Cálculo III - Lista de Exercícios 12 - UFF
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Cálculo III - Lista de Exercícios 12 - UFF


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Universidade Federal Fluminense
Instituto de Matema´tica e Estat\u131´stica
Departamento de Matema´tica Aplicada
Ca´lculo 3A \u2013 Lista 12
Exerc´\u131cio 1: Verifique o teorema de Gauss para o campo vetorial
\u2212\u2192
F (x, y, z) = = (x, y, z) no so´lido
W limitado pelas superf´\u131cies z = x2 + y2 e z = 4.
Soluc¸a\u2dco: O esboc¸o do so´lido W esta´ representado na figura que se segue.
x
y
z
S1
S2
D 2
2
4
\u2212\u2192n1
\u2212\u2192n2
Vemos que \u2202W = S1 \u222a S2, orientada positivamente. Logo,\u222b\u222b
S
\u2212\u2192
F · ~n dS =
\u222b\u222b
S1
\u2212\u2192
F · ~n1 dS +
\u222b\u222b
S2
\u2212\u2192
F · ~n2 dS .
Ca´lculo de
\u222b\u222b
S1
\u2212\u2192
F · ~n1 dS
Temos S1 : z = 4 = f(x, y), com (x, y) \u2208 D : x2 + y2 \u2264 4, ~n1 = ~k e
dS =
\u221a
1 + (fx)2 + (fy)2 dx dy =
\u221a
1 + 0 + 0 dx dy = dx dy .
Enta\u2dco: \u222b\u222b
S1
\u2212\u2192
F · ~n1 dS =
\u222b\u222b
D
(x, y, 4) · (0, 0, 1) dx dy = 4A(D) = 4\u3c0 · 22 = 16\u3c0.
Ca´lculo de
\u222b\u222b
S2
\u2212\u2192
F · ~n2 dS
Ca´lculo 3A Lista 12 180
Temos S2 : z = x
2 + y2 = g(x, y), com (x, y) \u2208 D : x2 + y2 \u2264 4. Um vetor normal a S2 e´ dado por\u2212\u2192
N = (\u2212gx,\u2212gy, 1) = (\u22122x,\u22122y, 1) que esta´ voltado para cima. Como ~n2 aponta para baixo enta\u2dco
~n2 =
(2x,2y,\u22121)\u221a
1+4x2+4y2
. Temos dS = \u2016\u2212\u2192N \u2016 dx dy =
\u221a
1 + 4x2 + 4y2 dx dy. Enta\u2dco:
\u222b\u222b
S2
\u2212\u2192
F · ~n2 dS =
\u222b\u222b
D
(
x, y, x2 + y2
) · (2x, 2y,\u22121) dx dy =
=
\u222b\u222b
D
(
2x2 + 2y2 \u2212 x2 \u2212 y2) dx dy = \u222b\u222b
D
(
x2 + y2
)
dx dy .
Passando para coordenadas polares, temos:\u222b\u222b
S2
\u2212\u2192
F · ~n2 dS =
\u222b\u222b
Dr\u3b8
r2 · r dr d\u3b8 =
\u222b
2
0
r3
\u222b
2pi
0
d\u3b8 dr = 2\u3c0
[
r4
4
]2
0
= 8\u3c0 .
Enta\u2dco: \u222b\u222b
S
\u2212\u2192
F · ~n dS = 16\u3c0 + 8\u3c0 = 24\u3c0 .
Por outro lado: \u222b\u222b\u222b
W
div
\u2212\u2192
F dV = 3
\u222b\u222b\u222b
W
dV = 3
\u222b\u222b
D
\u222b
4
x2+y2
dz dx dy =
= 3
\u222b\u222b
D
(
4\u2212 x2 \u2212 y2) dx dy = 3 \u222b\u222b
Dr\u3b8
(
4\u2212 r2) r dr d\u3b8 =
= 3
\u222b
2
0
(
4r \u2212 r3) \u222b 2pi
0
d\u3b8 dr = 6\u3c0
[
2r2 \u2212 r
4
4
]2
0
= 6\u3c0(8\u2212 4) = 24\u3c0 .
