Cálculo III - Lista de Exercícios 11 - UFF
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Cálculo III - Lista de Exercícios 11 - 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 11
Exerc´\u131cio 1: Seja o campo vetorial
\u2212\u2192
F (x, y, z) = (x\u2212 y, x+ y, z). Calcule o fluxo de \u2212\u2192F atrave´s de
S, orientada com \u2212\u2192n exterior a S se:
a) S : x2 + y2 = a2 com a > 0 e 0 \u2264 z \u2264 h;
b) S : x2 + y2 + z2 = a2 com a > 0.
Soluc¸a\u2dco:
a) O esboc¸o de S esta´ representado na figura que se segue.
x
y
z
S
\u2212\u2192n
a
a
h
Da teoria, temos no caso do cilindro x2 + y2 = a2, que o unita´rio normal exterior e´ da forma
\u2212\u2192n = (x, y, 0)
a
. Enta\u2dco o fluxo \u3c6 e´ dado por:
\u3c6 =
\u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS =
\u222b\u222b
S
(x\u2212 y, x+ y, z) · (x, y, 0)
a
dS =
=
1
a
\u222b\u222b
S
(
x2 \u2212 xy + xy + y2) dS = 1
a
\u222b\u222b
S
(
x2 + y2
)
dS =
=
a2
a
\u222b\u222b
S
dS = aA(S) = a(2piah) = 2pia2h .
Ca´lculo 3A Lista 11 165
b) O esboc¸o de S esta´ representado na figura a seguir.
x
y
z
S
\u2212\u2192n
a
a
a
Da teoria, temos no caso da esfera x2 + y2 + z2 = a2, que o unita´rio normal exterior e´ dado por
\u2212\u2192n = (x, y, z)
a
. Enta\u2dco fluxo e´ dado por:
\u3c6 =
\u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS =
\u222b\u222b
S
(x\u2212 y, x+ y, z) · (x, y, z)
a
dS =
=
1
a
\u222b\u222b
S
(
x2 \u2212 xy + xy + y2 + z2) dS =
=
1
a
\u222b\u222b
S
(
x2 + y2 + z2
)
dS =
a2
a
\u222b\u222b
S
dS = aA(S) =
= a (4pia2) = 4pia3 .
Exerc´\u131cio 2: Calcular o fluxo do campo vetorial de
\u2212\u2192
F = (x\u2212 y \u2212 4)\u2212\u2192i + y\u2212\u2192j + z\u2212\u2192k
atrave´s da semi-esfera superior de x2 + y2 + z2 = 1, com campo de vetores normais \u2212\u2192n tal que
\u2212\u2192n · \u2212\u2192k > 0.
Soluc¸a\u2dco: O esboc¸o de S esta´ representado na figura a seguir.
UFF IME - GMA
Ca´lculo 3A Lista 11 166
x
y
z
S
1
1
1
\u2212\u2192n
Como \u2212\u2192n · \u2212\u2192k > 0 enta\u2dco \u2212\u2192n aponta para cima e, portanto, \u2212\u2192n = = (x, y, z)
a
= (x, y, z) pois a = 1. O
fluxo e´ dado por: \u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS =
\u222b\u222b
S
(x\u2212 y \u2212 4, y, z) · (x, y, z) dS =
=
\u222b\u222b
S
(
x2 \u2212 xy \u2212 4x+ y2 + z2) dS =
=
\u222b\u222b
S
(
x2 + y2 + z2\ufe38 \ufe37\ufe37 \ufe38
= 1
\u2212xy \u2212 4x) dS = \u222b\u222b
S
(1\u2212 xy \u2212 4x) dS =
=
\u222b\u222b
S
dS \u2212
\u222b\u222b
S
(xy \u2212 4x) dS = A(S)\u2212
\u222b\u222b
S
(xy \u2212 4x) dS =
=
1
2
· 4pi · 12 \u2212
\u222b\u222b
S
(xy \u2212 4x) dS = 2pi \u2212
\u222b\u222b
S
(xy \u2212 4x) dS .
