
0
\u221a
1\u2212 v2
\u222b v
0
1 du dv =
\u222b 1
0
\u221a
1\u2212 v2 [u]v0 dv =
1
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
Figura 2: Esboc¸o da regia\u2dco R
\u222b 1
0
v
\u221a
1\u2212 v2 dv = \u22121
2
\u222b 1
0
\u22122v (1\u2212 v2)1/2 dv = \u22121
2
[(
1\u2212 v2)3/2
3/2
]1
0
= \u22121
3
(0\u2212 1) = 1
3
.
RESOLUC¸A\u2dcO DO EXERC´ICIO 6 c)
O dom\u131´nio de integrac¸a\u2dco R e´ definido por
R =
{
(x, y) \u2208 R2 : 0 \u2264 y \u2264 3 \u2227 y2 \u2264 x \u2264 9}
e esta´ representado na figura
Figura 3: Esboc¸o da regia\u2dco R
Para inverter a ordem de integrac¸a\u2dco temos que definir R como uma regia\u2dco verticalmente simples, isto e´,
com as condic¸o\u2dces
0 \u2264 x \u2264 9 \u2227 0 \u2264 y \u2264 \u221ax .
Podemos enta\u2dco escrever\u222b 3
0
\u222b 9
y2
y cos x2 dx dy =
\u222b 9
0
\u222b \u221ax
0
y cos x2 dy dx =
\u222b 9
0
cos x2
\u222b \u221ax
0
y dy dx =
\u222b 9
0
cos x2
[
y2
2
]\u221ax
0
dx =
\u222b 9
0
x
2
cos x2 dx =
1
4
\u222b 9
0
2x cos x2 dx =
1
4
[
sin x2
]9
0
=
1
4
sin 81 .
2
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
RESOLUC¸A\u2dcO DO EXERC´ICIO 9 c)\u222b 1
0
\u222b \u221a1\u2212x2
0
e
\u221a
x2 + y2 dy dx =
\u222b\u222b
D
e
\u221a
x2 + y2 dx dy onde
D =
{
(x, y) \u2208 R2 : 0 \u2264 x \u2264 1 \u2227 0 \u2264 y \u2264
\u221a
1\u2212 x2
}
.
D e´ representado graficamente por
Figura 4: Esboc¸o da regia\u2dco D
D pode ser definido de modo equivalente pelas condic¸o\u2dces
x2 + y2 \u2264 1 \u2227 x \u2265 0 \u2227 y \u2265 0 .
Fazendo a mudanc¸a para coordenadas polares, de acordo com as equac¸o\u2dces acima referidas, tem-se:
x2 + y2 \u2264 1 \u2212\u2192 r2 < 1 \u21d4 r < 1
x \u2265 0 \u2227 y \u2265 0 \u2212\u2192 0 \u2264 \u3b8 \u2264 pi2
,
e portanto, de acordo com a fo´rmula de mudanc¸a para coordenadas polares, tem-se:\u222b 1
0
\u222b \u221a1\u2212x2
0
e
\u221a
x2 + y2 dy dx =
\u222b\u222b
D\u2032
er r dr d\u3b8 ,
onde
D\u2032 =
{
(r, \u3b8) \u2208 R2 : 0 \u2264 r \u2264 1 \u2227 0 \u2264 \u3b8 \u2264 pi
2
}
.
Donde\u222b 1
0
\u222b \u221a1\u2212x2
0
e
\u221a
x2 + y2 dy dx =
\u222b pi/2
0
\u222b 1
0
er r dr d\u3b8 =
\u222b pi/2
0
[
er r \u2212
\u222b
er 1 dr
]1
0
d\u3b8 =
\u222b pi/2
0
[er r \u2212 er]10 d\u3b8 =
=
\u222b pi/2
0
(e\u2212 e)\u2212 (0\u2212 1) d\u3b8 =
\u222b pi/2
0
1 d\u3b8 =
pi
2
.
3
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
RESOLUC¸A\u2dcO DO EXERC´ICIO 19 b)
Seja
E =
{
(x, y, z) \u2208 R3 : y \u2265 x2 \u2227 x \u2265 y2 \u2227 0 \u2264 z \u2264 3} .
O so´lido E e´ limitado superiormente pela superf´\u131cie de equac¸a\u2dco z = 3 e inferiormente pela superf´\u131cie de
equac¸a\u2dco z = 0 .
A projecc¸a\u2dco, R , de E no plano XOY e´ dada por
R =
{
(x, y) \u2208 R2 : y \u2265 x2 \u2227 x \u2265 y2} .