Exerc´\u131cio 2: Calcule o fluxo do campo vetorial
\u2212\u2192
F atrave´s da superf´\u131cie aberta S, onde
\u2212\u2192
F (x, y, z) =
(xy2 + ey)
\u2212\u2192
i + (yz2 + sen2 x)
\u2212\u2192
j + (5 + zx2)
\u2212\u2192
k e S : z =
\u221a
4\u2212 x2 \u2212 y2, z \u2265 0, com \u2212\u2192n tendo
componente z positiva.
Soluc¸a\u2dco: O esboc¸o da superf´\u131cie aberta S esta´ representado na figura que se segue.
UFF IME - GMA
Ca´lculo 3A Lista 12 181
x
y
z
S
\u2212\u2192n
2
2
2
Para aplicar o Teorema de Gauss, devemos considerar a superf´\u131cie fechada S = S \u222a S1, onde S1 e´
dada por S1 : z = 0, (x, y) \u2208 D : x2 + y2 \u2264 4 com \u2212\u2192n1 = \u2212\u2212\u2192k e tambe´m temos que:
dS =
\u221a
1 + (zx)
2 + (zy)
2
dxdy =
\u221a
1 + 0 + 0 dxdy = dxdy .
Seja W o so´lido limitado por S. Logo, \u2202W = S.
x
y
z
S1
S
W
\u2212\u2192n1
2
2
2
Pelo Teorema de Gauss, temos\u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS +
\u222b\u222b
S1
\u2212\u2192
F · \u2212\u2192n1 dS =
\u222b\u222b\u222b
W
div
\u2212\u2192
F dV =
=
\u222b\u222b\u222b
W
(
y2 + z2 + x2
)
dV .
Passando para coordenadas esfe´ricas, temos:\u222b\u222b\u222b
W
(
x2 + y2 + z2
)
dV =
\u222b\u222b\u222b
W\u3c1\u3c6\u3b8
(
\u3c12
)
\u3c12 sen\u3c6 d\u3c1d\u3c6d\u3b8 =
=
\u222b pi/2
0
sen \u3c6
\u222b
2
0
\u3c14
\u222b
2pi
0
d\u3b8d\u3c1d\u3c6 = 2\u3c0
\u222b pi/2
0
sen \u3c6
[
\u3c15
5
]2
0
d\u3c6 =
=
64\u3c0
5
[
\u2212 cos \u3c6
]pi/2
0
=
64\u3c0
5
.
UFF IME - GMA
Ca´lculo 3A Lista 12 182
Ca´lculo de
\u222b\u222b
S1
\u2212\u2192
F · \u2212\u2192n1 dS
Temos: \u222b\u222b
S1
\u2212\u2192
F · \u2212\u2192n1 dS =
=
\u222b\u222b
S1
(
xy2 + ey, 0 + sen2 x, 5 + 0
) · (0, 0,\u22121) dS =
= \u22125A (S1) = \u22125 (\u3c022) = \u221220\u3c0 .
Logo, \u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS = 64\u3c0
5
+ 20\u3c0 =
164\u3c0
5
.
Exerc´\u131cio 3: Calcule
\u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS, onde
\u2212\u2192
F (x, y, z) = x
\u2212\u2192
i +
(\u2212 2y + ex cos z)\u2212\u2192j + (z + x2)\u2212\u2192k
e S e´ definida por z = 9 \u2212 (x2 + y2), 0 \u2264 z \u2264 5; z = 5, 1 \u2264 x2 + y2 \u2264 4 e z = 8 \u2212 3(x2 + y2),
x2 + y2 \u2264 1, com ~n exterior a S.
Soluc¸a\u2dco: A superf´\u131cie S na\u2dco e´ fechada e pode ser visualizada na figura que se segue.
x
y
z
\u2212\u2192n
\u2212\u2192n
\u2212\u2192n
3
3
5
8
Como div
\u2212\u2192
F = 1 \u2212 2 + 1 = 0, vamos usar o teorema de Gauss. Para isso, e´ necessa´rio fechar S
atrave´s da superf´\u131cie S1, porc¸a\u2dco do plano z = 0 com x
2 + y2 \u2264 9, orientada com \u2212\u2192n1 = \u2212\u2212\u2192k .