Ora, para calcular
\u222b\u222b
S
(xy \u2212 4x) dS devemos parametrizar S. Enta\u2dco temos que S : \u3d5(\u3c6, \u3b8) =
(sen \u3c6 cos \u3b8, sen \u3c6 sen \u3b8, cos\u3c6), com (\u3c6, \u3b8) \u2208 D :
{
0 \u2264 \u3c6 \u2264 pi/2
0 \u2264 \u3b8 \u2264 2pi . Tambe´m temos que dS =
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Ca´lculo 3A Lista 11 167
a2 sen \u3c6 d\u3c6d\u3b8 = sen\u3c6 d\u3c6d\u3b8. Logo:\u222b\u222b
S
(xy \u2212 4x) dS =
=
\u222b\u222b
D
(
sen2 \u3c6 sen \u3b8 cos \u3b8 \u2212 4 sen\u3c6 cos \u3b8) sen \u3c6 d\u3c6d\u3b8 =
=
\u222b\u222b
D
sen3 \u3c6 sen \u3b8 cos \u3b8 d\u3c6d\u3b8 \u2212 4
\u222b\u222b
D
sen2 \u3c6 cos \u3b8 d\u3c6d\u3b8 =
=
\u222b pi/2
0
sen3 \u3c6
\u222b
2pi
0
sen \u3b8 cos \u3b8 d\u3b8d\u3c6\u2212 4
\u222b pi/2
0
sen2 \u3c6
\u222b
2pi
0
cos \u3b8 d\u3b8d\u3c6 =
=
\u222b pi/2
0
sen3 \u3c6
[
sen2 \u3b8
2
]2pi
0
d\u3c6\u2212 4
\u222b pi/2
0
sen2 \u3c6
[
sen \u3b8
]2pi
0
d\u3b8 = 0 .
Portanto: \u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS = 2pi .
Exerc´\u131cio 3: Calcule
\u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS, onde \u2212\u2192F = \u2212z\u2212\u2192k e S e´ a parte da esfera x2 + y2 + z2 = 4 fora
do cilindro x2 + y2 = 1, \u2212\u2192n apontando para fora.
Soluc¸a\u2dco: A superf´\u131cie S esta´ ilustrada na figura a seguir:
x
y
z
S
\u221a
3
pi/6
1
2
2
2
1
\u3c6
\u221a
3 \u21d2 tg \u3c6 = 1\u221a
3
\u21d2 \u3c6 = pi
6
Uma parametrizac¸a\u2dco para S e´ dada por
S : \u3d5(\u3c6, \u3b8) = (2 sen\u3c6 cos \u3b8 , 2 sen\u3c6 sen \u3b8 , 2 cos\u3c6)
com (\u3c6, \u3b8) \u2208 D =
[
pi
6
,
5pi
6
]
× [0, 2pi] . Temos:
dS = a2 sen \u3c6 d\u3c6 d\u3b8
a=2
= 4 sen\u3c6 d\u3c6 d\u3b8 .
UFF IME - GMA
Ca´lculo 3A Lista 11 168
Como \u2212\u2192n e´ exterior a S, enta\u2dco
\u2212\u2192n = (x, y, z)
a
a=2
=
(x, y, z)
2
.
Assim: \u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS =
\u222b\u222b
S
(0, 0,\u2212z) · (x, y, z)
2
dS =
= \u2212
\u222b\u222b
S
z2 dS = \u2212
\u222b\u222b
D
4 cos2 \u3c6 · 4 sen\u3c6 d\u3c6 d\u3b8 =
= \u221216
\u222b
5pi/6
pi/6
\u222b
2pi
0
cos2 \u3c6 sen\u3c6 d\u3b8 d\u3c6 = 32pi
\u222b
5pi/6
pi/6
cos2 \u3c6 d(cos\u3c6) =
= 32pi
[
cos3 \u3c6
3
]5pi/6
pi/6
=
32pi
3
[
\u2212
(\u221a
3
2
)3
\u2212
(\u221a
3
2
)3]
=
= \u221232pi
3
· 3
\u221a
3
8
= \u22124pi\u221a3 .