Figura 5: R:projecc¸a\u2dco de E no plano XOY
Enta\u2dco o volume de E e´ dado pelo integral duplo
V (E) =
\u222b\u222b
R
3\u2212 0 dx dy =
\u222b\u222b
R
3 dx dy
A regia\u2dco R e´, simultaneamente, verticalmente e horizontalmente simples. Podemos, por exemplo, usar a
seguinte definic¸a\u2dco
R =
{
(x, y) \u2208 R2 : 0 \u2264 x \u2264 1 \u2227 x2 \u2264 y \u2264 \u221ax} ,
para escrever o integral estabelecido anteriormente, na forma iterada.
Tem-se enta\u2dco
V (E) =
\u222b 1
0
\u222b \u221ax
x2
3 dy dx = 3
\u222b 1
0
\u221a
x\u2212 x2 dx =
= 3
[
x3/2
3/2
\u2212 x
3
3
]1
0
= 3
[
2
3
\u2212 1
3
]
= 1 .
4
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
2 Integral Triplo
RESOLUC¸A\u2dcO DO EXERC´ICIO 22 a)
De
x+ y + z \u2264 1 \u2227 x \u2265 0 \u2227 y \u2265 0 \u2227 z \u2265 0
obte´m-se 0 \u2264 z \u2264 1\u2212 x\u2212 y ; assim a regia\u2dco E pode ser descrita como:
E =
{
(x, y, z) \u2208 R3 : (x, y) \u2208 R \u2227 0 \u2264 z \u2264 1\u2212 x\u2212 y} ,
com R = projXOY V definida por
R =
{
(x, y) \u2208 R2 : x+ y \u2264 1 \u2227 x \u2265 0 \u2227 y \u2265 0} = {(x, y) \u2208 R2 : 0 \u2264 x \u2264 1 \u2227 0 \u2264 y \u2264 1\u2212 x} .
Enta\u2dco \u222b\u222b\u222b
E
1
(x+ y + z + 1)3
dx dy dz =
\u222b\u222b
R
\u222b 1\u2212x\u2212y
0
1
(x+ y + z + 1)3
dz dxdy =
\u222b\u222b
R
[
(x+ y + z + 1)\u22122
\u22122
]z=1\u2212x\u2212y
z=0
dx dy =
\u222b\u222b
R
1
2
(x+ y + 1)\u22122 \u2212 1
8
dx dy =
=
\u222b 1
0
\u222b 1\u2212x
0
1
2
(x+y+1)\u22122\u2212 1
8
dy dx =
\u222b 1
0
[
\u22121
2
1
x+ y + 1
\u2212 1
8
y
]y=1\u2212x
y=0
dx =
\u222b 1
0
\u22121
4
\u2212 1
8
(x\u22121)+ 1
2
1
x+ 1
dx =
=
[
1
8
(1\u2212 x)2
2
\u2212 1
4
x+
1
2
ln |x+ 1 |
]1
0
= \u22121
4
+
1
2
ln 2\u2212 1
16
=
1
2
ln 2\u2212 5
16
.
RESOLUC¸A\u2dcO DO EXERC´ICIO 24 b)
Atendendo a que a regia\u2dco D e´ limitada por duas superf´\u131cies esfe´ricas centradas na origem, o sistema de
coordenadas mais conveniente para calcular o integral dado, e´ o sistema de coordenadas esfe´ricas.
Tem-se, enta\u2dco \u222b\u222b\u222b
E
1
(x2 + y2 + z2)
3
2
dV =
\u222b\u222b\u222b
E\u2032
1
(\u3c12)
3
2
\u3c12 sin \u3c6d\u3c1 d\u3c6 d\u3b8 ,
com E\u2032 definido por
E\u2032 = {(\u3c1, \u3b8, \u3c6) : 0 \u2264 \u3b8 \u2264 2pi \u2227 0 \u2264 \u3c6 \u2264 pi \u2227 2 \u2264 \u3c1 \u2264 3} .
Podemos, enta\u2dco escrever\u222b\u222b\u222b
D
1
(x2 + y2 + z2)
3
2
dV =
\u222b 2pi
0
\u222b pi
0
\u222b 3
2
1
\u3c1
sin \u3c6d\u3c1 d\u3c6 d\u3b8 =
\u222b 2pi
0
1 d\u3b8 .
\u222b pi
0
sin \u3c6d\u3c6 .
