3.3. MULTIPLE INTEGRALS 107
Example 3.42 Find A D
Z
1
1
e
x
2
=2
dx
:
Solution:
e solution works if you compute the square of the integral.
A
2
D
Z
1
1
e
x
2
=2
dx
2
D
Z
1
1
e
x
2
=2
dx
Z
1
1
e
x
2
=2
dx
D
Z
1
1
e
x
2
=2
dx
Z
1
1
e
y
2
=2
dy
Rename
D
Z
1
1
Z
1
1
e
x
2
=2
e
y
2
=2
dy dx
D
Z
1
1
Z
1
1
e
1=2.x
2
Cy
2
/
dA
Change to polar coordinates
D
Z Z
R
e
1=2.r
2
/
dA
D
Z Z
R
e
1=2.r
2
/
r dr d
e polar region in question is
0
r <
1
and
0
< 2
. Rebuild the integral with these
limits and we get:
A
2
D
Z
2
0
Z
1
0
r
e
r
2
=2
dr d
D
Z
2
0
d
Z
1
0
r
e
r
2
=2
dr
D
ˇ
ˇ
ˇ
ˇ
2
0
Z
1
0
r
e
r
2
=2
dr
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