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.

2