\u222b 3
2
1
\u3c1
d\u3c1 =
= 2pi . [\u2212 cos \u3c6]pi0 . [ln \u3c1]32 = 2pi . 2 . ln
3
2
= 4pi ln
3
2
.
5
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
RESOLUC¸A\u2dcO DO EXERC´ICIO 28 a)
Seja E a regia\u2dco de R3 definida pela condic¸a\u2dco
x2 + y2 + z2 \u2264 r2 , r > 0 .
O volume de E e´ dado pelo integral
V (E) =
\u222b\u222b\u222b
E
1 dV .
Fazendo uma mudanc¸a de varia´veis para coordenadas esfe´ricas, obte´m-se,
V (E) =
\u222b\u222b\u222b
E
1 dV =
\u222b\u222b\u222b
E\u2032
\u3c12 sin \u3c6d\u3c1 d\u3c6 d\u3b8 ,
com E\u2032 definido por
E\u2032 = {(\u3c1, \u3b8, \u3c6) : 0 \u2264 \u3b8 \u2264 2pi \u2227 0 \u2264 \u3c6 \u2264 pi \u2227 0 \u2264 \u3c1 \u2264 r} .
Enta\u2dco,
V (E) =
\u222b 2pi
0
\u222b pi
0
\u222b r
0
\u3c12 sin \u3c6d\u3c1 d\u3c6 d\u3b8 =
\u222b 2pi
0
1 d\u3b8 .
\u222b pi
0
sin \u3c6d\u3c6 .
\u222b r
0
\u3c12 d\u3c1 =
= 2pi . [\u2212 cos \u3c6]pi0 .
[
\u3c13
3
]r
0
= 2pi . 2 .
r3
3
=
4
3
pi r3 .
6
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
3 Integral Curvil´\u131neo
RESOLUC¸A\u2dcO DO EXERC´ICIO 43 c)
Figura 6: Curva C
A curva C pode ser definida pelas equac¸o\u2dces parame´tricas\uf8f1\uf8f4\uf8f2\uf8f4\uf8f3
x = t
y = t2
z = t4
, t \u2208 [ 0, 1 ] ,
e portanto \uf8f1\uf8f4\uf8f2\uf8f4\uf8f3
x\u2032(t) = 1
y\u2032(t) = 2t
z\u2032(t) = 4t3
.
Enta\u2dco \u222b
C
~F . d~r =
\u222b 1
0
~F (x(t), y(t), z(t)) .
(
x\u2032(t), y\u2032(t), z\u2032(t)
)
dt =
=
\u222b 1
0
(
t2 + t4, t+ t4, t+ t2
)
.
(
1, 2t, 4t3
)
=
\u222b 1
0
t2 + t4 + 2t2 + 2t5, 4t4 + 4t5 dt =
=
\u222b 1
0
3t2 + 5t4 + 6t5 dt =
[
t3 + t5 + t6
]1
0
= 1 + 1 + 1 = 3 .
RESOLUC¸A\u2dcO DO EXERC´ICIO 43 d)
Figura 7: Curva C
7
Ana´lise Matema´tica III \u2013 MIEM e LEGI Exerc´\u131cios Resolvidos
Seja C1 o segmento de recta de extremos (0, 0, 0) e (1, 0, 0) e C2 a curva parametrizada por\uf8f1\uf8f4\uf8f2\uf8f4\uf8f3
x = cos t
y = sin t
z = tpi
, 0 \u2264 t \u2264 pi .
Como C = C1 + C2 ,
\u222b
C
~F . d~r =
\u222b
C1
~F . d~r +
\u222b
C2
~F . d~r .
A curva C1 pode ser representada pelas equac¸o\u2dces parame´tricas
\uf8f1\uf8f4\uf8f2\uf8f4\uf8f3
x = t
y = 0
z = 0
, 0 \u2264 t \u2264 1 .
\uf8f1\uf8f4\uf8f2\uf8f4\uf8f3
x = t
y = 0
z = 0
=\u21d2
\uf8f1\uf8f4\uf8f2\uf8f4\uf8f3
x\u2032 = 1
y\u2032 = 0
z\u2032 = 0
e \uf8f1\uf8f4\uf8f2\uf8f4\uf8f3
x = cos t
y = sin t
z = tpi
=\u21d2
\uf8f1\uf8f4\uf8f2\uf8f4\uf8f3
x\u2032 = \u2212 sin t
y\u2032 = cos t
z\u2032 = 1pi
.
Enta\u2dco\u222b
C
~F ·d~r =
\u222b
C1