UFF IME - GMA
Ca´lculo 3A Lista 12 183
x
y
z
\u2212\u2192n1
S1
3
3
Seja W o so´lido limitado por S e S1. Como \u2202W = S \u222a S1 esta´ orientada positivamente, podemos
aplicar o teorema de Gauss. Temos enta\u2dco,\u222b\u222b
\u2202W
\u2212\u2192
F · \u2212\u2192n dS =
\u222b\u222b\u222b
W
div
\u2212\u2192
F dxdydz =
\u222b\u222b\u222b
W
0 dxdydz = 0
ou \u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS +
\u222b\u222b
S1
\u2212\u2192
F · \u2212\u2192n1 dS = 0 .
Mas \u222b\u222b
S1
\u2212\u2192
F · \u2212\u2192n1 dS =
=
\u222b\u222b
S1
(
x, \u22122y + ex cos 0, 0 + x2) · (0, 0, \u22121) dS =
= \u2212
\u222b\u222b
S1
x2 dS
onde S1 e´ dada por z = 0, com (x, y) \u2208 D : x2 + y2 \u2264 9, e, \u2212\u2192n1 = \u2212\u2212\u2192k . Logo:
dS =
\u221a
1 +
(
\u2202z
\u2202x
)2
+
(
\u2202z
\u2202y
)2
dxdy =
\u221a
1 + 0 + 0 dxdy = dxdy .
Enta\u2dco: \u222b\u222b
S1
\u2212\u2192
F · \u2212\u2192n1 dS = \u2212
\u222b\u222b
D
x2 dxdy (em coordenadas polares) =
= \u2212
\u222b
2pi
0
\u222b
3
0
r3 cos2 \u3b8 drd\u3b8 = \u22123
4
4
\u222b
2pi
0
cos2 \u3b8 d\u3b8 =
= \u22123
4
4
· 1
2
· 2\u3c0 = \u221281\u3c0
4
.
Logo: \u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS = 81\u3c0
4
.
UFF IME - GMA
Ca´lculo 3A Lista 12 184
Exerc´\u131cio 4: Calcule o fluxo do campo
\u2212\u2192
F (x, y, z) =
(
x3
3
+ y,
y3
3
,
z3
3
+ 2
)
atrave´s da superf´\u131cie S
do so´lido W definido por
W =
{
(x, y, z) \u2208 R3; x2 + y2 + z2 \u2265 1 , x2 + y2 + (z \u2212 2)2 \u2264 4 ,
z \u2265
\u221a
x2 + y2
}
com campo de vetores normais a S apontando para fora de W .
Soluc¸a\u2dco: A figura que se segue mostra o so´lido W .
x
y
z
\u2212\u2192n
\u2212\u2192n
\u2212\u2192n
W
1
2
4
Como estamos nas condic¸o\u2dces do teorema de Gauss, temos:\u222b\u222b
S=\u2202W
\u2212\u2192
F · \u2212\u2192n dS =
\u222b\u222b\u222b
W
div
\u2212\u2192
F dxdydz =
=
\u222b\u222b\u222b
W
(
x2 + y2 + z2
)
dxdydz =
=
\u222b\u222b\u222b
W\u3c1\u3c6\u3b8
\u3c12 · \u3c12 sen\u3c6 d\u3c1d\u3c6d\u3b8 =
\u222b\u222b\u222b
W\u3c1\u3c6\u3b8
\u3c14 sen \u3c6 d\u3c1d\u3c6d\u3b8
onde
W\u3c1\u3c6\u3b8 =
{
(\u3c1, \u3c6, \u3b8) \u2208 R3; 0 \u2264 \u3b8 \u2264 2\u3c0 , 0 \u2264 \u3c6 \u2264 \u3c0
4
, 1 \u2264 \u3c1 \u2264 4 cos\u3c6
}
.
