2. Evaluate the following double integrals:
(a)
±±
R
e
x
2
dA
,
R
is the region bounded by
y
=0
,y
=
x,
x
=1.
(b)
R
(
x
+
y
)
dA
,
R
is the region bounded by the two curves
y
=
√
x,
y
=
x
2
.
Preliminaries
Since it is not diﬃcult to deal with iterated integrals, we use the following theorem to change
a double integral into an iterated integral.
Fubini’s Theorem
THEOREM 2
Fubini’s Theorem (Stronger Form)
Let
ƒ
(
x
,
y
) be continuous on a region
R
.
1.
If
R
is defined by
with
and
continu-
ous on [
a
,
b
], then
2.
If
R
is defined by
with
and
continuous
on [
c
,
d
], then
6
R
ƒ
s
x
,
y
d
dA
=
L
d
c
L
h
2
s
y
d
h
1
s
y
d
ƒ
s
x
,
y
d
dx dy
.
h
2
h
1
c
…
y
…
d
,
h
1
s
y
d
…
x
…
h
2
s
y
d
,
6
R
ƒ
s
x
,
y
d
dA
=
L
b
a
L
g
2
s
x
d
g
1
s
x
d
ƒ
s
x
,
y
d
dy dx
.
g
2
g
1
a
…
x
…
b
,
g
1
s
x
d
…
y
…
g
2
s
x
d
,
Find the Limits of Integration
From the theorem, we know that, when faced with evaluating
R
f
(
x, y
)
dA
, we only need
to
fnd the limits oF integration
. For example, if the order of the integration is
dydx
,thenwe
use the following steps:
(a)
Sketch
. Sketch the region of integration and label the bounding curves.
(b)
Find the
y
-limits of integration
(Two functions of
x
)
. Imagine a vertical line
L
cutting
through
R
in the direction of increasing
y
.Ma
rkthe
y
-values where
L
enters and leaves.
These are the
y
-limits of integration and are usually functions of
x
(instead of constants).
That is, take a vertical intersection, and fnd the intersection points, where
you can fnd the
y
-limits oF integration.
(c)
Find the
x
-limits of integration
(Two constant numbers)
.C
h
o
o
s
e
x
-limits that include
all the vertical lines through
R
.
That is, fnd the range oF
x
in your sketch.
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