1
0
! −1−
√1−y2
−
√4−y2 f(x, y)dxdy + ! 1
! 0
−1+√1−y2 f(x, y)dxdy + ! 2
√4−y2 f(x, y)dxdy
Ωn = {(x, y, z) x ≥ 0 , y ≥ 0 , x2 + y2 ≤ n2 , 0 ≤ z ≤ n}
, !
Ωn sen (x2 + y2) dV = ! π/2
! n
0 sen (r2)rdzdrdθ = π
4n(1 − cos (n2)
5π/12
23π/160
Ωn = {(x, y, z) ∈ IR3 , 1
n2 ≤ x2 + y2 ≤ 1 , −1 ≤ z ≤ 1}, onde n ≥ 2.
e !
Ωn |Ln(
"x2 + y2)| dV ≤ 4
limn→∞ # 2π
# 1
1/n
−1
rLn(r)dzdrdθ = −π
θ.
!
D x dA = −π
a (D1)n = {(x, y) ∈ IR2 , 1 ≤ x2 + y2 ≤ n2}; n ≥ 2. Tem-se !
(D1)n |f|dA ≤
! 2π
1 r−3/2drdθ ≤ M
(D2)n = {(x, y) ∈ IR2 , 1
n2 ≤ x2 + y2 ≤ 1}. !
(D2)n |f|dA ≤ ! 2π
! 1
1/n r−1/2drdθ ≤ M˜
D3 = D1 ∪ D2
T −1 dada por u = xy, v = y
x . u = 1, u = 2, v = 1/2 e v = 2 e |JT −1 | = 2y
x , logo o |JT | = x
2y !
D(x2 + y2) dA = ! 2
! 2
1/2(u
2 + u
2v2 )dvdu = 9
4 . D = {(x, y) , 1 ≤ y ≤ e , 0 ≤ x ≤ Ln(y)}.
Dn = {(x, y) , 1 + 1
n ≤ y ≤
e , 0 ≤ x ≤ Ln(y)}. = !
Dn |f|dA ≤ 27 = limn→∞ !
Dn f dA = limn→∞ 1
2 (ee − e1+ 1
n ) = ee−e
2 .
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Answers & Comments
1
0
! −1−
√1−y2
−
√4−y2 f(x, y)dxdy + ! 1
0
! 0
−1+√1−y2 f(x, y)dxdy + ! 2
1
! 0
−
√4−y2 f(x, y)dxdy
Ωn = {(x, y, z) x ≥ 0 , y ≥ 0 , x2 + y2 ≤ n2 , 0 ≤ z ≤ n}
, !
Ωn sen (x2 + y2) dV = ! π/2
0
! n
0
! n
0 sen (r2)rdzdrdθ = π
4n(1 − cos (n2)
5π/12
23π/160
Ωn = {(x, y, z) ∈ IR3 , 1
n2 ≤ x2 + y2 ≤ 1 , −1 ≤ z ≤ 1}, onde n ≥ 2.
e !
Ωn |Ln(
"x2 + y2)| dV ≤ 4
limn→∞ # 2π
0
# 1
1/n
# 1
−1
rLn(r)dzdrdθ = −π
θ.
!
D x dA = −π
a (D1)n = {(x, y) ∈ IR2 , 1 ≤ x2 + y2 ≤ n2}; n ≥ 2. Tem-se !
(D1)n |f|dA ≤
! 2π
0
! n
1 r−3/2drdθ ≤ M
(D2)n = {(x, y) ∈ IR2 , 1
n2 ≤ x2 + y2 ≤ 1}. !
(D2)n |f|dA ≤ ! 2π
0
! 1
1/n r−1/2drdθ ≤ M˜
D3 = D1 ∪ D2
T −1 dada por u = xy, v = y
x . u = 1, u = 2, v = 1/2 e v = 2 e |JT −1 | = 2y
x , logo o |JT | = x
2y !
D(x2 + y2) dA = ! 2
1
! 2
1/2(u
2 + u
2v2 )dvdu = 9
4 . D = {(x, y) , 1 ≤ y ≤ e , 0 ≤ x ≤ Ln(y)}.
Dn = {(x, y) , 1 + 1
n ≤ y ≤
e , 0 ≤ x ≤ Ln(y)}. = !
Dn |f|dA ≤ 27 = limn→∞ !
Dn f dA = limn→∞ 1
2 (ee − e1+ 1
n ) = ee−e
2 .