Solution 4
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MATH2020A Advanced Calculus II, 2013-14 Assignment 4 Suggested Solution P.1062, Ex. 8 Solution: Notice the ellipse can be parameterized by (x, y, z)=(x, y, 6 - 2x - 3y) where x 2 + y 2 ≤ 2. Hence, area of ellipse = ZZ {x 2 +y 2 ≤2} q 1+ z 2 x + z 2 y dA = √ 14 ZZ {x 2 +y 2 ≤2} dA =2π √ 14. P.1071, Ex. 12 Solution: Let T : R 2 -{0} -→ R 2 be the transformation given by T (x, y)=(u, v)= 2x x 2 + y 2 , 2y x 2 + y 2 . Denote R be the region mentioned in the question, then T (R)= {(u, v): 1 3 ≤ u ≤ 1, 1 4 ≤ v ≤ 1}. ∂ (x, y) ∂ (u, v) = ∂ (u, v) ∂ (x, y) -1 = 2(y 2 -x 2 ) (x 2 +y 2 ) 2 -4xy (x 2 +y 2 ) 2 -4xy (x 2 +y 2 ) 2 2(x 2 -y 2 ) (x 2 +y 2 ) 2 -1 = - (x 2 + y 2 ) 2 4 . Hence, ZZ R 1 (x 2 + y 2 ) 2 dxdy = ZZ T (R) 1 (x 2 + y 2 ) 2 ∂ (x, y) ∂ (u, v) dudv = ZZ T (R) 1 4 dudv = 1 4 3 4 2 3 = 1 8 . P.1071, Ex. 19 Solution: Let (x, y, z)=(ρ cos θ sin φ, ρ sin θ sin φ, ρ cos θ). Denote L = Z ∞ -∞ Z ∞ -∞ Z ∞ -∞ p x 2 + y 2 + z 2 e -k(x 2 +y 2 +z 2 ) dxdydz. 1
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Transcript of Solution 4
MATH2020A Advanced Calculus II, 2013-14
Assignment 4 Suggested Solution
P.1062, Ex. 8 Solution:
Notice the ellipse can be parameterized by (x, y, z) = (x, y, 6− 2x− 3y) where x2 + y2 ≤ 2.
Hence,
area of ellipse =∫∫{x2+y2≤2}
(√1 + z2
x + z2y
)dA =
√14∫∫{x2+y2≤2}
dA = 2π√
14.
P.1071, Ex. 12 Solution:
Let T : R2 − {0} −→ R2 be the transformation given by
T (x, y) = (u, v) =(
2xx2 + y2
,2y
x2 + y2
).
DenoteR be the region mentioned in the question, then T (R) = {(u, v) :13≤ u ≤ 1,
14≤ v ≤ 1}.
∂(x, y)∂(u, v)
=(∂(u, v)∂(x, y)
)−1
=
∣∣∣∣∣∣∣∣2(y2−x2)(x2+y2)2
−4xy(x2+y2)2
−4xy(x2+y2)2
2(x2−y2)(x2+y2)2
∣∣∣∣∣∣∣∣−1
= − (x2 + y2)2
4.
Hence, ∫∫R
1(x2 + y2)2
dxdy =∫∫
T (R)
1(x2 + y2)2
∣∣∣∣∂(x, y)∂(u, v)
∣∣∣∣ dudv=∫∫
T (R)
14dudv =
(14
)(34
)(23
)=
18.
P.1071, Ex. 19 Solution:
Let (x, y, z) = (ρ cos θ sinφ, ρ sin θ sinφ, ρ cos θ). Denote
L =∫ ∞−∞
∫ ∞−∞
∫ ∞−∞
√x2 + y2 + z2e−k(x
2+y2+z2)dxdydz.
1
Then
L = limR→∞
∫∫∫{x2+y2+z2≤R}
√x2 + y2 + z2e−k(x
2+y2+z2)dV
= limR→∞
∫ R
0
∫ 2π
0
∫ π
0
(ρe−kρ2)(ρ2 sinφ)dφdθdρ
= limR→∞
4π∫ R
0
(ρ3e−kρ2)dρ
= limR→∞
−2πk
∫ R
0
ρ2de−kρ2
= limR→∞
−2πk
([ρ2e−kρ
2]R0−∫ R
0
e−kρ2dρ2
)
= limR→∞
−2πk
([ρ2e−kρ
2]R0
+1k
[e−kρ
2]R0
)
= limR→∞
−2πk
(R2e−kR
2+e−kR
2 − 1k
)=
2πk2.
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