8/16/2019 Calculo III - Versão 3
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3
ETAPA 3
Figura 1:
S A=∫0
1
∫ X
4
X
dydx=∫0
1
x− x
4
dx
S A= x
2
2 −
x2
8 [10=12−18=0,375u . a
sB=∫1
2
∫ x
4
1
x
dydx=∫1
21
x− x
4 dx
sB=ln x− x
2
8 [21=[ ln2−22
8 ]−[ ln 1−12
8 ]=0,318u .a
Área total é 0,693.
Figura 2
S A=4u .a
SB=∫1
44
x dx=4ln x [41
SB=4ln4−4ln1=5,545u .a
Área total é 38,18u. a .
8/16/2019 Calculo III - Versão 3
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4
ETAPA 4
Desafio A
∫1
4
4
2π . (4 √ x ) .√1+ 2√ x2
dx=2π ∫1
4
4
4 √ x .√1+ 4 x dx
2π ∫1
4
4
4 √ x .(1+4
x )=8 π ∫14
4
√ x+4
8π ∫1
4
4
( x+4 )1
2=( x+4 )
3
2
3
2[ 4
1
4
8π ( 83
2
3
2
−(14+4)
3
2
3
2 )=2π 3 . (128√ 2−17√ 17)Desafio B
V =∫0
π
2
π [ (2−sin x3 )2−(2−sin x )2 ]dx=π
(∫0
π
2
(2−sin x3 )2
dx+∫0
π
2
−(2−sin x )2dx
)
8/16/2019 Calculo III - Versão 3
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V=
x3
−sin¿ .2+22dx−(∫0
π
2
22+2.2. (−sin x )+(−sin x )
2
dx)¿
(−sin x3 )2+2.(¿¿ )
∫0
π
2
¿
π ¿
v=π
(∫0
π
2
sin x6dx−4∫
0
π
2
(sin x )3dx+4 x|π 20
−
(4 x|π 2
0
+4∫0
π
2
(−sin x ) dx+∫0
π
2
(−sin x )2dx
))v=π (( 5π 32 −83 + 4 π 2 )−( 4 π 2 −4+ π 4 ))=3,26