Exerc´\u131cio 4: Calcule o fluxo do campo
\u2212\u2192
F = \u2212x\u2212\u2192i \u2212 y\u2212\u2192j + 3y2z\u2212\u2192k sobre o cilindro x2 + y2 = 16,
situado no primeiro octante entre z = 0 e z = 5 \u2212 y com a orientac¸a\u2dco normal que aponta para o
eixo z.
Soluc¸a\u2dco: A superf´\u131cie S esta´ ilustrada na figura a seguir.
x y
z
S
C
4
4
5
5
\u2212\u2192n
Temos S : \u3d5(t, z) = (4 cos t , 4 sen t , z), com (t, z) \u2208 D :
{
0 \u2264 t \u2264 pi/2
0 \u2264 z \u2264 5\u2212 4 sen t . Ale´m disso,
dS = a dt dz
a=4
= 4 dt dz .
Como \u2212\u2192n aponta para o eixo z, enta\u2dco:
\u2212\u2192n = (\u2212x,\u2212y, 0)
a
=
(\u2212x,\u2212y, 0)
4
.
UFF IME - GMA
Ca´lculo 3A Lista 11 169
Portanto: \u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS =
\u222b\u222b
S
(\u2212 x,\u2212y, 3y2z) · (\u2212x,\u2212y, 0)
4
dS =
=
1
4
\u222b\u222b
S
(
x2 + y2
)\ufe38 \ufe37\ufe37 \ufe38
= 16
dS = 4
\u222b\u222b
S
dS = 4
\u222b\u222b
S
dS =
= 4
\u222b\u222b
D
4 dt dz = 16
\u222b pi/2
0
\u222b
5\u22124 sen t
0
dz dt =
= 16
\u222b pi/2
0
(5\u2212 4 sen t) dt = 16[5t+ 4 cos t]pi/2
0
=
= 16
(
5pi
2
\u2212 4
)
= 40pi \u2212 64 .
Exerc´\u131cio 5: Calcule
\u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS onde
\u2212\u2192
F (x, y, z) = xzey
\u2212\u2192
i \u2212 xzey\u2212\u2192j + z\u2212\u2192k
e S e´ a parte do plano x+ y + z = 1 no primeiro octante com orientac¸a\u2dco para baixo.
Soluc¸a\u2dco: O esboc¸o de S esta´ representado na figura a seguir.
x
y
z
S
1
1
1
\u2212\u2192n
x
y
D
1
1
y = 0
x+ y = 1
y = 1\u2212 x
A superf´\u131cie pode ser descrita por S : z = 1 \u2212 x \u2212 y = f(x, y), com (x, y) \u2208 D : 0 \u2264 x \u2264 1 e
0 \u2264 y \u2264 1\u2212x. Um vetor normal a S e´ dado por N = (\u2212fx,\u2212fy, 1) = (1, 1, 1) Como \u2212\u2192n aponta para
UFF IME - GMA
Ca´lculo 3A Lista 11 170
baixo enta\u2dco \u2212\u2192n = (\u22121,\u22121,\u22121)\u221a
3
. Temos que dS =
\u221a
1 + (fx)2 + (fy)2 dxdy =
\u221a
3 dxdy . Enta\u2dco:\u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS =
\u222b\u222b
S
(
xzey ,\u2212xzey , z) · (\u22121,\u22121,\u22121)\u221a
3
dS =
=
\u222b\u222b
S
(\u2212 xzey + xzey + z) · (\u22121,\u22121,\u22121)\u221a
3
dS =
=
\u222b\u222b
S
(
xzey \u2212 xzey \u2212 z)\u221a
3
dS =
\u222b\u222b
S
\u2212z\u221a
3
dS =
=
\u222b\u222b
D
d\u2212(1\u2212x\u2212y)\u221a
3
· \u221a3 dxdy =
\u222b\u222b
D
(\u22121 + x+ y) dxdy =
=
\u222b
1
0
\u222b
1\u2212x
0
(\u22121 + x+ y) dydx =
\u222b
1
0
[
\u2212 y + xy + y
2
2
]1\u2212x
0
dx =
\u22121
6
.
