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SECTION
16.2
Double Integrals over More General Regions
(ET Section 15.2)
881
Trying to evaluate the iterated integral in the original order of integration gives:
Z
9
0
Z
√
y
0
x
3
dx dy
(
3
x
2
+
y
)
1
/
2
=
Z
9
0
Ã
Z
√
y
0
x
3
(
3
x
2
+
y
)
1
/
2
dx
±
dy
The inner integral is much more difFcult to compute than the integrals involved in the previous computation.
44.
Z
1
−
1
Z
√
1
−
x
2
−
√
1
−
x
2
q
1
−
y
2
dydx
SOLUTION
The limits of integration determine the vertically simple region
D
described by the following inequalities:
−
1
≤
x
≤
1
,
−
p
1
−
x
2
≤
y
≤
p
1
−
x
2
x
−
1
−
x
2
≤
y
≤
1
−
x
2
y
1
−
1
The domain
D
(shown in the Fgure) is the unit circle. Since
f
(
x
,
−
y
)
=
f
(
x
,
y
)
and the domain is symmetric with
respect to the
x
axis, the double integral over the circle equals twice the integral over the upper half of the circle.
Similarly, since
f
(
−
x
,
y
)
=
f
(
x
,
y
)
(
f
does not depend on
x
)and
D
is symmetric with respect to the
y
axis, the double
integral over
D
is four times the integral over the part of the circle in the Frst quadrant. That is,
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This note was uploaded on 03/11/2008 for the course MATH 32B taught by Professor Rogawski during the Spring '08 term at UCLA.
 Spring '08
 Rogawski
 Integrals

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