Solutions for PSet 11
1. (11.28:14) We substitute u = x − y and v = x + y. the Jacobian of the
transformation (x, y) → (u, v) is
�
�
1 −1
1 1
This has determinant 2 and the vertices of the parallelogram S correspond to
−π < u < π, π < v < 3π. Thus:
� � � 3π � π
1 2 2
2
2
(x − y) sin (x + y) d x d y =
u sin v d u d v
S
π
−π 2
This simplifies to:
�
�3π � 3 �π
u
1 v sin(2v)
π4
−
=
2 2
3 u=−π
3
4
v=π
2. (11.28:16)
(a) We can apply Fubini’s Theorem:
� � �
−(x2 +y 2 )
e
dx dy =
R
r
−r
−x2
e
2
�
dx
r
−r
2
e−y d y = (I(r))2
2
(b) C1 ⊂ R ⊂ C2 and the function e−(x +y ) > 0 thus
� � � �
� �
−(x2 +y 2 )
−(x2 +y 2 )
e
d x d y−
e
dx dy =
R
C1
and
� �
C2
e
−(x2 +y 2 )
R\C1
� � � �
−(x2 +y 2 )
d x d y−
e
dx dy =
R
C2 \R
Combining the two results:
� � � �
� �
−(x2 +y 2 )
−(x2 +y 2 )
e
dx dy <
e
dx dy <
C1
R
thus proving the required statement.
1
e−(x
2 +y 2 )
dx dy > 0
e−(x
2 +y 2 )
dx dy > 0
C2
e−(x
2 +y 2 )
dx dy
(c) For a disc C of radius s
� �
�
−(x2 +y 2 )
e
dx dy =
C
�
2π
0
s2
2π
0
�
�
0
s
2
e−ρ ρ d ρ d ϑ =
�u=s2
1 −u
2
e d u = π −e−u u=0 = π(1 − e−s )
2
For the circles C1 (inscribing square of√side 2r) and C2 (circumscribing
square of side 2r), the radii are r and 2r respectively. Substituting to
(b) we get:
2
2
π(1 − e−r ) < I 2 (r) < π(1 − e−2r )
as r → ∞:
π ≤ lim I 2 (r) ≤ π
r→∞
�∞
√
2
Thus limr→∞ I(r) exists and equals −∞ e−x d x = π. In other words,
√
� ∞ −x2
e
d x = 2π .
0
2
MIT OpenCourseWare
http://ocw.mit.edu
18.024 Multivariable Calculus with Theory
Spring 2011
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