UFF IME - GMA
Ca´lculo 3A Lista 12 185
Enta\u2dco: \u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS =
\u222b
2pi
0
\u222b pi/4
0
\u222b
4 cos\u3c6
1
\u3c14 sen \u3c6 d\u3c1d\u3c6d\u3b8 =
=
\u222b
2pi
0
\u222b pi/4
0
sen \u3c6
[
\u3c15
5
]4 cos\u3c6
1
d\u3c6d\u3b8 =
=
1
5
\u222b
2pi
0
\u222b pi/4
0
(
45 cos5 \u3c6 sen\u3c6\u2212 sen\u3c6) d\u3c6d\u3b8 =
=
1
5
\u222b
2pi
0
[
\u22124
5
6
cos6 \u3c6+ cos \u3c6
]pi/4
0
d\u3b8 =
=
1
5
(
\u22124
5
6
(\u221a
2
6
)6
+
\u221a
2
2
+
45
6
\u2212 1
)\u222b
2pi
0
d\u3b8 =
=
2\u3c0
5
(
45
6
· 7
8
+
\u221a
2
2
\u2212 1
)
=
2\u3c0
5
(
43 · 7
3
+
\u221a
2
2
\u2212 1
)
=
=
2\u3c0
5
(
445
3
+
\u221a
2
2
)
=
\u3c0
15
(
890 + 3
\u221a
2
)
.
Exerc´\u131cio 5: Seja T o tetraedro de ve´rtices O = (0, 0, 0), A = (2, 0, 0), B = (0, 6, 0) e C =
(0, 0, 2). Sejam S a superf´\u131cie lateral de T constitu´\u131da pelas faces de T que na\u2dco esta\u2dco no plano xy
e
\u2212\u2192
F (x, y, z) = (3y + z, x+ 4z, 2y + x) um campo vetorial de R3. Calcule
\u222b\u222b
S
rot
\u2212\u2192
F · \u2212\u2192n dS, com a
normal exterior a S.
Soluc¸a\u2dco: A figura que se segue mostra o tetraedro T .
x y
z
A
B
C
\u2212\u2192n
2
2
6
Notemos que \u2202T = S \u222a S1 onde S1 e´ a porc¸a\u2dco do plano z = 0, limitada pelo tria\u2c6ngulo de ve´rtices
O, A e B. Considere \u2212\u2192n1 o vetor unita´rio normal a S1 igual a \u2212\u2212\u2192k .
Como \u2202T esta´ orientada positivamente, podemos aplicar o teorema de Gauss. Temos:\u222b\u222b
\u2202T
rot
\u2212\u2192
F · \u2212\u2192n dS =
\u222b\u222b\u222b
W
div rot
\u2212\u2192
F dxdydz = 0
UFF IME - GMA
Ca´lculo 3A Lista 12 186
x
y
z
A
B
\u2212\u2192n1
S1
2
6
pois div rot
\u2212\u2192
F = 0 (conforme observac¸a\u2dco importante) ou\u222b\u222b
S
rot
\u2212\u2192
F · \u2212\u2192n dS +
\u222b\ufffd\ufffd\ufffd
S1
rot
\u2212\u2192
F · \u2212\u2192n1 dS = 0 .
Temos:
rot
\u2212\u2192
F =
\u2223\u2223\u2223\u2223\u2223\u2223\u2223\u2223\u2223\u2223
\u2212\u2192
i
\u2212\u2192
j
\u2212\u2192
k
\u2202
\u2202x
\u2202
\u2202y
\u2202
\u2202z
3y + z x+ 4z 2y + x
\u2223\u2223\u2223\u2223\u2223\u2223\u2223\u2223\u2223\u2223
= (2\u2212 4, 1\u2212 1, 1\u2212 3) = (\u22122, 0, \u22122) .
Logo: \u222b\u222b
S1
rot
\u2212\u2192
F · \u2212\u2192n1 dS =
\u222b\u222b
S1
(\u22122, 0,\u22122) · (0, 0,\u22121) dS =
= 2
\u222b\u222b
S1
dS = 2A(S1) = 2 · 1
2
· 2 · 6 = 12 .
Portanto, \u222b\u222b
S
rot
\u2212\u2192
F · \u2212\u2192n dS = \u221212 .
Exerc´\u131cio 6: Seja a superf´\u131cie