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Math 293 Answers to Practice Prelim #1
Spring 2000
x
y
y = 4  x
2
4
2
1
.
R
2
0
R
4

x
2
0
xe
2
y
4

y
dy dx
. This is awkward to in
tegrate in the given order, so switch the order to
get
R
4
0
R
√
4

y
0
xe
2
y
4

y
dx dy
. Then after the ﬁrst inte
gration we have
R
4
0
e
2
y
2
dy
, which works out to be
e
8
4

1
4
.
2
.
R
1
0
R
√
1

x
2
0
e

(
x
2
+
y
2
)
dy dx
. The region of integration is the portion of
the unit disk lying in the ﬁrst quadrant, so in polar coordinates this integral
becomes
R
π/
2
0
R
1
0
e

r
2
rdrdθ
, which computes out to be
(

e

1
+1
2
)(
π
2
)
.
3
. In spherical coordinates the cone is given by the equation
φ
=
π
4
, so the
region below the cone and above the
xy
plane will have
π
4
≤
φ
≤
π
2
. The
region between the two spheres will have 3
≤
ρ
≤
4. Also,
x
2
+
y
2
=
ρ
2

z
2
.
So we get the integral
R
2
π
0
R
π/
2
π/
4
R
4
3
(
ρ
2

z
2
)
ρ
2
sin
φdρdφdθ
. Substituting
z
=
ρ
cos
φ
, this becomes
R
2
π
0
R
π/
2
π/
4
R
4
3
(
ρ
2

ρ
2
cos
2
φ
)
ρ
2
sin
φdρdφdθ
which
works out to be
(
2
π
)(
4
5

3
5
5
)(
1
√
2

1
3
·
2
√
2
)
.
4
. The vector ﬁeld
F
(
x, y
)=(
y
+
e
x
ln
y
)
i
+
e
x
y
j
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This homework help was uploaded on 01/21/2008 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 TERRELL,R
 Math, Differential Equations, Equations

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