prova gabarito cálculo 2
4 pág.

prova gabarito cálculo 2


DisciplinaCálculo II33.213 materiais868.393 seguidores
Pré-visualização1 página
Gabarito da Prova Final de Ca´lculo Diferencial e Integral III
Eng. Produc¸a\u2dco \u2013 06/12/2007
Prof. Rolci Cipolatti
1a Questa\u2dco: (2,0 pts)
Calcule as integrais duplas abaixo:
(a)
\u222b\u222b
D
dxdy
(x2 + y2)p
, sendo p < 1 e D =
{
(x, y) |x2 + y2 \u2264 4
}
;
(b)
\u222b\u222b
R
e(x+y)/(x\u2212y) dxdy, sendo R e´ o trape´zio cujos ve´rtices sa\u2dco os pontos
(1, 0), (2, 0), (0,\u22122) e (0,\u22121).
Soluc¸a\u2dco: (a) Usando coordenadas polares.\u222b\u222b
D
dxdy
(x2 + y2)p
=
\u222b 2pi
0
[\u222b 2
0
r dr
r2p
]
d\u3b8 = 2\u3c0
\u222b 2
0
r1\u22122pdr =
\u3c022(1\u2212p)
1\u2212 p
.
(b) Consideremos a mudanc¸a de varia´veis: u = x+ y e v = x\u2212 y. Enta\u2dco
\u2202(u, v)
\u2202(x, y)
=
\u2223\u2223\u2223\u2223 1 11 \u22121
\u2223\u2223\u2223\u2223 = \u22122 \u21d2 \u2202(x, y)\u2202(u, v) = \u22121/2
Assim,
I =
\u222b\u222b
R
e(x+y)/(x\u2212y) dxdy =
1
2
\u222b \u222b
G
eu/vdudv
Como a regia\u2dco G e´ do tipo I, temos
I =
1
2
\u222b 2
1
(\u222b v
\u2212v
eu/v du
)
dv =
1
2
\u222b 2
1
(
eu/v
1/v
\u2223\u2223\u2223\u2223v
\u2212v
)
dv =
e\u2212 e\u22121
2
\u222b 2
1
v dv =
3
4
senh(1).
2a Questa\u2dco: (3,0 pts)
(a) Calcule as coordenadas do centro de massa do so´lido W definido por
W =
{
(x, y, z) |x2 + y2 + z2 \u2264 R2, x \u2265 0, y \u2265 0, z \u2265 0
}
e cuja densidade volume´trica \u3c1(x, y, z) (em g/cm3) em cada ponto P \u2208 W e´ igual a` dista\u2c6ncia de
P ao eixo dos z\u2019s.
(b) Quando R = R(t) varia com o tempo, o centro de massa descreve uma curva ~\u3b3(t). Determine o
vetor velocidade em cada instante, supondo que R(t) = t2. Que tipo de movimento descreve o
centro de massa?
Soluc¸a\u2dco: (a) A densidade e´ \u3c1(x, y, z) =
\u221a
x2 + y2. Logo,
M =
\u222b\u222b\u222b
W
\u3c1(x, y, z) dx dy dz =
\u222b\u222b\u222b
W
\u221a
x2 + y2dx dy dz
Em coordenadas esfe´ricas, tem-se \uf8f1\uf8f4\uf8f2\uf8f4\uf8f3
x = \u3c1 sen\u3c6 cos \u3b8
y = \u3c1 sen\u3c6 sen \u3b8
z = \u3c1 cos\u3c6
de modo que
dx dy dz = \u3c12 sen\u3c6d\u3c1 d\u3c6 d\u3b8,
\u221a
x2 + y2 = \u3c1 sen\u3c6.
Nessas coordenadas o conjunto W toma a forma
W\u2dc = {(\u3c1, \u3b8, \u3c6) \u2208 R3 ; 0 \u2264 \u3c1 \u2264 R, 0 \u2264 \u3c6, \u3b8 \u2264 \u3c0/2}.
Portanto,
M =
\u222b\u222b\u222b
W\u2dc
\u3c1 sen\u3c6\u3c12 sen\u3c6d\u3c1 d\u3c6 d\u3b8
=
(\u222b pi/2
0
d\u3b8
)(\u222b pi/2
0
sen2 \u3c6d\u3c6
)(\u222b R
0
\u3c13 d\u3c1
)
=
\u3c0
2
×
\u3c0
4
×
R4
4
=
\u3c02R4
32
.
