MODEL ANSWERS TO HWK #10
(18.022 FALL 2010)
√
�
∞
�∞
√
2
2
(1) We want to show that
−∞ e−x dx = π. We show that I
=
0 e−x dx
=
2
π by proving
that
�
∞
�
∞
�
∞ �
∞
π
2
2
2
−x2
−y 2
I =
e dx
e dy =
e−(x +y ) dxdy = .
4
0
0
0
0
(i) The square of side a defined by 0 < x < a and 0 < y < a contains Qa , the quarter
circle of radius a, and thus:
�a
�
�
a �
a
� π� a
� �
−r2
�
2
π
−e
π
2
2
2
2
2
2
�
e−(x +y ) dxdy ≥
e−(x +y ) dxdy =
e−r rdrdθ =
=
(1 − e−a )
�
2
2
�
4
0
0
Qa
0
0
r=0
Taking the limit as a goes to ∞, we conclude that I 2 ≥ π4
.
(ii) The square of side a defined by 0 < x < a and 0 < y < a is contained in Q√2a , the
√
quarter circle of radius 2a, and thus:
�√2a
�
�
a �
a
� �
� π � √2a
−r2
�
2
π −e �
π
2
2
2
2
2
2
e−(x +y ) dxdy ≤
e−(x +y ) dxdy =
e−r rdrdθ =
=
(1 − e−2a )
�
2
2
�
4
0
0
Q√2a
0
0
r=0
2
π
.
4
Taking the limit as a goes to ∞, we conclude that I ≤
(2) (5.5.15)
�
5
�3
��
� 2π � 3
r ��
486
2
2 3/2
3
(x + y ) dA =
r rdrdθ = 2π
=
π
�
5
r=0
5
D
0
0
(3) (5.5.16)
�
a � √a2 −y2
x
2 +y 2
e
−a
�
π
2
a
�
dxdx =
− π2
0
0
�
2
�a
e
r ��
e
rdrdθ = π
�
2
�
r2
r=0
π
2
=
(e
a − 1)
2
(4) (5.5.20)
�
�
dA = 4
rose
�
π
2
sin 2θ
�
dA = 4
leaf
0
0
�
π
2
(sin 2θ)2
rdrdθ = 4
dθ
2
0
�
π
π
2
1 − cos 4θ
π 1
π
2
dθ = − (sin 4θ|
θ=0
= 2
=
2
2 4
2
0
(5) (5.5.21) The cardioid and the circle intersect at the points (0, 1) and (0, −1), and since we
want the area inside the cardioid and outside the circle, the bounds of integration for θ must
1
2
MODEL ANSWERS TO HWK #10 (18.022 FALL 2010)
be π/2 to 3π/2.
3π
2
�
π
2
1−cos θ
�
1
1
rdrdθ =
2
1
=
2
3π
2
�
π
2
1
((1 − cos θ) − 1 )dθ =
2
2
�
3π
2
(
π
2
2
�
3π
2
(cos2 θ − 2 cos θ)dθ
π
2
1 + cos 2θ
− 2 cos θ)dθ
2
π 1
3π
π
π
+ (sin 3π − sin π) − (sin
− sin ) = + 2
4 8
2
2
4
=
(6) (5.5.22)
2π
�
3θ
�
2π
�
9θ2
9
dθ =
2
2
rdrdθ =
0
0
0
�
�2π
θ3 ��
9 8π 3
=
= 12π 3
3 �θ=0 2 3
(7) (5.5.29)
���
2π
�
�
1
√
�
dV =
W
0
√
− 10−2r2
0
�
= −π
10−2r2
�
1
rdzdrdθ = 2π
√
2 10 − 2r2 rdr
0
�1
(10 − 2r2 )3/2 ��
�
3/2
r=0
√
√
4 2
=
π(5 5 − 8)
3
(8) (5.5.30)
���
�
2π
�
2
9−r2
�
dV =
W
�
0
0
�
3
(9r − r )dr = 2π
rdzdrdθ = 2π
0
2
0
�2
9r2 r4 ��
− �
= 28π
2
4 r=0
(9) (5.5.31)
�√
2
4 � 25−z
r
2
2
2
�
(2 + x + y )dV =
rdzdrdθ = 2π
r + �
dz
4 r=0
W
0
3
0
3
� 5
� 5
625 − 50z 2 + z 4
725 23 2 z 4
2
− z + )dz
=
(25 − z +
)dz =
(
4
4
2
4
3
3
656
=
π
5
���
�
2π
�
5
√
�
25−z 2
�
5
�
(10) (5.5.32) By symmetry reasons, the total volume of the solid is 16 times the volume of the
portion defined by z > 0, x > 0 and 0 < y < x: this sixteenth of the solid is bounded on the
bottom by the plane z = 0, on the sides by the planes y = 0 and y = x and by the cylinder
x2 + y 2 = a2 , and on top by x2 + z 2 = a2 . We write the integral in cylindrical coordinates:
MODEL ANSWERS TO HWK #10
�
π �
a
4
(18.022 FALL 2010)
√
�
Volume = 16
a2 −r2 cos2 θ
rdzdrdθ
0
0
0
�
π � a √
4
= 16
a2 − r2 cos2 θrdrdθ
0
0
�
�
π �
3 a
16
4
(a2 − r2 cos2 θ) 2 ��
−
=
� dθ
�
3 0
cos
2 θ
r=0
�
π
3
4
16
1 − sin θ
=
dθ
3 0
cos2 θ
�� π
�
�
π
4
4 sin3 θ
16
1
=
dθ −
dθ
2
2
3
0 cos θ
0 cos θ
�
�
� 1 2
π
16
t
−
1
=
(tan θ
|θ=0
4 +
√
dt
2
2
3
t
2
�
�
�
1
16
1
=
1 + √ 1 − 2 dt
2
3
t
2
�
� �1 �
�
16
1 �
=
1 + (t
|1t= √2 +
�
2
3
t t= √2
2
�
�
√
√
16
2
=
1 + 1 −
+ 1 −
2
3
2
√
= (16 − 8 2)a3
3
MIT OpenCourseWare
http://ocw.mit.edu
18.022 Calculus of Several Variables
Fall 2010
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