Exerc´\u131cio 6: Calcule
\u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS onde \u2212\u2192F (x, y, z) = x\u2212\u2192i + y\u2212\u2192j + 5\u2212\u2192k e S e´ a fronteira da regia\u2dco
delimitada pelo cilindro x2 + z2 = 1 e pelos planos y = 0 e x + y = 2 com a orientac¸a\u2dco positiva
(isto e´, \u2212\u2192n exterior a S).
Soluc¸a\u2dco: Para esboc¸ar S, fac¸amos uma inversa\u2dco nos eixos coordenados.
x
y
z
S1
S2
S3
1
1
2
2
\u2212\u2192n1
\u2212\u2192n2
\u2212\u2192n3
Temos que S = S1 \u222a S2 \u222a S3, orientada positivamente. Logo:\u222b\u222b
S
\u2212\u2192
F · \u2212\u2192n dS =
\u222b\u222b
S1
\u2212\u2192
F · \u2212\u2192n1 dS +
\u222b\u222b
S2
\u2212\u2192
F · \u2212\u2192n2 dS +
\u222b\u222b
S3
\u2212\u2192
F · \u2212\u2192n3 dS .
UFF IME - GMA
Ca´lculo 3A Lista 11 171
Ca´lculo de
\u222b\u222b
S1
\u2212\u2192
F · \u2212\u2192n1 dS
Temos S1 : y = 2 \u2212 x = f(x, z), com (x, z) \u2208 D : x2 + z2 \u2264 1. Logo, uma parametrizac¸a\u2dco de S1
e´ \u3d5(x, z) = (x, f(x, z), z) = (x, 2 \u2212 x, z), com (x, z) \u2208 D. Logo, \u3d5x = (1, fx, 0) = (1,\u22121, 0) e
\u3d5z = (0, fz, 1) = (0, 0, 1) donde
\u3d5x × \u3d5z =
\u2223\u2223\u2223\u2223\u2223\u2223\u2223\u2223
\u2212\u2192
i
\u2212\u2192
j
\u2212\u2192
k
1 fx 0
0 fz 1
\u2223\u2223\u2223\u2223\u2223\u2223\u2223\u2223 = (fx,\u22121, fz) = (\u22121,\u22121, 0) .
Logo, dS = \u2016\u3d5x × \u3d5z\u2016 dxdz =
\u221a
2 dxdz. Como \u2212\u2192n1 aponta para cima, enta\u2dco a componente y de \u2212\u2192n1
e´ positiva. Logo \u2212\u2192n1 = (1, 1, 0)\u221a
2
. Enta\u2dco:\u222b\u222b
S1
\u2212\u2192
F · \u2212\u2192n1 dS =
\u222b\u222b
D
(x, 2\u2212 x, 5) · (1, 1, 0)\u221a
2
· \u221a2 dxdz =
=
\u222b\u222b
D
2 dxdz = 2 · A(D) = 2pi .
Ca´lculo de
\u222b\u222b
S2
\u2212\u2192
F · \u2212\u2192n2 dS
Temos S2 : x
2+z2 = 1, com 0 \u2264 y \u2264 2\u2212x. Uma parametrizac¸a\u2dco de S2 e´: \u3d5(t, y) = (cos t, y, sen t),
com (t, y) \u2208 D1 : 0 \u2264 t \u2264 2pi e 0 \u2264 y \u2264 2\u2212 cos t. Temos
\u2212\u2192
N = \u3d5t × \u3d5y =
\u2223\u2223\u2223\u2223\u2223\u2223\u2223\u2223
\u2212\u2192
i
\u2212\u2192
j
\u2212\u2192
k
\u2212 sen t 0 cos t
0 1 0
\u2223\u2223\u2223\u2223\u2223\u2223\u2223\u2223 = (\u2212 cos t, 0,\u2212 sen t)
donde dS = \u2016\u2212\u2192N \u2016 dtdy = dtdy. Como \u2212\u2192n2 e´ exterior a S2 enta\u2dco \u2212\u2192n2 = (cos t, 0, sen t) . Logo:\u222b\u222b
S2
\u2212\u2192
F · \u2212\u2192n2 dS =
\u222b\u222b
D1
(cos