Por outro lado,
x =
\u222b\u222b\u222b
W
x\u3c1(x, y, z) dx dy dz =
\u222b\u222b\u222b
W
x
\u221a
x2 + y2dx dy dz,
que, em coordenadas esfe´ricas se escreve,
x =
\u222b\u222b\u222b
W\u2dc
\u3c1 sen\u3c6 cos \u3b8 \u3c1 sen\u3c6\u3c12 sen\u3c6d\u3c1 d\u3c6 d\u3b8
=
(\u222b pi/2
0
cos \u3b8 d\u3b8
)(\u222b pi/2
0
sen3 \u3c6d\u3c6
)(\u222b R
0
\u3c14 d\u3c1
)
=
(\u222b pi/2
0
cos \u3b8 d\u3b8
)(\u222b pi/2
0
(1\u2212 cos2 \u3c6) sen\u3c6d\u3c6
)(\u222b R
0
\u3c14 d\u3c1
)
=
(\u222b pi/2
0
cos \u3b8 d\u3b8
)(\u222b 1
0
(1\u2212 u2) du
)(\u222b R
0
\u3c14 d\u3c1
)
=
2R5
15
.
Observe pela simetria das fo´rmulas que y = x.
z =
\u222b\u222b\u222b
W\u2dc
\u3c1 cos\u3c6\u3c1 sen\u3c6\u3c12 sen\u3c6d\u3c1 d\u3c6 d\u3b8
=
(\u222b pi/2
0
d\u3b8
)(\u222b pi/2
0
sen2 \u3c6 cos\u3c6d\u3c6
)(\u222b R
0
\u3c14 d\u3c1
)
=
(\u222b pi/2
0
d\u3b8
)(\u222b 1
0
u2 du
)(\u222b R
0
\u3c14 d\u3c1
)
=
\u3c0R5
30
.
Assim, o centro de massa e´ o ponto
PM =
(
x
M
,
y
M
,
y
M
)
=
(
64R
15\u3c02
,
64R
15\u3c02
,
16R
15\u3c0
)
.
(b) Supondo R(t) = t2, tem-se:
\u2212\u2192\u3b3 (t) =
(
64R
15\u3c02
,
64R
15\u3c02
,
16R
15\u3c0
)
=
16t2
15\u3c02
(4, 4, \u3c0).
Logo o movimento e´ retil´\u131neo na direc¸a\u2dco do vetor (4, 4, \u3c0) e o vetor velocidade e´
d
dt
\u2212\u2192\u3b3 (t) =
32t
15\u3c02
(4, 4, \u3c0),
o que mostra que e´ um movimento uniformemente acelerado.
3a Questa\u2dco: (3,0 pts)
Seja C a curva parametrizada por ~\u3c3(t) = (cos t, sen t, cos 2t), 0 \u2264 t \u2264 2\u3c0 e
~F (x, y, z) =
(
ln(1 + x2)\u2212 y, e\u2212y
2
\u2212 z, sen(z2)\u2212 x
)
.
(a) Calcule
\u222e
C
~F · d~r usando o Teorema de Stokes.
(b) E´ poss´\u131vel calcular a integral de linha acima sem aplicar o Teorema de Stokes? Justifique sua
resposta.
Soluc¸a\u2dco: (a) Temos rot
\u2212\u2192
F = (1, 1, 1). Como cos 2t = cos2 t \u2212 sen2 t, ve\u2c6-se que a curva \u2212\u2192\u3c3 (t) esta´
contida na superf´\u131cie S dada por z = x2 \u2212 y2. Esta superf´\u131cie pode ser parametrizada por
\u2212\u2192
\u3a6 (x, y) =
(x, y, x2 \u2212 y2). Para que C seja o bordo de S, tomamos o dom\u131´nio de
\u2212\u2192
\u3a6 como o disco unita´rio D, isto e´,
D = {(x, y) ; x2 + y2 \u2264 1}. Desta forma, o vetor normal
\u2212\u2192
N (x, y), com orientac¸a\u2dco positiva relativamente
a` orientac¸a\u2dco da curva C, e´
\u2212\u2192
N (x, y) = (2x,\u22122y, 1).
Assim, do Teorema de Stokes, tem-se\u222e
C
\u2212\u2192
F · d\u2212\u2192r =
\u222b\u222b
S
rot
\u2212\u2192
F · d
\u2212\u2192
S =
\u222b\u222b
D
(1, 1, 1) · (2x,\u22122y, 1) dxdy =
\u222b\u222b
D
(2x\u2212 2y + 1) dxdy
Mas \u222b\u222b
D
2xdxdy = 2
\u222b 2pi
0
[\u222b 1
0
r cos \u3b8 rdr
]
d\u3b8 = 0,
\u222b\u222b
D
2y dxdy = 2
\u222b 2pi
0
[\u222b 1
0
r sen \u3b8 rdr
]
d\u3b8 = 0.
Logo, \u222e
C
\u2212\u2192
F · d\u2212\u2192r =
\u222b\u222b
D
dxdy = a´rea de D = \u3c0.
(b) O ca´lculo da integral de linha pode ser feito explicitamente, embora na definic¸a\u2dco do campo estejam
presentes func¸o\u2dces que na\u2dco possuem primitivas elementares, tais como e\u2212y
2
. De fato, se escrevermos
\u2212\u2192
F =
\u2212\u2192
G +
\u2212\u2192
H , com
\u2212\u2192
G (x, y, z) =
(
ln(1 + x2), e\u2212y
2
, sen(z2)
)
, e
\u2212\u2192
H (x, y, z) =
(
\u2212y,\u2212z,\u2212x
)
,
enta\u2dco rot
\u2212\u2192
G = (0, 0, 0). Logo,
\u2212\u2192
G e´ um campo gradiente e
\u222e
C
\u2212\u2192
G · d\u2212\u2192r = 0. Desta forma,
\u222e
C
\u2212\u2192
F · d\u2212\u2192r =
\u222e
C
\u2212\u2192
H · d\u2212\u2192r =
\u222b 2pi
0
(\u2212 sen t,\u2212 cos 2t,\u2212 cos t) · (\u2212 sen t, cos t,\u22122 sen2t) dt
que pode ser calculada explicitamente.
4a Questa\u2dco: (2,0 pts)
Calcule o fluxo
\u222b\u222b
S(
~F · ~n) dS, do campo ~F (x, y, z) = (x, y + ez
2
, z2 + 1), atrave´s da superf´\u131cie S =
{(x, y, z) | z = 1 \u2212 x2 \u2212 y2, z \u2265 0}, (com vetor normal apontando para fora). (Sugesta\u2dco: Use o Teorema
de Gauss, mas observe que a superf´\u131cie na\u2dco e´ fechada em z = 0.)
Soluc¸a\u2dco: Consideremos o so´lido
W = {(x, y, z) \u2208 R3 ; 0 \u2264 z \u2264 1\u2212 x2 \u2212 y2}.
Enta\u2dco a fronteira \u2202W de W e´ a superf´\u131cie S \u222a T , sendo T o disco {(x, y, z) ; z = 0, x2 + y2 \u2264 1}.
Pelo Teorema de Gauss, tem-se\u222b\u222b\u222b
W
div
\u2212\u2192
F dxdydz =
\u222b\u222b
\u2202W\u222aT
\u2212\u2192
F · d
\u2212\u2192
S =
\u222b\u222b
S
\u2212\u2192
F · d
\u2212\u2192
S +
\u222b\u222b
T
\u2212\u2192
F · d
\u2212\u2192
S
Como div
\u2212\u2192
F (x, y, z) = 2 + 2z, tem-se em coordenadas cil´\u131ndricas
\u222b\u222b\u222b
W
div
\u2212\u2192
F dxdydz =
\u222b\u222b\u222b
W
(2 + 2z) dxdydz =
\u222b\u222b
D
[\u222b 1\u2212r2
0
(2 + 2z) dz
]
rdrd\u3b8,
sendo D o disco unita´rio. Portanto,\u222b\u222b\u222b
W
div
\u2212\u2192
F dxdydz = 2\u3c0
\u222b 1
0
[
2(1\u2212 r2) + (1\u2212 r2)2 rdr
]
=
4\u3c0
3
.
Por outro lado,\u222b\u222b
T
\u2212\u2192
F · d
\u2212\u2192
S =
\u222b\u222b
T
(
\u2212\u2192
F · \u2212\u2192n ) dS =
\u222b\u222b
T
(x, y + 1, 1) · (0, 0,\u22121) dS = \u2212a´rea de T = \u2212\u3c0.
Assim, conclu´\u131mos \u222b\u222b
S
\u2212\u2192
F · d
\u2212\u2192
S =
4\u3c0
3
+ \u3c0 =
7\u3c0
3
.